[Ian Bell[_2_]]

I have been thinking a bit about Patrick's suggestion of using something like an EL84 in an SE
headphones amp. I was intrigued that he had built one using a 'cheap' output transformer so I have
looked at specs from various suppliers of such transformers. These transformers typically have a
primarily inductance of around 10 Henries (at 20 or 40mA) for a nominal 5K match. Now my transformer
rule of thumb is you want the primary inductance to be twice the plate load at the lowest frequency
of interest (say 20Hz). So for 5K ohms and 20Hz the inductance needs to be getting on for 40 Henries
so is not the bass response of these transformers going to suffer somewhat??

Cheers

Ian

[Ian Iveson]

"Ian Bell" wrote:

It likely depends on how close saturation is to 20Hz, *at
the signal level the OPT is likely to encounter*. If you're
willing to use feedback, and the signal is small, you may be
OK. An appropriate filter at the input should make
saturation impossible.

Patrick'll be along in a tick.

Twice the plate load? It *is* the anode load isn't it? 2 to
3 times ra, depending on whether you want power or fidelity
respectively, broadly speaking.

There is a tradition of not caring about head amps...as if
they are add-on extras to the main system. These days, they
quite often *are* the main system.

Ian

[Ian Bell[_2_]]

Ian Iveson wrote:
"Ian Bell" wrote:

It likely depends on how close saturation is to 20Hz, *at
the signal level the OPT is likely to encounter*. If you're
willing to use feedback, and the signal is small, you may be
OK. An appropriate filter at the input should make
saturation impossible.


One thing I was wondering was if I use a transformer with a nominal 16 ohm secondary but use it with
32 ohm headphones for example, then ra reflected into the secondary would be so low relative to the
load that the output could be considered essentially a voltage source. Equally the load reflected
to the plate would be four times higher so distortion would be lower?

Patrick'll be along in a tick.

Twice the plate load? It *is* the anode load isn't it? 2 to
3 times ra, depending on whether you want power or fidelity
respectively, broadly speaking.

Yes, but you are thinking in power amp terms alone and as I have said before I have no power amp
experience. My rule applies to low level transformers e.g. mic transformers and for broadly similar
reasons.


Cheers

Ian

There is a tradition of not caring about head amps...as if
they are add-on extras to the main system. These days, they
quite often *are* the main system.

Ian

[Patrick Turner]

On Sep 30, 8:00*am, Ian Bell wrote:
I have been thinking a bit about Patrick's suggestion of using something like an EL84 in an SE
headphones amp. I was intrigued that he had built one using a 'cheap' output transformer so I have
looked at specs from various suppliers of such transformers. These transformers typically have a
primarily inductance of around 10 Henries (at 20 or 40mA) for a nominal 5K match. Now my transformer
rule of thumb is you want the primary inductance to be twice the plate load at the lowest frequency
of interest (say 20Hz). So for 5K ohms and 20Hz the inductance needs to be getting on for 40 Henries
so is not the bass response of these transformers going to suffer somewhat??

Cheers

Ian

Most old radio OPS designed for a 6V6 operating as a beam tetrode with
no NFB have low Lp because they are designed to give from only 100Hz
upwards as the little speaker used just can't cope with any below
100Hz.

You are right about the 40H. If there is that much L and Ra was 2k2
with EL84 in triode and the RL was a lot higher or not even present,
then response will be -3dB when XLp = Ra or at 8.75Hz, so Lp could be
16H to give -3dB at 22Hz. FB can flatten the resonse further. The real
other problem is that if Lp is low, it is often because the core Afe
size is too small and there are not enough P turns to prevent core
saturation, ie, Bdc + Bac exceeding about 1.6 Tesla at an F that is
too high.
In other words, we want Fsat to be low as possible at the 1kHz
clipping signal voltage.

If one designs for Fsat at below 20Hz then the size of a 5 watt rated
SE OPT can be 4 times heavier than some silly bean counter designed
crap from an old radio. Some such old radios or old hi-fi sets do have
just enough Lp and freedom from saturation at too high an F but they
should be carefully measured before use.
Most old radi OPT just have one sec section wound over the top of a
single primary which gives a poor HF response.

But even if Fsat s at say 50Hz for full Vo, for phone use the Vo may
only be 1/4 full Vo max and Fsat will then be 12.5Hz which s
acceptable.

Patrick Turner.

[Patrick Turner]

On Sep 30, 9:58*am, flipper wrote:
On Wed, 29 Sep 2010 23:00:49 +0100, Ian Bell
wrote:

I have been thinking a bit about Patrick's suggestion of using something like an EL84 in an SE
headphones amp. I was intrigued that he had built one using a 'cheap' output transformer so I have
looked at specs from various suppliers of such transformers. These transformers typically have a
primarily inductance of around 10 Henries (at 20 or 40mA) for a nominal 5K match. Now my transformer
rule of thumb is you want the primary inductance to be twice the plate load at the lowest frequency
of interest (say 20Hz). So for 5K ohms and 20Hz the inductance needs to be getting on for 40 Henries
so is not the bass response of these transformers going to suffer somewhat??

Cheers

Ian

The short answer is yes.

In practice it depends on design and what you mean by 'suffer'.

Open loop the OPT response is what it is. Closed loop is more
complicated because FB will attempt to keep the low end response up;
but saturation is saturation so it can't make full power bandwidth. On
the other hand, 'music' is not a full power monotone and 'average
program level', including bass, is usually at least 8-10dB down even
for compressed 'rock'. So if you put in enough NFB and can muster 1W
of bass from an otherwise 10 watt amp it will 'sound' about right.

The other 'suffer' is you use up the FB dBs in compensating for the
output drop so they're not there to reduce distortion. But, then, the
human ear is not terribly sensitive to low frequency distortion.

Well, the LF harmonic distortion is rarely the problem. But the
intermodulation distortion is a big problem if it is excessive and
caused by an OPT saturating and then all higher F will sound muddied
up.

Patrick Turner.



The net result is you can get bass response that 'sounds' surprisingly
better than what 'just specs' would suggest.

[Patrick Turner]

On Sep 30, 11:47*am, "Ian Iveson"
wrote:
"Ian Bell" wrote:

It likely depends on how close saturation is to 20Hz, *at
the signal level the OPT is likely to encounter*. If you're
willing to use feedback, and the signal is small, you may be
OK. An appropriate filter at the input should make
saturation impossible.

Patrick'll be along in a tick.

That Patrick fella might say "try for as much Lp as possible, and make
sure Fsat at 1kHz clipping level with load of 4x Ra or higher is below
20Hz. "

Sometimes this means you must buy a 10W OPT rated for 10k load and
-1dB at 20Hz - 20kHz.

The use of a genteel little tube like EL84 in triode, or a 6A3 should
give real music with no bad sonics.

Patrick Turner.




Twice the plate load? It *is* the anode load isn't it? 2 to
3 times ra, depending on whether you want power or fidelity
respectively, broadly speaking.

There is a tradition of not caring about head amps...as if
they are add-on extras to the main system. These days, they
quite often *are* the main system.

Ian

[Ian Bell[_2_]]

Patrick Turner wrote:
On Sep 30, 8:00 am, Ian wrote:
I have been thinking a bit about Patrick's suggestion of using something like an EL84 in an SE
headphones amp. I was intrigued that he had built one using a 'cheap' output transformer so I have
looked at specs from various suppliers of such transformers. These transformers typically have a
primarily inductance of around 10 Henries (at 20 or 40mA) for a nominal 5K match. Now my transformer
rule of thumb is you want the primary inductance to be twice the plate load at the lowest frequency
of interest (say 20Hz). So for 5K ohms and 20Hz the inductance needs to be getting on for 40 Henries
so is not the bass response of these transformers going to suffer somewhat??

Cheers

Ian

Most old radio OPS designed for a 6V6 operating as a beam tetrode with
no NFB have low Lp because they are designed to give from only 100Hz
upwards as the little speaker used just can't cope with any below
100Hz.

You are right about the 40H. If there is that much L and Ra was 2k2
with EL84 in triode and the RL was a lot higher or not even present,
then response will be -3dB when XLp = Ra or at 8.75Hz, so Lp could be
16H to give -3dB at 22Hz. FB can flatten the resonse further. The real
other problem is that if Lp is low, it is often because the core Afe
size is too small and there are not enough P turns to prevent core
saturation, ie, Bdc + Bac exceeding about 1.6 Tesla at an F that is
too high.
In other words, we want Fsat to be low as possible at the 1kHz
clipping signal voltage.

If one designs for Fsat at below 20Hz then the size of a 5 watt rated
SE OPT can be 4 times heavier than some silly bean counter designed
crap from an old radio. Some such old radios or old hi-fi sets do have
just enough Lp and freedom from saturation at too high an F but they
should be carefully measured before use.
Most old radi OPT just have one sec section wound over the top of a
single primary which gives a poor HF response.

But even if Fsat s at say 50Hz for full Vo, for phone use the Vo may
only be 1/4 full Vo max and Fsat will then be 12.5Hz which s
acceptable.

Which is an interesting point. Say we have a SET OP transformer with 10H and a 16 ohm secondary.
One watt into 16 ohms need 4V rms whcih is plenty to drive the phones I am intersted in. So suppose
I connect a 32 ohm headphone to this 16 ohm winding and it requires no more than 2V rms (for 125mW)
I would expect lower distortion because of the lower level and higher reflected ac load. What would
happen to the bass response with a 10H primary?

Cheers

Ian

Patrick Turner.

[Ian Bell[_2_]]

Patrick Turner wrote:
On Sep 30, 11:47 am, "Ian
wrote:
"Ian Bell" wrote:

It likely depends on how close saturation is to 20Hz, *at
the signal level the OPT is likely to encounter*. If you're
willing to use feedback, and the signal is small, you may be
OK. An appropriate filter at the input should make
saturation impossible.

Patrick'll be along in a tick.

That Patrick fella might say "try for as much Lp as possible, and make
sure Fsat at 1kHz clipping level with load of 4x Ra or higher is below
20Hz. "

Sometimes this means you must buy a 10W OPT rated for 10k load and
-1dB at 20Hz - 20kHz.

The use of a genteel little tube like EL84 in triode, or a 6A3 should
give real music with no bad sonics.

Patrick Turner.



Tha I understand but what intrigued me was your earlier post about how you got perfectly
satisfactory results from a small radio OPT. What sort of Lp would that have had?

Cheers

Ian


Twice the plate load? It *is* the anode load isn't it? 2 to
3 times ra, depending on whether you want power or fidelity
respectively, broadly speaking.

There is a tradition of not caring about head amps...as if
they are add-on extras to the main system. These days, they
quite often *are* the main system.

Ian

[Ian Iveson]

Ian Bell wrote:

It likely depends on how close saturation is to 20Hz, *at
the signal level the OPT is likely to encounter*. If
you're
willing to use feedback, and the signal is small, you may
be
OK. An appropriate filter at the input should make
saturation impossible.

One thing I was wondering was if I use a transformer with
a nominal 16 ohm secondary but use it with 32 ohm
headphones for example, then ra reflected into the
secondary would be so low relative to the load that the
output could be considered essentially a voltage source.
Equally the load reflected to the plate would be four
times higher so distortion would be lower?

Patrick'll be along in a tick.

Twice the plate load? It *is* the anode load isn't it? 2
to
3 times ra, depending on whether you want power or
fidelity
respectively, broadly speaking.

Yes, but you are thinking in power amp terms alone and as
I have said before I have no power amp experience. My rule
applies to low level transformers e.g. mic transformers
and for broadly similar reasons.

10H is in shunt with load. Lower load makes shunt more
significant. Try sim. Should shift 0 to higher F. Use fb to
restore...what then?

Ian, who's English has been wrecked by Twitter.

[Ian Bell[_2_]]

Ian Iveson wrote:
Ian Bell wrote:

It likely depends on how close saturation is to 20Hz, *at
the signal level the OPT is likely to encounter*. If
you're
willing to use feedback, and the signal is small, you may
be
OK. An appropriate filter at the input should make
saturation impossible.

One thing I was wondering was if I use a transformer with
a nominal 16 ohm secondary but use it with 32 ohm
headphones for example, then ra reflected into the
secondary would be so low relative to the load that the
output could be considered essentially a voltage source.
Equally the load reflected to the plate would be four
times higher so distortion would be lower?

Patrick'll be along in a tick.

Twice the plate load? It *is* the anode load isn't it? 2
to
3 times ra, depending on whether you want power or
fidelity
respectively, broadly speaking.

Yes, but you are thinking in power amp terms alone and as
I have said before I have no power amp experience. My rule
applies to low level transformers e.g. mic transformers
and for broadly similar reasons.

10H is in shunt with load. Lower load makes shunt more
significant.

Yeah but, no but, my load is higher not lower.

Cheers

Ian

Try sim. Should shift 0 to higher F. Use fb to
restore...what then?

Ian, who's English has been wrecked by Twitter.

[Ian Iveson]

One thing I was wondering was if I use a transformer
with
a nominal 16 ohm secondary but use it with 32 ohm
headphones for example, then ra reflected into the
secondary would be so low relative to the load that the
output could be considered essentially a voltage source.
Equally the load reflected to the plate would be four
times higher so distortion would be lower?


Language. Lower load = higher resistance.

If you follow my logic, my point is clear. The load
presented by the primary of a transformer comprises:

1. Primary winding resistance
2. Leakage inductance in series with 1
3. Reflected secondary winding resistance in series with 1
and 2
4. Reflected secondary load in series with 1, 2 and 3
5. Primary inductance in shunt with 3 and 4

Think.

Ian

[Ian Bell[_2_]]

Ian Iveson wrote:
One thing I was wondering was if I use a transformer
with
a nominal 16 ohm secondary but use it with 32 ohm
headphones for example, then ra reflected into the
secondary would be so low relative to the load that the
output could be considered essentially a voltage source.
Equally the load reflected to the plate would be four
times higher so distortion would be lower?


Language. Lower load = higher resistance.

If you follow my logic, my point is clear. The load
presented by the primary of a transformer comprises:

1. Primary winding resistance
2. Leakage inductance in series with 1
3. Reflected secondary winding resistance in series with 1
and 2
4. Reflected secondary load in series with 1, 2 and 3
5. Primary inductance in shunt with 3 and 4

Think.


Nope, maybe I am just thick. You said "10H is in shunt with load. Lower load makes shunt more
significant."

Now you are saying "Language. Lower load = higher resistance."

Which means "Higher resistance makes shunt more significant"

That means to drive a higher resistance load I need MORE inductance in the primary.

Does not make sense to me.

Cheers

Ian

Ian

[Ian Iveson]

flipper wrote:

If I understood his meaning he has the relative effect
backwards.

For a given shunt inductance a lower load has better LF
response
(shunt less significant).

This is the first time for many years that the meaning of
the term "load" in this context has been challenged.

Generally, I have avoided the issue and used the less-easily
misunderstood term "load resistance" or just "resistance".
Lately, there has been so little discussion by amateurs that
the jargon of commercial operators, or "professionals" as
they prefer to call themselves, has erected barriers against
incursion. Now the self-styled professionals are getting
sloppy.

An open circuit represents zero load, right? So a high load
resistance is a low load.

I'll try to use more accessible language in the future.

Ian

[Ian Iveson]

Ian Bell wrote:

One thing I was wondering was if I use a transformer
with
a nominal 16 ohm secondary but use it with 32 ohm
headphones for example, then ra reflected into the
secondary would be so low relative to the load that
the
output could be considered essentially a voltage
source.
Equally the load reflected to the plate would be four
times higher so distortion would be lower?


Language. Lower load = higher resistance.

If you follow my logic, my point is clear. The load
presented by the primary of a transformer comprises:

1. Primary winding resistance
2. Leakage inductance in series with 1
3. Reflected secondary winding resistance in series with
1
and 2
4. Reflected secondary load in series with 1, 2 and 3
5. Primary inductance in shunt with 3 and 4

Think.


Nope, maybe I am just thick. You said "10H is in shunt
with load. Lower load makes shunt more
significant."

Now you are saying "Language. Lower load = higher
resistance."

Which means "Higher resistance makes shunt more
significant"

That means to drive a higher resistance load I need MORE
inductance in the primary.

Does not make sense to me.


I can't see why not. Oh well. I suggested you simulate.
Perhaps flipper can explain better.

Ian

[Patrick Turner]

On Oct 1, 7:58*am, Ian Bell wrote:
Patrick Turner wrote:
On Sep 30, 8:00 am, Ian *wrote:
I have been thinking a bit about Patrick's suggestion of using something like an EL84 in an SE
headphones amp. I was intrigued that he had built one using a 'cheap' output transformer so I have
looked at specs from various suppliers of such transformers. These transformers typically have a
primarily inductance of around 10 Henries (at 20 or 40mA) for a nominal 5K match. Now my transformer
rule of thumb is you want the primary inductance to be twice the plate load at the lowest frequency
of interest (say 20Hz). So for 5K ohms and 20Hz the inductance needs to be getting on for 40 Henries
so is not the bass response of these transformers going to suffer somewhat??

Cheers

Ian

Most old radio OPS designed for a 6V6 operating as a beam tetrode with
no NFB have low Lp because they are designed to give from only 100Hz
upwards as the little speaker used just can't cope with any below
100Hz.

You are right about the 40H. If there is that much L and Ra was 2k2
with EL84 in triode and the RL was a lot higher or not even present,
then response will be -3dB when XLp = Ra or at 8.75Hz, so Lp could be
16H to give -3dB at 22Hz. FB can flatten the resonse further. The real
other problem is that if Lp is low, it is often because the core Afe
size is too small and there are not enough P turns to prevent core
saturation, ie, Bdc + Bac exceeding about 1.6 Tesla at an F that is
too high.
In other words, we want Fsat to be low as possible at the 1kHz
clipping signal voltage.

If one designs for Fsat at below 20Hz then the size of a 5 watt rated
SE OPT can be 4 times heavier than some silly bean counter designed
crap from an old radio. Some such old radios or old hi-fi sets do have
just enough Lp and freedom from saturation at too high an F but they
should be carefully measured before use.
Most old radi OPT just have one sec section *wound over the top of a
single primary which gives a poor HF response.

But even if Fsat s at say 50Hz for full Vo, for phone use the Vo may
only be 1/4 full Vo max and Fsat will then be 12.5Hz which s
acceptable.

Which is an interesting point. Say we have a SET OP transformer with 10H and a 16 ohm secondary.
One watt into 16 ohms need 4V rms whcih is plenty to drive the phones I am intersted in. So suppose
I connect a 32 ohm headphone to this 16 ohm winding and it requires no more than 2V rms (for 125mW)
I would expect lower distortion because of the lower level and higher reflected ac load. What would
happen to the bass response with a 10H primary?

Its all very well to consider a "SET OP transformer with 10H and a 16
ohm secondary" but you have not mentioned the primary load nor type of
device driving the primary load.

If you have a single EL84 in SE class A1 triode then consider a
typical op point of Ea = +300Vdc and Ia = 30mAdc, and Ra = 2k0 approx..
Total RL + Rw for maximum PO = ( Ea / Ia ) - ( 2 x Ra ) =
( 300/0.03 ) - 4k0 = 6k0 and let us say Rw total is 10% of the totalo
RL so actual RL = 6k0 = 600R = 5k4.
Max PO = RL x ( Idc squared ) / 2 = 2.43W, and anode load Va = 0.707 x
5,400 x 0.03 = 114.5Vrms.
Your OPT has a 5k4 : 16 ohm ratio which gives ZR = 337.5:1 so the Ra
appears at the output as 2k0 / 337.5, plus the total Rw / 337.5 =
5.96R + 1.77 = 7.73R.

This is a not very good outcome unless we add at least about 12dB GNFB
from speaker connection to the driver tube cathode to reduce Rout to
about 2 ohms, giving a DF of 8 with 16 ohm load.

The response at low levels is given as the pole formed by Lp in
parallel to RA, which is Ra + RL in parallel. RDH4 spells it out.
Anyway, assuming the load is resistive at low F where LF cut off
occurs, then RA = 6,000 // 2k0 = 1k5, and 10H has XL = 1k5 at 23.9Hz,
not too bad. If the load around cut off was very high the low triode
Ra manages to keep the source R low so the Fco remains low.

If the sec load is higher than 16 ohms, say 32 ohms or 320 ohms, the
triode Ra remians the main dominant source resistance and Fco will
remain fairly low. My example presumes anode RL with 16 ohm sec =
total of 6k0, and this is 3 x Ra and maybe you get 4% THD at 2 Watts.
With anode RL = high as possible, ie, CCS, then THD will be about 1%
at the maximum Vo but because there is no Ia change there is no PO, so
THD/IMD will vary between about 4% and maybe 2% at maximum *useful*
Vos for RL above 6k0.

But should you use ther EL84 in pentode with Ea = 250V and Ia = 40mA,
Ra = maybe 50k0 and then the Fco is MUCH higher and more dominated by
total RL which may be 0.9 x 250/0.04 = 5k6, if in fact there is a an R
load present at Fco. With 10H and RA = approx 5k0, Fco = 79.6Hz.
The lower the Rsource the lower is the distortions caused by the damn
iron.
EL84 THD in SE mode at near clipping without any NFB anywhere is
usually well over 10%, and fulla odd order H.

Triodes are the gold standard for driving OPTs but one may completely
overcome the shortcomings of pentodes by using NFB in the form of
global, or local CFB or maybe UL plus global.


When I made a pair od SET amps for a customer a couple of years ago I
used 2 x 845 in parallel to get PO max = 60W.
The noise was only barely discernible with head held against a
speaker, and noise was estimated at 0.25mV.This illustrates that one
may use any power tube one wishes to use for either driving a speaker
*or phones*. With a luxury of a 60W output max to 5 ohms, or 17.3Vrms,
one might only want say 3Vrms max at the phones of 32 ohms which is
280mW. The R divider might have R across phones output = 2R2, and feed
R of 10R, so that noise is then reduced to 0.25 x ( 2.2/12.2 )
0.045mV, and if the average phones level was 0.2Vrms, then unweighted
SNR = 20 log 0.000045/0.2 which I guess = -72dB approx and possibly
acceptable, because the background noise of the recording venue is
likely to be higher if nothing else is.

So a single EL34 for phones would be OK, or a 1/2 of 6AS7, and so on,
providing you think it through. PP use of 6SN7 was common with a 20k:
16 OPT.

If one wants to drive down to 8 ohms without an OPT, I'd suggest a
single power mosfet in source follower mode which may be driven by CR
coupling from EL84 in triode. The triode needs a B+ of say +350V and a
CCS feed to anode so that Ea = 250V and Ia = 25mAdc. The EL84 will be
**extremely linear** while producing only up to say 3V into its very
high R loading which is merely the biasing resistors of the mosfet
gates.
Diode clamps would be needed to prevent excessive drive voltage
appearing across Vgate-source. The mosfet would have Ed at +20Vdc and
Idc = 0.5A, and a CCS sink taken to -20vdc, with load connected via
4,700uF.
But better would be to use NPN and PNP mosfets in a complementary pair
and biased for class A1 between the +/- 20Vdc rails.
Direct coupling to a load is then possible if DC offset is controlled.
The Pdd of each mosfet = 10W, and somewhat high to drive phones - 20W
total per channel.
But you can get about 9W of max PO into 18 ohms, 12.7Vrms pure class A
with THD approx 1%. But the noise problem may still be there and a
resistance divider needed. No phones should ever require more than say
1V, so R divider can be 22R + 2R2 is OK and probably class AB action
is OK so Id may be reduced to 250mAdc.
Rout from mosfet followers is about 2 ohms without any NFB, so the
Rout is about determined by the R divider lower resistance.
At 0.2Vrms from the divider, there may be 2V from the mosfets where
THD may be as low as 0.17%. Reducing the THD further means using
global loop NFB and more tube gain. The complementary pair of mosfets
will produce far more undistorted Vo than will one mosfet with
CCS.

Since an EL84 has a gain = approx 20 with CCS loading, then gain with
R divider will be slightly above unity, and about right for use if the
source is a CD player.

One's imagination may easily be employed to make a headphone amp much
simpler and better than the dreadful range of insipid bean counter
inspired designs currently available at excessive prices in hi-fi
stores.

Patrick Turner.



Cheers

Ian





Patrick Turner.- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

[Patrick Turner]

On Oct 1, 8:00*am, Ian Bell wrote:
Patrick Turner wrote:
On Sep 30, 11:47 am, "Ian
wrote:
"Ian Bell" wrote:

It likely depends on how close saturation is to 20Hz, *at
the signal level the OPT is likely to encounter*. If you're
willing to use feedback, and the signal is small, you may be
OK. An appropriate filter at the input should make
saturation impossible.

Patrick'll be along in a tick.

That Patrick fella might say "try for as much Lp as possible, and make
sure Fsat at 1kHz clipping level with load of 4x Ra or higher is below
20Hz. "

Sometimes this means you must buy a 10W OPT rated for 10k load and
-1dB at 20Hz - 20kHz.

The use of a genteel little tube like EL84 in triode, or a 6A3 should
give real music with no bad sonics.

Patrick Turner.

Tha I understand but what intrigued me was your earlier post about how you got perfectly
satisfactory results from a small radio OPT. What sort of Lp would that have had?

The "radio OPT" were from a Trio AM/FM stereo receiver, somewhat a
reasonable quality radio from about 1958, and the OPTs did have
sufficient Lp and low enough LL to give a wide enough BW. The Trio had
SE EL84 in pentode with 6AU6 drivers with 20dB NFB. I removed all the
audio circuits from the receiver to add tubes dedicated towards making
the darn set a pure AM/FM tuner of largely my own design.
The EL84 could get 4W at the OPT sec into 4,8or16R, so 8Vrms was
available to 16R, but this drops if triode mode is used, but it still
worked out OK, and I used global NFB.

Now many mantle radio sets with a single 6V6 output won't have a very
good OPT. The average radio audio circuit with 6V6 is **designed to
make 2W max** from the anode and OPT winding losses with very thin
wire might be 25% so expect 1.5W at clipping and BW 70Hz ( Fsat ) to
10kHz, -3dB. The typical LS has Z = 4 ohms, and OPT ratio might be 7k:
4.

Now 1.5W to 4 ohms = 2.44Vrms, and should phones only need 1V approx,
then Fsat might be at 30Hz.
NFB can straighten out the mess, if one obeys all the rules about
critical damping to achieve stability with low Lp and high LL.

The thing to be remembered is that a mantle radio set produces
terrible sound if turned up loud. It causes domestic bliss to cease,
and rolling pins, plates, and cooking pots to fly across the nation's
kitchens and dining rooms.

I recall an old movie where a TV set which must have had a single 6BM8
as the audio amp was flung out through a window of an appartment on
the 4th floor. Damn thing! Silence! Peace at last!

But at extremly low levels with head against the speaker, the THD/IMD
becomes quite low even with beam tetrode operation and even without a
stitch of NFB.

Patrick Turner.




Cheers

Ian





Twice the plate load? It *is* the anode load isn't it? 2 to
3 times ra, depending on whether you want power or fidelity
respectively, broadly speaking.

There is a tradition of not caring about head amps...as if
they are add-on extras to the main system. These days, they
quite often *are* the main system.

Ian- Hide quoted text -

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[Patrick Turner]

On Oct 2, 11:34*am, flipper wrote:
On Fri, 01 Oct 2010 23:51:42 +0100, Ian Bell
wrote:





Ian Iveson wrote:
Ian Bell wrote:

It likely depends on how close saturation is to 20Hz, *at
the signal level the OPT is likely to encounter*. If
you're
willing to use feedback, and the signal is small, you may
be
OK. An appropriate filter at the input should make
saturation impossible.

One thing I was wondering was if I use a transformer with
a nominal 16 ohm secondary but use it with 32 ohm
headphones for example, then ra reflected into the
secondary would be so low relative to the load that *the
output could be considered essentially a voltage source.
Equally the load reflected to the plate would be four
times higher so distortion would be lower?

Patrick'll be along in a tick.

Twice the plate load? It *is* the anode load isn't it? 2
to
3 times ra, depending on whether you want power or
fidelity
respectively, broadly speaking.

Yes, but you are thinking in power amp terms alone and as
I have said before I have no power amp experience. My rule
applies to low level transformers e.g. mic transformers
and for broadly similar reasons.

10H is in shunt with load. Lower load makes shunt more
significant.

Yeah but, no but, my load is higher not lower.

If I understood his meaning he has the relative effect backwards.

For a given shunt inductance a lower load has better LF response
(shunt less significant).

E.g. At 20Hz 10H is roughly 1250 ohms and, relative to the desired
load, 1250 in parallel with 10k (twice the nominal 5k load) is clearly
worse than in parallel with 5k, which is worse than in parallel with
2.5k (half the nominal 5k load).

Thing is, you can't usually accomplish that by simply buying a lower
reflected load transformer because the inductance will be
correspondingly less since the LF frequency response is usually the
same for the same 'quality level' OPT.

What you could do is get a 10k OPT, with likely a 20H primary, and
then, for example, put an 8 ohm load on the 16 ohm secondary, to halve
the load back to 5k, but leakage inductance correspondingly lowers HF
response (the no 'free lunch' axiom).

Unfortunately, with headphone impedances you're working in the other
direction, which aggravates the LF end.

Yes, but the Rsource in parallel with RL and Lp determine the Fco.

if the tube is low Ra, say EL34 in triode then Ra = say 1,300 ohms and
even with no load there at all the Fco is 1,300 / ( 10H x 6.28 ) =
20.7Hz.

An OPT made for a 5k RL may have Lp = 10H and little consideration has
been made for Rsource which might be assumed to be a high Ra pentode.
But also assumed is that NFB will be used.

Now 5k plus 10H gives a load = 3.5k at 80Hz.

So if you had a tube just cabable of driving 5k at 1kHz to say 5W,
then at 80Hz the load has dropped to 3.5k and the output appears
distorted because of the load reduction. It may be possible the OPT is
also saturating.

Suppose you had an OPT meant for 10k, then it'd be likely its LP would
be 20H, and the load will fall to 7k at 80Hz.
The same tube driving this OPT with 10k load will easily produce the
same voltage as for 5W into 5k and the load won't fall to cause
overloading at 80Hz. Such an OPT may be loaded to reflect an anode
load of 5k and the 20H gives a load reduction to 3k5 at 40Hz which is
better.

But loading of the 10k OPT may not reflect low load at all to the
anode and the response with pentode becomes a big arch because gain =
RL x gm.
Triode connection flattens the arch and so does NFB. if the OPT is
rated for 10W to 10k, its Fsat will be for a higher Va than for
a 5W OPT for 5k, so Fsat will be lower with the bigger OPT.

But the winding losses with phones may be negligible, and if P load =
10k for 4,8,16 ohms the the phones might be directly connected to the
4 ohm outlet wih no R divider needed because the OPT ratio 10k:4R =
50:1, and a good enough Vo reduction to lower noise.

I may have said an R divider is a useful thing to reduce noise for
headphones and many amps have such an arrangement. But having low
enough taps on the OPT do the same thing providing that the LL does
not increase so much that HF is cut because such a small portion of
the secondary is used and coupled to the primary. No tube amps I have
seen have a dedicated low Vo tap for headphones. Two resistances are
usually cheaper for the manufacturer.

It is crucial that OPT used for headphone amps and the PT used on the
same chassis be potted lest stray magnetic coupling spoil the outcome.
I recall despairing while trying to reduce hum noise on a MingDa
phones amp. The guys at MingDa also were worried because they tried to
mount flat metal plates between PT and OPT but this crummy fix was
ineffective and what was needed was total re-design of layout and
decent magnetic shielding with pots. GlowOne phone amps had similar
problems. The applied science of making really quiet phones amps with
tubes is still something to be learnt in parts of China.

Patrick Turner.



Cheers

Ian

*Try sim. Should shift 0 to higher F. Use fb to
restore...what then?

Ian, who's English has been wrecked by Twitter.- Hide quoted text -

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[Ian Bell[_2_]]

flipper wrote:
On Sat, 02 Oct 2010 11:03:10 +0100, Ian
wrote:

Ian Iveson wrote:
One thing I was wondering was if I use a transformer
with
a nominal 16 ohm secondary but use it with 32 ohm
headphones for example, then ra reflected into the
secondary would be so low relative to the load that the
output could be considered essentially a voltage source.
Equally the load reflected to the plate would be four
times higher so distortion would be lower?


Language. Lower load = higher resistance.

If you follow my logic, my point is clear. The load
presented by the primary of a transformer comprises:

1. Primary winding resistance
2. Leakage inductance in series with 1
3. Reflected secondary winding resistance in series with 1
and 2
4. Reflected secondary load in series with 1, 2 and 3
5. Primary inductance in shunt with 3 and 4

Think.


Nope, maybe I am just thick. You said "10H is in shunt with load. Lower load makes shunt more
significant."

Now you are saying "Language. Lower load = higher resistance."

Which means "Higher resistance makes shunt more significant"

That means to drive a higher resistance load I need MORE inductance in the primary.

Does not make sense to me.

It's because the inductance is in parallel with the load impedance.

With a 5k (or any other value) OPT you'd like it to be that same
impedance at all frequencies but it won't because the inductance
impedance, which is in parallel with the load impedance, falls in
value as the frequency decreases, so the parallel value also falls.

Let's pick a convenient reference point: effective impedance half of
nominal. That's going to occur at a frequency where the 10 Henry
impedance equals the OPT nominal impedance, I.E. 5k. If the nominal
OPT impedance is 10k then that (parallel impedance half of nominal) is
going to happen at a higher frequency.

Now, that was presuming the EL84 in pentode mode where the tube's
anode impedance is high and, so, not much of a factor. See Patrick's
post for triode mode.


Yes, you are of course right. I have a stinking headache today and I just cannot think straight.
Bigger L means higher Zl or lower load - duh??

Cheers

ian

Cheers

Ian

Ian

[Ian Bell[_2_]]

Ian Iveson wrote:
flipper wrote:

If I understood his meaning he has the relative effect
backwards.

For a given shunt inductance a lower load has better LF
response
(shunt less significant).

This is the first time for many years that the meaning of
the term "load" in this context has been challenged.

Generally, I have avoided the issue and used the less-easily
misunderstood term "load resistance" or just "resistance".
Lately, there has been so little discussion by amateurs that
the jargon of commercial operators, or "professionals" as
they prefer to call themselves, has erected barriers against
incursion. Now the self-styled professionals are getting
sloppy.

An open circuit represents zero load, right? So a high load
resistance is a low load.

Moot. An open circuit implies 'no' load which could equally imply infinite resistance.

Cheers

ian

I'll try to use more accessible language in the future.

Ian

[Patrick Turner]

On Oct 2, 11:09*pm, Ian Bell wrote:
Ian Iveson wrote:
flipper wrote:

If I understood his meaning he has the relative effect
backwards.

For a given shunt inductance a lower load has better LF
response
(shunt less significant).

This is the first time for many years that the meaning of
the term "load" in this context has been challenged.

Generally, I have avoided the issue and used the less-easily
misunderstood term "load resistance" or just "resistance".
Lately, there has been so little discussion by amateurs that
the jargon of commercial operators, or "professionals" as
they prefer to call themselves, has erected barriers against
incursion. Now the self-styled professionals are getting
sloppy.

An open circuit represents zero load, right? So a high load
resistance is a low load.

Moot. An open circuit implies 'no' load which could equally imply infinite resistance.

This is basic stuff.

An open circuit sure is NO LOAD PRESENT but load ohms is always the
result of an equation and always = V / I and if I = 0.0 Amps then the
load = V / 0 = infinite number of ohms = very high resistance value in
ohms.

A low load is always assumed to be a low ohm value load, usually in
comparison to a source resistance so if Rsource = 100 ohms then a load
of 10 ohms is a low load, and it is never a high load resistance. The
context helps determine the meaning.
A load of 1,000ohms is a high value load resistance - in comparision
to the Rsource of 100 ohms.

Patrick Turner.

Cheers

ian





I'll try to use more accessible language in the future.

Ian- Hide quoted text -

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[Patrick Turner]

On Oct 3, 10:35*am, flipper wrote:
On Sat, 2 Oct 2010 14:06:40 -0700 (PDT), Patrick Turner





wrote:
On Oct 2, 11:09*pm, Ian Bell wrote:
Ian Iveson wrote:
flipper wrote:

If I understood his meaning he has the relative effect
backwards.

For a given shunt inductance a lower load has better LF
response
(shunt less significant).

This is the first time for many years that the meaning of
the term "load" in this context has been challenged.

Generally, I have avoided the issue and used the less-easily
misunderstood term "load resistance" or just "resistance".
Lately, there has been so little discussion by amateurs that
the jargon of commercial operators, or "professionals" as
they prefer to call themselves, has erected barriers against
incursion. Now the self-styled professionals are getting
sloppy.

An open circuit represents zero load, right? So a high load
resistance is a low load.

Moot. An open circuit implies 'no' load which could equally imply infinite resistance.

This is basic stuff.

An open circuit sure is NO LOAD PRESENT but load ohms is always the
result of an equation and always = V / I and if I = 0.0 Amps then the
load = V / 0 = infinite number of ohms = very high resistance value in
ohms.

A low load is always assumed to be a low ohm value load, usually in
comparison to a source resistance so if Rsource = 100 ohms then a load
of 10 ohms is a low load, and it is never a high load resistance. The
context helps determine the meaning.
A load of 1,000ohms is a high value load resistance - in comparision
to the Rsource of 100 ohms.

With all due respect I dispute your contention that 'low load' is
"always" assumed to be an 'ohm value' and Iveson has a point that it's
context sensitive. For example, if one is clearly talking about power
then a 'low load' could be referring to 'low power' and, so, perhaps a
'high impedance'.

However, I also dispute his argument about "in this context" because
the issue at hand was an inductance, specifically it's impedance at
frequency, in parallel with 'the load' and I don't think of inductor
(or capacitor, or resistor, etc.) impedance in parallel with 'power'
but with other impedances (which may be why you consider it 'always'
ohms).

I'm sure it was blindingly obvious to his own mind's eye but we don't
have that 'context' and it's just a shame he decided to get all
indignant even though I specifically qualified with "If I understood
his meaning." I apparently did *not* 'understand his meaning' but, in
my own defense, that's because I'm not a mind reader.

Hmm, when we have an inductance involved as a load then you have its
varying REACTANCE which is measured in Ohms for a specific frequency
to be mixed up with context.

My mind likes to operate with serious engineering terms which have
evolved over many years and which don't include the vaguities of the
multitudes who like to talk in riddles, can't say what they mean and
don't mean what they say, and swan through life in a fog of brevity.

Thus I can't be popular with everyone, but I mean well.

Anything said abour amplifiers should be unambiguous, precise,
factual, and lead to better tube craft and the understanding of it all
and should not tempt the God Of Triodes to zap anyone, especially on a
religious day, a Sunday.

Now including the word "should" has probably made a little army of ppl
very upset.......

I had a nice bicycle ride this am but it left me rather tired and the
Typical Experience happened - I am able to gradually catch up to the
25 year olds in front of me along flat roads, but up hills they ride
away from me. Alas being 63 is a PIA, and teaches me that it is
impossible to maintain a low fat content % of my body which would then
enable me to keep up on hills. So it is with intellectual efforts; as
we age we gather fat in our thinking and sluggishness in our
incisiveness when examining some favourite subject, so we tend to
waffle on, and be in-precise, which I guess I am doing right now.

But I can laugh at all of me at least, both my antics on the road, and
in a discussion group, and I suggest all of you try to be un-serious
for a greater % of the time. Un-worry!

But I did catch up and pass one 25 yr who was passed by the other pair
and after a lot of effort. This indicates that anyone much older than
someone much younger has a relevant place in the tapestry of life, and
ageism is not a happy attitude.

I have to marvel at those with brighter minds than myself, and those
who can ride away from me.

Patrick Turner.




Patrick Turner.

Cheers

ian

I'll try to use more accessible language in the future.

Ian- Hide quoted text -

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[Ian Bell[_2_]]

Patrick Turner wrote:
On Oct 1, 7:58 am, Ian wrote:
Patrick Turner wrote:
On Sep 30, 8:00 am, Ian wrote:
I have been thinking a bit about Patrick's suggestion of using something like an EL84 in an SE
headphones amp. I was intrigued that he had built one using a 'cheap' output transformer so I have
looked at specs from various suppliers of such transformers. These transformers typically have a
primarily inductance of around 10 Henries (at 20 or 40mA) for a nominal 5K match. Now my transformer
rule of thumb is you want the primary inductance to be twice the plate load at the lowest frequency
of interest (say 20Hz). So for 5K ohms and 20Hz the inductance needs to be getting on for 40 Henries
so is not the bass response of these transformers going to suffer somewhat??

Cheers

Ian

Most old radio OPS designed for a 6V6 operating as a beam tetrode with
no NFB have low Lp because they are designed to give from only 100Hz
upwards as the little speaker used just can't cope with any below
100Hz.

You are right about the 40H. If there is that much L and Ra was 2k2
with EL84 in triode and the RL was a lot higher or not even present,
then response will be -3dB when XLp = Ra or at 8.75Hz, so Lp could be
16H to give -3dB at 22Hz. FB can flatten the resonse further. The real
other problem is that if Lp is low, it is often because the core Afe
size is too small and there are not enough P turns to prevent core
saturation, ie, Bdc + Bac exceeding about 1.6 Tesla at an F that is
too high.
In other words, we want Fsat to be low as possible at the 1kHz
clipping signal voltage.

If one designs for Fsat at below 20Hz then the size of a 5 watt rated
SE OPT can be 4 times heavier than some silly bean counter designed
crap from an old radio. Some such old radios or old hi-fi sets do have
just enough Lp and freedom from saturation at too high an F but they
should be carefully measured before use.
Most old radi OPT just have one sec section wound over the top of a
single primary which gives a poor HF response.

But even if Fsat s at say 50Hz for full Vo, for phone use the Vo may
only be 1/4 full Vo max and Fsat will then be 12.5Hz which s
acceptable.

Which is an interesting point. Say we have a SET OP transformer with 10H and a 16 ohm secondary.
One watt into 16 ohms need 4V rms whcih is plenty to drive the phones I am intersted in. So suppose
I connect a 32 ohm headphone to this 16 ohm winding and it requires no more than 2V rms (for 125mW)
I would expect lower distortion because of the lower level and higher reflected ac load. What would
happen to the bass response with a 10H primary?

Its all very well to consider a "SET OP transformer with 10H and a 16
ohm secondary" but you have not mentioned the primary load nor type of
device driving the primary load.

If you have a single EL84 in SE class A1 triode then consider a
typical op point of Ea = +300Vdc and Ia = 30mAdc, and Ra = 2k0 approx.
Total RL + Rw for maximum PO = ( Ea / Ia ) - ( 2 x Ra ) =
( 300/0.03 ) - 4k0 = 6k0 and let us say Rw total is 10% of the totalo
RL so actual RL = 6k0 = 600R = 5k4.
Max PO = RL x ( Idc squared ) / 2 = 2.43W, and anode load Va = 0.707 x
5,400 x 0.03 = 114.5Vrms.
Your OPT has a 5k4 : 16 ohm ratio which gives ZR = 337.5:1 so the Ra
appears at the output as 2k0 / 337.5, plus the total Rw / 337.5 =
5.96R + 1.77 = 7.73R.

This is a not very good outcome unless we add at least about 12dB GNFB
from speaker connection to the driver tube cathode to reduce Rout to
about 2 ohms, giving a DF of 8 with 16 ohm load.

The response at low levels is given as the pole formed by Lp in
parallel to RA, which is Ra + RL in parallel. RDH4 spells it out.
Anyway, assuming the load is resistive at low F where LF cut off
occurs, then RA = 6,000 // 2k0 = 1k5, and 10H has XL = 1k5 at 23.9Hz,
not too bad. If the load around cut off was very high the low triode
Ra manages to keep the source R low so the Fco remains low.

If the sec load is higher than 16 ohms, say 32 ohms or 320 ohms, the
triode Ra remians the main dominant source resistance and Fco will
remain fairly low. My example presumes anode RL with 16 ohm sec =
total of 6k0, and this is 3 x Ra and maybe you get 4% THD at 2 Watts.
With anode RL = high as possible, ie, CCS, then THD will be about 1%
at the maximum Vo but because there is no Ia change there is no PO, so
THD/IMD will vary between about 4% and maybe 2% at maximum *useful*
Vos for RL above 6k0.

But should you use ther EL84 in pentode with Ea = 250V and Ia = 40mA,
Ra = maybe 50k0 and then the Fco is MUCH higher and more dominated by
total RL which may be 0.9 x 250/0.04 = 5k6, if in fact there is a an R
load present at Fco. With 10H and RA = approx 5k0, Fco = 79.6Hz.
The lower the Rsource the lower is the distortions caused by the damn
iron.
EL84 THD in SE mode at near clipping without any NFB anywhere is
usually well over 10%, and fulla odd order H.

Triodes are the gold standard for driving OPTs but one may completely
overcome the shortcomings of pentodes by using NFB in the form of
global, or local CFB or maybe UL plus global.


When I made a pair od SET amps for a customer a couple of years ago I
used 2 x 845 in parallel to get PO max = 60W.
The noise was only barely discernible with head held against a
speaker, and noise was estimated at 0.25mV.This illustrates that one
may use any power tube one wishes to use for either driving a speaker
*or phones*. With a luxury of a 60W output max to 5 ohms, or 17.3Vrms,
one might only want say 3Vrms max at the phones of 32 ohms which is
280mW. The R divider might have R across phones output = 2R2, and feed
R of 10R, so that noise is then reduced to 0.25 x ( 2.2/12.2 )
0.045mV, and if the average phones level was 0.2Vrms, then unweighted
SNR = 20 log 0.000045/0.2 which I guess = -72dB approx and possibly
acceptable, because the background noise of the recording venue is
likely to be higher if nothing else is.

So a single EL34 for phones would be OK, or a 1/2 of 6AS7, and so on,
providing you think it through. PP use of 6SN7 was common with a 20k:
16 OPT.

If one wants to drive down to 8 ohms without an OPT, I'd suggest a
single power mosfet in source follower mode which may be driven by CR
coupling from EL84 in triode. The triode needs a B+ of say +350V and a
CCS feed to anode so that Ea = 250V and Ia = 25mAdc. The EL84 will be
**extremely linear** while producing only up to say 3V into its very
high R loading which is merely the biasing resistors of the mosfet
gates.
Diode clamps would be needed to prevent excessive drive voltage
appearing across Vgate-source. The mosfet would have Ed at +20Vdc and
Idc = 0.5A, and a CCS sink taken to -20vdc, with load connected via
4,700uF.
But better would be to use NPN and PNP mosfets in a complementary pair
and biased for class A1 between the +/- 20Vdc rails.
Direct coupling to a load is then possible if DC offset is controlled.
The Pdd of each mosfet = 10W, and somewhat high to drive phones - 20W
total per channel.
But you can get about 9W of max PO into 18 ohms, 12.7Vrms pure class A
with THD approx 1%. But the noise problem may still be there and a
resistance divider needed. No phones should ever require more than say
1V, so R divider can be 22R + 2R2 is OK and probably class AB action
is OK so Id may be reduced to 250mAdc.
Rout from mosfet followers is about 2 ohms without any NFB, so the
Rout is about determined by the R divider lower resistance.
At 0.2Vrms from the divider, there may be 2V from the mosfets where
THD may be as low as 0.17%. Reducing the THD further means using
global loop NFB and more tube gain. The complementary pair of mosfets
will produce far more undistorted Vo than will one mosfet with
CCS.

Since an EL84 has a gain = approx 20 with CCS loading, then gain with
R divider will be slightly above unity, and about right for use if the
source is a CD player.

One's imagination may easily be employed to make a headphone amp much
simpler and better than the dreadful range of insipid bean counter
inspired designs currently available at excessive prices in hi-fi
stores.

Patrick Turner.



Cheers

Ian





Patrick Turner.- Hide quoted text -

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A stunning post hat is just the answer I was looking for Patrick. I have saved that for detailed
digestion along with a cup of tea and RDH4.

Just one small point about damping factor. I am not clear how important this is for driving
headphones. You see many designs with a series output resistor of the same order as the headphone
load they are driving which means a damping factor of 1 or less. Must they therfrtoe sound awful or
is damping factor not sn issue with headphones.

Cheers

Ian

[Ian Iveson]

Not indignant. The mild invective was purely for amusement,
and offered, gratis, for you to share.

I'm sure if you check back through posts here you will find
that a high load has always been a low impedance, and a low
load is a high impedance.

You appear to have ignored the incontrovertible truth that
an open circuit is a low load.

Consider also "lightly loaded".

Consider also *looking it up*.

It's me that used the term, and I have clarified what I
meant by it. End of story. Time for you all to stop
whinging.

Ian

"flipper" wrote in message
...
On Sat, 2 Oct 2010 14:06:40 -0700 (PDT), Patrick Turner
wrote:

On Oct 2, 11:09 pm, Ian Bell
wrote:
Ian Iveson wrote:
flipper wrote:

If I understood his meaning he has the relative
effect
backwards.

For a given shunt inductance a lower load has better
LF
response
(shunt less significant).

This is the first time for many years that the meaning
of
the term "load" in this context has been challenged.

Generally, I have avoided the issue and used the
less-easily
misunderstood term "load resistance" or just
"resistance".
Lately, there has been so little discussion by
amateurs that
the jargon of commercial operators, or "professionals"
as
they prefer to call themselves, has erected barriers
against
incursion. Now the self-styled professionals are
getting
sloppy.

An open circuit represents zero load, right? So a high
load
resistance is a low load.

Moot. An open circuit implies 'no' load which could
equally imply infinite resistance.

This is basic stuff.

An open circuit sure is NO LOAD PRESENT but load ohms is
always the
result of an equation and always = V / I and if I = 0.0
Amps then the
load = V / 0 = infinite number of ohms = very high
resistance value in
ohms.

A low load is always assumed to be a low ohm value load,
usually in
comparison to a source resistance so if Rsource = 100 ohms
then a load
of 10 ohms is a low load, and it is never a high load
resistance. The
context helps determine the meaning.
A load of 1,000ohms is a high value load resistance - in
comparision
to the Rsource of 100 ohms.

With all due respect I dispute your contention that 'low
load' is
"always" assumed to be an 'ohm value' and Iveson has a
point that it's
context sensitive. For example, if one is clearly talking
about power
then a 'low load' could be referring to 'low power' and,
so, perhaps a
'high impedance'.

However, I also dispute his argument about "in this
context" because
the issue at hand was an inductance, specifically it's
impedance at
frequency, in parallel with 'the load' and I don't think
of inductor
(or capacitor, or resistor, etc.) impedance in parallel
with 'power'
but with other impedances (which may be why you consider
it 'always'
ohms).

I'm sure it was blindingly obvious to his own mind's eye
but we don't
have that 'context' and it's just a shame he decided to
get all
indignant even though I specifically qualified with "If I
understood
his meaning." I apparently did *not* 'understand his
meaning' but, in
my own defense, that's because I'm not a mind reader.


Patrick Turner.

Cheers

ian





I'll try to use more accessible language in the
future.

Ian- Hide quoted text -

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[Ian Bell[_2_]]

Just wanted to add a related question. In looking up SE transformers on the net many that quote
inductance do so at a specified dc current. So the difference between a 5W and a 10W transformer
might typically be the 5W has a 10H primary inductance measure at a dc current of 48mA whereas a 10W
type would have an inductance of 15H at a dc current of 64mA.

So my question is, how does the inductance vary with (dc) current. If it measures higher with lower
dc current then if I use a 10W instead of a 5W but with a lower dc current, do I get even higher
inductance because of the lower current?

Cheers

Ian

[Ian Iveson]

So my question is, how does the inductance vary with (dc)
current. If it measures higher with lower dc current then
if I use a 10W instead of a 5W but with a lower dc
current, do I get even higher inductance because of the
lower current?

This isn't a simple question, IIRC from longago posts. I
venture the following expecting a tongue-lashing.

The inductance varies with the slope of the BH curve. If you
look at one common, DC depiction of that curve, it appears
as a sigmoid, shallow at the zero crossing because the
magnetic domains exhibit what you might call stiction,
becoming steeper, and then shallow again as the majority of
domains reach their elastic limit, and some reach the yield
point.

The yielding dissipates power, resulting in hysteresis.
Consequently, for AC, the sigmoid becomes a loop, steepest
at the zero crossing, and shallow at both extremes of its
variation. What's more, any AC signal will generate a
sigmoid, with the magnetic bias affecting how flat or steep
the sigma-shaped loop is, according to where its centre is
on to the DC BH curve.

Break for ridicule.

Now, it seems from the DC curve that inductance should be
zero as B crosses the H axis. Much nonsense arises from this
interpretation. In fact, for pure AC the curve is steepest
there, both on the upward and downward journey.

However the sigmoid is flattish for small AC signals, so the
mean inductance is lowish. Then it gets steeper as the
signal increases, so the mean inductance rises, and then
begins to flatten out at its extremes and the inductance
falls again, as the core saturation approaches.

Still with me?

Add DC and small signals see a higher inductance, but the
average inductance for larger signals doesn't increase very
much, especially if the signal results in the core getting
to the sharper curve approaching saturation.

I would have thought that the great benefit of using a
transformer rated for high DC current, assuming the same
rated inductance, would be that it gives you greater
separation between roll-off and saturation. That means you
can use more feedback to extend its bass range at a given
max signal. Alternatively, some designs bias the SE
transformer and deliberately oversize it, to avoid the bass
disappearing at small signal levels. So they say, I think.

So, the answer is: yes and no.

Ian


"Ian Bell" wrote in message
...
Just wanted to add a related question. In looking up SE
transformers on the net many that quote inductance do so
at a specified dc current. So the difference between a 5W
and a 10W transformer might typically be the 5W has a 10H
primary inductance measure at a dc current of 48mA whereas
a 10W type would have an inductance of 15H at a dc current
of 64mA.


Cheers

Ian

[Ian Iveson]

The dialectic'll see us through.

"flipper" wrote in message
...
On Sun, 3 Oct 2010 20:56:34 +0100, "Ian Iveson"
wrote:

Not indignant. The mild invective was purely for
amusement,
and offered, gratis, for you to share.

Another 'context' problem I guess.

I'm sure if you check back through posts here you will
find
that a high load has always been a low impedance, and a
low
load is a high impedance.

I suppose you mean that for Patrick since it is he who
insists the
exact opposite is 'always' the case.

You appear to have ignored the incontrovertible truth that
an open circuit is a low load.

I 'ignored' it because your attitude didn't seem worth the
bother but
since you insist.

No one, well, except Patrick, disputed the explanation of
what you had
'meant'. The problem was the original ambiguity and things
being
'clear' after an explanation does not mean they were clear
before the
explanation.

Perhaps 'twitter speak' wasn't the best choice.

Consider also "lightly loaded".

Consider also *looking it up*.

It's me that used the term, and I have clarified what I
meant by it. End of story. Time for you all to stop
whinging.

You're the one whining about it but I suppose that's
simply more
'amusement'.

Hey, if you want a real 'amusement', you claimed 'twitter'
had ruined
your English and then argued it was perfectly clear and
obvious.
That's kinda funny, although I think the even better
amusement is you
deciding to jump on me again in a message where I defended
your
position.



Ian

"flipper" wrote in message
. ..
On Sat, 2 Oct 2010 14:06:40 -0700 (PDT), Patrick Turner
wrote:

On Oct 2, 11:09 pm, Ian Bell
wrote:
Ian Iveson wrote:
flipper wrote:

If I understood his meaning he has the relative
effect
backwards.

For a given shunt inductance a lower load has
better
LF
response
(shunt less significant).

This is the first time for many years that the
meaning
of
the term "load" in this context has been challenged.

Generally, I have avoided the issue and used the
less-easily
misunderstood term "load resistance" or just
"resistance".
Lately, there has been so little discussion by
amateurs that
the jargon of commercial operators, or
"professionals"
as
they prefer to call themselves, has erected barriers
against
incursion. Now the self-styled professionals are
getting
sloppy.

An open circuit represents zero load, right? So a
high
load
resistance is a low load.

Moot. An open circuit implies 'no' load which could
equally imply infinite resistance.

This is basic stuff.

An open circuit sure is NO LOAD PRESENT but load ohms is
always the
result of an equation and always = V / I and if I = 0.0
Amps then the
load = V / 0 = infinite number of ohms = very high
resistance value in
ohms.

A low load is always assumed to be a low ohm value load,
usually in
comparison to a source resistance so if Rsource = 100
ohms
then a load
of 10 ohms is a low load, and it is never a high load
resistance. The
context helps determine the meaning.
A load of 1,000ohms is a high value load resistance - in
comparision
to the Rsource of 100 ohms.

With all due respect I dispute your contention that 'low
load' is
"always" assumed to be an 'ohm value' and Iveson has a
point that it's
context sensitive. For example, if one is clearly
talking
about power
then a 'low load' could be referring to 'low power' and,
so, perhaps a
'high impedance'.

However, I also dispute his argument about "in this
context" because
the issue at hand was an inductance, specifically it's
impedance at
frequency, in parallel with 'the load' and I don't think
of inductor
(or capacitor, or resistor, etc.) impedance in parallel
with 'power'
but with other impedances (which may be why you consider
it 'always'
ohms).

I'm sure it was blindingly obvious to his own mind's eye
but we don't
have that 'context' and it's just a shame he decided to
get all
indignant even though I specifically qualified with "If
I
understood
his meaning." I apparently did *not* 'understand his
meaning' but, in
my own defense, that's because I'm not a mind reader.


Patrick Turner.

Cheers

ian





I'll try to use more accessible language in the
future.

Ian- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

[Ian Iveson]

PS I haven't distinguished between PP and SE, and obviously
for the latter you need some bias anyway. The important
point is that the steepness of the sigma depends on where
you are on the curve *and* on the signal amplitude. Small
signals will have flatter loops the closer you get to zero
bias, but a larger signal will result in a steeper loop at
that same low bias.

What I particularly want to attack is the notion that, for
PP, zero inductance at the crossing point leads to a kind of
crossover distortion. ********. The sigmoid is always
steepest at the crossing point. The problem for an unbiased
PP transformer is that the sigmoid is flatter for smaller
signals, and very flat indeed for infinitesimal signals, so
bass extension may be lost *for small signals only*.

Finally, back to your question. For highest max power you
should bias the transformer half way between zero and max
current, rather as you might a power valve for SE. That's
also likely to be the point for lowest distortion, happily.
Because of the gap, the BH curve is shallow and so B is more
linear wrt H, and where you are on the curve doesn't matter
so much.

Is it not an assumption of your example specs that bias will
be half the saturation current?

Ian

"Ian Iveson" wrote in
message ...
So my question is, how does the inductance vary with (dc)
current. If it measures higher with lower dc current then
if I use a 10W instead of a 5W but with a lower dc
current, do I get even higher inductance because of the
lower current?

This isn't a simple question, IIRC from longago posts. I
venture the following expecting a tongue-lashing.

The inductance varies with the slope of the BH curve. If
you look at one common, DC depiction of that curve, it
appears as a sigmoid, shallow at the zero crossing because
the magnetic domains exhibit what you might call stiction,
becoming steeper, and then shallow again as the majority
of domains reach their elastic limit, and some reach the
yield point.

The yielding dissipates power, resulting in hysteresis.
Consequently, for AC, the sigmoid becomes a loop, steepest
at the zero crossing, and shallow at both extremes of its
variation. What's more, any AC signal will generate a
sigmoid, with the magnetic bias affecting how flat or
steep the sigma-shaped loop is, according to where its
centre is on to the DC BH curve.

Break for ridicule.

Now, it seems from the DC curve that inductance should be
zero as B crosses the H axis. Much nonsense arises from
this interpretation. In fact, for pure AC the curve is
steepest there, both on the upward and downward journey.

However the sigmoid is flattish for small AC signals, so
the mean inductance is lowish. Then it gets steeper as the
signal increases, so the mean inductance rises, and then
begins to flatten out at its extremes and the inductance
falls again, as the core saturation approaches.

Still with me?

Add DC and small signals see a higher inductance, but the
average inductance for larger signals doesn't increase
very much, especially if the signal results in the core
getting to the sharper curve approaching saturation.

I would have thought that the great benefit of using a
transformer rated for high DC current, assuming the same
rated inductance, would be that it gives you greater
separation between roll-off and saturation. That means you
can use more feedback to extend its bass range at a given
max signal. Alternatively, some designs bias the SE
transformer and deliberately oversize it, to avoid the
bass disappearing at small signal levels. So they say, I
think.

So, the answer is: yes and no.

Ian


"Ian Bell" wrote in message
...
Just wanted to add a related question. In looking up SE
transformers on the net many that quote inductance do so
at a specified dc current. So the difference between a 5W
and a 10W transformer might typically be the 5W has a 10H
primary inductance measure at a dc current of 48mA
whereas a 10W type would have an inductance of 15H at a
dc current of 64mA.


Cheers

Ian

[Ian Bell[_2_]]

Ian Iveson wrote:
So my question is, how does the inductance vary with (dc)
current. If it measures higher with lower dc current then
if I use a 10W instead of a 5W but with a lower dc
current, do I get even higher inductance because of the
lower current?

This isn't a simple question, IIRC from longago posts. I
venture the following expecting a tongue-lashing.

The inductance varies with the slope of the BH curve. If you
look at one common, DC depiction of that curve, it appears
as a sigmoid, shallow at the zero crossing because the
magnetic domains exhibit what you might call stiction,
becoming steeper, and then shallow again as the majority of
domains reach their elastic limit, and some reach the yield
point.

The yielding dissipates power, resulting in hysteresis.
Consequently, for AC, the sigmoid becomes a loop, steepest
at the zero crossing, and shallow at both extremes of its
variation. What's more, any AC signal will generate a
sigmoid, with the magnetic bias affecting how flat or steep
the sigma-shaped loop is, according to where its centre is
on to the DC BH curve.

Break for ridicule.

Now, it seems from the DC curve that inductance should be
zero as B crosses the H axis. Much nonsense arises from this
interpretation. In fact, for pure AC the curve is steepest
there, both on the upward and downward journey.

However the sigmoid is flattish for small AC signals, so the
mean inductance is lowish. Then it gets steeper as the
signal increases, so the mean inductance rises, and then
begins to flatten out at its extremes and the inductance
falls again, as the core saturation approaches.

Still with me?

Add DC and small signals see a higher inductance, but the
average inductance for larger signals doesn't increase very
much, especially if the signal results in the core getting
to the sharper curve approaching saturation.

I would have thought that the great benefit of using a
transformer rated for high DC current, assuming the same
rated inductance, would be that it gives you greater
separation between roll-off and saturation. That means you
can use more feedback to extend its bass range at a given
max signal. Alternatively, some designs bias the SE
transformer and deliberately oversize it, to avoid the bass
disappearing at small signal levels. So they say, I think.

So, the answer is: yes and no.

Ian


"Ian wrote in message
...
Just wanted to add a related question. In looking up SE
transformers on the net many that quote inductance do so
at a specified dc current. So the difference between a 5W
and a 10W transformer might typically be the 5W has a 10H
primary inductance measure at a dc current of 48mA whereas
a 10W type would have an inductance of 15H at a dc current
of 64mA.


Cheers

Ian





Thanks for that. Not sure I understood it all. Magnetic theory I have always
found difficult.

From what I have read recently in RDH4 is seems that the more primary Henries
you have the better bass extension you get which makes sense sine the reatance
is in parallel with the load. Also the number of Henries you get depends on the
signal level being lowest at low signal levels and RDH4 recommends quoted
primary Henries should be measured at low levels. I have seem SET transformer
specs give values for primary inductance but not one mentions the level it is
measured at. I doubt it is safe to assume they are all made at low levels.

Cheers

Ian

[Ian Iveson]

Ian wrote:


Thanks for that. Not sure I understood it all. Magnetic
theory I have always found difficult.

Me too. Nightmare for SPICE!

From what I have read recently in RDH4 is seems that the
more primary Henries you have the better bass extension
you get which makes sense sine the reatance is in parallel
with the load. Also the number of Henries you get depends
on the signal level being lowest at low signal levels and
RDH4 recommends quoted primary Henries should be measured
at low levels. I have seem SET transformer specs give
values for primary inductance but not one mentions the
level it is measured at. I doubt it is safe to assume they
are all made at low levels.


It's not easy to explain, for me anyway, without diagrams. I
can't remember where I saw the explanation I'm trying to
convey.

Here's an article for a different material, with a less
wiggly loop than yours, but the principle remains the same:

http://www.bitechnologies.com/pdfs/6appnotes.pdf

Look for the diagram showing a "BH minor loop". You should
be able to see that the steepness of the minor loop will
increase with signal amplitude, as long as it doesn't come
close to saturation. This is true whether the core is biased
with DC or not. A shallower loop represents a lower
inductance.

Not easy to see with the depicted core material which is so
linear in its central region, but consider how the steepness
and shape of the minor loop will also depend on the bias,
for a core with more wiggly characteristics.

Note that the sigmoid loop for an unbiased core is steepest
at zero B. A common misconception is that dB/dH falls to
zero at zero B. That's because ppl look at the initial
magnetising curve, which starts at the origin (zero B and
zero H). The initial curve is indeed flat for small B, but
that's only for the first part of the first cycle of an AC
signal, assuming an unmagnetised core.

If I understood all this well, I might have simulated it by
now. As it is, so far, I can't deal with hysteresis. The
only SPICE models I have seen which do, do so only at a
given frequency and amplitude, which is OK for a SMPS, but
no good for audio.

Ian

[Ian Iveson]

Ian wrote:

An open circuit represents zero load, right? So a high
load
resistance is a low load.

Moot. An open circuit implies 'no' load which could
equally imply infinite resistance.

Why moot? No load = infinite resistance is the limiting
condition of what I said: a high load resistance is a low
load.

Ian

[Ian Bell[_2_]]

Ian Iveson wrote:
Ian wrote:

An open circuit represents zero load, right? So a high
load
resistance is a low load.

Moot. An open circuit implies 'no' load which could
equally imply infinite resistance.

Why moot? No load = infinite resistance is the limiting
condition of what I said: a high load resistance is a low
load.

Ian




Moot because zero is not equal to infinity.

Cheers

Ian

[Ian Bell[_2_]]

Ian Iveson wrote:
Ian wrote:


Thanks for that. Not sure I understood it all. Magnetic
theory I have always found difficult.

Me too. Nightmare for SPICE!

From what I have read recently in RDH4 is seems that the
more primary Henries you have the better bass extension
you get which makes sense sine the reatance is in parallel
with the load. Also the number of Henries you get depends
on the signal level being lowest at low signal levels and
RDH4 recommends quoted primary Henries should be measured
at low levels. I have seem SET transformer specs give
values for primary inductance but not one mentions the
level it is measured at. I doubt it is safe to assume they
are all made at low levels.


It's not easy to explain, for me anyway, without diagrams. I
can't remember where I saw the explanation I'm trying to
convey.

Here's an article for a different material, with a less
wiggly loop than yours, but the principle remains the same:

http://www.bitechnologies.com/pdfs/6appnotes.pdf


Can't seem to get this to download - currently says 1 day left to complete!!

Cheers

Ian

Look for the diagram showing a "BH minor loop". You should
be able to see that the steepness of the minor loop will
increase with signal amplitude, as long as it doesn't come
close to saturation. This is true whether the core is biased
with DC or not. A shallower loop represents a lower
inductance.

Not easy to see with the depicted core material which is so
linear in its central region, but consider how the steepness
and shape of the minor loop will also depend on the bias,
for a core with more wiggly characteristics.

Note that the sigmoid loop for an unbiased core is steepest
at zero B. A common misconception is that dB/dH falls to
zero at zero B. That's because ppl look at the initial
magnetising curve, which starts at the origin (zero B and
zero H). The initial curve is indeed flat for small B, but
that's only for the first part of the first cycle of an AC
signal, assuming an unmagnetised core.

If I understood all this well, I might have simulated it by
now. As it is, so far, I can't deal with hysteresis. The
only SPICE models I have seen which do, do so only at a
given frequency and amplitude, which is OK for a SMPS, but
no good for audio.

Ian

[Ian Iveson]

"Ian Bell" wrote in message
...
Ian Iveson wrote:
Ian wrote:

An open circuit represents zero load, right? So a high
load
resistance is a low load.

Moot. An open circuit implies 'no' load which could
equally imply infinite resistance.

Why moot? No load = infinite resistance is the limiting
condition of what I said: a high load resistance is a low
load.

Ian


Moot because zero is not equal to infinity.

A nonsequiter, surely? Neither of our propositions claims or
comes anywhere near implying that it is. Any statement that
were to claim so would be not just moot, but clearly false.

I'm saying that zero is the limiting condition of low.
Infinity is the limiting condition of high. I assume that
you agree. Where, exactly, in our sequence of logical
ruminations above, do you detect a difference between us?

Ian

[Ian Iveson]

oops. I'm coming from the future, and I've fallen a day
short. Look tomorrow, when it'll be yesterday for me, at
this rate.

It's a 100Kb PDF that will be what it says in its file name.
It was produced on one of those irksome Macintosh computers
but there's nothing strange about it.

I googled "minor BH loop". Lots of links demand money. That
one is the first for free. The PDF is aimed at inductors for
SMPS, but it starts from the beginning, so much is common.

http://www.bitechnologies.com/pdfs/6appnotes.pdf

Inductor manufacturers have similar literature on their
sites.

I just remembered a blinding flash of inspiration, somewhere
beyond Jupiter. The sigmoidal BH curve puts current, H,
along the horizontal axis. For unbiased AC, if the voltage
is at right angles to the current, that puts the flat
sections of the sigma at the zero voltage crossing point, so
in terms of voltage, the inductance will look a bit like
B=V^3, with an inflection at the origin. That's where the
"crossover distortion" idea comes from, I saw, approaching
Mars all those years ago.

Or will be soon.

Ian


"Ian Bell" wrote in message
...
Ian Iveson wrote:
Ian wrote:


Thanks for that. Not sure I understood it all. Magnetic
theory I have always found difficult.

Me too. Nightmare for SPICE!

From what I have read recently in RDH4 is seems that
the
more primary Henries you have the better bass extension
you get which makes sense sine the reatance is in
parallel
with the load. Also the number of Henries you get
depends
on the signal level being lowest at low signal levels
and
RDH4 recommends quoted primary Henries should be
measured
at low levels. I have seem SET transformer specs give
values for primary inductance but not one mentions the
level it is measured at. I doubt it is safe to assume
they
are all made at low levels.


It's not easy to explain, for me anyway, without
diagrams. I
can't remember where I saw the explanation I'm trying to
convey.

Here's an article for a different material, with a less
wiggly loop than yours, but the principle remains the
same:



Can't seem to get this to download - currently says 1 day
left to complete!!

Cheers

Ian

Look for the diagram showing a "BH minor loop". You
should
be able to see that the steepness of the minor loop will
increase with signal amplitude, as long as it doesn't
come
close to saturation. This is true whether the core is
biased
with DC or not. A shallower loop represents a lower
inductance.

Not easy to see with the depicted core material which is
so
linear in its central region, but consider how the
steepness
and shape of the minor loop will also depend on the bias,
for a core with more wiggly characteristics.

Note that the sigmoid loop for an unbiased core is
steepest
at zero B. A common misconception is that dB/dH falls to
zero at zero B. That's because ppl look at the initial
magnetising curve, which starts at the origin (zero B and
zero H). The initial curve is indeed flat for small B,
but
that's only for the first part of the first cycle of an
AC
signal, assuming an unmagnetised core.

If I understood all this well, I might have simulated it
by
now. As it is, so far, I can't deal with hysteresis. The
only SPICE models I have seen which do, do so only at a
given frequency and amplitude, which is OK for a SMPS,
but
no good for audio.

Ian

[Ian Bell[_2_]]

Ian Iveson wrote:
"Ian wrote in message
...
Ian Iveson wrote:
Ian wrote:

An open circuit represents zero load, right? So a high
load
resistance is a low load.

Moot. An open circuit implies 'no' load which could
equally imply infinite resistance.

Why moot? No load = infinite resistance is the limiting
condition of what I said: a high load resistance is a low
load.

Ian


Moot because zero is not equal to infinity.

A nonsequiter, surely? Neither of our propositions claims or
comes anywhere near implying that it is. Any statement that
were to claim so would be not just moot, but clearly false.

I'm saying that zero is the limiting condition of low.
Infinity is the limiting condition of high. I assume that
you agree. Where, exactly, in our sequence of logical
ruminations above, do you detect a difference between us?

Ian



OK. You said

"
Zero is the limiting condition of low.

An open circuit is zero load

No (zero) load is infinite resistance

Therefore zero = infinity

QED

Cheers

Ian

[Ian Iveson]

"Ian Bell" wrote in message
...
Ian Iveson wrote:
"Ian wrote in message
...
Ian Iveson wrote:
Ian wrote:

An open circuit represents zero load, right? So a
high
load
resistance is a low load.

Moot. An open circuit implies 'no' load which could
equally imply infinite resistance.

Why moot? No load = infinite resistance is the limiting
condition of what I said: a high load resistance is a
low
load.

Ian


Moot because zero is not equal to infinity.

A nonsequiter, surely? Neither of our propositions claims
or
comes anywhere near implying that it is. Any statement
that
were to claim so would be not just moot, but clearly
false.

I'm saying that zero is the limiting condition of low.
Infinity is the limiting condition of high. I assume that
you agree. Where, exactly, in our sequence of logical
ruminations above, do you detect a difference between us?

Ian



OK. You said

"
Zero is the limiting condition of low.

An open circuit is zero load

No (zero) load is infinite resistance

OK so far. I said those things.

Therefore zero = infinity

er...why? Wiki is quite good, but long-winded, on
syllogisms. I don't think you'll find that one in the list
of valid forms. I don't think you can rip nouns out, willy
nilly, with impunity.

QED

or not


Cheers

Ian

[Ian Bell[_2_]]

Ian Iveson wrote:
"Ian wrote in message
...
Ian Iveson wrote:
"Ian wrote in message
...
Ian Iveson wrote:
Ian wrote:

An open circuit represents zero load, right? So a
high
load
resistance is a low load.

Moot. An open circuit implies 'no' load which could
equally imply infinite resistance.

Why moot? No load = infinite resistance is the limiting
condition of what I said: a high load resistance is a
low
load.

Ian


Moot because zero is not equal to infinity.

A nonsequiter, surely? Neither of our propositions claims
or
comes anywhere near implying that it is. Any statement
that
were to claim so would be not just moot, but clearly
false.

I'm saying that zero is the limiting condition of low.
Infinity is the limiting condition of high. I assume that
you agree. Where, exactly, in our sequence of logical
ruminations above, do you detect a difference between us?

Ian



OK. You said

"
Zero is the limiting condition of low.

An open circuit is zero load


An open circuit (A) is zero load (B)

A = B

No (zero) load is infinite resistance


No (zero) load (B) is infinite resistance (C)

B = C

OK so far. I said those things.

Therefore zero = infinity


Therefore A = C

QED

Cheers

ian
er...why? Wiki is quite good, but long-winded, on
syllogisms. I don't think you'll find that one in the list
of valid forms. I don't think you can rip nouns out, willy
nilly, with impunity.

QED

or not


Cheers

Ian