I have now got the first channel of my KT88
50W monobloc amp up and running, powered
by a lab psu.

The B+ rails is at 500V. The amp draws 320mA
when driven at full power continuous
(20V into 8 Ohms) with a sine wave

I plan to use a SS full wave bridge in the psu.
and rate the mains transformer at 1.61 times
the DC current drawn by the amp.

It is highly unlikely that the amplifier will ever run
at full power, but classical music requires plenty of
headroom.

So, should I rate the transformer
at 1.61 x 320mA, or add some headroom to this,
or specify a lower rating for the winding?

Iain

"Iain M Churches" wrote in message
...
I have now got the first channel of my KT88
50W monobloc amp up and running, powered
by a lab psu.

The B+ rails is at 500V. The amp draws 320mA
when driven at full power continuous
(20V into 8 Ohms) with a sine wave

I plan to use a SS full wave bridge in the psu.
and rate the mains transformer at 1.61 times
the DC current drawn by the amp.

It is highly unlikely that the amplifier will ever run
at full power, but classical music requires plenty of
headroom.

So, should I rate the transformer
at 1.61 x 320mA, or add some headroom to this,
or specify a lower rating for the winding?

Iain

Wouldn't it be prudent to check the amp at the full spectrum of audio
frequencies? Current draw can widely vary with different frequencies. Just
my $.02.
west

"west" wrote in message
om...
Wouldn't it be prudent to check the amp at the full spectrum of audio
frequencies? Current draw can widely vary with different frequencies. Just
my $.02.
west

Driving white noise into a dummy load, I get the same current
requirement - 320mA.

Your $.02. much appreciated :-)

Iain

"Iain M Churches"

I have now got the first channel of my KT88
50W monobloc amp up and running, powered
by a lab psu.

The B+ rails is at 500V. The amp draws 320mA
when driven at full power continuous
(20V into 8 Ohms) with a sine wave


** Err - so you have built a power stage that is only 31 % efficient at
full power ??

( 50 / (500 x 0.32) = 0.31 )

Even pure, bloody class A is better.



I plan to use a SS full wave bridge in the psu.
and rate the mains transformer at 1.61 times
the DC current drawn by the amp.

It is highly unlikely that the amplifier will ever run
at full power, but classical music requires plenty of
headroom.


** Err - you have not supplied the range of DC supply current.



So, should I rate the transformer
at 1.61 x 320mA, or add some headroom to this,
or specify a lower rating for the winding?



** The AC tranny needs to be of 250 + VA rating or more to run that
electron sucking POS.


( 500 x 0.32 x 1.6 = 256 VA , plus heater current )



................. Phil

Iain M Churches wrote:

I have now got the first channel of my KT88
50W monobloc amp up and running, powered
by a lab psu.

The B+ rails is at 500V. The amp draws 320mA
when driven at full power continuous
(20V into 8 Ohms) with a sine wave

So it must be operating in mainly class B to make the 50 watts.
A KT88 would have Ia = 60 mA at idle, and Ea = 500v,
if you have fixed bias, which you should.
So two tubes in class A would draw 120mA if the poer was pure class A,
but for class AB, the input power rises, and the Ia
could double, but you have more again.

Its probably because the load the tubes are seeing at 50 watts
is about 3k, rather low imho.

I like to have the maximum power at clipping from a pair of KT88 at 54
watts at 4 ohms, and RL a-a is around 5k.
Then at 8 ohms the power falls to 34 watts, but with a huge class A %,
and the sound is good and it measures very well.



I plan to use a SS full wave bridge in the psu.
and rate the mains transformer at 1.61 times
the DC current drawn by the amp.

Allowing for 3 times the idle current is OK.
In practice, a good class AB amp rarely ever gets out of class A,
and spends 99.9% of its life with the power draw being nearly =
to the idle condition.




It is highly unlikely that the amplifier will ever run
at full power, but classical music requires plenty of
headroom.

Then use large caps to store a big charge.

No need to make the PT rated too high.

But it should have Bmax = 0.9 tesla or less, and have the core rated for

3 times the max VA, then copper rated for twice the VA,
then the PT will run quiet, and cool.
Use only GOSS, or have a toroidal especially wound, varnished and
potted.
Plitron make some nice gear.



So, should I rate the transformer
at 1.61 x 320mA, or add some headroom to this,
or specify a lower rating for the winding?

320 mA of draw is enormous for a pair of KT88.

OK, so the drive amp might use 20 mA, but that still leaves
300 mA, and that's 150 mA per tube, or more than twice what i
would use at idle, but some folks run
KT88 at 40 watts, or at 80 mA of plate and screen current.
Its too much.

The total power needs in total have to be examined.

Heater power :-
KT88, 2 x 6.3 x 1.8 = 22.68VA
input tubes, allow 3 x 6.3 x 0.6 = 11.34VA
idle power, B+, = 500v x 2 x 80 mA,
but allow for 500v x 2 x 160 mA = 160 VA,
Input stage power, 500v x 20 mA = 10VA
bias supply, 100v x 10 mA = 1 VA,

Total VA = 205VA.

Don't muck about, get a 300VA tranny which has very
adequate ratings.

This will mean a power tranny with 2.7" stack of 1.5" tongue E&I lams,
so round that up to a 3" stack.
Use only GOSS lams, lest the tranny run too warm.

The Bmax should be under 0.9 T for low losses, but most importantly
for quite op with a rectifier.

the best possible regulation with the tranny isn't needed, since the amp
will operate mostly
class A and its rare that you will ever seen the EA change more than a
volt or 3 even with a
high resistance tube rectifier.

Have a CLC filter before the CT, with series 470 uF x 350v rated caps,
so you have at least 235 uF, about 3H, 235 uF, and that should be OK.
If dcr of the choke is less than 30 ohms,
allow the loaded vrms of the B+ winding to be 0.74 tims the B+ proposed,

so for +500v, the winding is 370v.
I have often used a doubler, and a 190 volt winding with thicker wire.
Have a few taps on the winding at about 20vrms
postions to allow B+ = 528v, 500v, 472v, 444v and 416v,
so that if other tubes are selected such as EL34 or 6L6, you are not
stuck.

Then if you were sensible, and wanted more class A %
in the output power at the slight expense of maximum class AB power,
then you would have B+ = 416v, or use cathode bias with say B+ = 444v,
so Ea would be around 410v, and raise the Ia to 78 mA,
and retain the load value you have now, and settle for
the outcome with greater fidelity for the same amount of NFB.

Seen this?
http://www.turneraudio.com.au/htmlwe...0ulabinteg.htm

I know the SE triode tribe hates this sort of amp but I don't
care a hoot, I know it sounds fantabulistic.

If you are doing mono amp chassis, with
a separate PS, that PS can have 600VA tranny, which will not
be much larger than a 300VA, and then run an umbilical cable
to each amp chassis.

For the cables, use no less than a decent heavy duty mobile crane 5 way
cable with at least five 5 amp wires and very flexible.
I generally hard wire the cable into the amp and have an octal socket
to fit into a recessed socket hole on the PS, so the
accidental drag or yank on the cable won't **** up the cable or plug.
The 5 way wire allows for B+, B-, OV, and heaters.
On the octal socket&plug, 4 of the 8 pins should be used for
the high heater current, and that leaves 4 for the other things.
If you get some 7 way cable with each wire rated at 3 amps, heaters
and the rest can still be catered for without hot cable and losses.

There are military grade plugs and sockets, but I have never used them,
and they cost heaps.
Just make sure the male plug is on a cable from the amp, not the PS,
lest you have the PS turned on and exposed lethal B+ if you have
forgotten
to plug in a cable.
Its good idea to araldite plates to the plug, and screw it to the PS
chassis to
prevent a child pulling out a cable and sucking it before the B+
has subsided.

Patrick Turner.















Iain

Iain M Churches wrote:

"west" wrote in message
om...
Wouldn't it be prudent to check the amp at the full spectrum of audio
frequencies? Current draw can widely vary with different frequencies. Just
my $.02.
west

Driving white noise into a dummy load, I get the same current
requirement - 320mA.

I doubt white noise is the right stuff for a test like this;
certainly it is uneccessary.

Pink noise I have used, and the amp will start to clip
on peaks in the noise at a power output level well below what
is gained with a sine wave at clipping.

Pink noise is like a music signal, and again, busy rock and roll that has been
compressed
will give a power output about 1/2 the sine wave power.

Full range pink noise will also stimulate the OPT to saturate with large LF
signals,
and cause a knocking noise, and current pulses in the OPVs, and the HF part of
the noise will
perhaps saturate the amp with HF, depending on the amp quality.

A sine wave at 1 kHz is fine.

Patrick Turner.


Your $.02. much appreciated :-)

Iain

Phil Allison wrote:

"Iain M Churches"

I have now got the first channel of my KT88
50W monobloc amp up and running, powered
by a lab psu.

The B+ rails is at 500V. The amp draws 320mA
when driven at full power continuous
(20V into 8 Ohms) with a sine wave


** Err - so you have built a power stage that is only 31 % efficient at
full power ??

( 50 / (500 x 0.32) = 0.31 )

Even pure, bloody class A is better.

Agreed.
It's OK Phil. I have him thinking about the loading in another post.
I think he has RLa-a rather too low.

I also think that if there was some way to alter the load match on his OPTs,
he ought to investigate.

SS amps can be up to 78% efficient at full power, class aB.
But tube amps rarely more than 66%,
and class A pentode amps at about 40%.
Class AB UL at full power should reach about 50%.



I plan to use a SS full wave bridge in the psu.
and rate the mains transformer at 1.61 times
the DC current drawn by the amp.

It is highly unlikely that the amplifier will ever run
at full power, but classical music requires plenty of
headroom.


** Err - you have not supplied the range of DC supply current.

So, should I rate the transformer
at 1.61 x 320mA, or add some headroom to this,
or specify a lower rating for the winding?


** The AC tranny needs to be of 250 + VA rating or more to run that
electron sucking POS.

( 500 x 0.32 x 1.6 = 256 VA , plus heater current )

I said 300VA trannies are appropriate...

A 600VA tranny for both channels is better, since the larger tranny can have
better regulation
than a pair of trannies.

Patrick Turner.



................ Phil

"Patrick Turner" wrote in message
...


Iain M Churches wrote:

"west" wrote in message
om...
Wouldn't it be prudent to check the amp at the full spectrum of audio
frequencies? Current draw can widely vary with different frequencies.
Just
my $.02.
west

Driving white noise into a dummy load, I get the same current
requirement - 320mA.

I doubt white noise is the right stuff for a test like this;
certainly it is uneccessary.

I am wondering if the fig of 320mA is correct.
The Solartron psu has a single meter which can be switched
between voltage and current.

Perhaps I should check by measuring the voltage drop
across a series 1 Ohm resistor.

Iain

"Patrick Turner" wrote in message
...
I said 300VA trannies are appropriate...

A 600VA tranny for both channels is better, since the larger tranny can
have
better regulation
than a pair of trannies.

This amp is the first of a pair of monoblocs, so each will have its own
mains transformer.

Iain

I now have a little more info regarding the
prototype 50W amp.

The OPT has Ra-a of 4.5k
and rated at 100W.

The meter on the Solartron bench psu is difficult
to read. The voltage scale is clear, as it is printed
in black, but the ammeter scale in in red, and faded
over the years.

So, I put a 1R 5W resistor in series with the psu,
with a DVM across it.

The amp has fixed bias.
I have set each of the KT88's to draw 60mA idle.
The idle current is measured across a 33 Ohm
resistor to give 2V. (2/33 = 0.060)

Now here's the new bit. With a sine wave input
and driving at 50W (20V into 8 Ohms) the total
current drawn is 240mA (not 320mA as I thought
earlier)

So that's an efficiency of 50/(500 x 0.240) = 42%
Better?

So, to ask my question again, "At what current rating
should I specify the HT winding on the transformer secondary"

Thanks,

Iain

On Tue, 15 Mar 2005 10:49:41 +0200, "Iain M Churches" wrote:

I have now got the first channel of my KT88
50W monobloc amp up and running, powered
by a lab psu.

The B+ rails is at 500V. The amp draws 320mA
when driven at full power continuous
(20V into 8 Ohms) with a sine wave

Are you sure about that 320ma???

Sounds awful high to me! The last 50 watt amp I made ran at closer to 120ma...

Maybe your load is a bit off? Or your meter?

I plan to use a SS full wave bridge in the psu.
and rate the mains transformer at 1.61 times
the DC current drawn by the amp.

According to my Fairchild PS book, the transformer VA could be 2.16 PO with a
capacitive loaded bridge, no CT. Not sure what the peak current would be...
maybe you could calculate it.

It is highly unlikely that the amplifier will ever run
at full power, but classical music requires plenty of
headroom.

With music, you probably will fill the peak demands with energy from the filter
caps, so you may want to make these quite large. The average current from the
tranny will probably be low, I forget the percent that music requires. Of
course, it could depend on the music as well! I build guitar amps, and they
have a more stringent current requirement, since they run at 11 all the time!

So, should I rate the transformer
at 1.61 x 320mA, or add some headroom to this,
or specify a lower rating for the winding?

Iain


You realize you're talking about a 500ma transformer in a 50 watt amp!?!?

I have a huge Hammond here, the 475ma one, and I have it reserved for a 200 watt
stereo...

I'd suggest you run some more tests!

Good luck!

"Iain M Churches"

I have set each of the KT88's to draw 60mA idle.
The idle current is measured across a 33 Ohm
resistor to give 2V. (2/33 = 0.060)

Now here's the new bit. With a sine wave input
and driving at 50W (20V into 8 Ohms) the total
current drawn is 240mA (not 320mA as I thought
earlier)


** Is IMC measuring the voltage drop with an average responding meter or
a "true rms" one ??

And does he have the slightest idea of the difference ???





............ Phil

"Bob" wrote in message
...
Are you sure about that 320ma???

No. See my second post. I checked the current drawn with two other
meters (Fluke) and they give the total as 240mA.
(0.240V dropped across a 1 Ohm resistor)

Maybe your load is a bit off? Or your meter?

The load is a 100W wirewound.
The transformer Ra-a = 4.5k
With the new reading of 240mA, that makes the
efficiency about 42%, which is probably OK.

Patrick has suggested that the transformer should
be specificied at 3x idling current, which seems like
a good idea.

Iain

Iain M Churches wrote:

I now have a little more info regarding the
prototype 50W amp.

The OPT has Ra-a of 4.5k
and rated at 100W.

How do you work that out, bearing in mind
that your other statements about your amp were a bit wonky
to start with?



The meter on the Solartron bench psu is difficult
to read. The voltage scale is clear, as it is printed
in black, but the ammeter scale in in red, and faded
over the years.

So, I put a 1R 5W resistor in series with the psu,
with a DVM across it.

The amp has fixed bias.
I have set each of the KT88's to draw 60mA idle.
The idle current is measured across a 33 Ohm
resistor to give 2V. (2/33 = 0.060)

Now here's the new bit. With a sine wave input
and driving at 50W (20V into 8 Ohms) the total
current drawn is 240mA (not 320mA as I thought
earlier)

Where did you measure that? with 33 ohms between each k and 0V?

What allowance have you made for the signal voltage present
at each k?

What if you had a 10 ohm R in series with the CT, but before the
cap to 0V, so that plate current has virtually no signal present?

What of the screen dissipation?



So that's an efficiency of 50/(500 x 0.240) = 42%
Better?

Not including the screen current, you'd get a better calculated plate

Is the amp UL, if so, what part in the efficiency measure does the
screen play?

And BTW, at 50 watts, the B+ will have dropped,
and power input to the plates is less than 500v at idle,
so power input at 50 watts won't be 500 x 0.24 amps,
but maybe 470 x 0.24.

To know reality, be conscious of variations of the conditions
where it is measured.
IE, when conditions change always *expect* to have to make allowances
in your measurements.



So, to ask my question again, "At what current rating
should I specify the HT winding on the transformer secondary"

The rating can be for the maximum current draw condition
for the load that produces the most power.

Its not a strict design premise.

If you observe the wave output wave form you will see when clipping just
begins.
That's where thd = approx 2%.
That's where max PO is usually measured.

Then plot a graph
of power output vs load value using your dummy loads of
1,2,3,4,5,6,8,12,16, 24, and 32 ohms.

This graph will show a steep eise of PO from o ohms to say a peak at 4
ohms,
then a sag downwards towards zero PO as RL approaches very high value.

Then where is the peak in the PO?

Is it at 3 ohms, 5 ohms, 8 ohms?

Is it where you want it?

If you have speakers nominally 8 ohms, you should assume part
of the Z will be at 6 ohms or even lower, while the average is maybe 8
ohms
across the band.

If I know a speaker is said to be 8, I still set the amp up to ideally
suit a
5.6 ohm load, which is 1/2 way between 4 and 8 ohms.

Then any speaker between 3 and 16 ohms is fine.

There are almost no 16 ohm speakers around now
but where there are they are old and also usually sensitive, so
it doesn't matter if the amp is "set to a low Z match"
for such speakers.

Modern speakers are often nominally 6 to 8 ohms, often with a did
to maybe 4 ohnms, so you want your amp to cope,
and be able to give the extra current grunt without lots of distortion,
so the choice of 5.6 ohms is not absurd.

I usually set up my PP amps to make max PO into
about 3 ohms, below which PO rolls off if we keep to the 1% thd rule.
It means that the RLa-a has to be carefully chosen for that
and the turn ratio, and hence impedance ratio
depends on the B+, and class.
There is some reading material about EH6550, which are identical to KT88
at
http://www.turneraudio.com.au/htmlwe...50pographs.htm

At the KT88 UL schematic at
http://www.turneraudio.com.au/htmlwe...0ulabinteg.htm
I state I get 53 watts at 4 ohms, 35 watts at 8 ohms.
But there would be slightly more than 53 watts at 3 ohms.
The UL% = 50%, and the PO is less than pure beam tetrode.
But its OK, because AB tetrode has far higher thd and Rout, and I don't
need
80 watts.

In the UL amp of mine the RLa-a = approximately 9k when 8 ohms,
4.5k with 4 ohms, and so with 3 ohms, RLa-a = is used

For pure class A operation, and Pd at 30 watts for each tube,
Ia quiescent = 60 mA, Ea = 500v.

When you plot the load lines for a single tube
at 60 mA x 500v, and in tetrode or UL mode,
theoretical maximum class A is where RL = 7k approx.
The -ve and +ve V swings are around 450v either side of the 500v
quiescent point
which won't move when class A is used, because Pin stays the same for
class A.

Where you have a PP circuit, for pure class A, each of the tubes is
working
with a 7 k load, so the RLa-a is 14k, and
voltage swing is 900 peak volts a-a,
so we get 636 vrms, and that gives us 28.9 watts of pure class A.

Pin = 2 x 500 x 0.06 = 60 watts, so class A plate efficiency
( neglecting any effects the screens may have, since they are minor, )
is 100% x ( 28.9 / 60 ) = 48% which is near the theroetcal
maximum amount of class A ever available from any pure class A circuit.
Losses in the OPT of 10% would reduce the secondary load output power to

26 watts, so efficiency is 43%.
Plate efficiency is plate efficiency, and should be measured in the
plate circuit,
lest we measure plate efficiency wrongly by including the OPT.
So if you don't know the winding R, you have to have 10 ohm plate
current
sensors, and so the measuremts carefully.

Its easier to simply ascertain the general amp efficiency with OPT
losses included, which saves us having to measure distorted wave forms
in the
plate currents when examining class AB operation.

So, a surprisingly high load value is needed for pure class A if we sart
with Ea = 500v
and Ia = 60 mA for each tube.

As soon as we reduce Ia at idle, the RLa-a for pure class A has to be
increased,
and if Ia is increased, RLa-a can be reduced and still have all class A.

With all class A the circuit gain is highest, and applied NFB is the
greatest, so thd
will perhaps be 0.07% at 28 watts, not a bad result.
With a 14k a-a load, even triode will give 25 watts, and
we don't need so much global NFB and the Rout, thd will still be just as
low.

But nobody is much impressed with this low power hi-fi idea, they say
they must have 50 watts.
Or 100.

The tetrode plate resistance line on the KT88/6550 data sheet for where
Eg1 = 0V
is a curve that starts at 0V and 0amps, and rises nearly straight
to 0.35 amps at Ea = 80v, then swings down to near flat to pass through
0.4 amps at 500v.
The "corner" of the grapgh is the called the knee,
and the load lines for all proposed loads for AB1 should
intersect the Ra line to the left of the knee.

Let us suppose we have 5k a-a as a load for the UL amp with the KT88 and
500v and 60 mA.

Now to determine the v swing, imagine we had the amp working in class B.

The loadline condition can be drawn for each tube separately, since to
draw composite load line for both tubes is utterly confusing to a
beginner.
So I never do that.

Back to the single tube, biased for class B.

For class B, the tube being turned off by a -ve going Eg1, its as if it
was not
in the amp.
The one that is being turned on by the +ve Eg1 going voltage
Is connected to only 1/2 the primary, and thus sees a load which is 1/4
of the class A a-a load.

So if RL a-a was a nominal 5k, then the load for one tube in class B is
5,000 / 4 = 1,250 ohms.

Since the class B idle cindition has Ia = 0amps, we can draw
the loadline for the tube from the 0.0 amps axis and from the Ea = 500v
spot.

500v / 1,250 = 0.4 amps on the verical Ia axis, and we can dtaw a line
there.
It will intersect the Ra tube line at about
Ea = 80v and Ia = 0.35 amps.

Regardless of whatever class A % of power produced, there
cannot be any greater load voltage swing than indicated by the
class B load line for 1 tube, in this case, 500v - 80v, = 420 peakv

The same thing happens with two tubes in a PP circuit,
and the max swing for the 5 ka-a load is 2 x 840 peakv = 594v.

So the maximum power into 5k is ( 594 x 594 ) / 5,000 = 70.5 watts.

This is further spelled out down the page for the 6550 at my website
at the above url.

The initial pure class A watts can be calculated by
loadline analysis and plotting class A load lines on the same graph
for the class B situation, which BTW is *never actually used*
but unless one becomes quite used to imaginary conditions for tube
operation, and
models of perfect tubes, you won't learn anything fast.
There is always some plate current, and class B is impossible to arrange

without enormous amounts of unwanted X-over distortion.

If the load is increased from 5k, the PO will decrease, and if the load
is reduced to say
3k a-a, then the class B load line will become more vertical,
and pass through Eg1 = 0V at a point to the right of the knee in the
curve.
This usually then brings the load line into the region where Pd for the
tube
is greater than the limit, for 6550 its 42 watts, and a line should be
drawn on the plate curves
indicating this, and where load lines should never be placed.

If RLa-a = 3k, the class B load line = 750 ohms, and the class B
loadline runs from 500v to 0.666 amps, and well into the
"more than 42 watts Pd area".
The load swing for one tube will be from 500v down
to 200v, so 300peakv, so 600pv across 3k
and that gives 60 watts.

So the power peaks at some load value between
5k and 3k.

At that peak, say 4.5k, is where i would match the amp to a 3 ohm load.

With UL operation, i won't get the 70 watts for tetrode, and nor will
you.

For 100 watts from a pair of KT88/6550, even in tetrode mode, higher Ea
is needed, and/or class AB2 operation needed.
Then things get complex, the reliability suffers, and thd skyrockets.

Far better to use 4 output tubes if over 50 watts is expected
for hi-fi.

So where we have and OPT to give 4.5k to 3 ohms, the
ZR = 1,500:1, and to match to 5.6 gives 8.4k a-a, and for UL we get
48 watts, with the first 16 watts in class A.

For 8 ohms, the RL a-a becomes 12.0 k, and PO = 37 watts,
with the first 24 watts in pure class A.

I have found all other ways to match loads to tubes to be wrong for me.

If you use the above load matching technique and ideology, and you have
about 16 db of global NFB
when 8 ohms is connected, then expect Ro to be under 0.5 ohms,
and thd under 0.2% at 37 watts, and at 2 watts, expect about 0.02% thd,
and even slightly lower Rout.

Operation with triodes is possible with the same load values as I like
to use with UL.

As for the PT rating, follow the ideas in the last detailed post I sent;
I hate
having to repeat myself.

Always use a far bigger VA rating than the actual VA likely to be drawn.

Bean counters pair everything down to save a penny here and there
and everyfukkinwhere, but such folks could never find employment at this

establishment.

The last thing the diyer/hobbiest person needs is to allow
penny pinching ideologies to stuff the righteousness of a design.
So don't think like the guys in the office at Quad, Leak, Radford,
Dynaco et all;
Think BETTER.

Spread the wings of your mind and fly!.

Patrick Turner.















Thanks,

Iain

On Tue, 15 Mar 2005 10:49:41 +0200, Iain M Churches wrote:
So, should I rate the transformer
at 1.61 x 320mA, or add some headroom to this, or specify a lower rating
for the winding?

Generally speaking, power transformers are rated based on their
temperature rise at full load over a long period of continuous
use (like between 8 to 24 hours). For shorter periods or on peak demand
they can normally supply somewhat more than the rated current, how
much more depends on how conservatively they're rated, ie: what
temperature rise is considered acceptable at rated load.

320 ma is 160 watts of plate input power, sounds like plenty of
headroom for a 50 watt amp!

--
Ned Carlson Triode Electronics Chicago,IL USA
www.triodeelectronics.com

In message , Phil Allison
writes

"Iain M Churches"

I have set each of the KT88's to draw 60mA idle.
The idle current is measured across a 33 Ohm
resistor to give 2V. (2/33 = 0.060)

Now here's the new bit. With a sine wave input
and driving at 50W (20V into 8 Ohms) the total
current drawn is 240mA (not 320mA as I thought
earlier)


** Is IMC measuring the voltage drop with an average responding meter or
a "true rms" one ??

And does he have the slightest idea of the difference ???





........... Phil





He's using a sine wave, so if the meter is correctly calibrated it won't
matter.
Now with pink noise it's another thing altogether.
--
Chris Morriss

"Chris Morriss" wrote

** Is IMC measuring the voltage drop with an average responding
meter or
a "true rms" one ??

He's using a sine wave, so if the meter is correctly calibrated it
won't matter.

But he's measuring supply current.

Quite *where* he's measuring is not very clear.

Not much of all this is making sense to me. To optimise a mains
transformer, the load on the transformer itself must be known. Do we
know what kind of rectification is being used? I tend to miss stuff
if there's too much crap in a post.

Iain, have you tried the free power supply modelling program at

http://www.duncanamps.com/software.html?

cheers, Ian

"Chris Morriss"
Phil Allison

"Iain M Churches"

I have set each of the KT88's to draw 60mA idle.
The idle current is measured across a 33 Ohm
resistor to give 2V. (2/33 = 0.060)

Now here's the new bit. With a sine wave input
and driving at 50W (20V into 8 Ohms) the total
current drawn is 240mA (not 320mA as I thought
earlier)


** Is IMC measuring the voltage drop with an average responding meter
or
a "true rms" one ??

And does he have the slightest idea of the difference ???



He's using a sine wave,


* The DC supply current to a class AB tube amplifier is **NOT** sine
wave - dickhead.


so if the meter is correctly calibrated it won't matter.


** Better go work out what that wave shape is - dickhead.


Now with pink noise it's another thing altogether.


** **** off.




............... Phil

Patrick Turner wrote:

Iain M Churches wrote:

I now have a little more info regarding the
prototype 50W amp.

The OPT has Ra-a of 4.5k
and rated at 100W.

How do you work that out, bearing in mind
that your other statements about your amp were a bit wonky
to start with?



The meter on the Solartron bench psu is difficult
to read. The voltage scale is clear, as it is printed
in black, but the ammeter scale in in red, and faded
over the years.

So, I put a 1R 5W resistor in series with the psu,
with a DVM across it.

The amp has fixed bias.
I have set each of the KT88's to draw 60mA idle.
The idle current is measured across a 33 Ohm
resistor to give 2V. (2/33 = 0.060)

Now here's the new bit. With a sine wave input
and driving at 50W (20V into 8 Ohms) the total
current drawn is 240mA (not 320mA as I thought
earlier)

Where did you measure that? with 33 ohms between each k and 0V?

What allowance have you made for the signal voltage present
at each k?

What if you had a 10 ohm R in series with the CT, but before the
cap to 0V, so that plate current has virtually no signal present?

What of the screen dissipation?



So that's an efficiency of 50/(500 x 0.240) = 42%
Better?

Not including the screen current, you'd get a better calculated plate

Is the amp UL, if so, what part in the efficiency measure does the
screen play?

And BTW, at 50 watts, the B+ will have dropped,
and power input to the plates is less than 500v at idle,
so power input at 50 watts won't be 500 x 0.24 amps,
but maybe 470 x 0.24.

To know reality, be conscious of variations of the conditions
where it is measured.
IE, when conditions change always *expect* to have to make allowances
in your measurements.



So, to ask my question again, "At what current rating
should I specify the HT winding on the transformer secondary"

The rating can be for the maximum current draw condition
for the load that produces the most power.

Its not a strict design premise.

If you observe the wave output wave form you will see when clipping just
begins.
That's where thd = approx 2%.
That's where max PO is usually measured.

Then plot a graph
of power output vs load value using your dummy loads of
1,2,3,4,5,6,8,12,16, 24, and 32 ohms.

This graph will show a steep eise of PO from o ohms to say a peak at 4
ohms,
then a sag downwards towards zero PO as RL approaches very high value.

Then where is the peak in the PO?

Is it at 3 ohms, 5 ohms, 8 ohms?

Is it where you want it?

If you have speakers nominally 8 ohms, you should assume part
of the Z will be at 6 ohms or even lower, while the average is maybe 8
ohms
across the band.

If I know a speaker is said to be 8, I still set the amp up to ideally
suit a
5.6 ohm load, which is 1/2 way between 4 and 8 ohms.

Then any speaker between 3 and 16 ohms is fine.

There are almost no 16 ohm speakers around now
but where there are they are old and also usually sensitive, so
it doesn't matter if the amp is "set to a low Z match"
for such speakers.

Modern speakers are often nominally 6 to 8 ohms, often with a did
to maybe 4 ohnms, so you want your amp to cope,
and be able to give the extra current grunt without lots of distortion,
so the choice of 5.6 ohms is not absurd.

I usually set up my PP amps to make max PO into
about 3 ohms, below which PO rolls off if we keep to the 1% thd rule.
It means that the RLa-a has to be carefully chosen for that
and the turn ratio, and hence impedance ratio
depends on the B+, and class.
There is some reading material about EH6550, which are identical to KT88
at
http://www.turneraudio.com.au/htmlwe...50pographs.htm

At the KT88 UL schematic at
http://www.turneraudio.com.au/htmlwe...0ulabinteg.htm
I state I get 53 watts at 4 ohms, 35 watts at 8 ohms.
But there would be slightly more than 53 watts at 3 ohms.
The UL% = 50%, and the PO is less than pure beam tetrode.
But its OK, because AB tetrode has far higher thd and Rout, and I don't
need
80 watts.

In the UL amp of mine the RLa-a = approximately 9k when 8 ohms,
4.5k with 4 ohms, and so with 3 ohms, RLa-a = is used

For pure class A operation, and Pd at 30 watts for each tube,
Ia quiescent = 60 mA, Ea = 500v.

When you plot the load lines for a single tube
at 60 mA x 500v, and in tetrode or UL mode,
theoretical maximum class A is where RL = 7k approx.
The -ve and +ve V swings are around 450v either side of the 500v
quiescent point
which won't move when class A is used, because Pin stays the same for
class A.

Where you have a PP circuit, for pure class A, each of the tubes is
working
with a 7 k load, so the RLa-a is 14k, and
voltage swing is 900 peak volts a-a,
so we get 636 vrms, and that gives us 28.9 watts of pure class A.

Pin = 2 x 500 x 0.06 = 60 watts, so class A plate efficiency
( neglecting any effects the screens may have, since they are minor, )
is 100% x ( 28.9 / 60 ) = 48% which is near the theroetcal
maximum amount of class A ever available from any pure class A circuit.
Losses in the OPT of 10% would reduce the secondary load output power to

26 watts, so efficiency is 43%.
Plate efficiency is plate efficiency, and should be measured in the
plate circuit,
lest we measure plate efficiency wrongly by including the OPT.
So if you don't know the winding R, you have to have 10 ohm plate
current
sensors, and so the measuremts carefully.

Its easier to simply ascertain the general amp efficiency with OPT
losses included, which saves us having to measure distorted wave forms
in the
plate currents when examining class AB operation.

So, a surprisingly high load value is needed for pure class A if we sart
with Ea = 500v
and Ia = 60 mA for each tube.

As soon as we reduce Ia at idle, the RLa-a for pure class A has to be
increased,
and if Ia is increased, RLa-a can be reduced and still have all class A.

With all class A the circuit gain is highest, and applied NFB is the
greatest, so thd
will perhaps be 0.07% at 28 watts, not a bad result.
With a 14k a-a load, even triode will give 25 watts, and
we don't need so much global NFB and the Rout, thd will still be just as
low.

But nobody is much impressed with this low power hi-fi idea, they say
they must have 50 watts.
Or 100.

The tetrode plate resistance line on the KT88/6550 data sheet for where
Eg1 = 0V
is a curve that starts at 0V and 0amps, and rises nearly straight
to 0.35 amps at Ea = 80v, then swings down to near flat to pass through
0.4 amps at 500v.
The "corner" of the grapgh is the called the knee,
and the load lines for all proposed loads for AB1 should
intersect the Ra line to the left of the knee.

Let us suppose we have 5k a-a as a load for the UL amp with the KT88 and
500v and 60 mA.

Now to determine the v swing, imagine we had the amp working in class B.

The loadline condition can be drawn for each tube separately, since to
draw composite load line for both tubes is utterly confusing to a
beginner.
So I never do that.

Back to the single tube, biased for class B.

For class B, the tube being turned off by a -ve going Eg1, its as if it
was not
in the amp.
The one that is being turned on by the +ve Eg1 going voltage
Is connected to only 1/2 the primary, and thus sees a load which is 1/4
of the class A a-a load.

So if RL a-a was a nominal 5k, then the load for one tube in class B is
5,000 / 4 = 1,250 ohms.

Since the class B idle cindition has Ia = 0amps, we can draw
the loadline for the tube from the 0.0 amps axis and from the Ea = 500v
spot.

500v / 1,250 = 0.4 amps on the verical Ia axis, and we can dtaw a line
there.
It will intersect the Ra tube line at about
Ea = 80v and Ia = 0.35 amps.

Regardless of whatever class A % of power produced, there
cannot be any greater load voltage swing than indicated by the
class B load line for 1 tube, in this case, 500v - 80v, = 420 peakv

The same thing happens with two tubes in a PP circuit,
and the max swing for the 5 ka-a load is 2 x 840 peakv = 594v.

So the maximum power into 5k is ( 594 x 594 ) / 5,000 = 70.5 watts.

This is further spelled out down the page for the 6550 at my website
at the above url.

The initial pure class A watts can be calculated by
loadline analysis and plotting class A load lines on the same graph
for the class B situation, which BTW is *never actually used*
but unless one becomes quite used to imaginary conditions for tube
operation, and
models of perfect tubes, you won't learn anything fast.
There is always some plate current, and class B is impossible to arrange

without enormous amounts of unwanted X-over distortion.

If the load is increased from 5k, the PO will decrease, and if the load
is reduced to say
3k a-a, then the class B load line will become more vertical,
and pass through Eg1 = 0V at a point to the right of the knee in the
curve.
This usually then brings the load line into the region where Pd for the
tube
is greater than the limit, for 6550 its 42 watts, and a line should be
drawn on the plate curves
indicating this, and where load lines should never be placed.

If RLa-a = 3k, the class B load line = 750 ohms, and the class B
loadline runs from 500v to 0.666 amps, and well into the
"more than 42 watts Pd area".
The load swing for one tube will be from 500v down
to 200v, so 300peakv, so 600pv across 3k
and that gives 60 watts.

So the power peaks at some load value between
5k and 3k.

At that peak, say 4.5k, is where i would match the amp to a 3 ohm load.

With UL operation, i won't get the 70 watts for tetrode, and nor will
you.

For 100 watts from a pair of KT88/6550, even in tetrode mode, higher Ea
is needed, and/or class AB2 operation needed.
Then things get complex, the reliability suffers, and thd skyrockets.

Far better to use 4 output tubes if over 50 watts is expected
for hi-fi.

So where we have and OPT to give 4.5k to 3 ohms, the
ZR = 1,500:1, and to match to 5.6 gives 8.4k a-a, and for UL we get
48 watts, with the first 16 watts in class A.

For 8 ohms, the RL a-a becomes 12.0 k, and PO = 37 watts,
with the first 24 watts in pure class A.

I have found all other ways to match loads to tubes to be wrong for me.

If you use the above load matching technique and ideology, and you have
about 16 db of global NFB
when 8 ohms is connected, then expect Ro to be under 0.5 ohms,
and thd under 0.2% at 37 watts, and at 2 watts, expect about 0.02% thd,
and even slightly lower Rout.

Operation with triodes is possible with the same load values as I like
to use with UL.

As for the PT rating, follow the ideas in the last detailed post I sent;
I hate
having to repeat myself.

Always use a far bigger VA rating than the actual VA likely to be drawn.

Bean counters pair everything down to save a penny here and there
and everyfukkinwhere, but such folks could never find employment at this

establishment.

The last thing the diyer/hobbiest person needs is to allow
penny pinching ideologies to stuff the righteousness of a design.
So don't think like the guys in the office at Quad, Leak, Radford,
Dynaco et all;
Think BETTER.

Spread the wings of your mind and fly!.

Patrick Turner.


Thanks,

Iain

Anyone can end up with something by simply throwing money at it, perhaps
even quality. A real designer knows how to do that while keeping cost under
control. Working inside a cost target is something achieved only by
professionals. JLS

"John Stewart"

Anyone can end up with something by simply throwing money at it, perhaps
even quality.


** That would be a fair definition of what "boutique" audio businesses are
all about.


A real designer knows how to do that while keeping cost under
control. Working inside a cost target is something achieved only by
professionals.



** Didn't President Herbert Hoover once say

..... " an engineer can do for 50 cents what any damn fool can do for a
dollar " .





................ Phil

Snip what I said Iain said except :-


The last thing the diyer/hobbiest person needs is to allow
penny pinching ideologies to stuff the righteousness of a design.
So don't think like the guys in the office at Quad, Leak, Radford,
Dynaco et all;
Think BETTER.

Spread the wings of your mind and fly!.

Patrick Turner.


Thanks,

Iain

Anyone can end up with something by simply throwing money at it, perhaps
even quality. A real designer knows how to do that while keeping cost under
control. Working inside a cost target is something achieved only by
professionals. JLS

The hobbyist places no monetary value on his creative time,
which might be otherwise wasted by watching tele, going to the pub.
So should he resolve to never come down the lowest common denominator
like much of the "entry fi " junk of the past, then he sure can fly very well,
and cheaply.
He would only have the parts costs to consider.

He should not go overboard with overly expensive parts, lest he risk a possible
divorce,
should he never take his wife and childeren for a holiday and instead choose
silver wiring in the OPTs, which may not make much difference to the sound.
A sensible man with a sensibly constructed life but one where he isn't a high
wage earner
and who has spare time can cobble together a superb system which functions
identically
in comparison to whatever is cionsidered the best from CJ, ARC, Ongaku, etc.
Balance, and the middle path in life has its plusses, and dosn't imply
compromise
on audio parts has to be made, but may mean patience and slow progress,
until funds are saved.

Some how the theme in society's consciousness is to make everyone feel guilty
for having spare time, because the wowsers would say hobbies and time off work
is sloth,
and as evil as bonking the bosses daughter when you are married, which is lust,

or doing all manner of other things that could be considered to be vice-ridden
activities.
We are made to feel fearful about the future, threatened by terrorists, and
insecure,
and the only cure for our background emotional disquiet and anxiety is to pay
up
like a good little consumer and never say boo to authorities and big business.
We are expected to believe the relentless advertising, but I won't.
I have never been a slave to societal conventions, believing in give and take,
and society will take more than it gives, should you let it, and I don't,
and If I wanted to spend 6 years cycling, so be it, rather than building
someone else's
house, and if I chose to give up the challenge of this for what I consider
a lowly paid career in analog electronics, so be it.
In many other societies, I wouldn't have such freedoms.
Should my righhts be infringed, I will be there mixing it with the partisans.

So I have always had the right to snub mainstream products and culture.
I never go to Mc Donalds for food, and I have never had a need or desire
to buy a Hammond tranny, but for those who wish to, then good luck.
I cannot remember when I have eaten a sausage, for they are full of **** as AJ
pointed out.
Meat is USD $6 per Kg, (2.2lb) where I am in the supermarkets,
largely fat free, and $12 worth of *real* meat is enough per week for me, along
with a lot
of other fresh and non processed food stuffs. Sausages are fake nuitrition.
It might be cheaper to live some other way, but I don't care to.
So, the message is, self reliance, low debts, insistence on quality,
don't waste time on tele, ( I am amoung 1% of Oz ppl who does not watch TV )
don't drink more than a glass of wine a day, never smoke, and never waste
anything.
And don't marry a woman who has the opposite ideas about the basics.
I tried several times, but never minded when they left.
Self reliance needs stickatedness. Its extremely unromantic, by today's
absurd standards.

Between 23 and 26, I saved $11,000 from my pay as an employee, which is about
USD $125,000 in today's money.
When I bought a house I bought a small one which I then needed a small loan
to pay off a small balance, because I established high equity at once.
Since I was a builder then, it made sense that I construct additional rooms
to give me the room more suitable to enjoy myself in.
So unlike nearly everyone else, I finished up with a very comfy workshop
and house which I had fully paid for by age 37.
I worked my butt off between 18 and 38, to make damn sure it all happened.
And I constructed the additions to my house so I could rent out 1/2 the area to

give me a little extra $$ to fund time to dabble in hobby persuits, without
always worrying about
finding enough income.

I also built all my own furniture, and saved another 10 grand, today's value.

I have lived very cheaply, but very well, with few worries, and continue to do
so.
All the rest of those with less self reliance have paid a lot more for their
cost of living, because every
time you pay another to do something for you, you pay his profits, his taxes,
and
oncosts, and the interest on the loans needed to pay him, stamp duties, etc,
etc, etc.


The attitude transferred to the electronics for my music, and a hobby grew into
a business,
when i realised that at 50 I was not competitive with younger builders,
and that I was sick and tired of building after 32 years.
Ladders and barrow fulls of concrete began to cause episodes of
knee pain and illness, so I quit building.

I avoided paying twice for my life by avoiding an expensive divorce.
Ppl around me could allways have the freedoms they craved,
except the freedom to **** up my life.

While others may discuss cars and other OT matters, I'd prefer
to drift to life philosophy, because life choices will get you further than any
motor vehicle, imho.
Audio gives me an easier life than building, yet is far less well paid.
But I meet more people, and there is a personal reward.

One needs personal satisfaction from keeping ppl happy, not just oneself.
Work makes the soul thrive.....
If I'd had the expectation that I may have been valued by the offspring
I never had then I may have been seriously dissapointed, judging by
the dysfunction in family lives I see all around me, so I feel no guilt
about my choices, philosophy, or lifestyle.

I knew an old guy who built a CD player from scratch, with a gold plated double
sided board.
That's dedication to self reliance.

I don't have the time for everything I want to do, but lazing around isn't
on the list.

Patrick Turner.

"Ned Carlson" wrote in message
news
On Tue, 15 Mar 2005 10:49:41 +0200, Iain M Churches wrote:
So, should I rate the transformer

Generally speaking, power transformers are rated based on their
temperature rise at full load over a long period of continuous
use (like between 8 to 24 hours). For shorter periods or on peak demand
they can normally supply somewhat more than the rated current, how
much more depends on how conservatively they're rated, ie: what
temperature rise is considered acceptable at rated load.

320 ma is 160 watts of plate input power, sounds like plenty of
headroom for a 50 watt amp!

I posted a correction to this. The total current draw by one channel
B+ at full power is 240mA at 500V.


Iain