Hello,

I have too much HT in my amp because my power transformers are not
exactly what i need.
I have a GZ34 and then a CLC filter, the resulting HT is 480V and i
need 375V.
I have 0.2A current.

I have inserted a 500ohm resistor in serie after the CLC filter.
It is a lot of heat (20W). I use a big radiator and it is ok (around
60=B0C). I plan to isolate it thermically from the rest of the
amplifier.

My question is :
Does that 500ohm resistor has an impact sonically on the rest of the
amplifier?
I think it should not but i wanted to be sure.

Thank you,
Luc D.

wrote:

Hello,

I have too much HT in my amp because my power transformers are not
exactly what i need.
I have a GZ34 and then a CLC filter, the resulting HT is 480V and i
need 375V.
I have 0.2A current.

I have inserted a 500ohm resistor in serie after the CLC filter.
It is a lot of heat (20W). I use a big radiator and it is ok (around
60°C). I plan to isolate it thermically from the rest of the
amplifier.

My question is :
Does that 500ohm resistor has an impact sonically on the rest of the
amplifier?
I think it should not but i wanted to be sure.

Thank you,
Luc D.

You would be much better off putting the resistor before the CLC
filter, I think. With it after the filter you are raising the PSU
impedance considerably making it much more sensitive to load
variation - much bigger change in voltage with load change. Moving
the resistor won't lower the static impedance but will lower the
dynamic impedance for transients. Not such a problem for class 'A'
but it will alter the sound for 'AB'.

Alternatively add another big capacitor after the resistor. The
CLC filter plus resistor will protect the GZ34 so you can go
into the hundreds of uF.


best


Andy

Luc D.

It's a bit drastic, but how about removing the first C of the filter
connected to the GZ34! This will change the voltage from approximately the
peak transformer voltage to the average voltage ~305V in theory. You will
probably get a bit more than this. However, the hum on the h.t. will
increase. Is this a problem? Big advantage is no loss of power.

--
Kind regards

Graham Holloway
WPS/Accuphon Electronics
www.accuphon.co.uk

wrote in message
oups.com...
Hello,

I have too much HT in my amp because my power transformers are not
exactly what i need.
I have a GZ34 and then a CLC filter, the resulting HT is 480V and i
need 375V.
I have 0.2A current.

I have inserted a 500ohm resistor in serie after the CLC filter.
It is a lot of heat (20W). I use a big radiator and it is ok (around
60°C). I plan to isolate it thermically from the rest of the
amplifier.

My question is :
Does that 500ohm resistor has an impact sonically on the rest of the
amplifier?
I think it should not but i wanted to be sure.

Thank you,
Luc D.

wrote:

Hello,

I have too much HT in my amp because my power transformers are not
exactly what i need.
I have a GZ34 and then a CLC filter, the resulting HT is 480V and i
need 375V.
I have 0.2A current.

I have inserted a 500ohm resistor in serie after the CLC filter.
It is a lot of heat (20W). I use a big radiator and it is ok (around
60°C). I plan to isolate it thermically from the rest of the
amplifier.

My question is :
Does that 500ohm resistor has an impact sonically on the rest of the
amplifier?
I think it should not but i wanted to be sure.

If your amp is a pure class A amp, the B+ won't sag while
you use it, since the power input remains very nearly constant
up to clipping.
If the amp is class AB, then at high po the B+ will sag,
which may suit a guitar amp, since the rail drop will cause clipping
to occur at some low power, then give the note sustain.

But even in a hi-fi amp with class AB, the B+ voltage usually
doesn't move if only 3 watts is used out of the 30 max watts,
since 3watts is withing the class A amount produced by the amp.

The 500 ohm resistor added will only slightly add to the choke's
impedance, and thus slightly increase the filtering of the B+ rail
and thus give you less hum on the B+ and thus a cleaner sound.

Should you develop a fault, say a tube saturates and draws say 300mA,
then that will pull down the B+, and Ia for 3 of 4 tubes will get less,
but one may have thr 300mA, then the dissipation in the 500 ohms will
rise from
20 watts at 200 mA to say 80 watts at say 400mA, so if the
R consists of say 3 ten watt x 1,500 ohms in parallel, each
would have 27 watts in it and perhaps this will cause resistor failure,
and hopefully an OPT primary won't fry instead.

Using a 20 watt rated resistor with 20 watts may be quite safe,
and the R won't fail for 10minutes with twice the I.
meanwhile a tube may have 300mA x say 250v across it, = 75 watts,
and if its an EL34, sure, the tube will still melt down in that time.

Since you have a spare 100v of wanted B+, there is
a possibility of using an EL34 as a series pass element for a
regulated supply, so that the 20 watts of power you are wasting in the R
could be wasted in a regulator
tube.
It needn't be a state of the art reg. EL34 in triode cathode follower
mode
will have Ro of about 150 ohms, so a change of
Ia from 200mA to 300 mA would give a B+ sag of only 15v.
You'd need to fix the voltage at the grid with a voltage regulator
tube or string of 100v zeners.
And you'd need a heater supply of 6.3v at 1.6A biased at +400V.

I have an emitter follower regulator using a high voltage line output TV
transistor
on an insulated heatsink..
Its been in a PP amp for 5 years and I use this amp daily in my workshop

for testing speakers.
Ro is about 5 ohms only, ripple voltage is
less than 1mV. No choke is used, only silicon diodes.



Patrick Turner.



Thank you,
Luc D.

Andy Cowley wrote:

wrote:

Hello,

I have too much HT in my amp because my power transformers are not
exactly what i need.
I have a GZ34 and then a CLC filter, the resulting HT is 480V and i
need 375V.
I have 0.2A current.

I have inserted a 500ohm resistor in serie after the CLC filter.
It is a lot of heat (20W). I use a big radiator and it is ok (around
60°C). I plan to isolate it thermically from the rest of the
amplifier.

My question is :
Does that 500ohm resistor has an impact sonically on the rest of the
amplifier?
I think it should not but i wanted to be sure.

Thank you,
Luc D.

You would be much better off putting the resistor before the CLC
filter, I think. With it after the filter you are raising the PSU
impedance considerably making it much more sensitive to load
variation - much bigger change in voltage with load change. Moving
the resistor won't lower the static impedance but will lower the
dynamic impedance for transients. Not such a problem for class 'A'
but it will alter the sound for 'AB'.

Alternatively add another big capacitor after the resistor. The
CLC filter plus resistor will protect the GZ34 so you can go
into the hundreds of uF.

best

Andy

The use of resistors before the first C of the CLC
is another idea, and a good one, but finding the right value
is a bit tricky.

Since the AC current in such R is not a sine wave,
such R tend to get quite hot, and require a higher wattage to do the same
B+ lowering trick.

The regulation with such R is no better than with R
placed after the choke, with mainly only DC flow.

Peak charge currents to C1 will be reduced though, as well
as the DC working voltages on C1 and C2 and the choke.
Voltage ratings need to be kept constant, and yes, 470uF for
C2 is fine, or maybe 2 x 470 uF x 350v rated in series, since one should
always allow for the PSU to remain unloaded by tube Ia and for B+
to soar to its max, and still not cause the electros to blow up.

Patrick Turner.

Assuming you don't have an actual need to heat your house as a by-product,
just reduce the input capacitor to lower the voltage.

wrote in message
oups.com...
Hello,

I have too much HT in my amp because my power transformers are not
exactly what i need.
I have a GZ34 and then a CLC filter, the resulting HT is 480V and i
need 375V.
I have 0.2A current.

I have inserted a 500ohm resistor in serie after the CLC filter.
It is a lot of heat (20W). I use a big radiator and it is ok (around
60°C). I plan to isolate it thermically from the rest of the
amplifier.

My question is :
Does that 500ohm resistor has an impact sonically on the rest of the
amplifier?
I think it should not but i wanted to be sure.

Thank you,
Luc D.

Graham Holloway wrote:

Luc D.

It's a bit drastic, but how about removing the first C of the filter
connected to the GZ34! This will change the voltage from approximately the
peak transformer voltage to the average voltage ~305V in theory. You will
probably get a bit more than this. However, the hum on the h.t. will
increase. Is this a problem? Big advantage is no loss of power.

To get say +490v at C1 and at 0.2A he'd have about 410-0-410 vrms for the
PT HT secondary, if the AC to DC conversion is 1.0vrms for 1.2v of DC.
The only time one sees the maximum ACC conversion rate of 1:1.41
is when no load is used.
Even with Si diodes which have little DC R, the ratio is only 1:1.35.

410vrms in theory will give 410V x 0.89 V DC, = +365V, less the DC drop across
the choke
which may be 10v, so perhaps he'd get +355V only.
The choke value needs to be above the critical value of RL / 940,
which is ( 375 / 0.2 ) / 940 = 2H, and it needs to be a choke that won't hum
or vibrate
when excited by a huge increase in applied AC.

The value of the one C in an LC input can be any high value one wants, since
the charge current in the choke is never a high peak value like it is with a
CLC scheme.
A tenfold increase in C value would be needed to give the same low ripple
as at C2 of a CLC.

Yout option is mentionable, and doable, but there can be problems.

Patrick Turner.





--
Kind regards

Graham Holloway
WPS/Accuphon Electronics
www.accuphon.co.uk

wrote in message
oups.com...
Hello,

I have too much HT in my amp because my power transformers are not
exactly what i need.
I have a GZ34 and then a CLC filter, the resulting HT is 480V and i
need 375V.
I have 0.2A current.

I have inserted a 500ohm resistor in serie after the CLC filter.
It is a lot of heat (20W). I use a big radiator and it is ok (around
60°C). I plan to isolate it thermically from the rest of the
amplifier.

My question is :
Does that 500ohm resistor has an impact sonically on the rest of the
amplifier?
I think it should not but i wanted to be sure.

Thank you,
Luc D.

John Willoughby wrote:

Assuming you don't have an actual need to heat your house as a by-product,
just reduce the input capacitor to lower the voltage.

This is another way of reducing ther B+, and the by products are
poorer B+ reg, but then adding series R anywhere does that also.
The ripple at C1 will be much higher, and since the
attentuation factor of the ripple in L/C2 will stay about constant, ripple at
C2 will
also rise, maybe 5 times.
However, for best LC ripple reduction ZC1 should be at least ZL / 10.
So If L = 2H, ZL = 1,256 ohms at 100 Hz,
and C should be 125 ohms minimum, so C = 12.8 uF minimum.

If the value of C1 was reduced to say 3 uF, then L starts getting lots of AC
applied,
and you are going towards an LC filter, so to keep ripple
low as at present with the existing C2 = 40uF I suspect in there,
one may have to use 235uF, or a couple of 470uF in series, each bypassed with
100k.

I am not sure what value of C would be needed to get the B+ at +375V without
any added R values.

This can be worked out by trial and error though.
the cap type used for C1 can be a oil filled cap, or something polyester or
polypropylene,
and with a generous V rating, 630v at least.

Patrick Turner.







wrote in message
oups.com...
Hello,

I have too much HT in my amp because my power transformers are not
exactly what i need.
I have a GZ34 and then a CLC filter, the resulting HT is 480V and i
need 375V.
I have 0.2A current.

I have inserted a 500ohm resistor in serie after the CLC filter.
It is a lot of heat (20W). I use a big radiator and it is ok (around
60°C). I plan to isolate it thermically from the rest of the
amplifier.

My question is :
Does that 500ohm resistor has an impact sonically on the rest of the
amplifier?
I think it should not but i wanted to be sure.

Thank you,
Luc D.

Why not just use a bucking transformer? Less waste heat, doesn't kill the
regulation and won't cause HT ripple to increase.

"Jason R." wrote:

Why not just use a bucking transformer? Less waste heat, doesn't kill the
regulation and won't cause HT ripple to increase.

That works well sometimes, but in this case the drop needs to be 22%. You
could get part of that thru a smaller input C to the filter but perhaps not
enough. Dropping the line with a bucking transformer on the primary will drop
the heaters as well, perhaps too much in this case. JLS

wrote:

Hello,

I have too much HT in my amp because my power transformers are not
exactly what i need.
I have a GZ34 and then a CLC filter, the resulting HT is 480V

Sounds like your transformer HV is 400-0-400. Simply connecting the
rectifier system for choke input filter will get you 360 volts less a
few volts drop across the choke. The choke will need to be more than the
critical inductance to maintain current flow at all parts of the power
cycle. In this case that would be at least 3H.

See rectifier curves at
http://frank.pocnet.net/sheets/093/5/5AR4.pdf

John Stewart

and i
need 375V.
I have 0.2A current.

I have inserted a 500ohm resistor in serie after the CLC filter.
It is a lot of heat (20W). I use a big radiator and it is ok (around
60=B0C). I plan to isolate it thermically from the rest of the
amplifier.

My question is :
Does that 500ohm resistor has an impact sonically on the rest of the
amplifier?
I think it should not but i wanted to be sure.

Thank you,
Luc D.

Another easy solution is to find a surplus Jap to US line matching
transformer, and plug your amp into that. Get it at a surplus house for a
couple of bucks, and get great isolation to boot. That will lower your HT
by about 15%.


wrote in message
oups.com...
Hello,

I have too much HT in my amp because my power transformers are not
exactly what i need.
I have a GZ34 and then a CLC filter, the resulting HT is 480V and i
need 375V.
I have 0.2A current.

I have inserted a 500ohm resistor in serie after the CLC filter.
It is a lot of heat (20W). I use a big radiator and it is ok (around
60°C). I plan to isolate it thermically from the rest of the
amplifier.

My question is :
Does that 500ohm resistor has an impact sonically on the rest of the
amplifier?
I think it should not but i wanted to be sure.

Thank you,
Luc D.

A big thank's for all your well documented answers.

First i'm reassured for the sonics impacts.

I will try to put the resistor before the clc as it is a big (50w)
reostat. So i can adjust it to have the correct B+. Thank's Andy.

My amp is class AB.

My CLC is 8mfd 10H 8mfd

I don't want to diminue too much the first C because then i have to
increase a lot the second and the time constant is then too big.

I'll give a try at Patrick regulator option EL34 (when i'll have time),
thank's Patrick.

Luc.

wrote:

A big thank's for all your well documented answers.

First i'm reassured for the sonics impacts.

I will try to put the resistor before the clc as it is a big (50w)
reostat. So i can adjust it to have the correct B+. Thank's Andy.

My amp is class AB.

My CLC is 8mfd 10H 8mfd

I don't want to diminue too much the first C because then i have to
increase a lot the second and the time constant is then too big.

I'll give a try at Patrick regulator option EL34 (when i'll have time),
thank's Patrick.

Any old octal power tube will do. They will conduct
200mA at Ea = 100v, which is 20W.
6550 or KT88 are nice, but 6L6, 5881, etc are OK.

But for a simple regulator, a BU208 power transistor,
with MJE340 darlington connected will make a nice
darlington pair emitter follower, Ro very low.
But the secret is to make sure the base voltage is well filtered,
and the circuit is protected since bjts don't like excess voltages or
reverse
voltages, or short circuits which allow PS caps to discharge through them,
so one shoud have 50v across the bjt at 200mA, and a 250 ohm R in series
with the collector.
This means that if the output is shorted to 0V, the max current in the bjt
is only about 1.9 amps.

Patrick Turner.





Luc.

Yo, Luc:

Others have given you some good strategies already but before you even
start on those you should first implement the rectifier correctly and
make your power supply safe, each of which will drop some voltage. This
is an opportunity, not a difficulty!

First off, something often overlooked today is that you are supposed to
insert ballast resistors in the legs between the transformer ends and
the rectifier plates as a matter of course, in your case around 150-180
ohm per leg. See the spec sheets for the GZ34. Such ballasts are a good
thing sonically too as they stabilize the voltage to a certain extent.

Second, no amp is safe without a bleeder resistor to draw current from
charged power caps after the amp is switched off. Generally speaking
you should draw at least ten per cent as much current as you will use
but you can draw much more as a voltage regulation strategy, in both
senses of the word. The bigger the bleed you draw, the better the
voltage regulation, and of course the lower the voltage put out at the
end of the power filter. The bleed resistor value should be V/I =3D R
and the power dissipated in this resistor will be I x I x R =3D W, so
specify 2x or 3x W for safety.The bleed should be across the lines at
the end of the filter or, traditionally, after the last cap of the pi
or CLCL filter. More about this bleeder below where I mention choke
input filters.

First check your choke is the correct value. The choke inductance in
Henries should be at least:
Voltage/(Current x 940) for 50Hz supplies (Europe)
or
Voltage/(Current x 1130) for 60Hz supplies (US)
with the current specified in amps. The cap value should be at least
56/L in Henries. Generally modern "fast" amps use double the cap value
so determined and of course the inductor's value is rounded up to the
nearest convenient larger size, or even a much larger size.

At this point you can measure again and decide how much further voltage
to drop. You can drop this voltage with a series resistor and filter
cap, with values determined in the familiar manner by Colonel Ohm's
handy law and the filter constant you're working to, or by turning your
power supply into a choke input filter.

If you drop further voltage by turning your pi filter into a choke
input filter, the bleeder may well be essential to provide the
necessary standing current for the choke input filter.
To calculate the minimum bleed required as standing current, divide the
voltage by the inductance of the choke, and add at least ten per cent.
This doesn't actually matter so much because your rectifier is a slow
warm-up type and you don't use directly heated tubes so there is no
instant demand on the choke. But, since you are desperate to lose
voltage, the bigger the bleed current the better, since the voltage
dropped here is not just uselessly dissipated in heat as it is in a
series resistor.

Everything changes the sonic signature of your amp, if not always so
much that you can hear it. You will definitely hear a change to a choke
input filter and you may also hear the difference in loading a big
series resistor makes in the frequency response of the amp. The idea is
to give the power supply as low an impedance as realistically possible.
I loved Patrick's idea of shunting a power tube across your lines as a
regular; I've done this with 300B amps that were already exceptionally
fine and discovered an elevated dimension to their sound, though the
cost and complication soon becomes offputting (1).

HTH.

Andre Jute

(1) there are a few remarks about this on my netsite
http://members.lycos.co.uk/fiultra/ (go to JUTE ON AMPS and then to
KISS 114 and KISS 122).

wrote:
Hello,

I have too much HT in my amp because my power transformers are not
exactly what i need.
I have a GZ34 and then a CLC filter, the resulting HT is 480V and i
need 375V.
I have 0.2A current.

I have inserted a 500ohm resistor in serie after the CLC filter.
It is a lot of heat (20W). I use a big radiator and it is ok (around
60=B0C). I plan to isolate it thermically from the rest of the
amplifier.

My question is :
Does that 500ohm resistor has an impact sonically on the rest of the
amplifier?
I think it should not but i wanted to be sure.
=20
Thank you,
Luc D.

Patrick Turner wrote:

wrote:

A big thank's for all your well documented answers.

First i'm reassured for the sonics impacts.

I will try to put the resistor before the clc as it is a big (50w)
reostat. So i can adjust it to have the correct B+. Thank's Andy.

My amp is class AB.

My CLC is 8mfd 10H 8mfd

I don't want to diminue too much the first C because then i have to
increase a lot the second and the time constant is then too big.

I'll give a try at Patrick regulator option EL34 (when i'll have time),
thank's Patrick.

Any old octal power tube will do. They will conduct
200mA at Ea = 100v, which is 20W.
6550 or KT88 are nice, but 6L6, 5881, etc are OK.

If using 6L6/5881 family as a series regulator passer you will need two in
parallel for a reliable 200 ma. In any case, the heater supply will have to be
tied to the cathodes in order to avoid heater-cathode insulation failure. That
means another 6.3 volt winding independent of the others.

An IRF840 FET whose gate is tied to a series of zener diodes totaling your
required voltage plus about 4 volts is a better alternative. The IRF840 is
rated to 500 volts. You will need an adequate heatsink insulated from the
chassis for that one, same as you would when using a bipolar transistor as the
pass element.

Before doing any of that you should try the choke input filter alternative.
Simply unhook the positive lead from your first 8 mfd cap & reconnect it to
the positive lead on the second. Thats all! Since the output stage is
push-pull it automatically cancels any hum that may otherwise appear in the
output signal. At that point your B+ voltage should be close to what you want.
Choke input filtering provides very good power supply regulation, considerably
better than a capacitor input system. JLS

But for a simple regulator, a BU208 power transistor,
with MJE340 darlington connected will make a nice
darlington pair emitter follower, Ro very low.
But the secret is to make sure the base voltage is well filtered,
and the circuit is protected since bjts don't like excess voltages or
reverse
voltages, or short circuits which allow PS caps to discharge through them,
so one shoud have 50v across the bjt at 200mA, and a 250 ohm R in series
with the collector.
This means that if the output is shorted to 0V, the max current in the bjt
is only about 1.9 amps.

Patrick Turner.



Luc.

You could swap the GZ-34 for a 5R4. That will drop the voltage a fair
amount.

wrote in message
oups.com...
Hello,

I have too much HT in my amp because my power transformers are not
exactly what i need.
I have a GZ34 and then a CLC filter, the resulting HT is 480V and i
need 375V.
I have 0.2A current.

I have inserted a 500ohm resistor in serie after the CLC filter.
It is a lot of heat (20W). I use a big radiator and it is ok (around
60°C). I plan to isolate it thermically from the rest of the
amplifier.

My question is :
Does that 500ohm resistor has an impact sonically on the rest of the
amplifier?
I think it should not but i wanted to be sure.

Thank you,
Luc D.

" wrote:

Yo, Luc:

Others have given you some good strategies already but before you even
start on those you should first implement the rectifier correctly and
make your power supply safe, each of which will drop some voltage. This
is an opportunity, not a difficulty!

First off, something often overlooked today is that you are supposed to
insert ballast resistors in the legs between the transformer ends and
the rectifier plates as a matter of course, in your case around 150-180
ohm per leg. See the spec sheets for the GZ34. Such ballasts are a good
thing sonically too as they stabilize the voltage to a certain extent.

The series R would only limit peak charge currents in the GZ34 diodes.
Is not each diode only 120 ohms turned on when used with 400-0-400 vrms HT
?
The 150 ohms in series would allow C1 of a CLC input filter to be large as
possible
with greater reliability, and 60 uF is the largest C value mentioned in
my Miniwatt data book.
There is no mention of the series R though, and about every commercial amp
omits them, but they would make the GZ34 more reliable, last longer,
but wouldn't improve regulation since any series R in any PS reduces the
regulation.
Anyway, where the Ia draw is above 150mA, it is good practice to use some
series R in front of each anode.



Second, no amp is safe without a bleeder resistor to draw current from
charged power caps after the amp is switched off. Generally speaking
you should draw at least ten per cent as much current as you will use
but you can draw much more as a voltage regulation strategy, in both
senses of the word.

Wait a sec, something to discharge caps in a CLC type of supply does not
have to amount to 10% of the anode current.
That would amount to 20mA, and require 10 watts of dissipation.
Surely 3 mA would be enough, giving somebody enough time for caps to
discharge,
bearing in mind that after being on for some time, an amp rail drops fast
with Ia draining away most of it.
The real danger is where Si diodes are used and one switches on, then off
quickly,
and reaches in to work on the amp, only to find the caps have charged to
+560v
with 400-0-400 HT.
Yeow!!, but a small current drain will drop that V to less than lethal
levels fairly fast,
and if C was 100uF, and R was 160k, the time to get from 560v to 200v would
be 16 seconds,
so after 1/2 a minute during which time one would be aware of charged caps,
the
voltage would sag to 36v.

The only time 10% current drain is needed is with the choke input type
of PS, imho.

The bigger the bleed you draw, the better the
voltage regulation, and of course the lower the voltage put out at the
end of the power filter. The bleed resistor value should be V/I = R
and the power dissipated in this resistor will be I x I x R = W, so
specify 2x or 3x W for safety.The bleed should be across the lines at
the end of the filter or, traditionally, after the last cap of the pi
or CLCL filter. More about this bleeder below where I mention choke
input filters.

The regulation offered by a bleeder R drawing 10% of Ia would be very
marginal. The RL of the amp supply on the PS output = B+ / Ia,
or 375 / 0.2 = 1.875k ohms in this case, and a bleeder
of 18.75kohms won't make much difference if Rout is now say a total of
700 ohms with the 500 ohms he wants to use with the GZ34, choke, winding R
etc.





First check your choke is the correct value. The choke inductance in
Henries should be at least:
Voltage/(Current x 940) for 50Hz supplies (Europe)
or
Voltage/(Current x 1130) for 60Hz supplies (US)
with the current specified in amps. The cap value should be at least
56/L in Henries. Generally modern "fast" amps use double the cap value
so determined and of course the inductor's value is rounded up to the
nearest convenient larger size, or even a much larger size.

I assume this last recomendation applies to choke input filters only,
since one could build a CLC with
quite large C1, then the usual rule is that ZL = 10 x ZC1 and C2 value can
be well above C1 value.
If you have C1 = 470 uF, then ZL need only be say 50ohms, so an 80mH choke
will
do and if C2 was another 470uF, then the hum attenuatiuon factor would be
0.067, so hum would be reduced more than 10 times between C1 and C2 with
very little choking.
1.0 Henry would of course be what one would use, for resonance and "desire
for perfection" reasons.

But the figures above that you quote are right for choke input, or LC
filters.



At this point you can measure again and decide how much further voltage
to drop. You can drop this voltage with a series resistor and filter
cap, with values determined in the familiar manner by Colonel Ohm's
handy law and the filter constant you're working to, or by turning your
power supply into a choke input filter.

If you drop further voltage by turning your pi filter into a choke
input filter, the bleeder may well be essential to provide the
necessary standing current for the choke input filter.

The bleeder R need not be used when the output tubes start conducting
where the Ia is quite large. the R stops the B+ soaring to the full
1.41 x HT rms unloaded voltage, maybe +600v
In class B amps where idle current is below the bleeder current,
the bleeder R is a must.
But in class AB amps with a heavy Ia at idle, the bleeder
if used can be disconnected, and since the GZ34 gives a delayed turn on
perhaps it can be omitted entirely.



To calculate the minimum bleed required as standing current, divide the
voltage by the inductance of the choke, and add at least ten per cent.
This doesn't actually matter so much because your rectifier is a slow
warm-up type and you don't use directly heated tubes so there is no
instant demand on the choke. But, since you are desperate to lose
voltage, the bigger the bleed current the better, since the voltage
dropped here is not just uselessly dissipated in heat as it is in a
series resistor.

Everything changes the sonic signature of your amp, if not always so
much that you can hear it. You will definitely hear a change to a choke
input filter and you may also hear the difference in loading a big
series resistor makes in the frequency response of the amp. The idea is
to give the power supply as low an impedance as realistically possible.
I loved Patrick's idea of shunting a power tube across your lines as a
regular; I've done this with 300B amps that were already exceptionally
fine and discovered an elevated dimension to their sound, though the
cost and complication soon becomes offputting (1).

I mentioned that he use a power tube as a series element regulator,
not a shunt regulator. This is the most efficient use of voltage
headroom in this case.

To shunt reg in this case would mean drawing the usual 200mA for the output

tubes, and an extra 100mA in the shunt element, so 300 mA total.
Then when the Ia rises with class AB from 200 mA to 300mA, the shunt
regulator draws no current, so as current from the PS stays constant,
so does the B+.
Man, that's the hard way to get there. It means increasing the PS I draw,
and having 37 watts of dissipation in some shunt tube.

Shunt reg belongs only in preamps where Ia is small, imho.

Using Si diodes in the amp would give him about +540v to start with,
and then he could place a 6550 in the socket meant for the GZ34 as a series
pass
regulator, and maybe all he'd need would be a heater tranny since the 6550
filament needs biasing at the B+ voltage.

Patrick Turner.


HTH.

Andre Jute

(1) there are a few remarks about this on my netsite
http://members.lycos.co.uk/fiultra/ (go to JUTE ON AMPS and then to
KISS 114 and KISS 122).

wrote:
Hello,

I have too much HT in my amp because my power transformers are not
exactly what i need.
I have a GZ34 and then a CLC filter, the resulting HT is 480V and i
need 375V.
I have 0.2A current.

I have inserted a 500ohm resistor in serie after the CLC filter.
It is a lot of heat (20W). I use a big radiator and it is ok (around
60°C). I plan to isolate it thermically from the rest of the
amplifier.

My question is :
Does that 500ohm resistor has an impact sonically on the rest of the
amplifier?
I think it should not but i wanted to be sure.

Thank you,
Luc D.

John Stewart wrote:

Patrick Turner wrote:

wrote:

A big thank's for all your well documented answers.

First i'm reassured for the sonics impacts.

I will try to put the resistor before the clc as it is a big (50w)
reostat. So i can adjust it to have the correct B+. Thank's Andy.

My amp is class AB.

My CLC is 8mfd 10H 8mfd

I don't want to diminue too much the first C because then i have to
increase a lot the second and the time constant is then too big.

I'll give a try at Patrick regulator option EL34 (when i'll have time),
thank's Patrick.

Any old octal power tube will do. They will conduct
200mA at Ea = 100v, which is 20W.
6550 or KT88 are nice, but 6L6, 5881, etc are OK.

If using 6L6/5881 family as a series regulator passer you will need two in
parallel for a reliable 200 ma. In any case, the heater supply will have to be
tied to the cathodes in order to avoid heater-cathode insulation failure. That
means another 6.3 volt winding independent of the others.

I would have thought having 100v across a 6L6 at 200mA would be OK;
its only 20 watts dissipation. Naturally a spare 6550 would do better,
or a 6AS7G with a 12AX7 gain tube to drive it.



An IRF840 FET whose gate is tied to a series of zener diodes totaling your
required voltage plus about 4 volts is a better alternative. The IRF840 is
rated to 500 volts. You will need an adequate heatsink insulated from the
chassis for that one, same as you would when using a bipolar transistor as the
pass element.

I have used the far more rugged BU208.
But yeah, the heatsink needs to be insulated, or else use
two washers instead of just one.


Before doing any of that you should try the choke input filter alternative.
Simply unhook the positive lead from your first 8 mfd cap & reconnect it to
the positive lead on the second. Thats all! Since the output stage is
push-pull it automatically cancels any hum that may otherwise appear in the
output signal. At that point your B+ voltage should be close to what you want.
Choke input filtering provides very good power supply regulation, considerably
better than a capacitor input system. JLS

Now that we know he has a 10H choke, yeah, the choke input is worth a try.

If he hasn't quite got enough B+, a 1 uF or 2 uF cap cor C1 could be used.

He could have a MUCH BIGGER electro cap for C2, say a couple of 470 uF in series
with
100k across each, and the ripple with a choke input filter with no added trimmer
C1
cap would be about 150mV.

This is ok for a PP amp where common mode rejection is good during the
class A working.

Quad II with only 16 uF at the OPT CT and which is charged straight from the
rectifier
has about 17vrms of ripple voltage at the CT, which is quite attrocious.
0.15v is ok.

Patrick Turner.



But for a simple regulator, a BU208 power transistor,
with MJE340 darlington connected will make a nice
darlington pair emitter follower, Ro very low.
But the secret is to make sure the base voltage is well filtered,
and the circuit is protected since bjts don't like excess voltages or
reverse
voltages, or short circuits which allow PS caps to discharge through them,
so one shoud have 50v across the bjt at 200mA, and a 250 ohm R in series
with the collector.
This means that if the output is shorted to 0V, the max current in the bjt
is only about 1.9 amps.

Patrick Turner.



Luc.

Hello,

I have changed yesterday the place of the resistor and tried before the
CLC filter.
I need now a smaller resistor (about 340R).

It works ok and i think it is better than after the CLC filter.

I would like to try Patrick 6550 regulator option. Patrick maybe you
have a schemo somewhere ?

I could also try the LC filter. But i'm not sure of the time contant
(charging time) of a 470mfd. Couldn't that make my amplifier have a bad
bass response?

Thank's for that interesting thread.
Luc D.

Patrick Turner wrote:

John Stewart wrote:

Patrick Turner wrote:

wrote:

A big thank's for all your well documented answers.

First i'm reassured for the sonics impacts.

I will try to put the resistor before the clc as it is a big (50w)
reostat. So i can adjust it to have the correct B+. Thank's Andy.

My amp is class AB.

My CLC is 8mfd 10H 8mfd

I don't want to diminue too much the first C because then i have to
increase a lot the second and the time constant is then too big.

I'll give a try at Patrick regulator option EL34 (when i'll have time),
thank's Patrick.

Any old octal power tube will do. They will conduct
200mA at Ea = 100v, which is 20W.
6550 or KT88 are nice, but 6L6, 5881, etc are OK.

If using 6L6/5881 family as a series regulator passer you will need two in
parallel for a reliable 200 ma. In any case, the heater supply will have to be
tied to the cathodes in order to avoid heater-cathode insulation failure. That
means another 6.3 volt winding independent of the others.

I would have thought having 100v across a 6L6 at 200mA would be OK;
its only 20 watts dissipation.

The dissipation limit is OK but 200 ma is somewhat beyond the safe limit for 6L6
family cathode emission.

Naturally a spare 6550 would do better,
or a 6AS7G with a 12AX7 gain tube to drive it.



An IRF840 FET whose gate is tied to a series of zener diodes totaling your
required voltage plus about 4 volts is a better alternative. The IRF840 is
rated to 500 volts. You will need an adequate heatsink insulated from the
chassis for that one, same as you would when using a bipolar transistor as the
pass element.

I have used the far more rugged BU208.
But yeah, the heatsink needs to be insulated, or else use
two washers instead of just one.


Before doing any of that you should try the choke input filter alternative.
Simply unhook the positive lead from your first 8 mfd cap & reconnect it to
the positive lead on the second. Thats all! Since the output stage is
push-pull it automatically cancels any hum that may otherwise appear in the
output signal. At that point your B+ voltage should be close to what you want.
Choke input filtering provides very good power supply regulation, considerably
better than a capacitor input system. JLS

Now that we know he has a 10H choke, yeah, the choke input is worth a try.

With the circuit constants given the ripple will be less than 2 volts while the
filter resonance is at 12 Hz.
Should work very well with the exception of possible choke audible hum if there are
loose laminations. JLS

If he hasn't quite got enough B+, a 1 uF or 2 uF cap cor C1 could be used.

He could have a MUCH BIGGER electro cap for C2, say a couple of 470 uF in series
with
100k across each, and the ripple with a choke input filter with no added trimmer
C1
cap would be about 150mV.

This is ok for a PP amp where common mode rejection is good during the
class A working.

Quad II with only 16 uF at the OPT CT and which is charged straight from the
rectifier
has about 17vrms of ripple voltage at the CT, which is quite attrocious.
0.15v is ok.

Patrick Turner.



But for a simple regulator, a BU208 power transistor,
with MJE340 darlington connected will make a nice
darlington pair emitter follower, Ro very low.
But the secret is to make sure the base voltage is well filtered,
and the circuit is protected since bjts don't like excess voltages or
reverse
voltages, or short circuits which allow PS caps to discharge through them,
so one shoud have 50v across the bjt at 200mA, and a 250 ohm R in series
with the collector.
This means that if the output is shorted to 0V, the max current in the bjt
is only about 1.9 amps.

Patrick Turner.



Luc.

John Stewart wrote:

Patrick Turner wrote:

John Stewart wrote:

Patrick Turner wrote:

wrote:

A big thank's for all your well documented answers.

First i'm reassured for the sonics impacts.

I will try to put the resistor before the clc as it is a big (50w)
reostat. So i can adjust it to have the correct B+. Thank's Andy.

My amp is class AB.

My CLC is 8mfd 10H 8mfd

I don't want to diminue too much the first C because then i have to
increase a lot the second and the time constant is then too big.

I'll give a try at Patrick regulator option EL34 (when i'll have time),
thank's Patrick.

Any old octal power tube will do. They will conduct
200mA at Ea = 100v, which is 20W.
6550 or KT88 are nice, but 6L6, 5881, etc are OK.

If using 6L6/5881 family as a series regulator passer you will need two in
parallel for a reliable 200 ma. In any case, the heater supply will have to be
tied to the cathodes in order to avoid heater-cathode insulation failure. That
means another 6.3 volt winding independent of the others.

I would have thought having 100v across a 6L6 at 200mA would be OK;
its only 20 watts dissipation.

The dissipation limit is OK but 200 ma is somewhat beyond the safe limit for 6L6
family cathode emission.

I should have added to that the 6L6 family when triode connected as is the usual case
when operated as the series pass element in a regulator can deliver only 50 ma with a
100 volt drop. They do better as pentodes but that adds further complication to the
circuit since yet another off ground (common) auxiliary power supply is required for
the screen(s) connexion.

A triode such as the 6080/6AS7G family would be OK. IMO the added complication of a
vacuum tube regulator circuit is not justified at all. Besides the 6080 you will need a
gain tube such as a 6AU6 & a regulator VR tube such as an 0B2.

JLS

Naturally a spare 6550 would do better,
or a 6AS7G with a 12AX7 gain tube to drive it.



An IRF840 FET whose gate is tied to a series of zener diodes totaling your
required voltage plus about 4 volts is a better alternative. The IRF840 is
rated to 500 volts. You will need an adequate heatsink insulated from the
chassis for that one, same as you would when using a bipolar transistor as the
pass element.

I have used the far more rugged BU208.
But yeah, the heatsink needs to be insulated, or else use
two washers instead of just one.


Before doing any of that you should try the choke input filter alternative.
Simply unhook the positive lead from your first 8 mfd cap & reconnect it to
the positive lead on the second. Thats all! Since the output stage is
push-pull it automatically cancels any hum that may otherwise appear in the
output signal. At that point your B+ voltage should be close to what you want.
Choke input filtering provides very good power supply regulation, considerably
better than a capacitor input system. JLS

Now that we know he has a 10H choke, yeah, the choke input is worth a try.

With the circuit constants given the ripple will be less than 2 volts while the
filter resonance is at 12 Hz.
Should work very well with the exception of possible choke audible hum if there are
loose laminations. JLS

If he hasn't quite got enough B+, a 1 uF or 2 uF cap cor C1 could be used.

He could have a MUCH BIGGER electro cap for C2, say a couple of 470 uF in series
with
100k across each, and the ripple with a choke input filter with no added trimmer
C1
cap would be about 150mV.

This is ok for a PP amp where common mode rejection is good during the
class A working.

Quad II with only 16 uF at the OPT CT and which is charged straight from the
rectifier
has about 17vrms of ripple voltage at the CT, which is quite attrocious.
0.15v is ok.

Patrick Turner.



But for a simple regulator, a BU208 power transistor,
with MJE340 darlington connected will make a nice
darlington pair emitter follower, Ro very low.
But the secret is to make sure the base voltage is well filtered,
and the circuit is protected since bjts don't like excess voltages or
reverse
voltages, or short circuits which allow PS caps to discharge through them,
so one shoud have 50v across the bjt at 200mA, and a 250 ohm R in series
with the collector.
This means that if the output is shorted to 0V, the max current in the bjt
is only about 1.9 amps.

Patrick Turner.



Luc.

wrote:

Hello,

I have changed yesterday the place of the resistor and tried before the
CLC filter.
I need now a smaller resistor (about 340R).

It works ok and i think it is better than after the CLC filter.

I would like to try Patrick 6550 regulator option. Patrick maybe you
have a schemo somewhere ?

I have never used a tube series regulator; its always been a darlington
pair emitter follower,
and the SS gave far better regulation than any single cathode follower tube
could.

It was always easier to build the SS circuit on a small board and have a
small heatsink
about the size of a choke, which isn't needed or wanted when one has
an active series regulator, since whenever current is reduced in the choke
the voltage
input to the bjt collector circuit tries to rise, and may cause a bjt to go
into avalanche.





I could also try the LC filter. But i'm not sure of the time contant
(charging time) of a 470mfd. Couldn't that make my amplifier have a bad
bass response?

The charge up time is not important; you don't want or need it to be
instant.

The idea of using a pair of 350v rated 470 uF caps in series is to
create the effect of a low impedance supply for signal frequencies, so that
once the caps are charged
at +375V, any DC supply variations due to signal excursions into class AB
working region
will not much change the B+ instantly.
Since crescendos or guitar rifts have a small duty cycle, ie, they don't
last long like a continuous sine wave applied to the
amp then the B+ won't move fast as it **definately will** with 10H followed
by only 8 uF.

If you are using the amp for hi-fi, and if power at clipping = 30watts,
then this will be say 13.4 vrms into 6 ohms, and class AB, with a sine wave
test signal
and Ia to the two tubes may double above the quiescent value and B+ would
sag say
15% from 375V to 319V.
But with music, no peaks can be greater than 13.4 vrms, so the average
music voltage level will typically be only 2.68 vrms which ius a very loud
sounding 1.2 watts into 6 ohms.

Since the average power is only 1.2 watts, and the range of class A power
is the first 10 watts,
then the amp will spend its evenings with you only ever using its class AB
action on the occasional drumbeat
and transient.
Since these don't last long, the B+ will not sag, due to the large store of
energy in the
235 uF that I suggest you try.

If you want to understand the issue, just leave the C1 = 8 uF you have,
add your 500 ohms after the choke, and then use the 235 uF, ( 2 x 470uF in
series ).

The run some of your busiest music loudly, and use a peak and hold
digital voltage meter to record the largest signal peaks.
If your amp does 30 watts to 6 ohms, that's 13.4 vrms, which is 18.9 peak
volts.
So when you run some music up to that level, you know there ain't no more
volume to be had unless you like severe thd and imd.

While running up the music to such absurdly loud levels,
you may deafen yourself, so use a dummy resistor rated at 50 watts for the
load instead
of a speaker, and just to hear the sound a bit, use a 100ohm to 2 ohm
divider, and take the speaker drive
off the 2 ohms.

An oscilliscope is the right way to inspect what is happening in your amp
with regard to clipping.

While this test is done the B+ voltage shouldn't drop down more than about
5%.

But at normal listening levels, it won't drop much at all.

The impedance of an 8uF cap at 50 Hz is 398 ohms.
The 10H choke has an impedance at 50 Hz = 6,280 ohms, so the only
thing mainly holding the CT at a stable working B+ voltage is that 398 ohms
of the 8 uF.

The 8uF and 10H form a second order low pass filter, and it obeys all the
filter rules,
and the cut off frequency is at 17.8 Hz.

The load on the filter isn't anywhere near low enough to damp the resonant
behaviour of the LC filter, so everytime Ia changes with amp current
demands because of AB
operation, the PS will oscillate a bit at 17.8 Hz, which is not an ideal
situation.

But if we use 235 uF, the ZC = 3.38 ohms at 50 Hz and the filter
resonance is 3.28 Hz, and this is far less likely to get excited by the
amp demand behaviour.

Whenever the amp goes into class AB operation there is only signal current
in one half
of the OPT, so there is no common mode hum rejection that can only
occur while the amp is in class A.
So at low F when a large signal is present, the LF signal can modulate the
gain of the amp
at higher F; this is intermodulation by power supply rail hum, and isn't
desirable at all. But if the ZC at the CT is 3.4 ohms instead of 398 ohms,
then the
LF signal at the CT will be tiny, and the amount imd will be well reduced.

Anyway, try it, observe it, measure it, get to know what you are doing
about basic issues, and then the sound can only improove.

To make the 6550 regulator, you really don't need a choke because the
plate input impedance will be high if the tube is working in beam tetrode.
This would mean having a regulated grid voltage at about +395V, and having
a
screen supply fed by the from the 475V, and also having the screen bypassed
to the cathode
with about 100 uFchoke, and this way the Ro at the 6550 cathode
would be lowest, and the plate input impedance high at about 15 kohms,
so that where you have 55 vrms of ripple at C1 8uF cap, there will be
3.66 mA of ripple current into the 6550 anode circuit.
The Ro of the cathode will be 1 / gm = 1 / 0.01 = 100 ohms, so ripple
voltage at the cathode will
100 ohms x 3.66 mA = 0.366 volts.

Trouble is with such a regulator is that you still need a humungous cap at
the CT which
the cathode of the 6550 will feed, or else you will have 100 ohms
impedance at all signal F. Caps offer a falling Z as f rises, so at
1 kHz, 235 uF has Z = 0.677 ohms, a rather low Z indeed.

Anyway, the simplest option is to use the 20 watt resistor,
but change the C2 from 8 to 235 uF, and that will
give you a far firmer supply voltage which your amp needs
to work at its best.

I usually use use all silicon rectifiers with say 50 ohms in series from
the
HT CT to 0V, then C1 = 235 uF,
L = about 3H, then C2 = about 1,000 uF.

At 0.5 amps of DC, the ripple at the CT is in millivolts,
and LF signals don't cause any large signal to appear at the CT.

People say they hate electrolytic caps, and that electros ruin music,
but **all** the real hard evidence suggests the opposite.
I am also seen as a heretic in some hi-fi circles because I
don't think the brand of HV elcaps matter much either.
Generic cheap Jamiko 470 uF x 350v caps made in Taiwan
each with a 2 amp ripple current rating, and made to work damned hard in
switch mode PS will do the job in anyone's tube power amp excellently.
I don't believe that Black Gate, Cerafine, Elna, or any oiler caps or
huge lumbering polypropylene or any fancy-shmancy
brands have a monopoly at providing good music with tubes.
If in doubt, use 1uF poly caps rated at 1,000v for across the rails
at C1 and C2 and for other decoupled stages;
Elcaps have a rising Z at RF, and the poly caps need to bypass supply
rails at such F.

Elcaps made now are *far better* than the unreliable types made 40 years
ago
which gave elcaps such a horrid reputation.
Any PC or any other TV or complex HT reciever might have
50 electrolytics within, and the failure rates have to be
extremely low, lest customers loose faith in the industry.

Patrick Turner.








Thank's for that interesting thread.
Luc D.

John Stewart wrote:

Patrick Turner wrote:

John Stewart wrote:

Patrick Turner wrote:

wrote:

A big thank's for all your well documented answers.

First i'm reassured for the sonics impacts.

I will try to put the resistor before the clc as it is a big (50w)
reostat. So i can adjust it to have the correct B+. Thank's Andy.

My amp is class AB.

My CLC is 8mfd 10H 8mfd

I don't want to diminue too much the first C because then i have to
increase a lot the second and the time constant is then too big.

I'll give a try at Patrick regulator option EL34 (when i'll have time),
thank's Patrick.

Any old octal power tube will do. They will conduct
200mA at Ea = 100v, which is 20W.
6550 or KT88 are nice, but 6L6, 5881, etc are OK.

If using 6L6/5881 family as a series regulator passer you will need two in
parallel for a reliable 200 ma. In any case, the heater supply will have to be
tied to the cathodes in order to avoid heater-cathode insulation failure. That
means another 6.3 volt winding independent of the others.

I would have thought having 100v across a 6L6 at 200mA would be OK;
its only 20 watts dissipation.

The dissipation limit is OK but 200 ma is somewhat beyond the safe limit for 6L6
family cathode emission.

I agree, 200 mA is on the highside, but at 100V across the tube, Pda = 20 watts only.
A 6550 would be far better..

But why have a tube regulator when solid state acting as
compliant slaves to the tubes' every desire for current will be cheaper
and work better.

The trouble with any regulator is that it is something that has to be made to work
properly,
and self protect itself, so that blatting the output with an old screw driver to 0V
won't make the devices fail, and same goes for shorts on the
input side of the reg.

In short, if anyone has not built a few regulated supplies, its a very
steep learning curve to get **all** the details right.

There is no use me providing too much info; ppl must be prepared to
take the long hard hours to work all this stuff out themselves
like I did.





Naturally a spare 6550 would do better,
or a 6AS7G with a 12AX7 gain tube to drive it.



An IRF840 FET whose gate is tied to a series of zener diodes totaling your
required voltage plus about 4 volts is a better alternative. The IRF840 is
rated to 500 volts. You will need an adequate heatsink insulated from the
chassis for that one, same as you would when using a bipolar transistor as the
pass element.

I have used the far more rugged BU208.
But yeah, the heatsink needs to be insulated, or else use
two washers instead of just one.


Before doing any of that you should try the choke input filter alternative.
Simply unhook the positive lead from your first 8 mfd cap & reconnect it to
the positive lead on the second. Thats all! Since the output stage is
push-pull it automatically cancels any hum that may otherwise appear in the
output signal. At that point your B+ voltage should be close to what you want.
Choke input filtering provides very good power supply regulation, considerably
better than a capacitor input system. JLS

Now that we know he has a 10H choke, yeah, the choke input is worth a try.

With the circuit constants given the ripple will be less than 2 volts while the
filter resonance is at 12 Hz.

let me see now,
With a CLC, 8uF, 10H, 8 uF, Vr at C1 at 100Hz worst case = 55 v at 0.2 amps.
Yeah, I get 1.75 vrms of 100Hz ripple at C2.

With C2 = 235 uF with the choke ( and the added 500 ohms )
will give about 58 mV, and this is far better than your 2v, and Fo will be about
3.3 Hz, not 12Hz, or 17Hz as I would have calculated for 10H and 8 uF.



Should work very well with the exception of possible choke audible hum if there are
loose laminations. JLS

With a choke input filter, ripple V at the choke input is about 140vrms,
and so with 10H and 8 uF, Vr would be 4.45 vrms, no?

With 235 uF for C2, the Vr with just a choke would be about 0.15vrms.
Its only marginally worse than the 8u -10H - 35u set up.

The use of the 8 uF without ballast resistors in series with the tube diodes
is fine; the peak charge currents are very low with only 8 uF.
Far higher C values are possible with GZ34, so 8 uF is a doddle.


Patrick Turner.




If he hasn't quite got enough B+, a 1 uF or 2 uF cap cor C1 could be used.

He could have a MUCH BIGGER electro cap for C2, say a couple of 470 uF in series
with
100k across each, and the ripple with a choke input filter with no added trimmer
C1
cap would be about 150mV.

This is ok for a PP amp where common mode rejection is good during the
class A working.

Quad II with only 16 uF at the OPT CT and which is charged straight from the
rectifier
has about 17vrms of ripple voltage at the CT, which is quite attrocious.
0.15v is ok.

Patrick Turner.



But for a simple regulator, a BU208 power transistor,
with MJE340 darlington connected will make a nice
darlington pair emitter follower, Ro very low.
But the secret is to make sure the base voltage is well filtered,
and the circuit is protected since bjts don't like excess voltages or
reverse
voltages, or short circuits which allow PS caps to discharge through them,
so one shoud have 50v across the bjt at 200mA, and a 250 ohm R in series
with the collector.
This means that if the output is shorted to 0V, the max current in the bjt
is only about 1.9 amps.

Patrick Turner.



Luc.

John Stewart wrote:

John Stewart wrote:

Patrick Turner wrote:

John Stewart wrote:

Patrick Turner wrote:

wrote:

A big thank's for all your well documented answers.

First i'm reassured for the sonics impacts.

I will try to put the resistor before the clc as it is a big (50w)
reostat. So i can adjust it to have the correct B+. Thank's Andy.

My amp is class AB.

My CLC is 8mfd 10H 8mfd

I don't want to diminue too much the first C because then i have to
increase a lot the second and the time constant is then too big.

I'll give a try at Patrick regulator option EL34 (when i'll have time),
thank's Patrick.

Any old octal power tube will do. They will conduct
200mA at Ea = 100v, which is 20W.
6550 or KT88 are nice, but 6L6, 5881, etc are OK.

If using 6L6/5881 family as a series regulator passer you will need two in
parallel for a reliable 200 ma. In any case, the heater supply will have to be
tied to the cathodes in order to avoid heater-cathode insulation failure. That
means another 6.3 volt winding independent of the others.

I would have thought having 100v across a 6L6 at 200mA would be OK;
its only 20 watts dissipation.

The dissipation limit is OK but 200 ma is somewhat beyond the safe limit for 6L6
family cathode emission.

I should have added to that the 6L6 family when triode connected as is the usual case
when operated as the series pass element in a regulator can deliver only 50 ma with a
100 volt drop. They do better as pentodes but that adds further complication to the
circuit since yet another off ground (common) auxiliary power supply is required for
the screen(s) connexion.

I am not sure. The tube could be allowed to draw grid current.
The voltage supply to the grid must itself be well
regged and smoothed, which suggests a couple of 200V reg tubes,
or some seriesed 5 watt zener diodes with cap bypassing.

Its a fancy idea, series tube regulation, and although it is initially
fascinating, and doable, after all these years the only series tube regulator
I have ever built is in my bench top PSU, allowing
setable output voltages at 50v increments above 200v up to
550v.

Series regs are used in amps made by audio research Reference 600 types, and i don't
wish these complex amps on anyone.

In class AB tube power amps in 2005, there is simply
**no need** to actively series regulate any B+
supply anywhere, since silicon diodes and low dcr chokes and huge caps allow
excellent self regulation of the electrode voltages.
However, where somebody is doing a trick to reduce the B+, I can see
the attraction of the regulation.

For screen supplies in my amps I have shunt regulation and hu-mungous value
bypass caps of 470 uF.
Then if a screen decides to conduct too much current, the voltage supply
sags, since there is a dropping resistor there to allow this to occur,
once Ig2 becomes a little too much.
I set up the shunt reg to give excellent reg using a string of zeners
and so that with music taken up to clipping
Eg2 only drops a few v with 1/2 rated load on the output.
This means that about the same idle current in the reg must flow as the screen current
itself at idle,
so as Ig2 increases, less I flows in the zeners, but once all the current flows to the
screens,
the series R to the zeners and screens gets a voltage drop larger than the idle condition.
One should only regulate where one has to, and no more.

Where I have used a fixed EG2 supply about 100V lower than the plate B+,
I could have used a separate HT winding on the PT to develop the EG2,
and this would offer ample self regging with Si diodes etc, but also mean that
I have to worry about fried screens from too much Ig2, and no ability
for the Eg2 to sag if Ig2 in any one tube becomes too much.
So there is a case for a fusible resistor link to each screen.
But usually if Ig2 goes too high, Ia will also be too high, and hence Ik
will be high, and active protection is the answer.

With a series reg, the series tube damn well tries to keep Eg2 at the wanted fixed level,
and a screen and a tube may melt down as a result.
Same goes for plate supplies; regulation is all very well, but
its best to rig things that once Ia goes over say 300 mA when 200mA is the idle value,
the regulation should cease and the B+ be allowed to fall, and of course
active sensing of cathode current in each of the 4 output tubes should
tell the PSU to turn itself off if over current persists longer than say 5 seconds.

In 2005, ppl expect tube electronics to be reliable, or at least polite when they
get a fault.
I don't like phone calls where folks tell me that the amp I built just electrocuted
their daughter, and burnt the house down.


A triode such as the 6080/6AS7G family would be OK. IMO the added complication of a
vacuum tube regulator circuit is not justified at all. Besides the 6080 you will need a
gain tube such as a 6AU6 & a regulator VR tube such as an 0B2.

All true.
Ro for the 6AS7 is probably lower than a 6550.
1 / gm = 1 / 0.014 = 71 ohms for both halves of the tube.

Patrick Turner.



JLS

Naturally a spare 6550 would do better,
or a 6AS7G with a 12AX7 gain tube to drive it.



An IRF840 FET whose gate is tied to a series of zener diodes totaling your
required voltage plus about 4 volts is a better alternative. The IRF840 is
rated to 500 volts. You will need an adequate heatsink insulated from the
chassis for that one, same as you would when using a bipolar transistor as the
pass element.

I have used the far more rugged BU208.
But yeah, the heatsink needs to be insulated, or else use
two washers instead of just one.


Before doing any of that you should try the choke input filter alternative.
Simply unhook the positive lead from your first 8 mfd cap & reconnect it to
the positive lead on the second. Thats all! Since the output stage is
push-pull it automatically cancels any hum that may otherwise appear in the
output signal. At that point your B+ voltage should be close to what you want.
Choke input filtering provides very good power supply regulation, considerably
better than a capacitor input system. JLS

Now that we know he has a 10H choke, yeah, the choke input is worth a try.

With the circuit constants given the ripple will be less than 2 volts while the
filter resonance is at 12 Hz.
Should work very well with the exception of possible choke audible hum if there are
loose laminations. JLS

If he hasn't quite got enough B+, a 1 uF or 2 uF cap cor C1 could be used.

He could have a MUCH BIGGER electro cap for C2, say a couple of 470 uF in series
with
100k across each, and the ripple with a choke input filter with no added trimmer
C1
cap would be about 150mV.

This is ok for a PP amp where common mode rejection is good during the
class A working.

Quad II with only 16 uF at the OPT CT and which is charged straight from the
rectifier
has about 17vrms of ripple voltage at the CT, which is quite attrocious.
0.15v is ok.

Patrick Turner.



But for a simple regulator, a BU208 power transistor,
with MJE340 darlington connected will make a nice
darlington pair emitter follower, Ro very low.
But the secret is to make sure the base voltage is well filtered,
and the circuit is protected since bjts don't like excess voltages or
reverse
voltages, or short circuits which allow PS caps to discharge through them,
so one shoud have 50v across the bjt at 200mA, and a 250 ohm R in series
with the collector.
This means that if the output is shorted to 0V, the max current in the bjt
is only about 1.9 amps.

Patrick Turner.



Luc.

Patrick Turner wrote:

John Stewart wrote:

Patrick Turner wrote:

John Stewart wrote:

Patrick Turner wrote:

wrote:

A big thank's for all your well documented answers.

First i'm reassured for the sonics impacts.

I will try to put the resistor before the clc as it is a big (50w)
reostat. So i can adjust it to have the correct B+. Thank's Andy.

My amp is class AB.

My CLC is 8mfd 10H 8mfd

I don't want to diminue too much the first C because then i have to
increase a lot the second and the time constant is then too big.

I'll give a try at Patrick regulator option EL34 (when i'll have time),
thank's Patrick.

Any old octal power tube will do. They will conduct
200mA at Ea = 100v, which is 20W.
6550 or KT88 are nice, but 6L6, 5881, etc are OK.

If using 6L6/5881 family as a series regulator passer you will need two in
parallel for a reliable 200 ma. In any case, the heater supply will have to be
tied to the cathodes in order to avoid heater-cathode insulation failure. That
means another 6.3 volt winding independent of the others.

I would have thought having 100v across a 6L6 at 200mA would be OK;
its only 20 watts dissipation.

The dissipation limit is OK but 200 ma is somewhat beyond the safe limit for 6L6
family cathode emission.

I agree, 200 mA is on the highside, but at 100V across the tube, Pda = 20 watts only.
A 6550 would be far better..

But why have a tube regulator when solid state acting as
compliant slaves to the tubes' every desire for current will be cheaper
and work better.

The trouble with any regulator is that it is something that has to be made to work
properly,
and self protect itself, so that blatting the output with an old screw driver to 0V
won't make the devices fail, and same goes for shorts on the
input side of the reg.

In short, if anyone has not built a few regulated supplies, its a very
steep learning curve to get **all** the details right.

There is no use me providing too much info; ppl must be prepared to
take the long hard hours to work all this stuff out themselves
like I did.



Naturally a spare 6550 would do better,
or a 6AS7G with a 12AX7 gain tube to drive it.



An IRF840 FET whose gate is tied to a series of zener diodes totaling your
required voltage plus about 4 volts is a better alternative. The IRF840 is
rated to 500 volts. You will need an adequate heatsink insulated from the
chassis for that one, same as you would when using a bipolar transistor as the
pass element.

I have used the far more rugged BU208.
But yeah, the heatsink needs to be insulated, or else use
two washers instead of just one.


Before doing any of that you should try the choke input filter alternative.
Simply unhook the positive lead from your first 8 mfd cap & reconnect it to
the positive lead on the second. Thats all! Since the output stage is
push-pull it automatically cancels any hum that may otherwise appear in the
output signal. At that point your B+ voltage should be close to what you want.
Choke input filtering provides very good power supply regulation, considerably
better than a capacitor input system. JLS

Now that we know he has a 10H choke, yeah, the choke input is worth a try.

With the circuit constants given the ripple will be less than 2 volts while the
filter resonance is at 12 Hz.

let me see now,
With a CLC, 8uF, 10H, 8 uF, Vr at C1 at 100Hz worst case = 55 v at 0.2 amps.
Yeah, I get 1.75 vrms of 100Hz ripple at C2.

With C2 = 235 uF with the choke ( and the added 500 ohms )
will give about 58 mV, and this is far better than your 2v, and Fo will be about
3.3 Hz, not 12Hz, or 17Hz as I would have calculated for 10H and 8 uF.


Should work very well with the exception of possible choke audible hum if there are
loose laminations. JLS

With a choke input filter, ripple V at the choke input is about 140vrms,
and so with 10H and 8 uF, Vr would be 4.45 vrms, no?

There are several variables here & that is why the results for ripple calculations may
differ. I cheated & simply used a simulation program. The result I got was in rms volts
where other kinds of calculations may give the answer in the form of peak-to-peak volts.
Where you had used 50 Hz as the supply I assumed 60 Hz, probably because the program
defaulted at that & I looked no further. Some of the cook book formulae make simplifying
assumptions such as ignoring all but the first two or three terms in the expansion of the
ripple spectrum, so another source of differing results.

I assumed the first 8 mfd to be reconnected in parallel with the 2nd so that the
simulation had 16 mfd after the 10 H choke.

Beyond all that we don't know anything about winding resistance in the transformer or the
condition of the 5AR4. To cover that I put 75R into each of the rectifier plate leads. I
also stuck 50R into the choke.

Accurate results are not really needed. I just look for answers 'within an order of
magnitude', ie, is it one volt or is it 10 volts? There are too many unknowns to
calculate an accurate result. But a good estimate is useful.

For the amplifier in question an easy fix would increase the cap after the choke input to
100 mfd which are cheap & easily available these days.

Cheers, John Stewart

With 235 uF for C2, the Vr with just a choke would be about 0.15vrms.
Its only marginally worse than the 8u -10H - 35u set up.

The use of the 8 uF without ballast resistors in series with the tube diodes
is fine; the peak charge currents are very low with only 8 uF.
Far higher C values are possible with GZ34, so 8 uF is a doddle.

Patrick Turner.

Patrick Turner wrote:

John Stewart wrote:

John Stewart wrote:

Patrick Turner wrote:

John Stewart wrote:

Patrick Turner wrote:

wrote:

A big thank's for all your well documented answers.

First i'm reassured for the sonics impacts.

I will try to put the resistor before the clc as it is a big (50w)
reostat. So i can adjust it to have the correct B+. Thank's Andy.

My amp is class AB.

My CLC is 8mfd 10H 8mfd

I don't want to diminue too much the first C because then i have to
increase a lot the second and the time constant is then too big.

I'll give a try at Patrick regulator option EL34 (when i'll have time),
thank's Patrick.

Any old octal power tube will do. They will conduct
200mA at Ea = 100v, which is 20W.
6550 or KT88 are nice, but 6L6, 5881, etc are OK.

If using 6L6/5881 family as a series regulator passer you will need two in
parallel for a reliable 200 ma. In any case, the heater supply will have to be
tied to the cathodes in order to avoid heater-cathode insulation failure. That
means another 6.3 volt winding independent of the others.

I would have thought having 100v across a 6L6 at 200mA would be OK;
its only 20 watts dissipation.

The dissipation limit is OK but 200 ma is somewhat beyond the safe limit for 6L6
family cathode emission.

I should have added to that the 6L6 family when triode connected as is the usual case
when operated as the series pass element in a regulator can deliver only 50 ma with a
100 volt drop. They do better as pentodes but that adds further complication to the
circuit since yet another off ground (common) auxiliary power supply is required for
the screen(s) connexion.

I am not sure. The tube could be allowed to draw grid current.

That would usually mean a CF & further complication, although it could be done. I did it in a
few regulated power supplies early on. One of the circuits used a set of three 6N7 double
triodes as the passer. The CF was a 12B4.

Another was the 4500 volt, one ampere monster. It used a pair of 304TH's as the passer. The
driving CF was an 813.

The voltage supply to the grid must itself be well
regged and smoothed, which suggests a couple of 200V reg tubes,
or some seriesed 5 watt zener diodes with cap bypassing.

Yes, in the simplest of regulator circuits you would need to do that. All of the many
regulator circuits I built where active in that they had error amplifiers of more or less
sophistication.

Its a fancy idea, series tube regulation, and although it is initially
fascinating, and doable, after all these years the only series tube regulator
I have ever built is in my bench top PSU, allowing
setable output voltages at 50v increments above 200v up to
550v.

Series regs are used in amps made by audio research Reference 600 types, and i don't
wish these complex amps on anyone.

In class AB tube power amps in 2005, there is simply
**no need** to actively series regulate any B+
supply anywhere, since silicon diodes and low dcr chokes and huge caps allow
excellent self regulation of the electrode voltages.
However, where somebody is doing a trick to reduce the B+, I can see
the attraction of the regulation.

Of all the many amplifiers & regulated PS's I've built until recently none were ever connected
on the same chassis. I never saw the need to have a regulated supply for any of my amplifiers.
However I've used simple power FET regulators on the past two amplifier projects as a easy
means of getting soft start when using a HV standby circuit. The added benefits of this
approach were regulation & very good PS hum elimination. The circuits are relatively simple
when compared to vacuum tube regulators.

JLS

Patrick Turner wrote:

Since you have a spare 100v of wanted B+, there is
a possibility of using an EL34 as a series pass element for a
regulated supply, so that the 20 watts of power you are wasting in the R
could be wasted in a regulator
tube.
It needn't be a state of the art reg. EL34 in triode cathode follower
mode
will have Ro of about 150 ohms, so a change of
Ia from 200mA to 300 mA would give a B+ sag of only 15v.

Forgive me my ignorance, but are you saying that the B+ voltage could
be reduced by putting a tube in series with the B+ power supply, and
that the EL34 would only sag 15V going from 200mA to 300mA current
draw? Does that mean that the same strategy could be employed by
putting a diode tube like a Type 80 in series to be used in a guitar
amp application where you'd want sag, and would the decrease in B+
voltage caused by the 100 or so ohms of the tube at lower current
levels drop the voltage too low to be compensated for otherwise, other
than just adjusting the bias? Could a high wattage resistor be added
in parallel with the tube to achieve a voltage drop that was within
acceptable parameters?

Again, forgive my ignorance.

Patrick Turner.

--Bryan