PT and tube limitations

[Chuck H]

Given the issue, why not use it for a tube rectifier?

+1

I'm not sure how a bucker could be arranged on the bi-phase winding?

I thought the same thing. But I didn't chime in because I don't know that it can't be done.

[Enzo]

Idle thought, not really all that clear headed today, but on a CT winding, could be put the 5v winding in the grounded CT leg? Or would that buck one phase and boost the other? Never mind...

[g1]

I think he probably meant the bucking winding on the primary. With line voltages the way they are today, there shouldn't be much problem with reducing everything along with the HV.
The heater voltage will go down a bit, but the main benefit will be for the HV winding.

[pdf64]

Idle thought, not really all that clear headed today, but on a CT winding, could be put the 5v winding in the grounded CT leg? Or would that buck one phase and boost the other? Never mind...

Yes, that's what I think would happen, so the average should work out about the same but ripple would be worse, due to the relative magnitudes of the phases being mismatched.

[Randall]

"A 5F1 amp has a 10k resistor between the plates and screens of the power tubes. This drops the voltage a little for the plates, as well as (the primary purpose) reduces ripple. You could put a resistor in front of the plate node, the resistor simply "eats" some power and the plates see a lower voltage. This allows you to torture a PSU resistor while en route to torturing the signal."

There are two versions of the Herzog. The first has no filter caps at all, simply a 1K 1W between plate and screen, and 10K 1W between screen and preamp. The second version has four filter nodes, an unused one straight off the rectifier, with 470R 5W between it and plate, 10K 2W between plate and screen, and a 10K 1W between screen and preamp. All four have filter caps, but they are not labeled.

So, do I understand that the first 470 5W is to be increased in order to drop some voltage? How would a 2K 10 watt fare in this position?

[eschertron]

four filter nodes, an unused one straight off the rectifier

So that's the pi filter I spoke of, CRC before anything. The caps smooth the ripple, the resistor eats some voltage - but again the primary purpose is ripple reduction.

So, do I understand that the first 470 5W is to be increased in order to drop some voltage? How would a 2K 10 watt fare in this position?

Calculate the total power dissipated in this resistor. Add idle current of the tube in mA, plus the current through the 'speaker replacement' resistor (which equals current in series with the tubes through the OT) in mA, plus a few mA for the preamp. Then add a generous helping of overhead. Any resistor you choose must dissipate all that current. From the power equation, there's "P = I squared R" where I is the current you estimated, and R is the value you chose. The power rating of the resistor must be greater than "P". The voltage dropped across the resistor will be "P / I". All this is estimated, mind you.

edit: looking above, if 470/5W is safe, then since P and R are directly related in the power equation, you'd need a 2k/20W to maintain the same safety margin. Whether you need that big a resistor is what the math is for.

[Dave H]

Calculate the total power dissipated in this resistor. Add idle current of the tube in mA, plus the current through the 'speaker replacement' resistor (which equals current in series with the tubes through the OT) in mA, plus a few mA for the preamp. Then add a generous helping of overhead. Any resistor you choose must dissipate all that current.

I don't think that's right. All you need is the idle current of the 6V6 + preamp. It's class A, the power supply current draw will hardly change from idle to full output. It's normally about 50mA, that's 2.5W dissipation in a 1k resistor or 5W in a 2k resistor. 2k 10W should be OK.

[pdf64]

Yes, in class A, power dissipated in the load is power not dissipated by the plate, so HT draw stays fairly constant, see tables at top of p2 http://www.mif.pg.gda.pl/homepages/f...93/6/6V6GT.pdf
Idle plate current 45mA, max signal plate current 47mA, 4.5mA rising to 7mA for the screen grids.

The higher HTs normally used in guitar amps will probably force a 'cold of centre' bias, which may skew things a bit.

[eschertron]

Yes, in class A, power dissipated in the load is power not dissipated by the plate, so HT draw stays fairly constant, see tables at top of p2 http://www.mif.pg.gda.pl/homepages/f...93/6/6V6GT.pdf
Idle plate current 45mA, max signal plate current 47mA, 4.5mA rising to 7mA for the screen grids.

The higher HTs normally used in guitar amps will probably force a 'cold of centre' bias, which may skew things a bit.

Of course you two are correct about class A. Hadn't thought it all the way through. But when it clips, is it really still in class A? Probably a pedantic exercise anyway, the power to the load is small compared to the tube dissipation. So as long as a reasonable safety factor is figured, I'd disregard any possible deviation from class A.

[Enzo]

It is a class A amp in distortion. Kinda like my mom's sedan. If I try to beat some guy at the traffic light in it, it doesn't become a "race car". Or... if I feed 1 volt of signal into an amplifier, and turn it down so a half a volt comes out, is it no longer an amplifier, but instead an attenuator? And so on.

[J M Fahey]

No need to torture anything, at all.
Nor do it the complicated way, just add an extra RC filter network in the main +V output at the rectifier.

Double bonus feature: you'll have the proper +V as required by standard Champ/Herzog and to boot youŽll have less ripple ... very important in a single ended amp which does not self cancel it.

DidnŽt read *all* the thread, too tired and sleepy now, but in general the Math is simple:
suppose you have 450V and need 350V , also idle current consumption is, say, 50mA .

Then you need to add after rectification and first filter cap a (450V-350V)/0.05A=100/0.05=2000 ohm resistor .

And after that added resistor, a second main filter cap , which to the amplifier, will look like the first one, offering proper voltage and current to it ... and with much reduced ripple

Repeat with your needed voltage and current .

Dissipation?: V * I , in this case 100V * .05A=5W ... repeat with your values and use next higher dissipation for reliability.

[eschertron]

It is a class A amp in distortion. Kinda like my mom's sedan. If I try to beat some guy at the traffic light in it, it doesn't become a "race car". Or... if I feed 1 volt of signal into an amplifier, and turn it down so a half a volt comes out, is it no longer an amplifier, but instead an attenuator? And so on.

OK, I don't know about your mom's sedan :wink: but taking one 6L6 out of my Bassman makes it a class A amplifier?
In my limited understanding, when a tube goes into cutoff we call that class B, don't we? Not being argumentative, just confused.

[Enzo]

Pulling one side of your bassman does not make the other side conduct 360 degrees of signal.

Others may argue the point, and you DID say it was pedantic, but if I fall off my deck, I do not become an aviator just because I am now flying. My dictionary is not a hammer just because I smack something with it. My class A champ doesn;t turn into something else just because I can drive it to a point outside the design specs. Otherwise we can't label any amp. We can always drive it into some other condition if we try.

Otherwise mom's sedan is a race car, and if I take gas money for taking a friend to the airport it is a taxi, and if I pick up a bale of peat moss for the garden on the way home, it is a delivery truck, and if I line up some food containers on the trunk deck at the park it is now a picnic table. "I just bought a new 1968 Chevy picnic table." Isn't it a sedan? Only when I drive to the store.

[eschertron]

Enzo, I do not wish you to fall off your deck!

But to the point of this thread, my confusion is about power supply draw when the SE tube clips. I know that screen current goes up, but now I reason that anode current drops way off for that half cycle. I was thinking incorrectly.
I'm not in a position to test right now, so can we say with confidence that the champ does not draw more current when grossly overdriven?

[Enzo]

Yes. I think that we do not send a sine wave steadily through the amp leaving it to steadily clip. Any clipping is going to be peaks, and any current anomalies will be transient. If you could drive the tube into cutoff, then current would drop. If you could drive the grid into conduction, you might get some more current. But we are talking peaks either way.

How much capability does the Champ have to do that?


You don't want to sail off my deck, you don't want to ride in mom's car, man, pleasing you is no picnic.