Close PT voltage but big B+1 voltage difference

[andrew]

Help please. I am comparing two power transformers. One is 260vts 300ma. At the first node I get 290vts with a GZ34. The other is 275vts 150ma. and with the same GZ34 at the first node I get 360vts. Why would there be such a difference? Thank you!

[collinsamps]

Need more info to help you. I'm assuming that it's per leg voltage? What amp? Unloaded or loaded(tubes installed?). The first filter cap size? Typically you can expect 1.414 x one leg of the secondary AC on an unloaded source.

[andrew]

I win the #%^! prize tonight! Four tubes = less voltage on B+1 -two tubes = more voltage on B+1. With all four tubes in place the voltage is correct now. However, the amp sounded great with just two EL84s but is 360vt too much with only a 100ohm cathode resistor? Thanks for that reply Mr. Collins!

[collinsamps]

It depends on the class of operation and total current draw. Measure your cathode voltage and I can tell you whats going on.

Here's a good reference:

http://www.diyguitarist.com/Images/BiasChart-EL84.jpg

[Structo]

You will want at least a 500 ohm cathode resistor on the EL84's.

Measure the voltage on the + side of the resistor.
Divide that by the resistance, that is your current for both EL84's.
E/R=I

[Phil_S]

Structo: Please help me understand your recommendation for 500 ohms.

Maybe I do the math wrong?

Assume Vk=20, Va=360 and Rk=130. Vk/Rk=20/130=154mA. Divide by 4 tubes = 38.5mA per tube. Va-Vk=true plate voltage=340v. 340v * 38.5mA = ~13W per tube or 52W for all 4. At 15v on the cathode, 100 ohm is just about right. This is right around 100% of the tube rating, and is the appropriate point for a cathode biased amp, is it not?

If he splits the pairs (inner pair/outer pair; one R/C for each pair of cathodes) to make it easy to pull one pair for half power, then he would double the Rk value.

Phil

[Structo]

I missed the part about 4 EL84's.

I skimmed it (bad habit) and just caught the part about two tubes.

As far as what I meant, I have a 2 x 6V6 amp that is cathode biased.

I have 336 on the plates.

With a 560 ohm cathode resistor, I have 22 ma current on each tube which is about right according to my calculations.

Hmmm, I forgot about going 100% on a cathode bias, I may have to recalculate that.

But with 4 tubes you are right.

[Structo]

Hmmm, I re did my calculations.

I have Va 336 - Vk 24.6 = 311.4v for true Va.

Rk is 560 ohms so 24.6 / 560 = .044 A / 2 = 22 ma per tube.

311.4 x .022 A = 6.85 watts per 6V6.

Way too low if I am doing this right!
So say I want 12 watts per 6v6, that would be 12/311.4 right?

That works out to be 38.5ma per tube or 77ma for the pair?

Help!

[tweedeluxe]

Bias to 100% (actually around 90%) for Class A ONLY.

70% max anode dissipation would be about 8W.

There is a little more to it though - read up on load-lines if you haven't already: google steve bench for a thorough explanation or valvewizard for a simplified method (don't have the addresses saved and can't access search pages at work).

[Structo]

I found that Excel spreadsheet that DIY guitar has.
It goes through all the values from cathode resistor to screen current to arrive at the pertinent numbers.
http://www.diyguitarist.com/Misc/PlateD ... onCalc.xls

I'll have to figure out a way to measure my screen pins though.
The chassis is very tight.

So with a push/pull cathode biased amp you still shoot for 70%?

I thought the 100% figure was for any cathode bias.

Well I have more research to do.

[Phil_S]

Bias to 100% (actually around 90%) for Class A ONLY.

70% max anode dissipation would be about 8W.

There is a little more to it though ...

I suppose there is more than a little bit more to it.

It is my understanding that class of operation isn't the determining factor here. I've seen significant flame wars over "What is class A?", and I'm not looking to start something.

As I understand it, for cathode bias 100% is the target. That's because, to be very brief, cathode bias is self limiting. As you drive the tubes, cathode voltage rises, keeping the thing in balance. Sorry, this isn't said so well.

For fixed bias, 70% is the target because, as you drive a signal into the tube, your negative bias voltage on the grid is fixed. This causes the tube to move towards 100% and it isn't generally a good thing to push it past the design limit.

Anyhow, to the original poster, you've now got some decent info on selecting a cathode resistor.

[Structo]

Aiken says:

Cathode-biased class AB amps are usually exempt from the "70% rule", because their cathode voltage rises when a signal is applied, effectively reducing the bias, and shifting the amp further into class AB operation. This means you can bias them hotter than a normal fixed-bias class AB amp and the tubes will still survive. Having said that, you have to experimentally determine how hot you can bias them by finding out how far the bias shifts during signal flow.

[tweedeluxe]

As I understand it, for cathode bias 100% is the target. That's because, to be very brief, cathode bias is self limiting. As you drive the tubes, cathode voltage rises, keeping the thing in balance. Sorry, this isn't said so well.

100% is only the target if you are shooting for Class A, even if the self-limiting cathode-bias method allows you to push the 70% figure.

Structo - you should be able to figure out how hot you can push the tubes on paper by drawing out the load lines on the output tube data sheet graphs. It will be easier too since your amp is already built.

You may very well be able to safely run your output at 100% anode dissipation (Class A PP) - but it depends on the rest of the design, especially the output transformer.

Disclaimer: I am trying to figure out all this stuff myself so don't take my words as infallible truth!