Please check my work.

[Alan0354]

Hi, I am studying the RDH4 and I want to do a design example to verify whether I am doing in correctly. I make up some number to make things easier, but I use the true data of the 6V6GTA for the design.
I am designing a Class A push pull with B+=320V with cathode bias and I run the tube in triode mode. I am using quiescent current of 35mA per tube so it's 70mA total. As the cathode is about +20V above the grounded grid, so I use a 320V supply so I have 300V across the tube. I draw the load line in red from the quiescent point of 300V and 35mA. I choose the second point of the load line at Vg1=0V at 80mA to make sure it stay below the 14W. I might have violate the max power along the line when travel from the quiescent point to the max current. I am too lazy to plot the max power curve, I just assume that I am not violating the power. I show the calculation how I arrive the load impedance RL by using the slope of the load line in red. Being a push pull, I assume half of the primary is 2.625K. So I need an OT of primary of 5.2K or so.



This is an exercise, I don't know enough to optimize for distortion or something. Please give me advice what else I need to look out.



From the graph, Vg1=-22V. I calculate the cathode resistor to drop 22V with 70mA, it turn out to be 314ohm. My question is if I am designing for class AB and if I set the tail current very low ( say 10mA). For -25V, the cathode resistor would have to be 2.5K. But then I will have issue when I put in large signal where one tube will conduct more than the total current of the tail ( larger than 10mA). How can a cathode bias work for class AB?

Thanks

Alan

[Steve Conner]

If you want to make a cold biased Class-AB amp, cathode bias isn't a good place to start. You would use fixed bias, though nowadays you can cheat by putting a Zener diode in parallel with the cathode resistor. However, cold bias tends to sound pretty bad for guitar.

With hot bias, it works reasonably well. The voltage across the cathode resistor just increases a bit as the amp is driven beyond its Class-A region of operation.

[Alan0354]

If you want to make a cold biased Class-AB amp, cathode bias isn't a good place to start. You would use fixed bias, though nowadays you can cheat by putting a Zener diode in parallel with the cathode resistor. However, cold bias tends to sound pretty bad for guitar.

With hot bias, it works reasonably well. The voltage across the cathode resistor just increases a bit as the amp is driven beyond its Class-A region of operation.

Thanks for the reply. What is cold biased? Do you mean cathode bias only good for class A amp? Do you mean to put a 23V zener in parallel with the cathode resistor so if one tube is driven to pull more current the voltage at the cathode rise and the zener turns on and provide the extra current?

Thanks

[loudthud]

14W is an average rating, not instantaneus like with semiconductors. Check the photos I posted in this thread: X-Y Plots of Output Tubes

[Alan0354]

What is the difference between tubes and semi conductors like transistors?

[loudthud]

What is the difference between tubes and semi conductors like transistors?

In the thread I linked above, compare the first photo in the first attachment with the first photo in the second attachment. The first shows typical plate curves and the 30W dissipation line. The second shows how a guitar amp can stress the tube well beyond the 30W average curve. Transistors die in milliseconds under that much stress beyond maximum ratings. Don't worry if your load line crosses the 14W line. Tubes only care about the long term average.

[J M Fahey]

Back to your math, which you are trying to check here (this is previous to talking about "sound"), it's basically right, but transformer impedance is 4 times that of each plate; so it's roughly 11K and not 5K2 .
You can also see how unefficient are triodes compared to pentodes.
180V saturation voltage is a terrible waste.
A pentode would easily reach down to 50 or 60V at a much higher current.

[Alan0354]

Back to your math, which you are trying to check here (this is previous to talking about "sound"), it's basically right, but transformer impedance is 4 times that of each plate; so it's roughly 11K and not 5K2 .
You can also see how unefficient are triodes compared to pentodes.
180V saturation voltage is a terrible waste.
A pentode would easily reach down to 50 or 60V at a much higher current.

Thanks for the reply. From my calculation, load resistance should be 2.625K. This is for one side. The total impedance should be twice which is 5.2K. Why is it supposed to be 4 time?

How do you make the plate voltage of pentodes go down to 50V? Do you increase the load resistance so the plate voltage reach 50V at grid voltage reach 0V. Or you just drive the grid +ve? But for class A, are you supposed not to drive the grid +ve?

[J M Fahey]

From my calculation, load resistance should be 2.625K. This is for one side. The total impedance should be twice which is 5.2K. Why is it supposed to be 4 time?

You calculated (correctly) the CT to one side impedance , this is whay 1 plate sees. Fine.
When you go buy the transformer, you specify its "plate to plate" impedance.
Plate to plate you have twice the turns, also twice the *voltage*, which means 4X the *impedance* which is proportional to the *square* of the turns ratio.

How do you make the plate voltage of pentodes go down to 50V? Do you increase the load resistance so the plate voltage reach 50V at grid voltage reach 0V. Or you just drive the grid +ve? But for class A, are you supposed not to drive the grid +ve?

By using them as pentodes.
That's the function of the screen, which is held at quite a high voltage *all the time*.
Electrons are attracted by voltage difference (in Physics: "difference of (electric) potential").
When plate is at idle, fine, you have your 320V or whatever the PSU supplies.
But now you tied the screen to the plate, so it follows same voltages.
As soon as current starts passing, plate voltage follows, screen too (you tied both together) and electrons are less attracted.
Check that to get 80mA (nothing to write home about) you need 180V (which is a waste)
Can your plate go lower?
Yes it can, but at a *much* reduced current, so you lose one way or another.
Pentode plates can swing quite low, but electrons are still attracted by the screen, which will stay at, say, around 300V in your case.
Repeat your design using same voltage but pentode curves.
You will be surprised.
In fact, do it and post your results.

[Alan0354]

You calculated (correctly) the CT to one side impedance , this is whay 1 plate sees. Fine.
When you go buy the transformer, you specify its "plate to plate" impedance.
Plate to plate you have twice the turns, also twice the *voltage*, which means 4X the *impedance* which is proportional to the *square* of the turns ratio.
:

I don't quite get this. If I need 2.625K for each tube, I get a 5.25K primary transformer with center tap. So I get 2.625K from each side to the center tap. I don't worry about the turn ratio. I just worry about the impedance. The reason I adjusted the load line is to get about 2.6K so it is easy to find a 5.2K primary transformer. It doesn't make sense to get a 10.4K primary for this.

BTW, what is cold bias?

[J M Fahey]

I don't quite get this. If I need 2.625K for each tube, I get a 5.25K primary transformer with center tap.

No, it's not so.
Center tap impedance is 1/4 the full transformer primary Impedance.

I don't worry about the turn ratio. I just worry about the impedance.

Well, you should.
They are related, and impedance varies with the square of the turns ratio.
Basic transformer theory.
Please refresh it.
It won't help you jumping straight to, say, chapter 12 or 14 of RDH4 skipping the basic theory.

The reason I adjusted the load line is to get about 2.6K so it is easy to find a 5.2K primary transformer. It doesn't make sense to get a 10.4K primary for this.

Well, 10.4K primary is what you need to give each plate 2.6K.

Repeat: 2:1 turns ratio means 4:1 impedance ratio.
And full primary to center tap means , precisely, 2:1 *turns* ratio (since both halves are equal) so full primary impedance (the spec by which output transformers are specified and sold) is 4:1 the impedance seen by a single plate (what you calculated graphically).

I found this interesting, and loved the practical, hands on approach:
http://www.sipi.edu/acadprog/progstu..._Labs45-46.pdf

[Alan0354]

No, it's not so.
Center tap impedance is 1/4 the full transformer primary Impedance.

Well, you should.
They are related, and impedance varies with the square of the turns ratio.
Basic transformer theory.
Please refresh it.
It won't help you jumping straight to, say, chapter 12 or 14 of RDH4 skipping the basic theory.

Well, 10.4K primary is what you need to give each plate 2.6K.

Repeat: 2:1 turns ratio means 4:1 impedance ratio.
And full primary to center tap means , precisely, 2:1 *turns* ratio (since both halves are equal) so full primary impedance (the spec by which output transformers are specified and sold) is 4:1 the impedance seen by a single plate (what you calculated graphically).

I found this interesting, and loved the practical, hands on approach:
http://www.sipi.edu/acadprog/progstu..._Labs45-46.pdf

I think I got it. For a transform with center tap primary, if the total impedance is 10K across the two end, then the impedance from each end to the center tap is 2.5K because of the square relation of impedance to voltage.

I don't use transformer much, thanks.

Seems the most common primary of OT is about 5.2K or so, should I check and redraw the load line to 1.25K or so?

Alan

[J M Fahey]

Yes, that's it, congratulations.
Looks counterintuitive at first sight, I know.
As of transformers, check what's available at Hammond Mfg. - "Classic" Transformers & Enclosures

Up to 15W, their "Universal" one can be wired for up to 22000 ohms, go figure.
Hammond Mfg. - Universal Tube Output - Push-Pull Transformers - (125 Series)
Or redesign for one of their "Guitar " ones:
Hammond Mfg. - REPLACEMENT & UPGRADES - Tube Guitar Amplifier - Output Transformers

In fact, *maybe* it pays to first build a classic (Bassman, etc) where you have a certain working circuit, a layout is suggested, even chassis dimensions and up to a matching front panel or cabinet if needed.
Build it, tweak it, and that experience will help you for the next one.
Ans so on