Designing push-pull power amp stages

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Matador

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I'm trying to wrap my head around the design procedure for push-pull class AB1 power amp stages.

I'm thinking about how to design using a specific transformer:

Hammond 1650RA, 5K @ 100W

So if I understand correctly, in a push-pull configuration, each tube would see one half of the output impedance in the class A portion of the output characteristic, which it will see until one of the tubes cut's off, at which point the impedance drops by 1/4 (since turns ratio is half, impedance is 1/4).

Starting with a 7581A tube, and placing these AC load lines on the plate curves, assuming B+ of 450V, gives this result:

7581_power_amp.png


The data sheet suggests a bias point of -37V @ 55mA with a plate-to-cathode voltage of 450V, which is the green point on the class A load line in purple.

So this seems to line up exactly with the datasheet recommendations.  Peak current is roughly 300mA (where the class B load line crosses the 0V grid curve) at a plate voltage of about 70V, so power output (per RDH) is roughly ((B+-Vamin)*ipeak)/2, which in this case is ((450-70)*0.3)/2 = 57 watts for a pair of tubes, which again agrees almost exactly with the operating point specified in the datasheet.

What I can't wrap my head around is doing this with parallel tubes per side.  In theory, I should be able to consider push-pull arrangement of parallel tubes as the same as a plain push-pull arrangement of two tubes, with the group of tubes acting as a single composite tube.

If I use two 7581's in parallel, then with the same grid potential, plate-to-cathode potential, and the same screen potential, I should flow exactly twice the current.  Thus if I relabel the plot above with the currents doubled then the analysis above should be exactly the same, yes?

In other words, if I replot these load lines on a new 'composite' curve with the plate current doubled, I get this:

7581_parallel_power_amp.png


The new ipeak is 340mA, the new plate minimum is 40V, so output power is ((450-40V)*0.34)/2 = 70W.  Not double by any means!

In order to restore the look of the first image above, I need to halve the impedance of the output transformer.  This superimposes the original load lines on the first image, however the plate currents are doubled.  The power output would be ((450-70)*0.6)/2 = 114 watts for a quad of tubes.

In order to do this, I would need an output transformer rated for 2.5K primary impedance, or, I think I would attach an 8 ohm load to the 16 ohm tap, which should reflect half of the rated primary impedance, correct?

In the alternate, it looks like I could increase the B+ voltage to move the load lines to the left, and pick a different bias voltage.  Increasing the B+ up to 550V, gives a new idling point at -40V.  ipeak increases to 440mA, at a plate minimum of 50V, which is )(550-50)*0.44)/2 = 110 watts, roughly equal to the power output of the half-impedance transformer.

7581_parallel_550_power_amp.png


Am I thinking about this correct?  If there a pro/con for each approach?
 
You are correct that impedance will half with 2 parallel tubes.  However,  it is better to just start with the transformer you need with appropriate power and impedance.  Also be careful not to exceed the tube maximums,  in this case something like 550 plate volts is too much.
 
I see where you are trying to draw your way to an answer; too much work for me.

7581 is, for practical purpose, a 6L6GC. Have there been any quad-6L6GC amps ever made? Fender made a ton. Rated 80W-100W. Fender's standard load for two 6L6GC was 4k; for quad-6L6GC 2k.

A 5k 100W OT will need (5k/2k)^-2 = 1.58 times the B+. Fender used roughly 450-500V to get 100W in 2k; in 5k we need 500V*1.58 or 790V of B+.

I'm not sure if Hammond had a specific use in mind when they wound this 1650RA.

Working four tubes in 5k is like working two tubes half power in 10k. Are there any popular bottles rated 50W/pair in 10k?

What you probably want is 1650TA.
 
Of course, everything PRR says is spot on. 
One thing I would add is, in a Hi-Fi tube amp,  it's not unheard of to use a higher turns ratio transformer.  Some folks prefer it for the lower output impedance (damping) and the higher primary inductance for the low end extension. 
The quad of tubes can be set to pretty much loaf along unless you want a richer combined class A region. 
 
I am no expert on tube power amp design but I don't think you can draw  ac load lines like that as if they were dc ones. The dc load line for a transformer is basically a vertical line at the HT voltage. You pick your dc operating point on this line and then draw ac load lines a the correct angle through this point. And then you have to remember the plate can in theory swing up to twice the HT volts.

Cheers

Ian
 
ruffrecords said:
I am no expert on tube power amp design but I don't think you can draw  ac load lines like that as if they were dc ones. The dc load line for a transformer is basically a vertical line at the HT voltage. You pick your dc operating point on this line and then draw ac load lines a the correct angle through this point. And then you have to remember the plate can in theory swing up to twice the HT volts.

Cheers

Ian

Yes that's about it.  Intersect your quiescent voltage (Qv) bias point (which you picked based on the vertical DC load line) with an AC load line that starts to the right at about double the Qv and the zero current point.  If things don't fit beneath the safe operating area curve, go back and pick another bias point on the DC load line.  Rinse and repeat until you're happy.
 
ruffrecords said:
I am no expert on tube power amp design but I don't think you can draw  ac load lines like that as if they were dc ones.
Did I not do that?  I didn't draw the DC load line, but it extends up from B+ through the green dot.  The AC load lines are drawn from there.

I started with that Hammond because I had one, and a handful of other power transformers too.  I think I'm trying to wrap my head around what figures of merit can be used to a) evaluate which tubes might be appropriate for a given output transformer, and b) which output transformer best matches with a given tube.  I also selected the 7581A because it is half the price of the common 6L6 types that are available now.

Looking at PRR's suggestion: plotting the load lines on some common Fender and Marshall 50W and 100W combinations, I see most (if not all) of the class B load lines pass right through (or below) the knee in the plate curves for 0V grid bias, and skirt along the max dissipation line.  So that means if I know the impedance of 1/4th of a transformer, I can lay a ruler on different power tubes and see if the ruler is close to a knee at different supply voltages.

Take  the 5K Hammond again for example: a 1.25K load line seems to be right in the correct spot for a KT90 (which for a 500V supply needs to be around the 400mA area), since it's knee bends right around the 410mA current level.  The 6550/KT88 also do something similar.  It would seem the Hammond 1650RA was more inclined to support a pair of higher power tubes rather than a quad of lower power ones.

Which might work out well for me:  ultimately, I'm trying to do two channels in a 3U rack, each supporting in the 60-100W range.  If I can two two channels around 80W each with only 4 glass bottles that would be nice.
 
6550 has a condition for 5kct load and 100 Watts, at 600V of B+.

Since 6550 is a "big 6L6", you could hit the same condition with two-pair 6L6, except 600V on the plate of 6L6GC is very reckless.
 

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ruffrecords said:
Not with the 1K25 one.
Oh, I see.  I think the reasoning is that in Class B, when the tube stops conduction, the current falls to zero, and that half of the plate winding therefore sits at the plate supply voltage, since there can be no AC (or DC) drop across it.

This seems consistent with how other sites draw the composite class AB1 load lines, for example:

PushpullLoadline2.jpg


And this one for an AC30:

AC30_loadline.jpg


And a last one for KT90 in triode operation (the class B load line intersects with the 550V supply):

KT90triode2tubes.jpg


I would be curious to confirm that this is correct.
 
Matador said:
Oh, I see.  I think the reasoning is that in Class B, when the tube stops conduction, the current falls to zero, and that half of the plate winding therefore sits at the plate supply voltage, since there can be no AC (or DC) drop across it.

I would be curious to confirm that this is correct.

Yes, that's as I would do it too.  I draw the class B load line first for 1/4 value the transformer plate to plate impedance and starting at actual HT voltage.  I'm more familiar with Hi-Fi type amps so I usually aim more for the triode region above the knee but I have seen guitar amplifier load lines steered towards the knee itself and up against the max power curve as you said.

I then hold a straight edge from the same HT point corresponding to 1/2 the plate to plate Z and , keeping the same angle, move it up and shifted to the right to get the bias for turning the tube on.

Anyway, it sounds like you know exactly what to do and did it correctly on your load lines for a class AB composite
 
> that half of the plate winding therefore sits at the plate supply voltage

The voltage kicks-up above plate supply, of course, while current in the half-winding goes toward zero. The loadline is curved. It passes through the idle point.
 

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