AC / DC loadline - please clarify

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Val_r

Well-known member
Joined
Jun 5, 2004
Messages
306
Location
Naples, Italy.
Let's say I want to start establishing a quiescent point for a 6CG7, Class A push-pull, the transformer is a 6.6 K ohms plate to plate.

An output stage could be the one in the Altec 436C comp.

Let's say I want the tube work in the -6V grid voltage, 200 plate volts, 7.5mA plate current, so this should be my ac loadline

6CG7load.gif


Each tube should see a half of the plate to plate primary impedance, so 3,300 ohms (am I right?)

How do I set the correct supply voltage to apply at the C.T. of the transformer?

What would be the dc loadline?

How to correct design the working point of the tube?

Any reading to suggest?

Thank you.
 
Each tube should see a half of the plate to plate primary impedance, so 3,300 ohms (am I right?)

Actually, each tube will see 1/4 of the p-p anode load. So thats 1650 ohms per tube.

As for the rest of your questions, they are all somewhat related to what you want out of your amp.

Output effect? Then work/design your way backwards from there.
 
> so 3,300 ohms (am I right?)

1,650 ohms. Half the turns. A quarter the impedance.

> How do I set the correct supply voltage to apply at the C.T. of the transformer? What would be the dc loadline?

The DC loadline is the DC resistance of the transformer. It is often about 300 ohms. At 7.5mA it drops about 2V. You can draw it, or increase your raw supply voltage 2V, but it is normally utterly negligible.

> How to correct design the working point of the tube?

Cart before the horse? You already told us: " 6CG7, Class A... 6.6 K ohms plate to plate... I want the tube work in the -6V grid voltage, 200 plate volts, 7.5mA plate current...". You have already picked a working point. Is it good? It works. You could do different, maybe better.

Always plot your plate dissipation, and note the zero-grid line. You normally want to stay inside these lines.

The DC loadline for a few-hundred ohm DC resistance is like the green dotted line. Nearly vertical.

6CG7-PP.gif


The red dot is your proposed DC bias condition. 200V, 7.5mA. You might need 202V supply to allow for the DC drop in the transformer (who cares?). If you cathode-bias (and you should whenever you can), then you need 200V+2V+6V= 208V total supply. The cathode resistor takes 6V and 15mA, 400 ohms.

The AC load is 1,650 ohms. I've plotted this in red. We can get about 5mA 8 Volt(!) peak swing per plate. This is under 20 milliWatts output.

You could do a little better working at 140V and 8 or 9mA.

Tip: for best power and distortion with triodes, the load should be larger than the plate resistance. For low distortion single-ended, much larger. For push-pull, equal or a little larger. 6CG7 plate resistance is about 7K. We want a load rated about 28K (not 6K!) plate to plate.

The yellow line is rougly 25K plate to plate. We get a little more current swing, a lot more voltage swing. About 7mA and 50V peak swing. This comes to about 170 milliWatts. Almost 10 times more than we got at 6K load.

The most power you can get from a single Class A push-pull 6CG7 would be an idle point near 300V 8mA, and a load-line tangent to the plate dissipation curve. Violet dotted line. That is about 160K plate to plate, which is NOT practical for hi-fi audio. It would give about 700 milliWatts.

In practice, you usually load a small twin triode with 20K. Higher impedance is a little more power, but much smaller wire, and much more trouble with frequency response.
 
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