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Old 1st November 2007, 04:52 PM   #1
Sherman is offline Sherman  United States
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Question Equalizing + & - output voltage?

I've been playing with TubeCad trying different tubes, B+ etc. for an input stage and I've noticed that certain combinations of B+ and plate resistor give widely different positive and negative maximum voltage outputs.

Depending on the combination I can get the positive and negative Vo to be nearly equal or I can get the positive to be much greater than the negative. Now intuition says matching the positive and negative Vo would be a good thing but my intuition has been grossly wrong before!

My question is this-

Assuming the total swing is adequate to drive the next stage is there any benefit obtained from matching positive and negative Vo? For instance if I want to drive a P-P EL84 output stage I need only about a 16V swing. Does it make any audible (or other important) difference if my driver stage Vo is -32/+95 versus -47/+53?
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Old 2nd November 2007, 05:59 AM   #2
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Sherman,
That of course makes sense.

If +ve swing is being limited that means that you don't have enough current through the tube and so the tube is cutting off (0 current) and the output is being "clamped" by the rail voltage - that is zero volts drop across the anode load resistor.

Conversely:
If -ve swing is being limited that means that you have too much current through the tube and so the tube is saturating (its maximum current) and the output is being clamped by the Saturation voltage of the tube (the minimum voltage it can pull the anode down to at full current)

In a normal bog standard Common Cathode stage the anode voltage at idle should be at least half rail voltage and preferrably more like 2/3 rail voltage. This means that you can get equal signal voltage swings in each direction.

Remember that the voltage swing is actually voltage dropped across the anode load resitor due to current swing. You don't want to swing current too low since thats the least linear section of its anode current vs grid voltage curve and you dont want to swing current too high, not passed its ratings.

Given a particular rail voltage you then choose somewhere between 1/2 and 2/3 of that for an anode voltage at idle. Look at the Ia vs Vg1 curves on the tube data sheet for the tubes of interest and given the rail less anode voltage and teh operating current you get the value of load resistor required.

OR SIMPLY
keep messing with TUBCAD and adjust things for equal +ve and -ve swings and yoou'll end up wher the above lengthy prose would have led you.

Cheers,
ian
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Old 2nd November 2007, 11:51 AM   #3
Sherman is offline Sherman  United States
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Ian,

Thanks for that very clear explanation! It makes sense to me now.

In the circuit I've been playing with I'm able to match swings within a few volts so I'll build that configuration and put the meter and scope on it. I've become a big fan of the Glassware software. (I've now got all three- TubeCAD, SE AmpCAD and Push-Pull Calculator.) I can try a lot more stuff than I could possibly breadboard and sometimes, with help from people like you I learn something!

Thanks again,
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Old 2nd November 2007, 06:56 PM   #4
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All Ian says is true. But there are times when you might not want to match the swings.

If you don't need the entire output range, sometimes you can get less - or more - 2nd order distortion by skewing "off-center" a bit.

Yes, sometimes I might want _more_ distortion. I do this to cancel some 2HD from the next stage, or sometimes just because I prefer to have more 2HD than 3HD in the end result.

This all has to do with the general statement (true by the way) that "changing the operating point changes the sound".

By all means, start with what TubeCAD shows, but don't be afraid to experiment from there.

Pete
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