Class (A)B output stage: why 2 emitter resistors, why not 1? - Page 3 - diyAudio
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Old 13th December 2006, 01:32 PM   #21
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My original plan was to use a single emitter resisitor on the bottom half of a quasi-complementary stage. It would help equalize the top/bottom transconductance a bit. Now I am looking at keeping package count down and considering some sanken darlingtons as output stage. So I cant use a quasi-complementary output stage if I use darlingtons.

As a potentially interesting side effect, if you use a r/c bootstrap for voltage amp bias then the bootstrap has a gain closer to 1 and this should increase the output impedance of the bootstrap current source. Although I havent simulated this effect yet.

Admittedly I am suspicious that I havent seen other amps which take the bootstrap cap from the emitter of the N half instead of the output. Maybe there is something wrong with the idea.
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Old 13th December 2006, 01:43 PM   #22
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You are being rather conservative with your numbers. Let me try:
40 volts peak output into 4 ohms is 10 amps. 10 amps across an emitter resisitor of 0.22 is 2.2 volts.

So if an amp has +40-36v output swing then it can only handle arbitrary 36v signals before clipping. However a very similar design with +38-38v can handle abritraty 38v signals before clipping.

I certainly agree its not much and probably not audible. I guess its an esthetic point that if I paid for a +-40v power supply I would prefer that the amp handle 38v signals before clipping instead of clipping at 36v signals. Just optimizing the specs/money ratio.
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Old 13th December 2006, 02:48 PM   #23
ilimzn is offline ilimzn  Croatia
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Wether or not it is worth investigating symetrical clipping WRT ground 'position' between the +-V rails, depends on what size resistor and what current we are talking about. No-one stipulatid the single emitter resistor is going to be 0.1 ohm, nor that the output voltage is going to be 10V - considering it's a bridge, I expect a fair bit more.
And, since this is a bridge amp, clipping at the load will be simetrical either way, but perhaps not at the maximum possible voltage. I am also guessing that there is no real center tap tot he power supply. In such amps the 'ground reference' is established by a passive network, so 'moving it about' a few V comes near or at a net cost of zero.
Besides, it was just an idea and basically, an invitation to have deleveld look into it and see if it is prudent to wring out that last W of power.
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Old 13th December 2006, 06:10 PM   #24
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Originally posted by deleveld

You are being rather conservative with your numbers. Let me try:
40 volts peak output into 4 ohms is 10 amps. 10 amps across an emitter resisitor of 0.22 is 2.2 volts.

For 10 amps I would use 3 transistors in parallel with resistor in each emitter, it is the different story.
The Devil is not so terrible as his math model is!
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