Class AB amplifier

536

10:51:00

Power:
-4ohms 10W (left lamp)
-8ohms 7W (right lamp)
Efficiency: around 50% 
Frequency response: flat from 40Hz-20Khz
A classic multistage class ab amplifier. Im taking an average of 450mA drawn for 8ohms which means around 16W current draw for the 7W output power. That means around 50% efficiency for 8ohms (normal for class B). For 4ohms im taking an average of 750mA current draw so around 26W for the full 10W in 4ohms. Here you see that efficiency levels drop beneath 50%. Now i know it seems that the output is too low for a 35V power source. This is because the amp is optimised for lowest noise and THD levels by adding  the 100nf capacitive negative feedback. This negative feedback will also ensure that no self excitations will occur. If you want more ,,kick,, to it you just decrese the capacitors value to 47nf or even less for higher power, but be aware that the sound will somewhat deteriorate if you decrease the negative feedback and the consumption will increase so the parameters i have given will no longer apply with the new conditions.

published 11 years ago

hurz

11 years ago

No, thats the wrong way to a good Amp. But i like to see you made that by your own!

thebugger

11 years ago

Whats wrong with it. Except that ive sacrificed gain for quality?

hurz

11 years ago

several things ;-) The first stage is good for 1.5% THD. 2 voltage follower because of to high input impedance. Power stage is out of loop. Whats in the loop is what makes the lowest THD, the voltage followers. And the efficency of 50%, i have calculated 32% in your 8Ohm case.

thebugger

11 years ago

1.5Thd is not bad for such simple design. And what does out of the loop mean. The second stage is indeed a current amplifier with own impedance.

hurz

11 years ago

1.5%THD for the first transistor stage only! And the loop does not include the power stage. Its feedback comes from the input of the power stage and not its output. Anyway the loop would not help much because of its high distortion. A regulator must be highly linear to reduce overall THD. But yours Regulator/loopcontroller/preamp what ever you can call it, is highly unlinear! 1.5% is just the first stage overall this amp might has a THD of about 5%

hurz

11 years ago

@8Ohm i see Pin=17.1Watt and Pout=3.4Watt which gives an efficency of 19.8%. You seems to calculated an average Pin against a peak power out. But instead 7W peak you have to calculate with 3.4Watt rms. So your "around 50%" is after your calculation more around 40%. And calculated correctly around 20%.

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