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Topic: Is this an appropriate circuit to drive my 8 ohm speaker using PWM? (Read 6 times) previous topic - next topic

s3n4te

Hi,

I was wondering if the circuit below will be able to drive a 0.2 W speaker?



The capacitor is used to remove the DC offset from the PWM output, and the voltage divider is used to reduce the vpp to 1 V. The 10 kOhm resistor is to reduce current to the OPAMP, and 20 kOHM is to multply the input by 2 so that the output wil be 2 V peak to peak. The transistors in the feedback loop helps to supply the required 125 mA to the speaker.

I don't have an scope to test this out, so I'm wondering if someone can verify that this circuit will not blow up.

Thanks,
s3n4te



retrolefty

Looks like it will function OK to me. I would use a audio taper pot on the input signal to give one a volume control. Keep in mind that there will be some 'cross-over' distortion in your output when the signal passes through 0vdc and neither transistors are conducting. 'Hi-Fi' class A/B amps usually apply a little dc bias such the both transistors are conducting a few milliamps with zero signal so that there is a smooth transistion between positive and negative cross-over. However as you are feeding this amp stage square waves anyway it's not like you won't have terrible odd order harmonic distortion anyway. Now if you did some low-pass filtering before the input stage and fixed the cross-over bias you might get musical quality tones through the thing.

Lefty


s3n4te

Would feedback cause the output to compensate for deadzone problem?

retrolefty

#3
Mar 01, 2011, 01:01 am Last Edit: Mar 01, 2011, 01:03 am by retrolefty Reason: 1

Would feedback cause the output to compensate for deadzone problem?


No. Recall your transistor basics. The emitter/base junction will only conduct when the applied voltage is greater then around .6vdc, so there is a time while the signal is moving from 0 to .6v that the transistor can't conduct. This is the source of the so called cross-over problem as one transistor stops conducting and the other starts conducting.

Any basic description/drawing of class AB amplifiers will explain the DC bias set-up where neither transistor is allowed to completely turn off no matter what the signal value is. In essence the amplifier runs in class A (at a low continuous current flow) until the signal is high enough to 'take control' of the conduction transistor and so enters class B operation.

Lefty

Grumpy_Mike

Why have you drawn 12v batteries all over the place?
You just need a +12 rail and a -12v rail.

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