# Help me understand how the opamp negative feedback reduces crossover distortion?

Discussion in 'Homework Help' started by Oser, Sep 27, 2013.

1. ### Oser Thread Starter New Member

Sep 27, 2013
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Hey everyone!

I've been searching for a couple of days for an explanation of how the negative feedback on the circuit below reduces the cross over distortion.

I understand the overall function of the push pull output and how the opamp will adjust it's output until the power output is the correct value to equalise the two opamp inputs.

I can't quite wrap my head around the part when both transistors are off. I've ran simulations and know that the output of the opamp " rushes through" the dead zone. But I cant find an explanation as to how it knows which way to drive it's output to compensate. All I've managed to find is, "the opamp is magic and just knows what to do to reach equilibrium".

If anyone could help it would be much appreciated. Thanks!

Feb 17, 2009
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3. ### Jony130 AAC Fanatic!

Feb 17, 2009
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And how op amp reduces crossover distortion?
Simple op amp has a very large open loop voltage gain. And that means that even small changes in input voltage cause a relatively large change in the output voltage. For example if op amp has a open loop gain equal to 1000V/V, 1mV change in the input voltage will result 1V change at the output voltage.
And in your amplifier circuit, when both transistors are off, op amp will work in open loop. So small change in input will result much large change in X node voltage.
And this large change in X node voltage will turn-on the transistor. So from the outside world it looks as if there was no crossover region.
Our example amplifier will reduce crossover region from +/-0.7V to +/-0.7mV.
And it's all thanks large open loop voltage gain.

Last edited: Sep 27, 2013
4. ### #12 Expert

Nov 30, 2010
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Consider the idea that, if you used a sine wave and sent it through a full wave bridge rectifier to get a bunch of positive half cycles, this circuit would faithfully produce a bunch of negative half cycles as its output. The op-amp does not know whether you are going to tell it to do half cycles or a normal sine wave, so it does not guess which way to go.

The fact of this matter is that the sine wave (at the input) keeps moving without any consideration for whether the transistors are on, off, broken, or even missing. You (personally) can see that, as the input voltage changes from positive toward zero, the voltage at NODEX will be changing from negative toward zero. It is the sine wave moving past zero (and becoming a negative voltage) that tells the op-amp to suddenly skip across the dead band of the transistors and start driving Q1.

The "other" circuit, the one with the feedback connection at NODEX, is the circuit this is being compared to. When the output voltage is taken from NODEX for use as negative feedback, the op-amp is only going to respond to its own output voltage and the result at NODEX will be a sine wave that is just like the input sine wave (except the polarity is reversed). If you apply that sine wave (at NODEX) to the transistors, the final output to RLOAD will have a flat spot around the zero crossing point. The op-amp still has negative feedback, but it is not being fed back from RLOAD.

By moving the feed back point to RLOAD you include the transistors in the negative feedback path and eliminate the flat spot. From this, I say that negative feedback does not decrease the distortion. Negative feedback taken from the proper place will reduce distortion.

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5. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Change V1 to a DC source and watch what happens at the op-amp output as V1 varies for small excursions about zero volts. For example go in 0.1V steps from +1V to -1V and this should provide some useful insights.

6. ### Oser Thread Starter New Member

Sep 27, 2013
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Thanks for the replies everyone! I'm happy to say it finally clicked for me last night whenever I was thinking about before I fell asleep. For some reason I had it in my head there would be a time whenever both transistors are off but I can see now that the only time the transistors will be off is whenever Vin is exactly 0. At other times the opamp will bias the transistors correctly to match even the smallest +/- Vin.

Now that I see how it works, it seems silly why I was having trouble with it in the first place but oh well I guess that's how these things go.

Thanks again!