Why has an audio amplifier negative feedback?

Here is a basic 10W audio amplifier - sorry for the European signs: Why does the amplifier in this application need negative feedback? What would happen if I'd connect pin 2 of IC1 to ground?

Typically, OpAmps on their own have a very large gain, typically tens of thousands.

So, take the setup you suggest: In this case we are using the basic gain of the amplifier:

Vout = gain x (Vp - Vn)

So a very small voltage difference at the inputs will appear at the output multiplied by 100,000. A tiny input signal of 1mV peak to peak will come out at 100V peak to peak! Well, this is impossible and the output will simply clip at the voltage rails (the voltage will not exceed the op-amps voltage supply). This will sound horrible if this is an audio amp.

Now let's take another example: In this case we have maximum negative feedback. What happens now? Imagine we start with the input signal at 0v, and the output at 0v too. When the input signal rises by 1mV, the OpAmp wants to increase its output to 100v. So the output starts to rise rapidly. But the output is connected to the negative input! So as soon as the output reaches 1mV, the two inputs are at the same voltage, and the OpAmp is happy. The output voltage stops rising. In this situation, the output will always match the input. This is known as a voltage follower.

Now for an actual amplifier: In this circuit I've created a potential divider between the output and the negative input. The voltage at the negative input will always be half the voltage at the output. So now, when the positive input rises to 1mV, the OpAmp's output will need to rise to 2mV before the two inputs are at the same voltage. This set up is an amplifier with a gain of 2, because the output will always be twice the input voltage.

It's a 10W audio amp, not an op amp. Negative feedback is required to set the gain at a known, fixed value.

"Fundamentally, all electronic devices (e.g. vacuum tubes, bipolar transistors, MOS transistors) exhibit some nonlinear behavior. Negative feedback corrects this by trading unused gain for higher linearity (lower distortion). An amplifier with too large an open-loop gain, possibly in a specific frequency range, will additionally produce too large a feedback signal in that same range. This feedback signal, when subtracted from the original input, will act to reduce the original input, also by "too large" an amount. This "too small" input will be amplified again by the "too large" open-loop gain, creating a signal that is "just right". The net result is a flattening of the amplifier's gain over all frequencies (desensitising)."

• Sorry for 'opamp'. But I don't quite understand your answer. How does negative feedback set the gain at a fixed value? – user17592 Feb 19 '13 at 17:50
• amplifiers commonly require negative feedback for stable operation – vicatcu Feb 19 '13 at 18:13

First let's take a look at what's feedback: It's portion of the output that system takes into account when determining next output. From systems point of view it would look something like this: So let's say that our amplifier is trying to multiply the input signal X by a factor of say 10. So ideally it would try to set Y=10X. The system will have a component which amplifies the signal and a component that takes feedback into account.

In case the feedback is positive we'll have an amplification "force" that will try to increase the output signal and the feedback "force" that will again try to amplify the signal. The result of that is that we'll get signal that tends to go to infinity (or in our practical example to as high value as the amplifier can output). Even if we null the input signal, the positive feedback will result in the increase of output signal.

In case of negative feedback, we'll have an amplification "force" that will try to amplify the signal and the feedback "force" that will try to attenuate the output signal. The gain of whole system would be set at the point at which they cancel out. This way, when there's no input signal, the feedback will terminate the output signal.