Negative Feedback In Amplifier Circuits

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NEGATIVE FEEDBACK IN AMPLIFIER CIRCUITS

Negative Feedback In Amplifier Circuits

Negative Feedback In Amplifier Circuits

A feedback amplifier is one in which the output signal is sampled and fed back to the input to form an error signal that drives the amplifier. (Jaeger 2006 If we connect the output of an op-amp to its inverting input and apply a voltage signal to the noninverting input? we find that the output voltage of the op-amp closely follows that input voltage):

As Vin increases? Vout will increase in accordance with the differential gain. However? as Vout increases? that output voltage is fed back to the inverting input? thereby acting to decrease the voltage differential between inputs? which acts to bring the output down. What will happen for any given voltage input is that the op-amp will output a voltage very nearly equal to Vin? but just low enough so that there's enough voltage difference left between Vin and the (-) input to be amplified to generate the output voltage. (Palumbo 2002)

The circuit will quickly reach a point of stability (known as equilibrium in physics)? where the output voltage is just the right amount to maintain the right amount of differential? which in turn produces the right amount of output voltage. (Sedra 2004) Taking the op-amp's output voltage and coupling it to the inverting input is a technique known as negative feedback? and it is the key to having a self-stabilizing system (this is true not only of op-amps? but of any dynamic system in general). This stability gives the op-amp the capacity to work in its linear (active) mode? as opposed to merely being saturated fully "on" or "off" as it was when used as a comparator? with no feedback at all.

Because the op-amp's gain is so high? the voltage on the inverting input can be maintained almost equal to Vin. Let's say that our op-amp has a differential voltage gain of 200?000. If Vin equals 6 volts? the output voltage will be 5.999970000149999 volts. This creates just enough differential voltage (6 volts - 5.999970000149999 volts = 29.99985 µV) to cause 5.999970000149999 volts to be manifested at the output terminal?(Jaeger 2006) and the system holds there in balance. As you can see? 29.99985 µV is not a lot of differential? so for practical calculations? we can assume that the differential voltage between the two input wires is held by negative feedback exactly at 0 volts. ...
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