Effects of Negative Feedback on Op-Amp Impedances



Negative feedback affects the input and output impedances of an op-amp.

Let’s examine the effect on both the noninverting and the inverting amplifiers.

i) Impedances of a noninverting amplifier.

a) Input impedance

Consider the picture shown below.

Assume there is a small differential voltage, Vd, between the two inputs.

Thus we are assuming the input impedance is not infinite (or the input current to be zero).

The input voltage can be expressed as

Vin = Vd + Vf

Let B = Ri/(Ri + Rf). Then BVout = Vf. Thus,

Vin = Vd + BVout

But Vout @ AolVd (i.e. the open loop gain times the differential voltage). So:

Vin = Vd + B AolVd = (1 + AolB)Vd

Now substitute IinZin for Vd to get

Vin = (1 + AolB) IinZin

Where Zin is the open-loop input impedance of the op-amp (i.e. without feedback connections).

Thus we get that the overall input impedance of a closed-loop noninverting amplifier configuration is:

Zin(NI) = Vin/Iin = (1 + AolB) Zin

Therefore, the input impedance of the noninverting amplifier with negative feedback is much greater than the internal input impedance of the op-amp itself (i.e. without feedback).

a) Output impedance

Consider the following circuit.

Apply Kirchhoff’s law to the output circuit to get:

Vout = AolVd - ZoutIout

The differential input is Vd = Vin – Vf.

Assume AolVd >> ZoutIout. Then,

Vout @ Aol(Vin­ – Vf)

Substitute BVout­ for Vf to get

Vout @ Aol(Vin – BVout)

= AolVin – AolBVout

AolVin = Vout + AolBVout

@ (1 + AolB)Vout

The output impedance of the noninverting amplifier is Zout(NI) = Vout/Iout.

Substitute IoutZout(NI) for Vout:

AolVin = (1 + AolB) IoutZout(NI)

Or

AolVin/Iout = (1 + AolB) Zout(NI)

The term on the left is the internal output impedance of the op-amp (Zout) because, without feedback, AolVin = Vout. Thus,

Zout = (1 + AolB) Zout(NI)

Zout(NI) = Zout/(1 + AolB)

Thus the output impedance of the noninverting amplifier is much less than the internal output impedance, Zout, of the op-amp itself.

Example.

a) Determine the input and output impedances of the amplifier shown below. The op-amp data sheet gives Zin = 2 M?, Zout = 75 ?, and Aol = 200000.

b) Find the closed-loop voltage gain.

Solution

a) The attenuation, B, of the feedback circuit is

B = Ri/(RI + Rf) = 10 k?/230 k? = 0.0435

Zin(NI) = (1 + AolB)Zin = [1+(200000)(0.0435)](2 M?)

= (1+8700)(2 M?) = 17.4 G?

Zout(NI) = Zout/(1+AolB) = 75 ?/(1 + 8700) = 8.6 m?

b) Acl(NI) = 1/B = 1/0.0435 @ 23.0

i) Impedances of a voltage-follower.

The formulas are the same as in the noninverting case, but the attenuation, B, is unity.

Thus:

Zin(VF) = (1 + Aol)Zin

Zout(VF) = Zout/(1 + Aol)

Example

The same op-amp as in the previous circuit is used in a voltage-follower configuration. Determine the input and output impedances.

Solution

Using the equations above, we get:

Zin(VF) = (1 + Aol)Zin = (1 + 2000000)(2 M?)

@ 400 G?

Zout(VF) = Zout/(1 + Aol) = 75 ?/(1+200000) = 375 ?V

iii) Impedances of an inverting amplifier.

From the circuit shown below, it is easy to see that

Zin(I) @ Ri

The expression for the output impedance can be shown to be exactly the same as for the noninverting case:

Zout(I) = Zout/(1 + AolB)v

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