Effects of Negative Feedback
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)
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.
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.
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.
Zin(VF) = (1 + Aol)Zin
Zout(VF) = Zout/(1 + Aol)
The same op-amp as in the previous circuit is used in a voltage-follower configuration. Determine the input and output impedances.
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)