Effects of Negative Feedback on Op-Amp Impedances – Negative feedback affects the input and output impedances of an op-amp

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, V_d, 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

V_in = V_d + V_f

– Let B = R_i/(R_i + R_f). Then BV_out = V_f. Thus,

V_in = V_d + BV_out

– But V_out @ A_olV_d (i.e. the open loop gain times the differential voltage). So:

V_in = V_d + B A_olV_d = (1 + A_olB)V_d

– Now substitute I_inZ_in for V_d to get

V_in = (1 + A_olB) I_inZ_in

– Where Z_in 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:

Z_in(NI) = V_in/I_in = (1 + A_olB) Z_in

– 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:

V_out = A_olV_d - Z_outI_out

– The differential input is V_d = V_in – V_f.

– Assume A_olV_d >> Z_outI_out. Then,

V_out @ A_ol(V_in­ – V_f)

– Substitute BV_out­ for V_f to get

V_out @ A_ol(V_in – BV_out)

= A_olV_in – A_olBV_out

A_olV_in = V_out + A_olBV_out

@ (1 + A_olB)V_out

– The output impedance of the noninverting amplifier is Z_out(NI) = V_out/I_out.

– Substitute I_outZ_out(NI) for V_out:

A_olV_in = (1 + A_olB) I_outZ_out(NI)

– Or

A_olV_in/I_out = (1 + A_olB) Z_out(NI)

– The term on the left is the internal output impedance of the op-amp (Z_out) because, without feedback, A_olV_in = V_out. Thus,

Z_out = (1 + A_olB) Z_out(NI)

Z_out(NI) = Z_out/(1 + A_olB)

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

Example.

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

b) Find the closed-loop voltage gain.

Solution

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

B = R_i/(R_I + R_f) = 10 k?/230 k? = 0.0435

Z_in(NI) = (1 + A_olB)Z_in = [1+(200000)(0.0435)](2 M?)

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

Z_out(NI) = Z_out/(1+A_olB) = 75 ?/(1 + 8700) = 8.6 m?

b) A_cl(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:

Z_in(VF) = (1 + A_ol)Z_in

Z_out(VF) = Z_out/(1 + A_ol)

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:

Z_in(VF) = (1 + A_ol)Z_in = (1 + 2000000)(2 M?)

@ 400 G?

Z_out(VF) = Z_out/(1 + A_ol) = 75 ?/(1+200000) = 375 ?V

iii) Impedances of an inverting amplifier.

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

Z_in(I) @ R_i

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

Z_out(I) = Z_out/(1 + A_olB)