Feedback can be arranged to subtract a voltage or a current from the input (these are sometimes called series feedback and shunt feedback, respectively). The non-inverting amplifier configuration we’ve been considering, for instance, subtracts a sample of the output voltage from the differential voltage appearing at the input, whereas the feedback scheme in Figure 2B subtracts a current from the input. The effects on input impedance are opposite in the two cases: voltage feedback multiplies the open-loop input impedance by 1+AB, whereas current feedback reduces it by the same factor. In the limit of infinite loop gain the input impedance (at the amplifier’s input terminal) goes to infinity or zero, respectively. This is easy to understand, since voltage feedback tends to subtract signal from the input, resulting in a smaller change (by the factor AB) across the amplifier’s input resistance; it’s a form of bootstrapping. Current feedback reduces the input signal by bucking it with an equal current.
Series (Voltage) Feedback
Let’s see explicitly how the effective input impedance is changed by feedback. We illustrate the case of voltage feedback only, since the derivations are similar for the two cases. We begin with a differential amplifier model with (finite) input resistance as shown in Figure 1. An input Vin is reduced by BVout, putting a voltage Vdiff =Vin−BVout across the inputs of the amplifier. The input current is therefore
giving an effective input impedance
In other words, the input impedance is boosted by a factor of the loop gain plus one. If you were to use the circuit of Figure 2B to close the feedback loop around a differential amplifier whose native input impedance is 100 kΩ and whose differential gain is 104, choosing the resistor ratio (99:1) for a target gain of 100 (in the limit of infinite amplifier gain), the input impedance seen by the signal source would be approximately 10MΩ, and the closed-loop gain would be 99.66.
Shunt (Current) Feedback Look at Figure 2A. The impedance seen looking into the input of a voltage amplifier with current feedback is reduced by the feedback current, which opposes voltage changes at the input.67 By considering the current change produced by a voltage change at the input, you find that the input signal sees a parallel combination of (a) the amplifier’s native input impedance Ri and (b) the feedback resistor Rf divided by 1+A. That is,
(see if you can prove this). In cases of very high loop gain (e.g, an op-amp), the input impedance is reduced to a fraction of an ohm, which might seem bad. But in fact this configuration is used to convert an input current into an output voltage (a “transresistance amplifier”), for which a low input impedance is a good characteristic.
By the addition of an input resistor (Figure 2B) the circuit becomes an “inverting amplifier,” with input resistance as shown. You can think of this (particularly in the high-loop-gain limit) as a resistor feeding a current-to-voltage amplifier. In that limit Rin approximately equals R1 (and the closed-loop gain approximately equals −R2/R1). It is a straightforward exercise to derive an expression for the closed-loop voltage gain of the inverting amplifier with finite loop gain. The answer is
where B is defined as before, B=R1/(R1+R2). In the limit of large open-loop gain A, G=1−1/B (i.e., G=−R2/R1).
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