Figure 1, shows the voltage shunt feedback amplifier using OP-AMP.

Figure 1

The input voltage drives the inverting terminal, and the amplified as well as inverted output signal is also applied to the inverting input via the feedback resistor Rf. This arrangement forms a negative feedback because any increase in the output signal results in a feedback signal into the inverting input signal causing a decrease in the output signal. The non-inverting terminal is grounded. Resistor R1 is connected in series with the source.

The closed loop voltage gain can be obtained by, writing Kirchoff's current equation at the input node V2.


The negative sign in equation indicates that the input and output signals are out of phase by 180. Therefore it is called inverting amplifier. The gain can be selected by selecting Rf and R1 (even < 1).

Inverting Input at Virtual Ground

In the Figure 1, shown earlier, the non-inverting terminal is grounded and the- input signal is applied to the inverting terminal via resistor R1. The difference input voltage Vd is ideally zero, (Vd = Vo/ A) is the voltage at the inverting terminals (V2) is approximately equal to that of the non-inverting terminal (V1). In other words, the inverting terminal voltage (V1) is approximately at ground potential. Therefore, it is said to be at virtual ground.


Input Resistance with Feedback

To find the input resistance Miller equivalent of the feedback resistor Rf, is obtained, i.e. Rf is splitted into its two Miller components as shown in Figure 2.

Figure 2

Therefore, input resistance with feedback Rif is then


Output Resistance with Feedback

The output resistance with feedback Rof is the resistance measured at the output terminal of the feedback amplifier. The output resistance can be obtained using Thevenin's equivalent circuit,shown in Figure 3.


io = ia + ib

Since Ro is very small as compared to Rf + (R1 || R2)

Therefore, i.e. iia

Vo = Ro io + AVd
VV1 – V2 = 0 - BVo

Similarly, the bandwidth increases by (1+AB) and total output offset voltage reduces by (1+AB).

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