**Voltage Shunt Feedback Amplifier Circuit:**

Voltage Shunt Feedback Amplifier Circuit is also known as **shunt-derived shunt fed feedback** or **voltage shunt inverse feedback**. Here, a small portion of the output voltage is coupled back to the input voltage, as shown in Fig. 19.10 (b). Since the feedback network shunt both the input and output of the amplifier, both of the input and output impedances are reduced by a factor 1/(1 + βA).

Figure 19.23 gives the circuit of a common-emitter amplifier stage with a resistor R_{f }providing feedback from collector to base. Here the feedback is proportional to the output voltage V_{out} and feedback current I_{f} gets added in shunt with the input. Thus this circuit forms the case of **voltage-shunt inverse feedback amplifier**.

From circuit shown in Fig. 19.23.

Feedback current,

From Eqs. (19.12) and (19.50) and assuming I_{s} ≈ I_{f}

Thus the transresistance equals the negative of the feedback resistance from output to input of the transistor and is stable if R_{f} is a stable resistance. Both the input and output resistances are reduced because the feedback network shunts both the input and output of the amplifier. Assuming input impedance to be zero, voltage gain with feedback.

Thus if resistors R_{f }and R_{s} are stable elements, then voltage gain is stable (independent of the transistor parameters, the temperature or supply voltage variations).

**Amplifier Without Feedback:**

Referring to second topology in Table 19.4, the input circuit of the amplifier without feedback is obtained by setting V_{out} = 0 (shorting the output node). Thus connecting resistor R_{f} from base to emitter of the transistor. The output circuit is obtained by setting V_{in} = 0 (shorting the input node), thus connecting R_{f} from collector to emitter. Thus, the amplifier circuit without feedback is depicted in Fig. 19.24. Since the feedback signal is a current, the source is represented by a Norton’s equivalent with I_{s} = V_{s}/R_{s}.

The feedback signal is the current I_{f} in the feedback resistor R_{f} which is the output circuit.

From Fig. 19.24,

If the amplifier is deactivated by reducing h_{fe} to zero, a current I_{f} passes through the β network (the resistor R_{f}) from input to output and this current is

The output current I_{out }with the amplifier activated is

So the condition that the forward transmission through the feedback network can be neglected is I_{out} >> I_{f}

Since this voltage gain is at least unity, this inequality is easily satisfied by selecting (R_{s} + R_{f}) >> R_{C}.