Voltage Shunt Negative Feedback:

Figure 1 shows the block diagram of voltage shunt negative feedback. Since voltage shunt is mentioned, output is connected in parallel configuration because of voltage sampling and input is connected in parallel mixing. It decreases both input and output impedance.

Figure 1: Block diagram of Voltage Shunt Negative Feedback

1. The input signal $I_S$ is applied as a input, $I_f$ is the feedback current and $V_O$ is the output voltage.
2. $A_f$ indicate relationship between $V_O$ and input current $I_S$. The feedback current is given by $I_f = β V_O$.
3. Resistance amplifier is called so because output is voltage and input is current, hence ratio gives us resistance gain.
4. The sample of output voltage is applied as a input to feedback network which feeds back the output signal to the input.
5. The difference of input signal and feedback signal gets amplified by the resistance amplifier.

Derivation for Rif, Rof and Af:

1. Rif (Input Resistance with feedback):

$Rif = \frac{Vin}{I_S} = \frac{Vin}{Iin + I_f}$

Since, $I_f = \beta V_O$

$Rif = \frac{Vin}{Iin + \beta V_O}$

But, $A = \frac{V_O}{Iin}$

$V_O = A \times Iin$

$Rif = \frac{Vin}{Iin+ (\beta \times A \times Iin)}$

$Rif = \frac{Vin}{Iin(1+A\beta)}$

$Rif = \frac{Ri}{1+A\beta}$

Due to negative feedback, Input impedance decreases by a factor 1+Aβ.

2. Rof (Output Resistance with feedback):

Figure 2 shows the equaivalent circuit for Rof.

Figure 2: Equivalent Circuit for Rof

For Rof :

i. Open the input current source.

ii. Remove output load resistance.

iii. Connect an imaginary current source that delivers current $I_O$.

Hence $Zof = \frac{V_O}{I_O}$

Apply KVL for negative feedback,

$V_O - I_O R_O - A \times Iin = 0$

But, $Iin = I_S - I_f$

Set $I_S = 0$ for Zof

$Iin = - I_f = -\beta V_O$

$V_O + A \beta V_O = I_O R_O$

$V_O (1+A\beta) = I_O R_O$

$\frac{V_O}{I_O} = Rof = \frac{R_O}{1+A\beta}$

Due to negative feedback, output impedance decreases by a factor 1+Aβ.

3. Af (Resistance gain with feedback ):

$A_f = \frac{V_O}{I_S} = Aif$

Apply KCL at input side,

$Iin = I_S - I_f$

$I_S = Iin + I_f = Iin + \beta V_O$

$A_f = \frac{V_O}{Iin+\beta V_O}$

Divide numerator & denominator by Iin,

$A_f = \frac{\frac{V_O}{Iin}}{1+\beta \frac{V_O}{Iin}}$

Since, $A = \frac{V_O}{Iin}$

$A_f = \frac{A}{1+\beta A}$

Due to negative feedback, gain decreases by a factor 1+Aβ.