Consider the shunt-shunt feedback amplifier shown in the figure below

The transistor is biased at $I_{c} = 3.94$ mA, then the value of output resistance $R_{0}$ is equal to

68.9 Ω

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560 Ω

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1 k Ω

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1.1 k Ω

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Solution

The correct option is A $68.9 Ω$
.

$v_{0} = (i_{b} + \frac{r_{π} i_{b}}{R_{2}}) . R_{1} + \frac{r_{π} i_{b}}{R_{2}} \times R_{2} = i_{b} (1 + \frac{r_{π}}{R_{2}}) . R_{1} + i_{b} r_{π}$
Applying KCL at node A is equal to
$I_{0} = β i_{b} + i_{b} + \frac{r_{π} i_{b}}{R_{2}}$
$I_{0} = i_{b} (β + 1 + \frac{r_{π}}{R_{2}})$
$\frac{v_{0}}{I_{0}} = \frac{(1 + \frac{r_{π}}{R_{2}}) R_{1} + r_{π}}{β + 1 + \frac{r_{π}}{R_{2}}}$
$= \frac{(1 + \frac{657}{700 \times 10^{3}}) \times 6.3 \times 10^{3} + 657}{1 + 100 + \frac{657}{700 \times 10^{3}}} = 68.9 Ω$

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