Question

Consider the voltage reference circuit shown in the figure below:

If the zener diode has a breakdown voltage $V_z=5.6V$ and a dynamic small signal resistance $r_d=100\Omega ,$ then the value of output resistance $R_0$ is equal to
(Assume $V_T=26$ mV)

• A. $19.11\Omega$
• B. $25.33\Omega$
• C. $9.84\Omega$
• D. $5.74\Omega$

Answer 1

The correct option is D $5.74\Omega$



$\begin{array}{cc}{V}_E& =V_z-V_{BE}\\ & =5.6-0.7=4.9V\\ \therefore I_E& =\frac{4.9}{1k}=4.9mA\\ \therefore r_{\pi }& =\frac{\beta V_T}{I_E}=\frac{200\times 26\times {10}^{-3}}{4.9\times {10}^{-3}}\\ r_{\pi }& =1.06k\Omega \end{array}$
$\text{Drawing, the small signal equivalent circuit, we get,}$


$\begin{array}{cc}{I}_0& =\frac{V_0}{1.06k+0.099}\\ I_0& =(1+\beta )I_b+\frac{V_0}{1k\Omega }\\ I_0& =\frac{-(1+\beta ).V_0}{1.06k\Omega +0.099}+\frac{V_0}{1k\Omega }\\ R_0& =\frac{V_0}{I_0}=\frac{1}{\frac{(1+\beta )}{1.06k\Omega +0.099}+\frac{1}{1k\Omega }}\\ R_0& \approx 5.74\Omega \end{array}$