Question

Consider the circuit shown in the figure below.


$\text{The voltage regulator circuit has a Zener diode with forward cut-in voltage}{V}_g=0.7\text{V and the zener breakdown voltage}{V}_z=4.8\text{V. The value of knee current is}{i}_{z(\text{min}}=5\text{mA and the maximum power rating of zener diode is 0.48 W.}\text{Then the maximum value of power that can be dissipated in the load resistance}\text{and the maximum value of resistance}{R}_L\text{respectively are}$

Answer 1

$\text{To find maximum power dissipation across the resistance, we have to find the maximum current}$ flowing in the load resistance $R_L$

$i_{L\text{(max)}}=i_{\text{in}}-i_{z\text{(min)}}$

$=\frac{6.3-4.8}{12}-5\times {10}^{-3}=(125-5)\times {10}^{-3}=120\text{mA}$

$\therefore P_{\text{(max)}}=(120\times {10}^{-3})\times 4.8$

$=0.576W$

$\text{Maximum value of resistance}{R}_L,\text{we have to find}{i}_{L\text{min}}$

$i_{R\text{(min)}}=i_{\text{in}}-i_{z\text{(max)}}$

$=125\times {10}^{-3}-\frac{0.48}{4.8}=125\times {10}^{-3}-100\times {10}^{-3}=25\text{mA}$

$\therefore R_{L\text{(max)}}=\frac{4.8}{25}\times {10}^3=192\Omega$