Question

Consider the circuit shown in the figure below:


In the circuit, if the maximum power rating of the zener diode is equal to 400 mW, then

• A. the maximum possible value of $I_Z=40mA$
• B. the minimum possible value of $I_L=1mA$
• C. the maximum possible value of $RL=1k\Omega$
• D. the value of $I_S=50mA$

Answer 1

Answer: (A), (C), (D)

the value of $I_S=50mA$



Applying KCL at node 'A', we get,
$I_S=I_Z+I_L$
$I_{Z(max)}=\frac{P_{Z(max)}}{V_Z}=\frac{400\times {10}^{-5}}{10}=40mA$
now, $I_S=\frac{V_S-V_Z}{R_S}=\frac{20-10}{100}=\frac{10}{200}=50mA$
$\therefore I_{L(min)}=I_S-I_{Z(max)}$
$=(50-40)\times {10}^{-3}$
$I_{L(min)}=10\times {10}^{-3}=10mA$
$\therefore R_{L(max)}=\frac{V_Z}{I_{L(min)}}=\frac{10}{10\times {10}^{-5}}=1k\Omega$