Question

The figure represents a voltage regulator circuit using a Zener diode. The breakdown voltage of the Zener diode is 6V and the load resistance is $R_L=4k\Omega$. The series resistance of the circuit is $R_i=1k\Omega$. If the battery voltage $V_B$ varies from $8V$ to $16V$, what are the minimum and maximum values of the current through Zener diode ?

• A. $0.5mA;6mA$
• B. $0.5mA;8.5mA$
• C. $1.5mA;8.5mA$
• D. $1mA;8.5mA$

Answer 1

Answer: (B)

At $V_B=8V$
$i_L=\frac{6\times {10}^{-3}}{4}=1.5\times {10}^{-3}A$
$i_R=\frac{8-6\times {10}^{-3}}{1}=2\times {10}^{-3}A$
$\therefore i_{zenerdiode}=i_R-i_{load}$
$=0.5\times {10}^{-3}A$
At $V_B=16V$
$i_L=1.5\times {10}^{-3}A$
$i_R=\frac{(16-6)\times {10}^{-3}}{1}=10\times {10}^{-3}A$
$\therefore i_{zenerdiode}=i_R-i_L$
$=8.5\times {10}^{-3}A$