Question

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

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

Answer 1

Answer: (B)

At $V_{B} =8V $
$ i_{L} = \frac{6\times10^{-3}}{4} =1.5\times10^{-3} A $
$i_{R} = \frac{8-6 \times10^{-3}}{1} = 2\times10^{-3} A $
$ \therefore i_{\text{zener} \, \text{diode}} = i_{R} -i_{\text{load}} $
$ = 0.5 \times10^{-3} A$
At $ V_{B} = 16 V$
$ i_{L} = 1.5 \times10^{-3} A $
$ i_{R} = \frac{\left(16-6\right)\times10^{-3}}{1} = 10\times10^{-3} A $
$\therefore i_{\text{zener}\; \text{diode}} = i_{R} - i_{L} $
$ = 8.5 \times10^{-3}A $