Question

The cut in voltage of both zener diode $D_z$ and diode $D$ shown in figure is 0.7 V, while break down voltage of $D_z$ is 3.3 V and reverse break down voltage of D is 50 V. Value of peak output current $\left(I_0\right)$ is

• A. 3.3 mA in both positive and negative half cycles
• B. 4 mA in both positive and negative half cycles
• C. 3.3 mA in positive half cycle and 1.4 mA in negative half cycle
• D. 4 mA in positive half cycle and 5 mA in negative half cycle

Answer 1

The correct option is D 4 mA in positive half cycle and 5 mA in negative half cycle

Case 1: (positive half cycle)
$V_0<4V$


Zener diode does not turn on,
$\therefore V_0=\frac{1}{1+1}{V}_i$
$\therefore$ Using voltage division
$V_0=5\sin \omega tV$
$I_0=\frac{5}{1}\sin \omega tmA$
For peak positive value of input, zener diode will be ON, So


$V_0=3.3+0.7=4V$
$\therefore I_0=\frac{4}{1}mA;$ peak value of output current $=4mA$
For negative value of input, zener diode does not turn ON, so


$V_0=-\frac{1}{1+1}\times 10\sin \omega t$
$=-5\sin \omega t$
$I_0=\frac{-5}{1}\sin \omega tmA$
peak value of output current $=-5mA$