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  4. A zener diode of voltage VZ(=6 V ) is used to maintain a constant voltage across a load resistance RL(=1000 Ω ) by using a series resistance Rs(=100 Ω). If the e.m.f. of source is E(=9 V ), what is the power being dissipated in Zener diode?

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Q. A zener diode of voltage $V_Z(=6 V )$ is used to maintain a constant voltage across a load resistance $R_L(=1000\, \Omega$ ) by using a series resistance $R_s(=100\, \Omega)$. If the e.m.f. of source is $E(=9 \,V )$, what is the power being dissipated in Zener diode?

Semiconductor Electronics: Materials Devices and Simple Circuits

Solution:

Here, $E =9 V ; V _{ z }=6 ; R _{ L }=1000 W$ and $R _{ s }=100 W$
Potential drop across series resistor
$V = E - V _{ Z }=9-6=3 \,V$
Current through series resistance $R_S$ is
$I =\frac{ V }{ R }=\frac{3}{100}=0.03 \,A$
Current through load resistance $R_L$ is
$I _{ L }=\frac{ V _{ Z }}{ R _{ L }}=\frac{6}{1000}=0.006\, A$
Current through Zener diode is
$I_Z=I-I_L=0.03-0.006=0.024 \,amp$.
Power dissipated in Zener diode is
$P_Z=V_Z I_Z=6 \times 0.024=0.144$ Watt