Figure shows a $D C$ voltage regulator circuit, with a Zener diode of breakdown voltage $=6 \mathrm{

Question

Figure shows a $D C$ voltage regulator circuit, with a Zener diode of breakdown voltage $=6 \mathrm{~V}$. If the unrcgulated input voltage varies between 10 V to 16 V , then what is the maximum Zener current?

Professor PC Thomas Classes · Answer

(a): In given circuit, $R_S=2 \mathrm{k} \Omega, R_L=4 \mathrm{k} \Omega$ $I_Z=$ Maximum current through zener diode. At maximum input voltage, maximum current will flow through zener diode. At breakdown voltage, Voltage through zener diode $=V_Z=$ constant Since zener diode and $R_L$ are in parallel, \ $ \begin{aligned} & 	herefore \quad V_L=V_Z=6 \mathrm{~V} 	ext { and } \ & I_L=\frac{6}{4000} \mathrm{~A}=\frac{6}{4} \mathrm{~mA} \ & V_{	ext {input }}=16 \mathrm{~V} 	ext { (given) } \end{aligned} $ \  Potential drop at $R_S=10 \mathrm{~V}$\ $ \left[\because V_{	ext {in }}=V_Z+V_Sight] $ \ $ 	herefore \quad I_S=\frac{10}{2 \mathrm{k} \Omega}=5 \mathrm{~mA} $ \  Now, $I_Z=\left(I_S-I_Light)$ or $I_Z=5 \mathrm{~mA}-\frac{6}{4} \mathrm{~mA} \Rightarrow I_Z=3.5 \mathrm{~mA}$ Voltage across $R_L=10 \mathrm{~V} 	herefore I_L=\frac{10}{5}=2 \mathrm{~mA}$ and $I=\frac{24-10}{1}=14 \mathrm{~mA}$ Now $I=I_Z+I_L \Rightarrow I_Z=14-2=12 \mathrm{~mA}$ So power across zener diode $=V I=10 	imes 12=120 \mathrm{~mW}$