Question

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

• A. $1\text{mA};8.5\text{mA}$
• B. $0.5\text{mA};6\text{mA}$
• C. $0.5\text{mA};8.5\text{mA}$
• D. $1.5\text{mA};8.5\text{mA}$

Answer 1

The correct option is C $0.5\text{mA};8.5\text{mA}$

Given:
The breakdown voltage of the Zener diode $=6\text{V}$
Load resistance $R_L=4\text{k}\Omega$
Series resistance of the circuit $R_i=1\text{k}\Omega$
If the battery voltage
$(V_B{)}_{min}$ $=8\text{V}$
$(V_B{)}_{max}$ $=16\text{V}$

For voltage,$V_B=8\text{V}$
voltage across $R_i,V_{Ri}=$Power supply-Zener voltage
$V_{Ri}=8-6=2\text{V}$
Now current
$I_{Ri}=\frac{V_{Ri}}{R_i}=\frac{2}{1\times {10}^3}=2\text{mA}$

Voltage across $R_L$
$V_{RL}=$Zener voltage
$V_{Ri}=6\text{V}$
Now current
$I_{RL}=\frac{V_{RL}}{R_L}=\frac{6}{4\times {10}^3}=1.5\text{mA}$
Current flowing through zener diode
$I_Z=I_{Ri}-I_{RL}=2-1.5=0.5\text{mA}$

For voltage,$V_B=16\text{V}$
Voltage across $Ri$
$V_{Ri}=16-6=10\text{V}$
Now current
$I_{Ri}=\frac{V_{Ri}}{R_i}=\frac{10}{1\times {10}^3}=10\text{mA}$

Voltage across $R_L$
$V_{RL}=$Zener voltage
$V_{Ri}=6\text{V}$
$I_{RL}=1.5\text{mA}$
Current flowing through zener diode
$I_Z=I_{Ri}-I_{RL}=10-1.5=8.5\text{mA}$

Hence option $\left(C\right)$ is correct.