Question

When a zener diode is used as a regulator with zener voltage of $10V$, nearly five times the load current passes through the zener diode. What should be the series resistance for the zener diode. If load resistance is $2k\Omega$ and the unregulated voltage supplied is $16V$.

• A. $500\Omega$
• B. $100\Omega$
• C. $200\Omega$
• D. $800\Omega$

Answer 1

Answer: (C)

According to the question, the circuit diagram of zener diode as a voltage regulator is shown below,

Given, voltage supply, $V_s=16V$, voltage across
zener diode, $V_Z=10V$, current passes through the
zener diede, $I_Z=5I_L$ and load resistance $R_L=2k\Omega$
The current through load resistance is
$I_L=\frac{\text{ voltage }\left(V_Z\right)}{\text{ resistance }\left(R_L\right)}$
$=\frac{10}{2\times 10^3}$
$=5\times 10^{-3}A...(i)$
The current through in the resistance in series
$I=I_Z+I_L$
$=5I_L+I_L$
$=6I_L$
$=6\times 5\times 10^{-3}$ [From Eq.(i)]
$=3\times 10^{-2}A$
Also,$I=\frac{V_S-V_Z}{R_S}$
$\Rightarrow R_S=\frac{V_S-V_Z}{I}...(ii)$
substituting the values in Eqs. (ii), we get
$R_S=\frac{16-10}{3\times 10^{-2}}=\frac{6}{3\times 10^{-2}}=200\Omega$
Thus, the series resistance for zener diode is $200\Omega$.