Question 210: Semiconductors and Logic Gates

Question

A voltage regulating circuit consisting of Zener diode, having break-down voltage of 10 V and maximum power dissipation of 0.4 W , is operated at 15 V . The approximate value of protective resistance in this circuit is $\_\_\_\_$ $\Omega$.

Accepted Answer

The series protective resistance limits the current flowing through the Zener diode so that it does not exceed the power rating.
The Zener diode maintains a constant voltage $\left(V_zight)$ across load resistance $R_L$. The excess voltage from the source ( $V_{	ext {in }}-V_z$ ) is dropped across the resistor $R_s$.
The Zener breakdown voltage $\mathrm{V}_{\mathrm{Z}}=10 \mathrm{~V}$
Maximum power dissipation $\mathrm{P}_{	ext {max }}=0.4 \mathrm{~W}$
Using the power formula $\mathrm{P}=\mathrm{V} 	imes \mathrm{I}$.
$$ I_{\max }=\frac{P_{\max }}{V_z} $$
$$\Rightarrow $$ $$ I_{\max }=\frac{0.4 \mathrm{~W}}{10 \mathrm{~V}}=0.04 \mathrm{~A} $$
The resistor is connected in series with the Zener diode. According to Kirchhoff's Voltage Law (KVL), the sum of voltage drops must equal the supply voltage.
$V_{i n}=V_{R s}+V_z$
$$\Rightarrow $$ $V_{R s}=V_{i n}-V_z$
$$\Rightarrow $$ $$ V_{R s}=15 \mathrm{~V}-10 \mathrm{~V}=5 \mathrm{~V} $$
To protect the diode, we choose the resistance such that the current flowing through it does not exceed $\mathrm{I}_{\max }$.
$$ \mathrm{R}_{\mathrm{s}}=\frac{\mathrm{V}_{\mathrm{Rs}}}{\mathrm{I}_{\mathrm{max}}} $$
$$\Rightarrow $$ $\mathrm{R}_{\mathrm{s}}=\frac{5}{0.04} \Omega=125 \Omega$

Semiconductors and Logic Gates

A voltage regulating circuit consisting of Zener diode, having break-down voltage of 10 V and maximum power dissipation of 0.4 W , is operated at 15 V . The approximate value of protective resistance in this circuit is \_\_\_\_ \Omega.

Integer / Numeric Answer Type