## Q. 2.SP.30

The Zener diode in the voltage-regulator circuit of Fig. 2-45 has a constant reverse breakdown voltage $V_Z = 8.2 \text{V}$, for $75 \text{mA} ≤ i_Z ≤ 1 \text{A}$.    If $R_L = 9 Ω$, size $R_S$ so that $v_L = V_Z$ is regulated to (maintained at) $8.2 \text{V}$ while $V_b$ varies by ±10 percent from its nominal value of $12 \text{V}$.

## Verified Solution

By Ohm’s law
$i_L = \frac{v_L}{R_L} = \frac{V_Z}{R_L} = \frac{8.2}{9} = 0.911 \text{A}$

Now an application of KVL gives
$R_S = \frac{V_b – V_Z}{i_Z + i_L}$             (1)

and we use (1) to size $R_S$ for maximum Zener current $I_Z$ at the largest value of $V_b$:
$R_S = \frac{(1.1)(12) – 8.2}{1 + 0.911} = 2.62 \Omega$

Now we check to see if $i_Z ≥ 75 \text{mA}$ at the lowest value of $V_b$:
$i_Z = \frac{v_b – v_Z }{R_S} – i_L = \frac{(0.9)(12) – 8.2}{2.62} – 0.911 = 81.3 \text{mA}$

Since $i_Z > 75 \text{mA}, v_Z = V_Z = 8.2 \text{V}$ and regulation is preserved.