Topic: Analog Circuits
What could be the output current rating of following shunt regulator?
(The circuit shows a 10 V source connected to a 50 $\Omega$ resistor, which is in series with a parallel combination of a 5 V, 0.4 W Zener diode and a load $I_L$.)
- A. $0<I_{L}<100 \text{ mA}$
- B. $20 \text{ mA}<I_{L}<100 \text{ mA}$
- C. $0<I_{L}<50 \text{ mA}$
- D. $10 \text{ mA}<I_{L}<100 \text{ mA}$
Answer: B. $20 \text{ mA}<I_{L}<100 \text{ mA}$
Explanation: Given Zener power $P_{Z} = 0.4 \text{ W}$ and Zener voltage $V_{Z} = 5 \text{ V}$.
The maximum Zener current is $I_{Z(max)} = \frac{P_{Z}}{V_{Z}} = \frac{0.4}{5} = 0.08 \text{ A} = 80 \text{ mA}$.
The total current supplied by the source is $I = \frac{V_{in} - V_{Z}}{R_{s}} = \frac{10 - 5}{50} = \frac{5}{50} = 0.1 \text{ A} = 100 \text{ mA}$.
To maintain regulation without exceeding the Zener's power rating, the load current $I_{L}$ must satisfy $I = I_{Z} + I_{L}$ where $I_{Z} \le 80 \text{ mA}$.
Thus, $100 \text{ mA} \le 80 \text{ mA} + I_{L} \Rightarrow I_{L} \ge 20 \text{ mA}$.
When $I_{Z} = 0 \text{ mA}$, the maximum load current is $I_{L} = 100 \text{ mA}$.
Therefore, $I_{L}$ lies in the range of 20 mA to 100 mA.