Suppose the Zener diode in Figure is ideal and has breakdown voltage of $10V$. What is the maximum current through the Zener diode?
To find the maximum current through the Zener diode, we first need to understand the circuit provided. We have a Zener diode with a breakdown voltage of 10 \, V, a series resistor of 820 \, \Omega, and a variable supply voltage ranging from 20 \, V to 40 \, V.
Step-by-step calculation:
- The Zener diode will start conducting when the supply voltage exceeds the breakdown voltage of 10 \, V. Therefore, the Zener maintains a constant output voltage of 10 \, V.
- To find the maximum current through the Zener diode, use the maximum supply voltage, which is 40 \, V.
- Apply Kirchhoff’s Voltage Law (KVL) around the loop: V_{supply} = V_{R} + V_{Z} where V_{supply} = 40 \, V and V_{Z} = 10 \, V.
- Calculate the voltage across the series resistor (V_{R}): V_{R} = V_{supply} - V_{Z} = 40 \, V - 10 \, V = 30 \, V.
- Use Ohm's Law to find the current (I_{Z}) through the Zener diode: I_{Z} = \frac{V_{R}}{R} = \frac{30 \, V}{820 \, \Omega} \approx 36.6 \, mA.
Therefore, the maximum current through the Zener diode is 36.6 \, mA, matching option B.
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