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A zener diode, having breakdown voltage equal to 15 V, is used in a voltage regulator circuit shown in figure. The current through the diode is
A
20 mA
B
5 mA
C
10 mA
D
15 mA
Detailed Solution
In the circuit the 20 V supply is connected through a 250 $\Omega$ series resistor to the zener diode, and a 1 k$\Omega$ load is in parallel with the zener.
The zener is in breakdown, so the voltage across it and across the load is 15 V.
Voltage across the series resistor = 20 − 15 = 5 V
Current through the series resistor: $I = \frac{5}{250} = 0.02$ A = 20 mA
Current through the load: $I_L = \frac{15}{1000} = 0.015$ A = 15 mA
Current through the zener diode: $I_Z = I - I_L = 20 - 15 = 5$ mA
The zener is in breakdown, so the voltage across it and across the load is 15 V.
Voltage across the series resistor = 20 − 15 = 5 V
Current through the series resistor: $I = \frac{5}{250} = 0.02$ A = 20 mA
Current through the load: $I_L = \frac{15}{1000} = 0.015$ A = 15 mA
Current through the zener diode: $I_Z = I - I_L = 20 - 15 = 5$ mA