A Zener diode with a breakdown voltage of 12 V is used in a voltage regulator circuit. It is connected in parallel with a load resistor R_L = 600 . A series resistor R = 200 connects the combination to a DC supply of 20 V . The current flowing through the Zener diode is ________ mA .

Step-by-step solution

The voltage across the load resistor and the Zener diode is the Zener breakdown voltage, V_Z = 12 V . The voltage drop across the series resistor R is: V_R = V_i - V_Z = 20 - 12 = 8 V The total current from the supply, which flows through the series resistor, is: I_S = V_R R = 8 200 = 0.04 A = 40 mA The current flowing through the load resistor R_L is: I_L = V_Z R_L = 12 600 = 0.02 A = 20 mA Applying Kirchhoff's current law, the current through the Zener diode is: I_Z = I_S - I_L = 40 - 20 = 20 mA Answer: 20