Question

The reverse breakdown voltage of a Zener diode is 5.6 V in the given circuit.

The current I_z through the Zener is ______.

Options

• 10 mA

• 17 mA

• 15 mA

• 7 mA

Answer 1

The reverse breakdown voltage of a Zener diode is 5.6 V in the given circuit.

The current I_z through the Zener is 10 mA.

Explanation:

Given: A Zener diode connected as described in the circuit below has a reverse breakdown voltage of V_z = 5.6V.

To find: The Zener diode's current is I_z.

The voltage drop across the Zener diode is as follows:

V_z = 5.6V

Voltage drop across the R₂ = 800Ω resistor:

V_z = 5.6V

Current through the R₂ = 800Ω resistor:

I₂ = `V_z/R_2 = 5.6/800` = 7mA

The voltage drop across the R₁ = 200Ω resistor is as follows:

V₁ = V - V_z = 9 - 5.6 = 3.4V

Current through the R₁ = 200Ω resistor: 

`I_1 = V_1/R_1 = 3.4/200 = 17` mA

Current I_z through the Zener diode:

`I_z = I_1 - I_2 = 17 - 7 = 10` mA