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

Engineering

Physics

PN Junction Diode

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 :

7 mA

17 mA

15 mA

10 mA

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9 = Vz + V_R1

Vz = 5.6 V

V_R1 = 9 – 5.6

V_R1 = 3.4

$I_{R_{t}} = \frac{V_{R_{1}}}{R} = \frac{3 .4}{200}; I_{R_{t}} = 17 mA$

V₂= VR₂ = IR₂ (R₂)

$\frac{5 .6}{800} = I_{R_{2}}; I_{R_{2}} = 7 mA$

I_z = (17 – 7)mA = 10 mA