Semiconductor Electronics Part 2

The circuit diagram is shown below The supply voltage, Vs=24V
Zener voltage, Vz=10V
The value of supply voltage is greater than the Zener voltage.
Hence, the voltage across the load resistance of 5 kΩ will be equal to Zener voltage as they are connected in parallel combination.
Current IL flowing through the load resistance can be calculated as
IL=

10

5×103

=2×10−3A=2‌mA
Since, the voltage drop across 5 kΩ is 10 V, then the voltage dropacross 1 kΩ resistance, V′ = 24 − 10 = 14 V
Current flowing through 1 kΩ resistance can be calculated as
I=

V′

1×103

=

14

1×103

=14×10−3A=14‌mA
From circuit diagram, I=IZ+IL
Iz=I−IL=14−2=12‌mA
Power across Zener diode is
P=Vz×Iz=10×12=120‌mW
Thus, the power across the Zener diode is 120 mW.