Consider a Zener diode connected in reverse bias with a 15 V battery and a 10 kΩ resistor in series. If the breakdown voltage of the diode is 4.5 V , then find the current flowing through the Zener in μA :

Question

Consider a Zener diode connected in reverse bias with a 15 V battery and a 10 kΩ resistor in series. If the breakdown voltage of the diode is 4.5 V , then find the current flowing through the Zener in μA : (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

I=(V textin - Vz/Rs) =(15 - 4 . 5/10 × 103) =(10 . 5/104) =1.05× 10- 3A ∴ I=1.05 mA=105 Î¼A

Q. Consider a Zener diode connected in reverse bias with a $15 V$ battery and a $10 k Ω$ resistor in series. If the breakdown voltage of the diode is $4.5 V$ , then find the current flowing through the Zener in $μ A$ :

$I = \frac{V _{in} - V _{z}}{R _{s}}$
$= \frac{15 - 4.5}{10 \times 1 0 ^{3}}$
$= \frac{10.5}{1 0 ^{4}}$
$= 1.05 \times 1 0^{- 3} A$
$∴ I = 1.05 m A = 105 μ A$

Q. Consider a Zener diode connected in reverse bias with a 15V battery and a 10kΩ resistor in series. If the breakdown voltage of the diode is 4.5V , then find the current flowing through the Zener in μA :

I=Rs​Vin ​−Vz​​ =10×10315−4.5​ =10410.5​ =1.05×10−3A ∴I=1.05mA=105μA

Q. Consider a Zener diode connected in reverse bias with a $15 V$ battery and a $10 k Ω$ resistor in series. If the breakdown voltage of the diode is $4.5 V$ , then find the current flowing through the Zener in $μ A$ :

$I = \frac{V _{in} - V _{z}}{R _{s}}$
$= \frac{15 - 4.5}{10 \times 1 0 ^{3}}$
$= \frac{10.5}{1 0 ^{4}}$
$= 1.05 \times 1 0^{- 3} A$
$∴ I = 1.05 m A = 105 μ A$