Manufacturers supply a zener diode with zener | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given:Zener voltage $V_z = 5.6 \, V$Maximum power dissipation $P_{z \max} = \frac{1}{4} \, W = 0.25 \, W$Input voltage $V_{in} = 10 \, V$The maxim

Vedclass · Accepted Answer

$98.56$

Vedclass · Answer

(A) Given:Zener voltage $V_z = 5.6 \, V$Maximum power dissipation $P_{z \max} = \frac{1}{4} \, W = 0.25 \, W$Input voltage $V_{in} = 10 \, V$The maximum current $I_z$ that the zener diode can handle is given by:$P_{z \max} = V_z 	imes I_z$$0.25 = 5.6 	imes I_z$$I_z = \frac{0.25}{5.6} \, A$In the circuit, the current through the series resistance $R_s$ is $I_s = I_z$. The voltage drop across the resistance $R_s$ is:$V_{R_s} = V_{in} - V_z = 10 \, V - 5.6 \, V = 4.4 \, V$Using Ohm's law for the resistance $R_s$:$R_s = \frac{V_{R_s}}{I_s} = \frac{4.4}{I_z} = \frac{4.4}{(0.25 / 5.6)}$$R_s = \frac{4.4 	imes 5.6}{0.25} = 4.4 	imes 5.6 	imes 4 = 98.56 \, \Omega$Thus, the minimum value of the resistance $R_s$ is $98.56 \, \Omega$.

Similar Questions

What will be the current flowing through the $6 \text{ k}\Omega$ resistor in the circuit shown, where the breakdown voltage of the Zener diode is $6 \text{ V}$?

In the given circuit,the current through the Zener Diode is close to.......$mA$.

$A$ reverse biased Zener diode when operated in the breakdown region works as

$A$ $9\,V$ stabilized voltage supply is required to run a car stereo system from the car's $12\,V$ battery. $A$ Zener diode with $V_z = 9\,V$ and $P_{max} = 0.27\,W$ is to be used as the voltage regulator. The load resistance is $450\,\Omega$. The value of the series resistor $R_s$ is....$\Omega$.

An ideal Zener diode with a breakdown voltage of $3 \text{ V}$ is reverse-biased with an input voltage $V_i = 5 \text{ V}$. The magnitude of the voltage difference between points $B$ and $A$ is: (in $\text{ V}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module