Question

In the given circuit, the current through the Zener diode is

• A. $2.5mA$
• B. $3.3mA$
• C. $5.5mA$
• D. $6.7mA$

Answer 1

The correct option is B

$3.3mA$

Step 1: Given Data

Resistance of the resistor is $R_1=500\Omega$ and resistance of the another resistor is $R_2=1500\Omega$

Voltage of the battery is $V=15V$

Voltage of the Zener diode is $V_Z=10V$

Let the voltage across the resistor $R_1$ be $V_{R_1}$.

Step 2: Zener diode

1. In a reverse-biased p-n junction, a small amount of reverse saturation current flows which is almost independent of the reverse-bias voltage.
2. However, if the reverse voltage exceeds a certain critical value the current suddenly rises to a very large value.
3. This is referred to as the breakdown in diodes.
4. Unless the diode is specially designed it may damage the diode.
5. Diodes that have adequate power dissipation capabilities to operate in the breakdown region are called breakdown diodes or Zener diodes.

Step 3: Calculate the Voltage

Let the voltage across the resistor $R_1$ be $V_{R_1}$.

According to Kirchoff's law

$V_{R_1}=V-V_Z$

Upon substituting the values we get,

$V_{R_1}=15-10$

$=5V$

Step 4: Calculate the Current

Therefore according to Ohm's law, the current across $R_1$ is

$I_{R_1}=\frac{5V}{500\Omega }$

$=0.01A$

Similarly, the current through $R_2$ can be given as

$I_{R_2}=\frac{10V}{1500\Omega }$

$=0.067A$

Step 5: Calculate the net current

Hence, the current passing through the Zener diode can be given as

$I_{R_1}-I_{R_2}=0.01-0.067$

$=0.0033A=3.3mA$

Hence, the correct answer is an option (B).