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

The current IThis question was previously asked in

JEE Mains Previous Paper 1 (Held On: 08 Apr 2019 Shift 1)

Option 1 : 10 mA

JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)

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90 Questions
360 Marks
180 Mins

**Concept: **

The current under reverse bias is essentially voltage independent upto a critical reverse bias voltage, known as **breakdown voltage**.

**Zener diode** is specially designed to work under the breakdown region. In zener diode, when the applied reverse voltage reaches the breakdown voltage, then there is a large change in current. But after breakdown voltage, zener voltage remains constant in the large current through the zener diode. This helps the zener diode to regulate the supply voltages. That is why zener diode acts as a voltage regulator.

**Calculation:**

The circuit is given in the detailed manner in the image below:

The voltage drop is:

\(9={{\text{V}}_{\text{Z}}}+{{\text{V}}_{{{\text{R}}_{1}}}}\)

From question, we know that,

V_{Z }= 5.6 V

On substituting this value in above equation, we get,

\(9=5.6+{{\text{V}}_{{{\text{R}}_{1}}}}\)

Now, the current passing through R_{1} is:

\({{\text{I}}_{{{\text{R}}_{1}}}}=\frac{{{\text{V}}_{{{\text{R}}_{1}}}}}{\text{R}}\)

\(\Rightarrow {{\text{I}}_{{{\text{R}}_{1}}}}=\frac{3.4}{200}\)

\(\therefore {{\text{I}}_{{{\text{R}}_{1}}}}=0.017\text{ }\!\!~\!\!\text{ A}=17\text{ }\!\!~\!\!\text{ mA}\)

Now, the voltage across Zener diode V_{Z} is:

\({{\text{V}}_{\text{z}}}={{\text{V}}_{{{\text{R}}_{2}}}}={{\text{I}}_{{{\text{R}}_{2}}}}\left( {{\text{R}}_{2}} \right)\)

\(\Rightarrow {{\text{I}}_{{{\text{R}}_{2}}}}=\frac{{{\text{V}}_{\text{z}}}}{{{\text{R}}_{2}}}\)

\(\Rightarrow {{\text{I}}_{{{\text{R}}_{2}}}}=\frac{5.6}{800}\)

\(\therefore {{\text{I}}_{{{\text{R}}_{2}}}}=0.007\text{ }\!\!~\!\!\text{ A}=7\text{ }\!\!~\!\!\text{ mA}\)

Now, the current passing through Zener diode is:

\({{\text{I}}_{\text{z}}}={{\text{I}}_{{{\text{R}}_{1}}}}-{{\text{I}}_{{{\text{R}}_{2}}}}\)

⇒ I_{Z }= (17 – 7) mA