The figure represents a voltage regulator circuit using a Zener diode. The breakdown voltage of the Zener diode is 6 V and the load resistance is, R_{L} = 4 kΩ. The series resistance of the circuit is R_{i} = R = 1 kΩ. If the battery voltage V_{B} varies from 8 V to 16 V, what are the minimum and maximum values of the current through the Zener diode?

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JEE Mains Previous Paper 1 (Held On: 10 Apr 2019 Shift 2)

Option 3 : 0.5 mA; 8.5 mA

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

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**Concept:**

**Zener diod****e:**

It can operate continuously without being damaged in the region of reverse biased

Functions:

1) It acts as a voltage regulator

2) In forward biasing it acts as an ordinary diode.

The current flowing through the Zener diode is given by the following formula

\({{\rm{I}}_{\rm{Z}}} = {{\rm{I}}_{{{\rm{R}}_{\rm{i}}}}} - {{\rm{I}}_{{{\rm{R}}_{\rm{L}}}}}\)

**Calculation:**

**When the supply voltage is 8 V:**

The voltage across R_{i} is:

V_{Ri} = Power supply – Zener voltage

⇒ V_{Ri} = 8 – 6

∴ V_{Ri} = 2 V

Now, the current through R_{i} is:

\({{\rm{I}}_{{{\rm{R}}_{\rm{i}}}}} = \frac{{{{\rm{V}}_{{\rm{Ri}}}}}}{{{{\rm{R}}_{\rm{i}}}}}\)

\(\Rightarrow {{\rm{I}}_{{{\rm{R}}_{\rm{i}}}}} = \frac{2}{{1000}}\)

∴ I_{Ri} = 2 mA

The voltage across R_{L} is:

V_{RL} = Zener voltage

∴ V_{RL} = 6 V

Now, the current through R_{L} is:

\({{\rm{I}}_{{{\rm{R}}_{\rm{L}}}}} = \frac{{{{\rm{V}}_{{{\rm{R}}_{\rm{L}}}}}}}{{{{\rm{R}}_{\rm{L}}}}}\)

\(\Rightarrow {{\rm{I}}_{{{\rm{R}}_{\rm{L}}}}} = \frac{6}{{4000}}\)

\(\therefore {{\rm{I}}_{{{\rm{R}}_{\rm{L}}}}} = \frac{3}{2}{\rm{\;mA}}\)

The current flowing through the Zener diode is:

\({{\rm{I}}_{\rm{Z}}} = {{\rm{I}}_{{{\rm{R}}_{\rm{i}}}}} - {{\rm{I}}_{{{\rm{R}}_{\rm{L}}}}}\)

\(\Rightarrow {{\rm{I}}_{\rm{Z}}} = 2 - \frac{3}{2}\)

\(\Rightarrow {{\rm{I}}_{\rm{Z}}} = \frac{{4 - 3}}{2} = \frac{1}{2}\)

∴ I_{Z }= 0.5 mA

**When the supply voltage is 16 V:**

The voltage across R_{i} is:

\({{\rm{V}}_{{{\rm{R}}_{\rm{i}}}}} = {\rm{Power\;supply}} - {\rm{Zener\;voltage}}\)

\(\Rightarrow {{\rm{V}}_{{{\rm{R}}_{\rm{i}}}}} = 16 - 6\)

\(\therefore {{\rm{V}}_{{{\rm{R}}_{\rm{i}}}}} = 10{\rm{\;V}}\)

Now, the current through R_{i} is:

\({{\rm{I}}_{{{\rm{R}}_{\rm{i}}}}} = \frac{{{{\rm{V}}_{{\rm{Ri}}}}}}{{{{\rm{R}}_{\rm{i}}}}}\)

\(\Rightarrow {{\rm{I}}_{{{\rm{R}}_{\rm{i}}}}} = \frac{{10}}{{1000}}\)

\(\therefore {{\rm{I}}_{{{\rm{R}}_{\rm{i}}}}} = 10{\rm{\;mA}}\)

The voltage across R_{L} is:

\({{\rm{V}}_{{{\rm{R}}_{\rm{L}}}}} = {\rm{Zener\;voltage}}\)

\(\therefore {{\rm{V}}_{{{\rm{R}}_{\rm{L}}}}} = 6{\rm{\;V}}\)

Now, the current through R_{L} is:

\({{\rm{I}}_{{{\rm{R}}_{\rm{L}}}}} = \frac{{{{\rm{V}}_{{{\rm{R}}_{\rm{L}}}}}}}{{{{\rm{R}}_{\rm{L}}}}}\)

\(\Rightarrow {{\rm{I}}_{{{\rm{R}}_{\rm{L}}}}} = \frac{6}{{4000}}\)

\(\therefore {{\rm{I}}_{{{\rm{R}}_{\rm{L}}}}} = \frac{3}{2}{\rm{\;mA}}\)

The current flowing through the Zener diode is:

\({{\rm{I}}_{\rm{Z}}} = {{\rm{I}}_{{{\rm{R}}_{\rm{i}}}}} - {{\rm{I}}_{{{\rm{R}}_{\rm{L}}}}}\)

\(\Rightarrow {{\rm{I}}_{\rm{Z}}} = 10 - \frac{3}{2}\)

\(\Rightarrow {{\rm{I}}_{\rm{Z}}} = \frac{{20 - 3}}{2} = \frac{{17}}{2}\)

∴ I