GATE 2022 IN Q.58: solution and answer

GATE 2022 · INQ.58 · 2 marks · NAT

In the circuit shown, $R_{1} = 100 k Ω$ and $R_{2} = 1 k Ω$ . If the base-to-emitter voltage of the npn BJT is $0.7 V$ and the collector-to-emitter voltage is $5.2 V$ , the $β$ (current gain) of the BJT is ________ (round off to two decimal places)

GATE 2022 IN question 58: Collector-feedback bias circuit with an npn BJT. — Figure from the GATE 2022 IN paper. It is part of the question.

Answer and solution

ANSWER

The current gain of the BJT is $β \approx 150.11$ .

Official answer: 148.00 to 152.00Source: Official answer key, GATE 2022 IN (IIT Kharagpur)

Assumptions: dc analysis, $V_{B E} = 0.7 V$ , emitter grounded ( $V_{E} = 0$ ). So the collector voltage is $V_{C} = V_{C E} = 5.2 V$ and the base voltage is $V_{B} = V_{B E} = 0.7 V$ .
Ohm's law across $R_{2}$ (from the 12 V rail to the collector node): $R_{2}$ carries the collector current plus the current fed back to the base through $R_{1}$ , so $I_{R_{2}} = I_{C} + I_{B} = \frac{12 - 5.2}{1 k Ω} = 6.8 mA$
Ohm's law across the feedback resistor $R_{1}$ (collector to base). All of the base current flows through it: $I_{B} = \frac{V _{C} - V _{B}}{R _{1}} = \frac{5.2 - 0.7}{100 k Ω} = 45 μ A = 0.045 mA$
KCL at the collector node: $I_{C} = I_{R_{2}} - I_{B} = 6.8 mA - 0.045 mA = 6.755 mA$
Check the region: $V_{C B} = V_{C} - V_{B} = 5.2 - 0.7 = 4.5 V > 0$ , so the collector-base junction is reverse-biased and the base-emitter junction is forward-biased. The BJT is in the active region, so $I_{C} = β I_{B}$ holds.
Current gain: $β = \frac{I _{C}}{I _{B}} = \frac{6.755 mA}{0.045 mA} = 150.11$

KEY IDEA

This is collector-feedback bias. Because the emitter is grounded, $V_{C E}$ gives the collector voltage directly. $R_{1}$ runs from collector to base, so the voltage across it fixes $I_{B} = (V_{C} - V_{B E}) / R_{1}$ . $R_{2}$ carries $I_{C} + I_{B}$ , which is the same as $I_{E}$ . Subtract $I_{B}$ from the current in $R_{2}$ to get $I_{C}$ , then take the ratio.

WATCH OUT FOR

The usual slip is to treat the whole 6.8 mA in $R_{2}$ as $I_{C}$ , which gives $β = 6.8/0.045 = 151.11$ . That lands inside the answer range by luck, but it ignores that $R_{2}$ also supplies the base current through $R_{1}$ . The other common error is to use $V_{C E} = 5.2 V$ as the drop across $R_{1}$ and forget that the base end of $R_{1}$ sits at $0.7 V$ .

From the GATE 2022 IN paper. Every question of this paper · Download the full paper · All 13 Biasing questions, with notes

GATE 2022 IN Q.58: solution and answer

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