GATE 2024 IN Q.24: solution and answer

GATE 2024 · INQ.24 · 1 mark · NAT

Figure shows an amplifier using an NMOS transistor. Assume that the transistor is in saturation with device parameters, $μ_{n} C_{o x} = 250 μ A/V^{2}$ , threshold voltage $V_{T} = 0.65 V$ and $W / L = 4$ . Ignore the channel length modulation effect. The drain current of the transistor at the operating point is ________ $μ$ A (rounded off to nearest integer).

GATE 2024 IN question 24: Common-source NMOS amplifier with drain-to-gate feedback bias. — Figure from the GATE 2024 IN paper. It is part of the question.

Answer and solution

ANSWER

The drain current at the operating point is $I_{D} = 500 μ A$ , with $V_{GS} = V_{D S} = 1.65 V$ .

Official answer: 498 to 502Source: Official answer key, GATE 2024 IN (IISc Bengaluru)

DC analysis: both $C_{large}$ capacitors are open circuits at DC. What remains is the 3.3 V supply, the 3.3 k $Ω$ drain resistor, the 100 k $Ω$ drain-to-gate resistor and the MOSFET, whose source is grounded.
The MOSFET gate draws no DC current, so no current flows in the 100 k $Ω$ resistor and there is no voltage across it. So $V_{G} = V_{D}$ , which means $V_{GS} = V_{D S}$ because the source is grounded. Since $V_{D S} = V_{GS} > V_{GS} - V_{T}$ , the device is indeed in saturation, as the question says.
Saturation square law, with channel-length modulation neglected: $I_{D} = \frac{1}{2} μ_{n} C_{o x} \frac{W}{L} (V_{GS} - V_{T})^{2} = \frac{1}{2} (250 μ A/V^{2}) (4) (V_{GS} - 0.65)^{2} = 500 μ A/V^{2} (V_{GS} - 0.65)^{2}$
KVL from the supply through the drain resistor. All of $I_{D}$ flows in the 3.3 k $Ω$ resistor, because the feedback resistor carries none: $V_{GS} = V_{D} = 3.3 V - (3.3 k Ω) I_{D}$
Let $x = V_{GS} - 0.65$ (in volts). Substituting the square law into the KVL equation: $3.3 - 0.65 - x = (3.3 k Ω) (500 μ A/V^{2}) x^{2} = 1.65 x^{2} \Rightarrow 1.65 x^{2} + x - 2.65 = 0$
Quadratic formula: $x = \frac{- 1 \pm 1 + 4 ( 1.65 ) ( 2.65 )}{2 ( 1.65 )} = \frac{- 1 \pm 18.49}{3.3} = \frac{- 1 \pm 4.3}{3.3}$ The negative root $x \approx - 1.61 V$ would make $V_{GS} < V_{T}$ . The transistor would then be off and the square law would not apply, so that root is rejected. That leaves $x = 1 V$ , so $V_{GS} = 1.65 V$ .
$I_{D} = 500 μ A/V^{2} \times (1 V)^{2} = 500 μ A$ . Check with the drain resistor: $(3.3 - 1.65) V /3.3 k Ω = 0.5 mA = 500 μ A$ , which agrees.

KEY IDEA

The gate draws no DC current, so the 100 kΩ feedback resistor has no voltage across it. This forces $V_{GS} = V_{D S}$ , which keeps the transistor in saturation (a diode-connected bias). Solve the square law and the drain-resistor KVL together, then keep only the root with $V_{GS} > V_{T}$ .

WATCH OUT FOR

One common slip is including a voltage drop across the 100 kΩ resistor, or letting it share current with the drain resistor, when it carries no DC current. Another is forgetting the factor $\frac{1}{2}$ in the square law, which gives $k = 1000 μ A/V^{2}$ and the wrong current. A third is keeping the negative root of the quadratic, which puts $V_{GS}$ below $V_{T}$ where the square law does not hold.

From the GATE 2024 IN paper. Every question of this paper · Download the full paper · All 13 Biasing questions, with notes

GATE 2024 IN Q.24: solution and answer

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