GATE 2025 EC Q.63: solution and answer

GATE 2025 · ECQ.63 · 2 marks · NAT

An ideal p-n junction germanium diode has a reverse saturation current of 10 μA at 300 K.

The voltage (in Volts, rounded off to two decimal places) to be applied across the junction to get a forward bias current of 100 mA at 300 K is ______.

(Consider the Boltzmann constant $k_{B} = 1.38 \times 1 0^{- 23} J/K$ and the charge of an electron $e = 1.6 \times 1 0^{- 19} C$ .)

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Answer and solution

ANSWER

V ≈ 0.24 V.

Official answer: 0.23 to 0.24Source: Official answer key, GATE 2025 EC (IIT Roorkee)

Diode law in forward bias: $I = I_{S} (e^{V / V_{T}} - 1)$ , with $V_{T} = \frac{k _{B} T}{q}$ .
Compute $V_{T} = \frac{1.38 \times 1 0 ^{- 23} \times 300}{1.6 \times 1 0 ^{- 19}} = 0.025875 V$ .
Since $I = 100 mA ≫ I_{S} = 10 μ A$ (ratio $1 0^{4}$ ), the $- 1$ term is negligible: $I \approx I_{S} e^{V / V_{T}}$ .
So $V = V_{T} ln (\frac{I}{I _{S}}) = V_{T} ln (\frac{100 \times 1 0 ^{- 3}}{10 \times 1 0 ^{- 6}}) = V_{T} ln (1 0^{4})$ .
$ln (1 0^{4}) = 4 ln (10) = 9.21034$ .
$V = 0.025875 \times 9.21034 = 0.238317 V$ .
Rounded to two decimal places: $V \approx 0.24 V$ .

KEY IDEA

In strong forward bias, $I ≫ I_{S}$ , so the diode equation $I = I_{S} (e^{V / V_{T}} - 1)$ reduces to $I \approx I_{S} e^{V / V_{T}}$ , giving $V = V_{T} ln (I / I_{S})$ with $V_{T} = k_{B} T / q$ .

WATCH OUT FOR

Rounding $ln (1 0^{4})$ too early (e.g. using $ln 10 \approx 2.3$ instead of $2.302585$ ) or misplacing a decimal in $V_{T}$ shifts the final answer's second decimal place, since the true value (0.2383 V) sits close to the 0.235 rounding boundary.

From the GATE 2025 EC paper. Every question of this paper · Download the full paper · All 6 P-N junction questions, with notes

GATE 2025 EC Q.63: solution and answer

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