GATE 2020 EE Q.31: solution and answer

GATE 2020 · EEQ.31 · 2 marks · MCQ

A non-ideal diode is biased with a voltage of $- 0.03 V$ , and a diode current of $I_{1}$ is measured. The thermal voltage is $26 mV$ and the ideality factor for the diode is $15/13$ . The voltage, in V, at which the measured current increases to $1.5 I_{1}$ is closest to:

(A)
$- 0.02$
(B)
$- 0.09$
(C)
$- 1.50$
(D)
$- 4.50$

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Answer and solution

ANSWER

With $η V_{T} = 30 mV$ , $I_{1} = I_{S} (e^{- 1} - 1) = - 0.632 I_{S}$ . Setting $I_{S} (e^{V /0.03} - 1) = 1.5 I_{1} = - 0.948 I_{S}$ gives $e^{V /0.03} = 0.0518$ , so $V \approx - 0.089 V \approx - 0.09 V$ . The answer is option (B).

Official answer: BSource: Official answer key, GATE 2020 EE (IIT Delhi)

Use the Shockley diode equation with ideality factor $η$ : $I = I_{S} (e^{V / (η V_{T})} - 1)$ . The bias is only about one $η V_{T}$ below zero, so the $- 1$ term is not negligible and must be kept.
Effective thermal voltage: $η V_{T} = \frac{15}{13} \times 26 mV = 30 mV = 0.03 V$ .
At $V = - 0.03 V$ : $\frac{V}{η V _{T}} = \frac{- 0.03}{0.03} = - 1$ , so $I_{1} = I_{S} (e^{- 1} - 1) = - 0.6321 I_{S}$ (a reverse current).
Required current: $1.5 I_{1} = 1.5 \times (- 0.6321 I_{S}) = - 0.9482 I_{S}$ . Set $I_{S} (e^{V /0.03} - 1) = - 0.9482 I_{S}$ .
Solve: $e^{V /0.03} = 1 - 0.9482 = 0.0518$ , so $\frac{V}{0.03} = ln (0.0518) \approx - 2.96$ .
Hence $V \approx - 2.96 \times 0.03 V \approx - 0.089 V \approx - 0.09 V$ , which is option (B). Check that a solution exists: in reverse bias the current magnitude can only approach $I_{S}$ , and $1.5 \times 0.632 I_{S} = 0.948 I_{S}$ is still less than $I_{S}$ .

KEY IDEA

At a reverse bias of only about one $η V_{T}$ , keep the full Shockley equation $I = I_{S} (e^{V / (η V_{T})} - 1)$ . The reverse current saturates toward $- I_{S}$ , so raising its magnitude by 50% (from $0.632 I_{S}$ to $0.948 I_{S}$ ) takes a lot more reverse voltage, and $V$ ends up about three times the original bias.

WATCH OUT FOR

Dropping the $- 1$ term and writing $I \approx I_{S} e^{V / (η V_{T})}$ . That approximation holds only for forward bias well above $η V_{T}$ . Using it here gives $V = - 0.03 + 0.03 ln 1.5 \approx - 0.018 V \approx - 0.02 V$ , which is distractor (A). It moves the voltage toward zero even though the reverse current is supposed to grow. A second slip is using $V_{T} = 26 mV$ without multiplying by the ideality factor.

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GATE 2020 EE Q.31: solution and answer

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