GATE 2018 EC Q.37: solution and answer

GATE 2018 · ECQ.37 · 2 marks · NAT

A dc current of $26 μ A$ flows through the circuit shown. The diode in the circuit is forward biased and it has an ideality factor of one. At the quiescent point, the diode has a junction capacitance of $0.5 nF$ . Its neutral region resistances can be neglected. Assume that the room temperature thermal equivalent voltage is $26 mV$ .

For $ω = 2 \times 1 0^{6} rad/s$ , the amplitude of the small-signal component of diode current (in $μ A$ , correct to one decimal place) is ________.

GATE 2018 EC question 37: A single series loop. — Figure from the GATE 2018 EC paper. It is part of the question.

Answer and solution

ANSWER

The amplitude of the small-signal diode current is about 6.4 µA.

Official answer: 6.2 to 6.6Source: Official answer key, GATE 2018 EC (IIT Guwahati)

Small-signal model of the diode. At the Q-point the forward-biased diode is modelled as its incremental (dynamic) resistance $r_{d}$ in parallel with the junction capacitance $C_{j} = 0.5 nF$ given in the question. The neutral-region (series) resistance is neglected, as stated. For the ac analysis the dc battery $V$ is replaced by a short circuit, which leaves the source $5 sin (ω t) mV$ in series with the diode model and the $100 Ω$ resistor.
Incremental resistance with ideality factor $η = 1$ and $I_{D} = 26 μ A$ : $r_{d} = \frac{η V _{T}}{I _{D}} = \frac{26 mV}{26 μ A} = 1 k Ω$
Capacitive reactance at $ω = 2 \times 1 0^{6} rad/s$ : $X_{C} = \frac{1}{ω C _{j}} = \frac{1}{( 2 \times 1 0 ^{6} ) ( 0.5 \times 1 0 ^{- 9} )} = \frac{1}{1 0 ^{- 3}} = 1 k Ω$
Diode impedance, with $r_{d}$ in parallel with $- j X_{C}$ : $Z_{D} = \frac{( 1000 ) ( - j 1000 )}{1000 - j 1000} = \frac{- j 1000}{1 - j} = \frac{- j 1000 ( 1 + j )}{2} = (500 - j 500) Ω$
Total loop impedance, adding the series $100 Ω$ : $Z = (600 - j 500) Ω, ∣ Z ∣ = 60 0^{2} + 50 0^{2} = 610000 \approx 781.0 Ω$
Phasor amplitude of the current. It is a single series loop, so the whole ac current flows through the diode (as the sum of its $r_{d}$ and $C_{j}$ branch currents): $∣ i_{d} ∣ = \frac{5 mV}{781.0 Ω} \approx 6.40 μ A$ Check on the small-signal assumption: the ac voltage across the diode is $∣ Z_{D} ∣ ∣ i_{d} ∣ = (707 Ω) (6.40 μ A) \approx 4.5 mV$ , well below $V_{T} = 26 mV$ , so the linear model is reasonable.
Correct to one decimal place, the amplitude of the small-signal diode current is $6.4 μ A$ .

KEY IDEA

For small signals, a forward-biased diode acts as its dynamic resistance $r_{d} = η V_{T} / I_{D}$ in parallel with its junction capacitance. With the dc source shorted, the ac current is the source amplitude divided by the magnitude of the complex loop impedance, which is $(r_{d} ∥ C_{j})$ in series with the resistor.

WATCH OUT FOR

Students often ignore $C_{j}$ and take $5 mV / (1000 + 100) Ω \approx 4.5 μ A$ . Another mistake is to add $r_{d}$ and $X_{C}$ in series, or to add the magnitudes $∣ Z_{D} ∣ + 100$ instead of adding the complex impedances first. Some also report only the current through $r_{d}$ and leave out the current through the capacitance, although the question asks for the total diode current.

From the GATE 2018 EC paper. Every question of this paper · Download the full paper · All 25 Diode circuits questions, with notes

GATE 2018 EC Q.37: solution and answer

Answer and solution

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