GATE 2024 IN Q.57: solution and answer

GATE 2024 · INQ.57 · 2 marks · NAT

In the figure shown, the diode current is given by $I_{D} = I_{S} e^{\frac{α V _{D}}{T}}$ . $V_{D}$ is the diode voltage in volts, $T$ is the absolute temperature in Kelvin, $α = 1.16 \times 1 0^{4} K/V$ , and $I_{S} = 1 0^{- 15} A$ is the saturation current. The dc current source, opamp and the resistors are ideal, and are assumed to be temperature independent. The change in the output voltage ( $V_{o u t}$ ) per Kelvin change in temperature is ________ mV (rounded off to one decimal place).

GATE 2024 IN question 57: A 250 nA dc current source hangs from V_DD and drives its current downward into a node (call it A). — Figure from the GATE 2024 IN paper. It is part of the question.

Answer and solution

ANSWER

The output voltage rises by about 10.0 mV for every 1 K rise in temperature.

Official answer: 9.5 to 10.5Source: Official answer key, GATE 2024 IN (IISc Bengaluru)

Diode bias. The ideal 250 nA source forces $I_{D} = 250 nA$ through the diode. Its anode is at the top node and its cathode is at the grounded rail, so this current forward-biases it. The op-amp's $+$ input is tied to the anode, so it sits at $V_{D}$ .
Invert the diode law. From $I_{D} = I_{S} e^{α V_{D} / T}$ : $V_{D} = \frac{T}{α} ln (\frac{I _{D}}{I _{S}})$
Op-amp stage (ideal op-amp, negative feedback, virtual short). The feedback resistor $5 R$ runs from $V_{o u t}$ to the $-$ input, and $R$ runs from the $-$ input to ground. So $V_{-} = V_{o u t} \frac{R}{R + 5 R} = V_{+} = V_{D}$ . This is a non-inverting amplifier: $V_{o u t} = (1 + \frac{5 R}{R}) V_{D} = 6 V_{D} = \frac{6 T}{α} ln (\frac{I _{D}}{I _{S}})$
Temperature derivative. $I_{D}$ comes from an ideal source, and $I_{S}$ , $α$ and the resistor ratio are given as temperature independent. So only the explicit $T$ changes: $\frac{d V _{o u t}}{d T} = \frac{6}{α} ln (\frac{I _{D}}{I _{S}})$
Evaluate the logarithm: $ln (\frac{250 \times 1 0 ^{- 9}}{1 0 ^{- 15}}) = ln (2.5 \times 1 0^{8}) = ln 2.5 + 8 ln 10 = 0.9163 + 18.4207 = 19.337$
Result: $\frac{d V _{o u t}}{d T} = \frac{6 \times 19.337}{1.16 \times 1 0 ^{4}} V/K = 1.0002 \times 1 0^{- 2} V/K \approx 10.0 mV/K$
Sign and sanity check. The result is positive. With $I_{S}$ held constant and a fixed current, $V_{D}$ is proportional to absolute temperature (the PTAT principle), so $\frac{d V _{D}}{d T} = \frac{V _{D}}{T}$ . At $T = 300 K$ this model gives $V_{D} = \frac{300}{1.16 \times 1 0 ^{4}} \times 19.337 \approx 0.50 V$ , a realistic silicon diode drop, and $\frac{d V _{D}}{d T} \approx 1.67 mV/K$ . The gain of 6 turns that into $10.0 mV/K$ . The answer does not depend on which temperature is chosen.

KEY IDEA

A fixed current in a diode whose saturation current is held constant gives $V_{D} = \frac{T}{α} ln (I_{D} / I_{S})$ , which is proportional to absolute temperature. The non-inverting amplifier multiplies it by $1 + 5 R / R = 6$ , so $d V_{o u t} / d T = \frac{6}{α} ln (I_{D} / I_{S})$ .

WATCH OUT FOR

(1) Using the familiar $- 2 mV/^{\circ} C$ rule for a real diode. That rule comes from $I_{S}$ rising steeply with temperature, and this question fixes $I_{S}$ at a constant, so here the voltage rises with temperature. (2) Taking the gain as $5 R / R = 5$ , the inverting-amplifier gain, when the circuit is non-inverting with gain $1 + 5 = 6$ ; this gives 8.3 mV. (3) Dropping the $8 ln 10$ term or using $lo g_{10}$ in place of $ln$ .

From the GATE 2024 IN paper. Every question of this paper · Download the full paper · All 25 Diode circuits questions, with notes

GATE 2024 IN Q.57: solution and answer

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