GATE 2024 IN – Question 57
The diode in the circuit shown obeys $I_D=I_S e^{\alpha V_D/T}$, where $V_D$ is in volts, $T$ is absolute temperature in kelvin, $\alpha=1.16\times10^4$ K/V, and $I_S=10^{-15}$ A. The ideal DC current source, opamp, and resistors are temperature independent. The change in output voltage per kelvin is ______ mV/K (one decimal place).
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Correct answer: 9.5 to 10.5
Explanation
At constant current, $V_D=(T/\alpha)\ln(I_D/I_S)$, so $dV_D/dT=\ln(250\times10^{-9}/10^{-15})/11600$ V/K. The amplifier gain is $1+5R/R=6$.
Thus $dV_{out}/dT=6\ln(2.5\times10^8)/11600\approx0.0100$ V/K, or **10.0 mV/K**.