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GATE 2015 EC – Question 52

Analog Circuits · BJT and MOSFET Amplifiers · 2 marks · Numerical answer

In the circuit shown, $I_1 = 80$ mA and $I_2 = 4$ mA. Transistors $T_1$ and $T_2$ are identical. Assume that the thermal voltage $V_T$ is 26 mV at 27 °C. At 50 °C, the value of the voltage $V_{12} = V_1 - V_2$ (in mV) is ________.

Two diode-connected transistors $T_1$ and $T_2$ with their emitters grounded, biased by current sources $I_1$ and $I_2$. $V_{12} = V_1 - V_2$ is the difference of their base-emitter voltages.

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Correct answer: 83.5 to 84.0

Explanation

Each diode-connected transistor has $V = V_T\ln\frac{I}{I_S}$, so $V_{12} = V_T\ln\frac{I_1}{I_2}$. At 50 °C, which is 323 K, the thermal voltage is $26 \times \frac{323}{300} = 27.99$ mV. Then $V_{12} = 27.99 \times \ln 20 = 27.99 \times 2.996 = 83.9$ mV.