Home › Previous year questions › GATE 2016 EC (EC2) › Question 24GATE 2016 EC (EC2) – Question 24 Analog Circuits · BJT and MOSFET Amplifiers · 1 mark · Numerical answer
Resistor $R_1$ has been adjusted so that $I_1=1$ mA. Transistors Q1 and Q2 are perfectly matched and have very high current gain, so their base currents are negligible. $V_{CC}=6$ V and $kT/q=26$ mV. The value of $R_2$ in ohms for $I_2=100\ \mu$A is ________.
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Show answer and explanation Correct answer: 570 to 610
Explanation The matched devices obey $I_C\propto e^{V_{BE}/V_T}$. Their bases are common, while Q2 has emitter voltage $I_2R_2$ above Q1's grounded emitter.
Therefore $V_{BE1}-V_{BE2}=V_T\ln(I_1/I_2)=26\ln10=59.867$ mV. This equals the voltage across $R_2$.
Divide by 100 μA to obtain **$R_2=598.67\ \Omega$**, within the official 570–610 Ω range.
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