GATE 2014 EC (EC3) – Question 22
The PNP transistor shown has |V_BE| = 0.7 V and β = 50. R_B = 100 kΩ. For output V₀ = 5 V, find R_C (kΩ).
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Correct answer: 1.04 to 1.12
Explanation
The emitter is at10V. With |VBE|=0.7V the base is at9.3V, giving IB=9.3/100kΩ=93µA through RB to ground. In active operation IC=βIB=50×93µA=4.65mA. The collector resistor has5V across it, so RC=5/4.65mA=**1.07527kΩ**. Emitter-collector headroom5V supports active operation.