Question

For the BJT, β=∞, VBEon=0.7V VCEsat=0.7V. The switch is initially closed. At t=0, it is opened. At which time the BJT leaves the active region?

(a) 20ms

(b) 50ms

(c) 60ms

(d) 70ms

This question was posed to me in a job interview.

I'm obligated to ask this question of Transistor Switching Times topic in section Transistor Characteristics of Electronic Devices & Circuits

Answer 1

Right answer is (b) 50ms

The best I can explain: At t < 0, the BJT is OFF in cut off region. IB=0 as β=∞, so IC=IE. When t > 0, switch opens and BJT is ON. The voltage across capacitor increases. From the input loop, -5-VBE-I(4.3K)+10=0 and gives I=1mA. IC1=1-0.5=0.5mA. VC1=0.7+4.3+10=-5V. IC1=C1dVC1/dt. From this equation, we get t=50ms.