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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?

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? Correct Answer 50ms

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.

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