A BJT current source is given in the figure. Assume the Si-PNP transistor to operate in the active region. The value of current I in mA is

A BJT current source is given in the figure. Assume the Si-PNP transistor to operate in the active region. The value of current I in mA is A single-phase AC distribution line supplies two single-phase loads as shown in the figure below. The impedances of line segments A-B and B-C are j0.25 Ω and j0.35 Ω respectively. The voltage drop from A to C is

Right Answer is:

2.15

SOLUTION

For given BJT current source operating in active region

IE = IC + IB

As the value of β is not given we assume that β is very large

Therefore, IB = 0

⇒ I = IE = IC

Applying the KVL loop at the input side through the Zener diode as it regulates the voltage across the emitter resistance.

VZ −IERE − VBE = 0

∵ For PNP transistor potential barrier is −0.7 V

5 − IE × 2 − 0.7 = 0

2 × 103 × IE = 4.3

IE  = 2.15 mA

∵ The value of β is not given we assume that β is very large

∴ IE = IC = I = 2.15 mA

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