1) For the circuit shown in figure, calculate emitter current(I<E>),Collector current(I<C>),base current(I<B>) and the collector emitter voltage.(Beta=current gain=[I<C>]/[I<B>]=50).Neglect the voltage drop across base emitter terminal.
I solved it in the following way:
Applying Kirchhoff voltage law to the loop SBEPS,
25+5-{(I<B>)R1 = 0
I<B> = 30/R1
= 30/200 m Amp
= 0.15 m Amp
But the answer given in my book is 0.1 m Amp.
I took the voltage across the loop as 25 + 5=30 volt because, the emitter connected to -5 volt is similar to the emitter being connected to the negative terminal of a 5 volt battery. Similarly, the base connected to +25 volt, is similar to the base being connected to the positive terminal of the 25 volt battery. So, as we traverse along the loop SBEPS, we would find that we move from negative terminal to positive terminal of the 2 sources and hence both positive.
If I take the voltage across the loop as 25 – 5=20 volt, I would get the book answer. Which is right?

 

The voltages add, as you would expect. From teh base source to emitter is 30 volts, from the collector source to emitter is 25 volts. I believe your result for base current is correct.

 

Thanx for the guidance.

 

Ib = (25 - (-5)) / 200k = 0.15mA
Ic = 50 * Ib = 7.5mA
therefore, 7.5mA = (20 - Vce - (-5)) / 2k
Vce = 10V
Ie = Ic + Ib
Ie = 7.65mA

Hope that's right. Haven't done transistors in a year.