# Common emitter amplifier question

Discussion in 'Homework Help' started by midnightblack, Jun 9, 2012.

1. ### midnightblack Thread Starter Member

Feb 29, 2012
31
0
Hi,

I am stuck on part (a) of this question:

I am not sure where I am going wrong exactly:

I calculate the voltage at the base of the transistor to be at 20*(43/(43+330) since the two resistors act as potential dividers.

This voltage is 2.31V. (Vb=2.31V)

Now if we assume the transistor to be on, the Vbe=0.7. So Ve=2.31-0.7=1.61V

If there is 1.61V across the 1.5kΩ resistor, the current must be I=V/R=1.61/1.5k=1.07mA. This is current Ie.

Ic=αIe. Where α=β/(β+1) and Ie=1.07mA ==> Ic=(200/201)*1.07m=1.06mA

I don't see any mistaking in my method but the answer is given as Ic=0.946mA.

Can someone explain?

2. ### panic mode Senior Member

Oct 10, 2011
1,330
305
that is your problem. you have neglected base current so base voltage you got is wrong (it is higher than real value).

3. ### panic mode Senior Member

Oct 10, 2011
1,330
305
Vb = Vbe + Ie*Re

you can relate Ib and Ie using known beta value.
you can also use KCL for base circuit - current flowing through 330k splits into two currents - base of transistor and 43k.

the other way is to use Thevenin and replace base resistors

4. ### midnightblack Thread Starter Member

Feb 29, 2012
31
0
Less current would flow through the 43k resistor if we take it elsewhere so the voltage at that point drops right?

So I took the Thévenin equivalent and got the correct answer.

Thank you!

5. ### panic mode Senior Member

Oct 10, 2011
1,330
305
you are welcome