Hi,

Calculate Ib? Vb?

The answer given for Ib is 25uA, so the base voltage is 0.8V (33k x Ib).

But how is 25uA found?
Hfe is given, but I also need to find Ic?
Ib is 25uA when Ic is 1mA.

 

Using

33k*IB+VBE+(1+HFE)*IB*8.2k=10

33k*IB+VBE+(41)*IB*8.2k=10

369.2k*IB=10-VBE

IB=(10-VBE)/369.2k=9.3/369.2k=25.2uA with VBE=0.7V

 

Hfe is given, but I also need to find Ic?
Ib is 25uA when Ic is 1mA.

hfe is defined as the ratio of Ic/Ib.

 

Thanks t_n_k.

You multiplied base current by emitter resistor, IB*8k2.
Is this because IE = IB + IC?

 

That's correct - the voltage drop across the emitter resistance is RE*IE. As to the base voltage VB keep in mind that in this case the base current is flowing from ground potential into the base via the 33kΩ resistor.

So VB=-25.2uA*33kΩ=-0.83V