# Biasing BJT – exact analysis

Lots of students forget how to analyse this bjt amplifier. This amplifier is also called CE (Common emitter) amplifier. It is also called Class A amplifier. To analyse this amplifier, starting from DC, we need to simplify it by removing all AC components such as AC signal input and capacitors.

After that you will have this circuit: To make use of Thevenin’s Equivalent circuit, this BJT biased circuit is further simplified as follow: Now the circuit is ready to be analysed by using KVL for the base circuit and then using the transistor’s amplfiication property to obtain the amplfied current IC.

Eth=VCC*R2/(R1+R2)=15V*30k/(130k+30k)=2.8125V
Rth=R1*R2/(R1+R2)=30k//130k=30K*130K/(130K+30K)=24.375K

If b=150 which means IC/IB=150, then
KVL at the base:
Eth=VRth+0.7+VRE=IBRth+0.7+(IB+IC)RE=IBRth+0.7+(IB+150IB)RE
=IB(Rth+151RE)+0.7
Solving for IB: IB=(Eth-0.7)/(Rth+151RE)=(2.8125-0.7)/(24.375K+151*1.2K)=0.5729uA

Now after finding IB, IC=150*IB with transistor helping to increase IC with the gain 150, provided of course the transistor is in active region which is not cut-off nor saturated. IC=1.4mA

Everything else can be solved after this: IE=IC+IB, VRC=IC*RC, VRE=IE*RE and using KVL to find VCE=VCC-VRC-VRE. VCE found to be 8.5268V which can be used as a check to see whether the BJT is actually in active region because if not, the VCE found can be negative which is impossible. In this case VCE shows that BJT is actually in active region.