# Common Emitter Amplifier DC analysis

I have a standard common emitter amplifier biased using a voltage divider. I'm given that RE = 500 Ω, IC = 2.5mA, β=100, Vcc=15V and the coefficient of thermal stability K=10 (RE = K * Rth / (b+1) where Rth = R1//R2). I'm asked to calculate every current, voltage and resistor. I have calculate pretty much everything. IB=25μA, IC=IE=2.5mA (supposing IB<<IC), R1=38.55K, R2=5.81K, IR1=IR2=0.34mA (supposing IR1>>IB). VB=1.96V and VE=1.26V.

I'm unable to calculate with the data that I have VC and RC.

Also the Early voltage is -60V but I believe it is to be used later on for ac analysis.

• You have the Ic current, so Rc= ~ (Vcc-1.26)/2 /IC = 2.75 kOhm. Aug 1 at 9:57
• @Antonio51 You are calculating VC=(Vcc-VE)/2. Why is this? Aug 1 at 10:37
• Because Vc "must" be at the mean of Vcc and VE for "maximum" change of Vc. Aug 1 at 10:39
• @Pavlos You have a degree of freedom to choose. You are right to ask why Vc is to be set at any specific place. What things are affected by either raising it or lowering it? Aug 1 at 11:00
• @Pavlos Do you exist? Can you respond? Sep 2 at 6:50

Assuming an unstated (in your question) $$\V_{_\text{BE}}=700\:\text{mV}\$$, then: simulate this circuit – Schematic created using CircuitLab

With your $$\K=10\$$ factor, you should be able to find that $$\X\approx 13.4\%\$$. The rest, except for $$\R_{_\text{C}}\$$, falls out as shown above.

Assuming you want to set the collector voltage to somewhere in the neighborhood of $$\8\:\text{V}\$$ then $$\R_{_\text{C}}\approx 2.7\:\text{k}\Omega\$$.

That's all there is.

An here is the practical "result" ... with VEarly = 60 V. With VEarly = 100 V, there is a "little" change. And the TRansient should be this 