# 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. Commented Aug 1, 2023 at 9:57
• @Antonio51 You are calculating VC=(Vcc-VE)/2. Why is this? Commented Aug 1, 2023 at 10:37
• Because Vc "must" be at the mean of Vcc and VE for "maximum" change of Vc. Commented Aug 1, 2023 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? Commented Aug 1, 2023 at 11:00
• @Pavlos Do you exist? Can you respond? Commented Sep 2, 2023 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