How would I calculate Vce and Ic from a given DC transistor gain?
With BJTs there is no real point trying to use math with complicated formulas to uncover precise voltage levels. Reason: there are two many things that are uncertain and even simulators are lucky to get close so, all that can be realistically done is make an estimate.
- Whatever voltage is at the base (set by the potentiometer and the 150 kΩ) will appear at the emitter but reduced by about 0.7 volts.
- So let's say the emitter voltage (as determined by the base voltage) is 1 volts (implying 1.7 volts at the base).
- That means that there will be a emitter current of 1/470 amps = 2.13 mA.
- Given that about 1% of that emitter current is base current you can pretty much say that all of the 2.13 mA flows into the collector.
- 2.13 mA through 2.2 kΩ (collector resistor) produces a volt drop of 4.68 volts.
- Hence, the collector voltage relative to ground will be 10 volts (the supply) minus 4.68 volts = 5.32 volts.
Vc = 5.32 volts and Ic = 2.13 mA
Of course, if you set the pot to a different position this will change but at least you have the mechanism for working this out yourself.