- #36
shayaan_musta
- 209
- 2
sophiecentaur said:That diagram makes things easier, thanks.
Re provides DC feedback which sets the current through the transistor. 'Looking into the base', that 220Ω will appear as approximately 220βΩ (say 22k), which will appear in parallel with R1. Almost negligible but it changes the effective value of R1 to about about 2.4k, giving a base voltage of about +1.45V (do you see the potential divider chain?). That produces an current through Re of (1.45-0.7)220 or 3mA (the 0.7V is the diode drop across the be junction). This 3mA , going through Rc gives 6V across it (the DC value).. Leaving the collector at 12V above ground.
Because of the capacitor Ce there is no AC feedback so the AC gain is not defined but, if you assume again the β is 100, then this is the best you can do. A reasonable voltage swing about the collector standing volts would be +-5V as you say. The actual voltage gain will depend on the value of RL but assume the Load is ten times the valus of Rc and you can 'ignore it'. So an input of just 100mV peak to peak will appear as a voltage swing of 10V peak to peak at the collector.
Hello sophiecentaur!
Thanks for this.
But my answer is still there where I was started it newly.
Kindly help me out with three basics questions i.e.
firstly, why coupling capacitors, bypass capacitor and emitter resistor are used?
secondly, what do coupling capacitors, bypass capacitor and emitter resistor do?
thirdly, what would happen with output response if they will not at there desired place?
Its a request that help me out with this. So that I could design a simplest CE amplifier.
Thanks a lot.