Hello,
I have a question about designing a common emitter amplifier. I know the amp will invert the input so a gain of 20 is what I am looking for as whatever goes in, will come out inverted and multiplied by 20.
Part of my coursework mentions that a rule-of-thumb is that the collector resistor (Rc) over the emitter resistor (Re) will give the voltage gain.
So, example:
Rc = 5.6k,
Re = 1k,
5.6k / 1k = 5.6 Av
I am struggling with the coursework a little, it starts by selecting a supply voltage and choosing a quiescent current of 1mA. So, if the supply is 9 volts, the output would be half the supply, 4.5 volts. So, 4.5 / 0.001 = 4.5k or 4.7k (real value resistor). Then if the same current is going through the emitter and another rule of thumb is that the voltage across this resistor is 8% - 10% of the supply. So, 10% of 9 volts is 0.9 volts, find resistor by 0.9 / 0.001 = 900 or real value 910 ohms.
with 4.7k over 910 = 5.16
I want to work backwards. starting with 20 as a gain. if I keep the 9 volt supply and 1mA quiescent i will have 4.7k still. rearrange some stuff to get 20 / 4.7k to get 4.25 milliohms. thats a small resistor!!!?
I was hoping there was a way to do this. The instructions in my course are really basic and only show this one way.
any advice would be great!

thanks
simon

 

rearrange some stuff to get 20 / 4.7k to get 4.25 milliohms. thats a small resistor!!!?

Try 4.7k / 20 => 220 ohms.

 

Again=alfa*Requiv_load/(Re+re) Requiv_load=Rc||RLoad for RLoad=inf ==> Requiv_load=Rc
Ic=1mA ==>re~25.8Ohm, alfa=Beta/(Beta+1), Beta=100 ==> alfa=0.99