How Do You Design a CE Amplifier with a Gain of 10 Using a 2N3906 Transistor?

[Sculptured]

Alright, I've been stuck on this problem for my electronics class for far too long.

" Design a CE amplifier with a gain of 10 using a 2n3906 transistor and a 20 V power supply. Calculate the required resistor and capacitor values for the circuit, assuming a minimum frequency of 10 Hz."

It turns out that there are four unknown resistors and one capacitor (if I'm thinking correctly). The voltage divider, the collector resistor, the emitter resistor, and the capacitor parallel to the emitter resistor. With such little information, at least for me, I'm completely lost on where to even begin to relate these values. Everything I've thought up so far has constantly interwoven the unknown values, given me new ones, and never enough equations to solve them all. Any help at all is appreciated in even getting started down the right path.

 

[pooface]

Well 2N3906 is a BJT PNP transistor right?

So gather your BJT PNP transistor formulas and work backwards from the gain of 10.
This looks sounds like a voltage divider bias with a full emitter bypass.

I don't remember the formulas by heart but make a list and I'll see what I can add to it from my notes.

 

[CEL]

The collector resistor is related to the gain. The emitter resistor is calculated in order that the voltage drop across it is much greater than the base-emitter voltage. With a 20 V power supply you can make the voltage drop around 5 V.
Calculate the base current from the collector current and the transistor's parameters. Calculate the base voltage from the emitter voltage and the base-emitter voltage.
The voltage divider must provide the base voltage. In order for the base current not to be significant, make the current through the divider 10 to 20 times greater.