Gee, thanks for the help! So how would you make the transistor so it's not near saturation and put it in its linear region?You said you want to learn how to design transistor circuits from the basic config. first.
Lets look at your circuit and analyze the values.
1. If you eleminated RE than voltage at the base would be 0.7v. because the VBE shunts the R2 resistor.
2.If you eliminated RE than emitter volatage woiuld be zero volts, because emitter is connected to ground.
2a.By using the RE than base voltage will be 4.5v.
so lets put RE back in.
3. since you have given 25Ma. for collector current. Then you should now calculate the emitter resistor. :3.8v. / 25Ma. = approx. 150 ohms.
from there you have a choice of how much volt. drop do you want across the transistor (VCE), or what voltage do you want at the collector with respect to ground. (VC)
Ussually for a class A amp the voltage at the collector (VC), is around
VCC/2. In this case 4.5V.
4. choosing VC=4.5v. then 4.5v. is dropped across the collector resistor.
so VCC-VC=VRC. can be also calculated as VRC=VCC/2 ALWAYS for a class A amp. And VCC/2=VC
so now RC=VC/IC, which would be 4.5v./25Ma.=180 ohms.
There are many ways to design a simple class A amp. but what dictates a design is the parameters given known as constraints.
Now due to the amount of base voltage gives a VE of 3.8 volts, which means that 0.7v. dropped across the transistor, so the transistor is near saturation, which would not work as a amplifier.
So values for VB. would have to be redetermined so as to get a VCE that would put the transistor in it's linear region.