I have a BJT problem that requires me to find the Vin and the resulting Vout(Ve). Given values are the Vp=5V, Rc=1kΩ, Rb=1kΩ, Rout=1k(located at the emitter), β=100. At full saturation, is Vce always 0.2V?

 

Let us know how far have you been able to get on your own?

hgmjr

 

Let me try. My answers WILL be wrong cos I am just studying this too.

Ic = Vp-Vcesat / 1000 = 4.8mA

βIb = Ic

so Ib = 48μA

Vin - Vbesat /1000 = Ib

Vin = 0.248V

I dunno how to find Vout. Neither do I know the purpose of the 1000Ω at the emitter. I also need help too

 

Let me try. My answers WILL be wrong cos I am just studying this too.

Ic = Vp-Vcesat / 1000 = 4.8mA

βIb = Ic

so Ib = 48μA

Vin - Vbesat /1000 = Ib


Vin = 0.248VI dunno how to find Vout. Neither do I know the purpose of the 1000Ω at the emitter. I also need help too

I'm not sure myself, but here is what I came up with:

VP=Vin where the two circles may be used to denote a connection.

so if VP=Vin
and if Vce sat. =0.2V.
then Ic=[(VP-Vce(sat.)) / (RC+RE)] where Ic will flow through both resistors and Q1 to ground as it is a series circuit.
IB would be (Ic / Beta)
and Vout would be [(Ic+Ib) X RE] where the combo of both Ic and Ib flow through RE.

I believe the RE at the emitter is for bias and temp. stability, as well as if the output is coupled through a capacitor then RE will allow a negative going signal to be produces as well as a positive signal. Without RE then the only signal output would be positive after a 0.7v. drop due to the Vbe, which would give a distorted output.


As I said I'm only guessing at this solution.

Nope my guess is wrong too.
Because RE and RC have circles also.....

Somebody can tell us how to do it.