How would you find base current for a class A amplifier with no base resistor?

 

For me the question is a little cryptic - are you able to elaborate further?

 

I don't have the exact numbers with me, but basically it was a voltage divider biased transistor. I have the two biasing resistor values and VCC, but that's it. I was asked for base current, but couldn't figure out how to get their answer.

 

A schematic would be helpful.

hgmjr

 

This sin't the exact one, but this is basically what I was given.

 

Were you given the Beta of the transistor as part of the problem statement? If so what is it?

Were you also given the Vbe to assume in the problem statement? If so, what is it?

hgmjr

 

Vbe was .6 if I remember correctly, but that was all I they gave.

 

You actually do have a base resistor embodied in the resistor divider R1 and R2. Calculate the Thevenin equivalents of the base voltage divider.

hgmjr

 

It's just R1 in parallel with R2 right? At least that's what I did, but it didn't work out.

 

That is because you probably overlooked that you have to use Thevenin's equivalent voltage as well. Not 12V.

hgmjr

 

I used the drop on R2 for the voltage. Is that correct?

 

Yes. The voltage across R2 is the Thevenin equivalent voltage for the network consisting of R1 and R2. Was your answer off by much?

hgmjr

 

Yeah, over double! I just took the voltage across R2 divided by the parallel combination of the two biasing resistors. Oh well...I don't know what else I could have done. Thanks for helping. At least I know I'm not crazy now!

 

Did you consider the input impedance of the base? Can you post your calculation that yielded the incorrect answer? I should be able to spot where you went off the rail.

hgmjr

 

I think that R1=4.7K and R2=1k with 9Vdc. I found Vr2 to be 1.579V. I divided that by 824.56 to get 1.91 mA. The correct answer was supposed to be around .7 mA

 

You forgot to subtract the value of Vbe from Vth before you divided.

hgmjr

 

That still leaves me with 1.187 mA . I know the answer was supposed to be .7 because I've been going over and over it.

 

If I divide Vbe by Rb I get close, but why would I do that?

 

is there an emitter resistor or is the emitter tied to ground?


hgmjr

 

There is a resistor, but I wasn't given the value.