NPN with a floating emitter

Thread Starter


Joined Oct 18, 2011
The question given is to solve for the collector current. The emitter is floating and the collector is biased at 5V with respect to the base.

The givens are Is = 10fA, BF = 100, and BR = 1

I understand the problem, I am just having trouble figuring out how to set it up properly. My assumptions are that iE = 0, so therefore Vce = 0

using the equation
ic = Is*e^((VBE/VT)(1+ VCE/VA))

Then VBE would be 5V, VT = 0.025, and VCE = 0?

Since iE is equal to zero then I know that iE = (BF +1)iB = 0, so then iB = 0 which does not make sense because ic = BF*iB and ic is what we are solving for.

Any help would be appreciated. Thanks in advance.


Joined Oct 26, 2011
Yes, assuming Big I is quiescent current and Little i is incremental current.
In this setup, the transistor base-collector junction is simply a reversed biased diode.
Hope I am not getting you into trouble with your prof.

A more difficult question is: What is the emitter voltage? a little trivia that I cannot answer--would have to measure it myself