# NPN with a floating emitter

Discussion in 'Homework Help' started by psycoadam, Nov 17, 2011.

Oct 18, 2011
19
0
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.

2. ### jimkeith Active Member

Oct 26, 2011
539
99
Note that in this case, Ib = Ic = Is
Simple series circuit problem

Oct 18, 2011
19
0
is big IB = iB? and Ic = ic?

4. ### jimkeith Active Member

Oct 26, 2011
539
99
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