# Simple question on BJT

Discussion in 'General Electronics Chat' started by yuhao.zhou, Mar 11, 2013.

1. ### yuhao.zhou Thread Starter New Member

Mar 11, 2013
1
0
Hi, new to this forum. Hope I post in the right place...

Simple question here about the output characteristic of BJT (npn type), why the collector current (IC) will become 0 when collector voltage (compare to emitter) VCE becomes 0 even when base current is not 0?

In this case I thought current would flow from base to collector and to emitter equally, because are both pn are forward biased. Is there any reason that IB is not 0 while IC should be 0???

2. ### TheComet Member

Mar 11, 2013
88
12
Hi,

What have you got connected to the collector?

With an NPN transistor, Base-Emitter has the same characteristics as a normal SI-Diode. If you drive the base (and I'm assuming you have a resistor to limit current), you can also measure 0.7V on VBE, just like with a diode.

The function of a transistor is to amplify current. If you apply 1mA to base, and the transistor has a current gain of hfe = 100, you can expect 100mA flowing into the collector.

HOWEVER, this only works if you are able to supply 100mA to the collector. Let's say you limit the current going into the collector to 5mA. You're still pumping 1mA into the base, but there will be 5mA flowing into the collector, regardless (because you're limiting the current).

So because VCE = 0V (this can only happen if no current is flowing), obviously Ic = 0A as well.

This is incorrect, the current flows from Base to Emitter, and from Collector to Emitter, where Ic = Ib * hfe.

TheComet

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3. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
If you are talking about saturation, where Vce≈0, then Ic=0 only if there is no current path (such as a resistor) from the collector to a positive supply voltage.

4. ### crutschow Expert

Mar 14, 2008
13,473
3,361
I did a simple simulation in LTspice of a 2N2222 with emitter and collector grounded and 1mA into the base (which I think is the condition the OP is talking about since he said the collector voltage is zero). The Operating Point analysis gave:

Ic = .954mA
Ie = .046mA

So it would seem that the current does not divide evenly between the base-emitter junction and base-collector junction under those conditions but rather most flows through the base collector junction. That may be because the base-collector junction has a lower impedance that the base-emitter junction.

5. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
It's probably mostly due to the fact that the C-B junction area is much larger than the C-E junction area. Doping is probably also a factor.

Here's a couple of cross-sectional images of NPNs: