Hi all,

I'm having a little trouble understanding the BJT. If we can model the BJT as two diodes that are back to back, where the Base-Collector is reversed biased and the Base-Emitter is forward biased, how can any current flow through the Base-Collector? I understand that electrons are swept through the BC, but I was always under the impression that no current besides leakage flows through a reversed-biased diode.

Can someone please give me a very basic explanation to how come in this case a reverse bias can have current running through it?

 

It's not a great model. However, that model should not indicate B - C current. All the base current goes through the B - E junction.

Oue Ebook has some different models in it - http://www.allaboutcircuits.com/vol_3/chpt_4/4.html

 

So I'm a little confused then... If there is a current going into the base, does a greater current come out of the emitter? If that's so, where is the extra current coming from since current cannot flow between base and collector?

One more thing... I think there might be an error inside the link you just sent me. When you run the experiment on SPICE for the second time, you get a graph that holds flat at 7.5 mA and then you say it's 100 times the base current. The previous paragraph says that your base current is kept at 20 uA... Is this a typo?

 

A transistor is not just two diodes connected together.
A small base to emitter current causes a much higher collector to emitter current.

 

In the case of a common emitter setup, all the current goes through the emitter - from the base and from the collector. The ratio of the base current to the collector current is the gain.

The two diode model may be used because you read two PN junctions when checking with a meter - base to emitter and base to collector. It does not explain BJT operation.