From: Joseph Legris
X-Mailer: Mozilla 4.77 [en]C-SYMPA (Win95; U)
Subject: Re: how does transistor work?
Date: Mon, 04 Nov 2002 11:35:37 -0500
NNTP-Posting-Date: Mon, 04 Nov 2002 11:35:47 EST
Organization: Bell Sympatico
> I have read some books and I still dont understand.
> Let's take NPN transisor for discussion- (respectively the Collector, Base,
> and common emitter configuration.
> once the Vbe > 0.7 V, the P-N junction is forward biased then,
> (I forgot which is donor/acceptor ion, but i know p-type majority carrier is
> +ve and n-type is -ve)
> the -ve charged ion on P-type material go to the anode and the +ve charged
> ion in n-type material go to cathode
> the depletion region gets smaller, and as the ions flows, there is current!!
> but then
> for the Vcb to be forward biased, the P-N juntion is reverse biased, then
> I dont understand what is influence by a larger depletion region.
> In another case for PNP(Emitter, Base Collector respectively) with common
> base configuration.
> if E-B is forward biased, then current flows, but the book says there is
> injection of minority carrier into the N-type material
> I dont quite understand the consequence.
> Thx for help
Considering just the E-B area, (NPN or PNP, it doesn't matter), when it
is forward biased, conduction is mainly by majority carriers. This mean
that holes and electrons meet each other near the junction and
recombine. Meanwhile, new holes and junctions are being created by the
voltage applied to the E-B. The result is a continuous E-B current.
Now, consider the C-B junction all by itself. It is reverse biased, so a
depletion region forms around the C-B junction, which means that there
are very few majority carriers on each side. Now, suppose some minority
carriers were added into the depletion region on either side of the C-B
junction (again, it doesn't matter if we are talking about NPN or PNP).
For example, if some electrons were put into the P-side of the depletion
region, they would be attracted to the N-side because it is reverse
biased. Or, if some holes were created in the N-side of the depletion
region they would be attracted to the P-side. The result in either case
is a flow of current because when the minority carriers cross the
junction they become majority carriers on the other side and continue
moving, right out of the depletion region.
In short, a reversed biased C-B junction will conduct a current if
minority carriers are injected into either side of depletion region -
electrons on the P side or holes on the N side.
The E-B junction can do this. When it is forward biased, majority
carriers from the emitter pass into the base where they become minority
carriers. Some of them recombine in the base and become part of the E-B
current, but most of them are attracted into the collector where they
become majority carriers once again. If the B-E voltage is changed, the
number of minority carriers going into the base changes and so does the
resulting collector current.