Re: VBC of npn transistor in common collector mode


"Eeyore" <rabbitsfriendsandrelations@xxxxxxxxxxxxxxxxxxxxxx> wrote in
message news:44AD3067.28D655D3@xxxxxxxxxxxxxxxxxxxxxxxxx


John B wrote:

For a typical npn bipolar transistor, such as the 2N2222A, we can expect
a
DC voltage drop of about 0.65 V, from base to emitter ("V BE"). Such is
the
case of the ubiquitous "common emitter" configuration.

Suppose we flip the transistor, and use it "backwards." Let's call
this,
"common collector." Correct me if I'm wrong, but that sounds like the
accurate description of the configuration, because we are grounding the
collector, driving the base with a controlling current, and attaching a
pulled-up load to the emitter.

The more common name for this is an emitter follower.
I disagree. Emitter follower has a different topology. There is nothing
"below" the collector, of my flipped transistor, other than an attached
ground.
If this npn transistor were to be suddenly replaced by a pnp, then I see an
emitter follower. Of course, the controlling signal at the base would then
have to be referenced to the upper rail, instead of to ground, as my npn is.


This is an inverting amplifier...sort of.

Absolutely not. Common emitter is inverting though.
It inverts voltage. When the base voltage rises, more current flows down
into the emitter from the upper rail, through a load, and out through the
collector to ground. The voltage at the emitter pin goes down, not up.

An emitter follower is non-inverting. This is not an emitter follower.


Beta can be expected to be "miserable," with a value of less than 1,
instead
of roughly 100. So amplification is "in the eye of the beholder."
Granted.

No - wrong. Beta hasn't changed and this configuration does indeed have
current
gain.

It's Beta reverse...as attested to by so many other posters in this thread.
It's drastically different from Beta forward.

Can anyone advise on the forward voltage drop of between base, and
ground?
(That is, "V BC") I expect that the collector is very lightly doped,
compared to the emitter, so the barrier should be far less, leading to a
lower threshold voltage to overcome.

Not much difference really. I can't ever recall seeing it specced in this
configuration. The c-b junction is always reverse biased anyway.
Not to be argumentative, but a c-b junction is forward biased in a saturated
circuit that employs common emitter. Very common situation.

John B


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