No. They are a transconductance device because applying a voltage across
the base emitter junction injects carriers from the emitter to the base
*region*. This charge essentially *all* flows out of the collecter, not
the base terminal.
---
Not no. From:
http://searchsmallbizit.techtarget.com/sDefinition/0,,sid44_gci214200,00.html
"Transconductance is an expression of the performance of a bipolar
transistor or field-effect transistor (FET). In general, the larger
the transconductance figure for a device, the greater the gain
(amplification) it is capable of delivering, when all other factors
are held constant.
Formally, for a bipolar device, transconductance is defined as the
ratio of the change in collector current to the change in base voltage
over a defined, arbitrarily small interval on the
collector-current-versus-base-voltage curve. For an FET,
transconductance is the ratio of the change in drain current to the
change in gate voltage over a defined, arbitrarily small interval on
the drain-current-versus-gate-voltage curve.
The symbol for transconductance is gm. The unit is the siemens, the
same unit that is used for direct-current (DC) conductance.
If dI represents a change in collector or drain current caused by a
small change in base or gate voltage dE, then the transconductance is
approximately:
gm = dI / dE
As the size of the interval approaches zero -- that is, the change in
base or gate voltage becomes smaller and smaller -- the value of dI /
dE approaches the slope of a line tangent to the curve at a specific
point. The slope of this line represents the theoretical
transconductance of a bipolar transistor for a given base voltage and
collector current, or the theoretical transconductance of an FET for a
given gate voltage and drain current."
---
This is not an accurate description of the bipolar transistor. This
description is more relevant to operation of the mosfet. The npn
junction simply does not act like a slap of N type. If it did, base
current would be huge.
---
yes, were it not for the current limiting resistance external to the
base the base current could become huge. After all, the base-emitter
diode is just that, a forward biased diode operating on the far side
of the VI knee.
The intent, in both devices, is the same. That is to cause a
non-conductive region in a semiconductor to become conductive. In a
MOSFET it's accomplished by treating the channel like the plate of a
capacitor and making it _seem_ like it's composed of the same material
as the drain and the source by influencing the charge distribution in
it using the gate metalization as the other plate of the capacitor,
while in a BJT it's accomplished by forcing dynamic charge into the
base ["base region" if you like
] and using that charge flow to make
it seem like the base region material is becoming more and more like
the emitter and collector material as the base current increases.
---
Indeed it is.
This is not too much detail at all. Its can't get any simpler. vbe
controls the collector/emitter current. End of story.
---
Hardly. Here this newbie asks "What makes a BJT different from a
FET?" and you reply "If you put a voltage across the base and emitter
terminals of a BJT current will flow between the collector and
emitter, while if you put a voltage across the gate and source
terminals of a FET current will flow between the drain and the
source." So, while your description may be true, its utter simplicity
leads the newb to think they're the same same thing with differently
named terminals.
Here is my original exchange with Skeleton Man:
<QUOTE>
so if I'm to understand correctly.. a bi-polar will pass current between
collector and emitter when a voltage is applied to the base ?
---
Essentially, yes. But, the voltage applied to the base must force
charge through the base-emitter junction before collector current can
flow.
---
and a fet will do
a simmilar thing only doesn't require a current ? (at whichever terminal
corresponds to a base on a bipolar)
---
Yes, but it still requires current to charge the gate capacitance.
However, once that capacitor is charged up, current can flow through
the drain-to-source channel with no further current required into the
gate.
<END QUOTE>
Do you have a problem with that?
---
No. No. No. It most certainly doesn't.
---
Yes. you're right. That was poorly stated. See my original reply,
above, to the OP for clarification.
---
Referring to the bipolar as "a current controlled device" causes never
ending confusion that is a bloody nightmare to correct. This is a case
in point. You yourself are trying to put forward the idea that that idea
has merit. It doesn't.
---
The problem which arises here, I think, is that the change in base
voltage required to affect a change in collector current is so tiny
that it becomes easier to consider what happens on the other side of
the change in base voltage. That is, the collector-to emitter current
change due to the base-to-emitter current change.
