I have done my research and much of what you guys are saying makes
sense!! Thanks!
BTW: How do you set the voltage at the emitter so that it is 0.7 v or
so less than the voltage at the base? I know that the base voltage is
set my some sort of potential divider resistor setup. But how do you
ensure that the emitter will be less than the base?
Well, this tells me about where you are at. I think I've been there,
too. But it's been a long time and I may not remember how I was led
out of it.
For the moment, just imagine that you have placed two batteries in
this way:
: (2)
: ,----------,
: | |
: | |
: | |
: (1) |/c Q1 | +
: ,----| -----
: | |>e --- B2
: + | | -----
: ----- | (3) ---
: --- B1 | | 10V
: ----- \ |
: --- / R1 |
: | 5V \ |
: | / |
: | | |
: | | |
: '------+----------'
: (gnd)
First off, before I go any further with the above circuit, note that I
added a (gnd) at the bottom. It doesn't mean anything except to note
where my "reference point" is at. There are three nodes in the
diagram other than (gnd) and these are (1), (2), and (3). If I talk
about the voltage at (1) as being +5V, you should keep in mind that I
am talking about the voltage at (1) compared with (gnd). Voltages are
always measured with respect to somewhere. And by common agreement,
we learn to designate one particular node as being special and calling
it (gnd). Which one makes the most sense to label that way will
depend on things and it will take some experience before you will
always pick the same place to label that way that a professional
might. But it doesn't change the circuit if you put the (gnd) label
somewhere a professional wouldn't -- it just means that folks
(including you) may be a bit confused when you talk about some voltage
here or there. But I could just as well have labeled the node above
called (2) as (gnd) and relabeled (gnd) as (15), for example. The
labels are entirely arbitrary and mean nothing as far as the circuit
itself goes. It's just inside our heads and nowhere else.
Okay. So trace out the above circuit in your mind until you agree
with me that there are four places of common connections between
components. We have 4 parts in the circuit; two batteries, one
resistor, and one transistor. They are tied together with wire and
since all places along a wire are considered to be at the same
potential, we can mentally consider everywhere that only wire exists
as being the same "point." I just labeled those points as (1), (2),
(3), and (gnd). I hope you agree with that much.
Now, what is the potential at (1)? It's +5 (with respect to (gnd), of
course.) That's because there is a battery there and its job is to
make sure that one end of it is at a fixed potential away from the
other end. In this case, the plus (+) side is tied to the base of Q1,
so the base of Q1 _must_ be at +5V because the battery will make sure
of this fact. That's its job.
Hopefully, we may agree that (1) will be at +5V (with respect to gnd.)
This means Q1's base is at +5. Now think about the effective diode
between Q1's base and Q1's emitter.. Ignoring the rest of the
circuit, that looks about like this:
: (1)
: ,------,
: | |
: | ---
: | \ /
: | V
: | ---
: | |
: + | |
: ----- | (3)
: --- B1 |
: ----- \
: --- / R1
: | 5V \
: | /
: | |
: | |
: '------'
: gnd
If you ignore R1 for a moment and think of it as just a wire, I think
you can see that the diode will be forward biased. This means it will
allow the flow of current, readily. Without R1 (with it shorted out
by a wire), the voltage across this Q1's base-to-emitter diode would
be the full 5V and a lot of current would flow. Way too much, and it
is almost certain that Q1 would self-destruct. Now, if you re-imagine
R1 included, what happens? Well, as more and more current is
considered to be flowing through the diode _and_ R1, R1 starts
dropping more and more voltage according to I*R1. Eventually, of
course, all of the 5V would be used up by R1. In other words, if you
imagine that "I" is large enough, the product of I*R1 would be exactly
5V. When that happens, there would be no voltage left to exist to
forward bias the diode. And no current (not much) would be able to
flow. So that would mean that that much "I" is too much to be
reasonable. So we are sure that "I" must be less (<) than this. In
other words, I*R1 < 5 or written another way, I < (5/R1).
But how much less?? Well, it turns out that the value of "I" will
automatically find itself about right when I = (5 - .7) / R1. In
other words, the diode itself "wants" to see about .7 volts across it
in order to work in the forward conduction region. It might be a
little bit less than .7 volts or a little more than .7 volts. But not
a lot different. This is because even a small increase in voltage
across it (say, about 0.06V) means a multiplication by 10 of the
current through it. So you can see that being off even by a tenth of
a volt one way or another might mean changes in current spanning a
factor of 100 or 1000 times. So diodes will tend to find their
forward drop to be pretty close to 0.7 volts most of the time. I see,
in transistors, between 0.6V and perhaps 0.9V depending on their use.
But quite often very close to the .7V.
So if it is true (and I tell you it is) that the diode, forward biased
as it is, will drop about .7V then we can calculate from that fact
that the voltage at node (3) should be .7V less than +5V or about
+4.3V. And it's true, it will be about that.
Now, before I continue, let me redraw the circuit and hope that you
still agree with me that it is the same circuit:
: (2)
: ,----------,
: | |
: | |
: | |
: (1) |/c Q1 | +
: ,----| -----
: | |>e --- B2
: + | | -----
: ----- | (3) ---
: --- B1 | | 10V
: ----- \ |
: --- / R1 |
: | 5V \ gnd
: | /
: | |
: gnd |
: gnd
I just broke the displayed wires, but conceptually this is the same
circuit. In practice, of course, you'd have to wire those ends
together. But for understanding the circuit, you don't need all that
wire. And in fact, it will tend to confuse you a little because
you'll start wondering how various components attached to it might
affect things. Breaking the wire and just labeling each end the same
is my way of forcing you to stop thinking that way and to watch for
the important stuff and ignore stuff that doesn't matter.
Now you were asking how we can know (or guarantee, somehow) that no
matter what happens in the circuit that node (3) will still be about
+4.3V? Well, it turns out that if you disconnect B2 entirely from the
circuit that it will be true and that if you hook up B2 it will still
be true. In fact, if you change the voltage for B2 from 10V to 20V,
it will still be true. This will remain just magic to you until you
really think more closely about things. But it does work that way,
close enough for now, anyway.
The only thing that matters here is that the base-emitter diode of Q1
is forward biased by the base-emitter circuit, which includes B2 and
R1, and that this is the only important detail in knowing that the
voltage at the emitter will be about 4.3V or 0.7V less than the base.
If you don't connect up the Q1 collector and leave B2 out of the
circuit, then all of the current flowing through R1 in order to
produce a voltage of 4.3V across it (I*R1 = 4.3V) will come from B1
via Q1's base. In other words, the base will be supplying all of the
current equal to (4.3V/R1). But if you now connect up B2 to Q1's
collector and include it as part of the circuit, then most of that
current (4.3V/R1) will actually come from the collector and B2 instead
of from B1 via the base. In other words, the base current will
suddenly and sharply decline to a tiny percent of its original value.
This is because Q1 will start conducting collector current to replace
the base current. R1's current will still be the same, 4.3/R1. But
most of it, about 99% of it, will come from the collector instead of
the base, when B2 is added and hooked up to the collector of Q1.
Does all that make sense, still?
Jon