Beginner's question: NPN common-emitter amplifier setup?

I have seen schematics with a resistor hooked up between the emitter
and ground. In a book by Stan Gibilisco (I can't remember the name off
hand) it stated that although the emitter was grounded from the signal
that there was still a voltage due to the resistor. Is there a voltage
because:

When the collector draws current from the emitter, that current passes
through the resistor creating a voltage drop? I don't think that
polarities are correct.

If this question doesn't even make sense to you guys, please let me
know.

Glenn

 

A bipolar resistor can be thought of as a current controlled switch. A
small current flowing into (out of) the base of an NPN (PNP) transistor
causes a larger current to flow in the collector - emitter. While the
collector current is not quite equal to the emitter current (due to the
base current) for easy analysis they are assumed to be since the
difference is usually very small. The ratio between base current and
the collector current is called the transistor Beta (current gain) and
is usually specified as a minimum value according to the datasheet.

If on the resistor to which you are referring, one end connects to
ground and the other end connects to the emitter, any current flowing
through the emitter will flow through the resistor, bringing the
emitter voltage above ground. In turn, the base voltage will be about
..6V - .7V above the emitter voltage, assumig the transistor is on. If
enough current is flowing, such that the collector voltage drops to
about or below the base voltage, the transistor will enter saturation
which means that an increase in base current will produce little, if
any increase in collector - emitter current. In this mode of
operation, the CE voltage is typically about .2V to .3V. This assumes,
that there is a collector resistor, which is generating a voltage drop
from the supply to the collector terminal. A circuit of this
configuration will exhibit an electrical "gain" equivalent to Rc / Re.
The idea is to design the circuit to be dependant on the resistors
rather than the inherent properties of the transistor which can vary
significantly

Sometimes a capacitor is placed in parallel with the emitter resistor.
This is done so that at DC the emitter resistor is present to "bias"
the transitor into a certain region of operation, while providing a
high gain to AC signals. Note also, that the emitter has an inherent
resistance associated with it that is proportional to the emitter
current.

 

Take a look at http://www.fncwired.com/TransistorExample/

Brian

 

Some of it makes sense. You are referring to:

: +V
: |
: |
: |
: |/c Q1
: IN-----|
: |>e
: |
: |
: |
: \
: / R1
: \
: /
: |
: |
: gnd

In this case, when collector current flows out the emitter and through
R1, a voltage is generated across it. Assuming that the base voltage
is high enough that Q1 doesn't otherwise turn off, this voltage moves
the emitter voltage closer to the collector voltage.

But the usual 'view' of this is the other way around. One starts with
the base voltage, drops it mentally by a presumed Vbe between the base
and emitter (say, 0.7V, for example), assumes that this will be the
voltage at the emitter, computes the current that would then flow
through R1 based on that applied voltage, and then concludes that this
is approximately the collector current -- less a small portion for the
base current.

Jon

 

Have a look at http://homepage.ntlworld.com/g.knott/elect225.htm

 

Yes of course. First principles.

Uh ?

An emitter R simply results in a degree of local feedback. It wil help
stabilise DC operating conditions and will also reduce AC voltage gain (
unless the emitter R is bypassed with a cap ).

Graham

 

That is exactly what happens. The emitter current is the sum of the
collector current and the normally much smaller base current, so it is
a way to produce a voltage related (mostly) to collector current that
is applied to one of the transistor's input terminals.

Makes perfect sense. Just remember that the input signal that
controls the transistor is the voltage between base and emitter. So
as you apply a signal to the base, the emitter resistor applies an
output (collector) current generated voltage that opposes it, acting
as negative feedback. This makes the output current much more
accurately proportional to the base voltage variations. It also
raises the base impedance and the collector impedance. And while it
lowers the gain (as all negative feedback does) it extends the
frequency range that flat gain occurs. And it helps reduce the
effects of temperature (increased current gain and lower base to
emitter voltage) on the bias situation.

 

Thanks for all the replies, everyone. I will definitely check out the
websites that have been suggested.

When you talk about current flowing through the emitter, is this
current coming from ground? I think that current cannot flow from
collector to emitter. If so, wouldn' t the voltage be higher at the
bottom of the resistor (closer to ground)? That voltage got dropped
across the resistor, leaving less voltage at the top.

I'm confused, but a little less so thanks to everyone.

Glenn

PS: the books I've been reading don't seem to address this very
clearly.

 

Current will flow out of the base and into the emitter in a PNP
transistor and yes, if the emitter were at ground, 0 volts, the base
would be below this potential. PNP transistors have their place when
the signals you want to work with are negative in amplitude such as an
AC coupled amplifier, and when you wnat to switch something by bringing
the voltage down rather than up, such as an active low signal.

The emitter current will be a combination of collector and base
currents. The collector-emitter current concept defies normal / basic
thinking and is a function of semiconductor physics. The application
of base current injects charge carriers into the C-E junction called
the depletion region. If the right type of charger carrier is injected
the depletion region gets narrower an if the other type is injected it
gets wider. When the depletion region is narrower and voltage is
apllied to C-E pair charge carriers are able to be swept across the
depletion region - hence current flows. The amount of current that is
able to flow depends on the width of the depletion region, hence more
base drive more current can flow in the C-E. The ratio of base
(current) drive to collector - emitter drive is the transistors Beta.

 

You have to decide early in your career if you're going to use electron
flow, which hams, techs, and other people who live in the real world
use, or "conventional" flow, which academecians, scientists, and other
visionaries use. ;-)

They're indistinguishable, except all of the signs are reversed.

Conventional flow flows in the direction of the arrow at PN junctions:
one in a diode, and the emitter/base junction in BJTs (bipolar junction
transistors). Electrons flow in the opposite direction to the arrow.

The conventional current in an electron beam goes upstream. ;-)

With a negative-ground supply, the electrons come out the ground, flow
up through your circuit, and get sucked into the positive terminal of
the battery, where they get circulated by the force of the chemical
reaction.

All you have to do is change your point of view, and it's exactly the
same as positive current flowing out of the positive terminal of the
battery, down through the load, and to ground.

Conventional current uses the right-hand rule, and electron flow uses
the left-hand rule, but save that for when you start to get into
electromagnetics. ;-)

Hope This Helps!
Rich

 

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?

Glenn

 

Getting a bipolar transistor to do what you want (within
its capabilities) is all about controlling b-e current,
and there are numerous was to do it, so there is no one
single answer to your question "how do you ensure that
the emitter is less than the base"?

But there is an implication in your question that needs
to be addressed. Your question implies that you may think
the base must always be ~.7 volts higher than the emitter.
You need to think of it a little differently.

The amount of b-e current controls the amount of c-e
current. It could be that you want *no* c-e current,
in which case you would not want the emitter to be ~.7
volts more negative than the base.

What you want to do is control the b-e current and not
have the base *always* ~.7 volts higher than the emitter.
The reason .7 volts is an important number is that
there will be no c-e current until the base voltage is
raised to about .7 volts more positive than the emitter
voltage in the NPN transistor. In a PNP, the base would
need to be about .7 volts more negative than the emitter
for there to be c-e current.

Ed

 

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