Can you recommend a few ? I would like an intuitive treatment along
with the math. The problem is that while I realize that Ib NEEDs to be
something by design, I dont see how that is achieved, mathematically.
Especailly because they ignore the Ib in calculating the voltage
Vb=Ve+.7.
You got a lot of information in this thread -- take some time and go through it.
But here's my own hobbyist viewpoint for DC operation of a typical common
emitter design...
I(C):
-----
First off, you need to select the collector current for the transistor. I'd
love to hear sophisticated views about this, but one way I do that is to just
look at the data sheet -- some of its specifications will specify an I(C) to
give it the appearance of being a "good" part. It's a good bet that you want to
operate your transistor around this area, without more meaningful reasoning to
pick another value. For example, in the Motorola data sheet for the 2N2222A,
they specify a number of the characteristics at 10mA, 15mA, and 20mA. That's
your first clue about what you want. (It's probably on the high end of what you
want, actually, unless you really need the performance.)
Later in the sheet, there are some nice curves (you want to find these, often)
for DC current gain versus I(C) and also V(CE) versus I(B) with several I(C)
curves (it's the collector saturation chart.) The 2N2222A from Motorola shows
the DC current gain dropping off starting around 30-40mA -- that's a broad
suggestion that you probably don't want to go a lot higher.
Other curves to look at might be the turn-on and turn-off times versus I(C)
[higher I(C) generally means 'faster'], but that usually isn't your problem for
audio amplifiers, for example. These numbers are often in nanoseconds. Another
curve is the noise figure versus source resistance for various I(C). Depending
on your source, it might push you one way or another. Though for hobby work, I
don't use this much, either.
As a guide, I tend to imagine that higher I(C) is for "higher operating
frequency and/or gain (up to a point.)" Lower I(C) is better for lower
dissipation. Noise figures and source resistance might push you to one side or
the other, depending.
So, let's say you are using a 2N2222(A) and without much into what's driving the
circuit. No need for the higher end I(C) values at this point. You can see
that the curves are given for 1mA and even somewhat lower I(C) and so you settle
on a working I(C) of 1mA. Probably just fine for most uses. Go with it.
Once you've figured your 'quiescent' I(C), or at least got your first hack at
it, then it's time to figure out your R(C).
R(C):
-----
Your collector load should be chosen to center the V(OUT) [the side of R(C)
connected to the collector of your 2N2222, for example] at about 1/2 of your
positive rail voltage (in this NPN case.) Let's say the upper voltage is +10V,
so this means about +5V. At a quiescent I(C) of 10mA, this means 5V/1mA = 5k
ohms.
R(E):
-----
For setting the DC operating point of your basic common emitter design, before
even considering an RC in the emitter leg, it's the general idea to figure the
voltage across R(E) at about 1V to add some temperature stability in the circuit
against V(BE) variations with I(C) over temperature, to help reduce the impact
of variations in the internal NPN's r(e) on gain, and to reduce distortion
caused by the Early effect. -- through the use of negative feedback.
I generally use 1V as a reasonable guideline. I try not to go lower, but higher
is fine. With this and knowing that I(C) is 1mA and that I(E) is roughly equal
to I(C) [when operating normally, anyway], you can figure that R(E) should be
1V/1mA = 1k ohm.
(Sometimes, I make modest adjustments above 1V, though.)
V(B):
-----
Well, for a normally operating NPN, your base voltage is going to be a little
more than 0.6V above your V(E). (I usually like to first guess at 0.65V.)
Since we arbitrarily decided to set V(E) = 1V earlier, that's easy to figure.
It's 1V. So the base voltage should be 1.65V, let's say.
Thevenin R for the biasing divider:
-----------------------------------
Setting R(E), by the way, sets your input R for the transistor, given the
current gain or beta of the NPN to about beta*R(E). You'll want the Thevenin
equivalent of the biasing divider to be about a tenth of this value. The gain
of this 2N2222A appears to vary from about 50-60 to over 200, over the
temperature range of -55 C to +125 C.
'Stiffer' (lower resistance) is better, but there's no need to go crazy. So
let's go with a gain figure of 100 (it's 150 hFE at +25 C) for this transistor
to cover our needs for now. That's an Rth(Divider) of (1/10)*100*R(E) = 10*1k =
10k ohms.
Resistor Dividers:
------------------
The resistors look like:
10V
|
\
/ R2 8.35V
\
|
+-----> to NPN base
|
\
/ R1 1.65V
\
|
gnd
We're ready to compute the two resistor values. Assuming, for the moment, that
the current going into (or out of) of the base of the NPN is negligible, the
current through R2 and the current through R1 are the same. Also, it's kind of
clear that R2's voltage is 8.35V, the difference between the 10V rail and the
V(B) we decided on, above. R1's voltage is 1.65V, of course. So:
I1 = I2, and,
1.65V = R1 * I1
8.35V = R2 * I2
so,
1.65V R1 * I1 R1
----- = ------- = --
8.35V R2 * I2 R2
or,
R2 = (8.35V/1.65V) * R1 = 5.061 * R1
Rth (mentioned before) is the parallel equivalent, or R1*R2/(R1+R2) = 10k ohms,
as already said. As R2 is more than 5 times larger than R1, R1 is going to be
close to our Rth of 10k, but a little higher. Let's substitute:
R1 * 5.061 * R1 / (R1 + 5.061 * R1) = 10kOhm
5.061 * R1^2 / ((1 + 5.061) * R1) = 10kOhm
5.061 * R1 / (1 + 5.061) = 10kOhm
5.061 * R1 = 10kOhm * (1 + 5.061)
or,
R1 = 10kOhm * (1 + 5.061) / 5.061
R1 is almost 12k. R2 is a bit more than 60k. For relatively standard values,
12k works for R1. But let's go ahead and pick 56k for R2. That will pull up
the base a little to about 1.76V and thus, V(E), to about 1.11V but that's okay.
We've got enough margin to work with that.
This pretty much sets up the DC conditions for the amplifier.
---------
Some Notes (gain and such):
With 1.11V as V(E) and the fact that we shouldn't let V(CE) get smaller than
about 1V, or so, this means that V(C) can't go below 2.11V. In quiescent state,
we've designed it to set at 5V at quiescent I(C) = 1mA. This means that the
output swing cannot be more than about +/-2.9V around that 5V point -- or from
2.1V to 8.9V. Just something to be careful about.
I haven't said much of gain at this point. It's all been about setting the DC
operating points. The gain is -R(C)/(R(E)+r(e)). I'll get back to where this
comes from, but....
The r(e) value comes from the Ebers-Moll equation and as a general guide is
about (25mV/I(C)). The 25mV comes from the (k*T/q) part of the Ebers-Moll
equation, when T = 25 C. In the above case, we chose 1mA, so this means that
r(e) is 25 Ohms, roughly speaking. Our R(E) resistor is 1k, by comparison, and
that makes r(e) not terribly important in computing the gain mentioned above.
R(C) = 5k
R(E) = 1k
r(e) = 25
so,
gain = G = -5k / (1k + 25) ~= -4.88
Not terribly impressive, gain-wise. That's the price of R(E)'s negative
feedback helping you elsewhere. (But there are things you can do to improve
that, if this is used for AC signals instead of just DC-only.)
Variations in V(C) are caused by variations in I(C).
V(C) = 10V - I(C)*R(C)
or,
I(C) = (10V - V(C)) / R(C)
Since V(C) is 5V at I(C) = 1mA, by design, and it will be 2.1V at (10-2.1)/5k or
1.58mA and it will be 7.9V at (10-7.9)/5k or 0.42mA, this is the range of I(C)
currents. I already mentioned that r(e) depends on I(C) and we can now compute
the variations on r(e) as going from 25mV/0.42mA to 25mV/1.58mA or from about 60
Ohms down to about 16 Ohms. Plugging that back into the gain estimate, we get:
gain = G = -5k / (1k + 60) ~= -4.72
gain = G = -5k / (1k + 25) ~= -4.88
gain = G = -5k / (1k + 16) ~= -4.92
As you can see, the gain variations over the reasonable I(C) values stays close
to the same value and this is a good thing as it reduces distortion. It's a low
gain, but at least it's not widely varying. If you want to see wide variations,
but with high gain, just think about what would happen with relatively small
R(E) values. r(e) would then drive the variations much more strongly,
distorting the signal unless you kept I(C) in a much tighter range of variation
than can be used above.
Now, getting back to the meaning of G and, I think, perhaps some answers about
that NPN base lead and what it is doing...
The G (gain) mentioned above is a unitless variety -- simply comparing
variations of V(B) with V(C). Because V(BE) (the difference between V(B) and
V(E)) is fairly constant in normal situations, lifting V(B) by a slight amount
will also lift V(E) by the same slight amount. When V(E) is lifted, though,
this means that the current through R(E) changes, too. This current is about
the same current flowing in I(C) and this current causes a voltage drop in R(C).
How much it does works like this:
V(SUPPLY) = 10V
V(BE) assumed a constant 1.65V, for now.
V(E) = V(B) - V(BE)
I(E) = V(E) / R(E)
I(C) ~= I(E)
V(C) = V(SUPPLY) - I(C)*R(C)
therefore,
V(C) = V(SUPPLY) - (V(E) / R(E)) * R(C) = V(SUPPLY) - V(E) * (R(C)/R(E))
Since the supply voltage is a constant, changes in V(C) are about -R(C) / R(E),
as you can see. Actually, there is an effective emitter resistance called r(e),
already mentioned, that adds to R(E). That's where I got my earlier equation.
In the above case, if we wanted to support +/-2.9V variations on V(C), this
would imply +/-2.9V/4.88 or about +/-0.6V variations on V(B). Frankly, that's a
lot. But at least you can see that a -.6V reduction from 1.76V as the quiescent
point is at least possible.
How are the voltage variations on V(B) impressed? Well, by loading it either
towards the 10V rail or ground. That gets back to how you intend to drive it,
though.
There are many improvements. For the kind of AC amplification usually used for
audio, for example, there is a series-RC (actually, there are several
topologies, but that's for later) that can be put in parallel with R(E) to, in
effect, maintain the DC bias points calculated in the above method but also to
change the effective combined-R(E) and reduce it significantly at some design
frequency so that the gain is increased to something lots better. Also, there
is something called "bootstrapping" that feeds back some of the emitter voltage
signal to the middle of the resistor divider and uses a resistor from that point
going into the base and with the base more directly coupled to the preceding
driver circuit that can greatly reduce the loading on the prior stage.
Really, if you can manage it, get a notebook and a meter and a few parts and a
power supply or two and wire up some of the various circuits. Take good notes
and make lots of measurements as you think about things. It will really help
out. Or, you can get a free simulator like LTSpice from linear.com and simply
play with circuits and see what that does (though it's not quite the same thing
and can't replace some of the value of actually wiring up a circuit.)
And again, I'm only a hobbyist type and have absolutely no professional design
experience. I'm not offering anything I say as gospel or based on a thorough
understanding of existing theory or based on well-practiced knowledge. It's
just my current take on things.
Jon