# common emitter configuration- voltage divider biasing.

Discussion in 'Electronic Basics' started by Jenny, Jul 26, 2004.

1. ### JennyGuest

I think I kind of SEE the equations, but the concepts behind the
voltage divider biasing does not quite enter my head. I am muddled
hopelessly i think.
V1= R1(R1+R2) Vb= R2(R1+R2) this correct ?

Why is the base current to be negligible ? I read this everywhere. Is
the base emitter junction like a high resistance that current that
comes down the R1 resistor (the first resistor in the voltage diver)
goes almost all through R2 ?

Also why should I chose R1 to make the current through it at least 10
times base current ?

Please try to explain if you could, with scenarios that show WHY. If
you can show what will happen if it is NOt the case that would
probably sink in easier. I have looked though som many posts and
we-sites, I think I am missing something very simple and obvious like
an elephant in the living room.

Thanks
JG

2. ### Rheilly PhoullGuest

MinTinkit you bin needin to show the circuit to us so we can be usin de
numbers of tings shown dere.

3. ### Fred BartoliGuest

No, it can't : voltages can't be equal to ohms squared.
This **probably** should be (depending on the schematics ):

V1 = Vsupply R1/(R1+R2)
Vb = Vsupply R2/(R1+R2)

What you want to be fixed by design is the base voltage. Current gain (beta)
of transistor is highly dependant on temperature, manufacturing parameters
and bias conditions.
If you "design" the base current a big part of the bias network current,
then the base voltage will change a lot with temperature, which transistor
A good figure is to make base current 10 times smaller than the bias network
one.

4. ### Joe McElvenneyGuest

Hi,
If V1 is the bias supply voltage, with R1 as the top resistor in the
divider chain and R2 as the bottom one, then -

VB = V1 * R2 / (R1 + R2) ; a simple voltage divider.

Now most small-signal transistors have a high ratio of collector to
base current (hFE) which is often around 100 or so. If then you bias
one of these devices so as to produce a collector current of 5mA, the
base current only needs to be 100th of that, say 50uA.

Sooo..., if you design the bias network to take about ten times the
base current (0.5mA = 10 * 50uA) from the supply line, any variations
in device characteristics, in temperature or in phases of the moon,
will have minimal effect on the bias voltage. In other words the bias
will be 'stiffer' and you may use the same circuit again and again
without laying awake at night.

Cheers - Joe

5. ### John PopelishGuest

Without seeing your schematic, I can only guess.
The base current is not assumed to be negligible. It is a design
requirement of this approach that the base current be negligible
compared to the divider current. You have to calculate the base
current and select divider resistances so that this requirement is
met, if the circuit is function as expected. So you start with the
expected collector current and the minimum current gain spec of the
transistor, and work backwards to the maximum expected base current.
Then you calculate the divider pair so that this worst case base
current will not upset the divider voltage by more than a little. And
voila, the assumption that the base current is negligible is met.
If the resistors a are the result of design and not chance, yes.
(snip)

Because otherwise, the design assumption that the divider produces a
specified voltage, regardless of gain variations in the transistor
(with different transistors and with changes in temperature) is not a
safe assumption.

If you have a source of transistors that have a very narrow range of
current gains (or are willing to live with more bias point variation),
you can take the range of expected base currents into specific account
and use much higher base divider resistances, saving power and raising
the amplifier input resistance, increasing effective gain. But you
can not then make the simplifying assumption that the voltage divider
output is unaffected by the base current, but have to calculate the
range of effect that the expected range of base currents will have on
the divider voltage. The ultimate extreme if this case is to
eliminate the lower half of the divider and feed base current through
a single resistor. This gives the most efficient form of the divider
(highest possible resistance for R1 and infinite resistance for R2),
but also the least stable.

Designers analyze alternatives.

6. ### JennyGuest

V1= R1(R1+R2) Vb= R2(R1+R2) this correct ?
ofcourse! sleepy and way too late at night, yes I meant R1 DIVIDED
BY(R1+R2) ANd I forgot the source voltage, sorry about that.
I get that beta is a function of temperature, doping and many other
things- being intrinsic to every transistor.
I am under the impression that one assumes the current that one wants
the Ic to be, and works backwards to get the base current required to
make that happen, Ib=Ic/beta. After that - because we know what Ib is
needed to make the required Ic, we figure out how to bias the base, to
enable such an Ib to flow...

So far so good. this is what I do not get.

For this circuit,
Vcc +
|
|_
| | Rc with Ic that we want
|_|
|
|
|
/C ------------o Vout
_____Ib___|
| \E->
| .7v | hfe=say 100
| |
=== Vb Re
= |
| |
______________________GND___o

In the above circuit, if I assume decide Ic to be X, Ib required will
be Ic/100. Ok. Now how to make Ib what we want it to be ? Attach a
voltage source to it.

Many questions:
What happens if I have a Vb of say 4V and
a] there is no Re, just emitter connected to ground. What happens to
Ib, Ie ?

b] Vb is 4v and Ib and Re is connected between Emitter and ground.

Vb=V drop acess Re + .7, and they say Ib will be small, why ? is it
just a characteristic of the Transistor for base current to be small,
what happens when a large voltage is applied ?

Next : this is the voltage divider circuit, because we can apply a
voltage to the base in anyway either through a new power source like
previously or through using a voltage divider resistors

Vcc +
|--------------|
| |
| | | Rc with Ic that we want
| | R1 | |
| | |
| |
| |
| /C ------------o Vout
|___Ib___|
| \E->
| .7v | hfe=say 100
| | |
| | R2 Re
| |
| |
|____________|__________GND___o

Let us ignore the capcitance stuff that is missing for now.
In the above circuit,
IR1 =IR2 +Ib. Ok that is fine. what decides how much current coming
down IR1 is siphoned off into Ib ? Is it related to Re ? How about if
there was no re and a direction connection to the ground.

It does appear that I am seeing it understanding it piecemeal, but not
really GETTING the whole picture.
why this number ?

Thanks

7. ### Joe McElvenneyGuest

Hi,

You know the current through 'Re', because it is almost the same (99%) of the
collector current, so from that you can work out the voltage across it. Add just
0.7V and you then have the bias voltage required.
ZAP - transistors won't stand that kind of voltage across the base/emitter
junction. Remember that current through a diode junction increases exponentially
with voltage and so it would be quickly destroyed. You can apply a fair reverse
bias though but not too much because the junction is very thin and breaks down
easily.

It isn't a magic number like 'pi', just a handy rule of thumb although the
bigger within reason, the better. Invent your own if you wish and keep it a
secret then no one will ever know how you get your voltage divider values.

Cheers - Joe

8. ### Ian BellGuest

No, that's the way it needs to be designed for dc stability. As a rule you
would aim for about 10 times the base current to flow thru the voltage
divider.
Like I said for dc stability.
You really need the math to appreciate it but as a beginning the design aim
is a stable collector current. This is related to base current by the hfe
parameter which unfortunately varies over a very large range between
different samples of transistor of the same part number. So whatever we do
to set the collector current has to be essentially independeant of the base
current. With a voltage divider the way to do this is to make the base
current small compared to the divider current.

Ok?

Ian

9. ### JennyGuest

Its clear till this point.
How do I do that ? By worst case base current do you mean that for a
given beta, Ic and Rc, Ibmax or Ibworst-case is Ic/beta ? and Ib can
never go above that ? What decides Ib ? If it is the voltage applied
to base, then wont higher voltage mean higher base current ?
I see the dots bUT not the connection. Sorry if I am being very duh-
can you explain like to a child ? :-\ now I am beginning to thik may
not understand how the voltage divider works.

So what is the resistance which decides Ib ? how total current in the
voltage divider circuit can be calculated and how do I know how much
will go through Ib ?

10. ### Ian BellGuest

Depends on how Vb is provided. Assuming it is a divider with a design
current 10 times the expected Ib then the transistor will turn hard on
(saturate), Vce will be about 0.2V and Rc and possibly the transistor may
get hot.
Emitter volts is 4 - Vbe ~ 3.3V, emitter current ~3.3/Re, collector volts
~Vcc-Rc/Ie.
It is hfe times smaller than Ic that's all. You *design* the divider so its
current is large compared to the expected Ib.
This is not a large signal amplifier so don't do that.
Ib is determined by the collector current required.
As before, probably a saturated transistor.

IAn

PS. There ar eplenty of elementary texts on this. you should read one.

11. ### andyGuest

One way to think of it is as a built in negative feedback loop which is
made when you put a transistor in this configuration.

The transistor won't conduct until Vbe > about 0.7V, then Ic increases
exponentially as Vbe goes above that. I.e. big changes in Ic only mean
small changes in Vbe, so Vbe will stay around 0.7V.

Say Re=1kOhm, Vcc=12 v, and R1=R2=10kOhm.
When you switch on, Vbe is 6V, so the transistor is fully on by a big
margin. This makes current flow in Rc and Re, increasing Ve. But as Ve
gets close to Vb, the transistor will start to shut off, and the current
will reduce until it reaches a stable operating point. Because of the
exponential relation between Vbe and Ic, this will be when Vbe is around
0.7-1v. I.e. it stabilises with Ve=6-0.7=5.3V, which gives
Ie=5.3/Re=5.3mA, Ib=Ie/hFE (roughly), and Ic=Ie-Ib. (This isn't quite
right because Hfe is Ic/Ib, i think) This will all happen very quickly.

Then think about what happens if there is a small up or down fluctuation
in Ic, and you'll see how it stabilises itself.

12. ### John PopelishGuest

Jenny wrote:
(snip)
The collector current and beta. Use the specified lowest bets to find
the highest expected base current.
Yes, but that would also change the collector current away from the
desired operating point.

Do you know how to calculate the effect on a voltage divider voltage
when a given current is pulled from it? Have you studies Thevenin's
equivalent, yet? This makes it very simple. It converts the divider
into an equivalent voltage source in series with a resistor. So the
voltage shift caused by the base current is just that current times
the equivalent resistance in the Thevenin's model of the divider.
That voltage shift will also be seen on the emitter resistor (because
the base to emitter junction changes voltage drop only slightly for
fairly substantial current changes). From this voltage change, you
can calculate the shift in the collector current (emitter current
times beta/(1+beta)). And that current change causes a change in the
drop across the collector resistor.
The voltage divider produces an intermediate voltage but also has
resistance in series with that voltage. Lets say we have a divider
made up of a 10k resistor to +10 volts and a 5k resistor to zero
volts. this produces a 3.33 volt divider voltage if nothing else is
connected (no other current reaches the divider node). But if you
short the divider node to ground, that short carries 1 milliamp (since
this is what current passes through the 10k resistor from the 10 volt
supply and all of it would rather go through the short to zero volt
and none of it will go through the 5k to zero volts, because that
would require voltage across that resistor. So the divider looks like
a 3.33 volt supply that has 3.33volts/1 milliamp)= 3.3k resistance in
series with it.

Another way to find this effective divider resistance is to take the
parallel combination of the two resistors (product over sum or
reciprocal of the sum of the reciprocals). So back to the base
current. If you needed to have the base at about 3.33 volts (to
produce the desired emitter voltage which produced the desired
collector current which produced the desired collector resistor drop.)
this pair of resistors might suffice. Lets also say that the
collector current is 10 mA, and the minimum beta is 100. So the
highest base current for this operating point is 0.1 mA (10 mA /
100).

This .1 ma passing through the equivalent resistance of the divider
will produce a 3300 ohm *.0001 amp = .33 volt sag in the divider
output. Another way to think of this base current is to picture it as
a resistance to zero volts. 3 volts divided by .0001 amp would be a
resistance of 30,000 ohms (though the base will not strictly follow a
resistive character over large voltage swings)

Is this an acceptable sag? Go back and figure out how this will
affect the collector voltage. Remember that this was a low gain worst
case, so higher gain transistors would sag the divider less. Is this
variation within specs? I don't have the specs, so I can't say. But
choosing lower values for the divider resistors would cause less
variation with different gain transistors, and higher values would
cause more sag. You should start with some idea what is acceptable
variation if you are choosing the resistors. If you are just
analyzing someone else's design, go through the calculations for the
lowest and highest gain cases and see how much things move around.
Only collector operating current and beta.

13. ### andyGuest

I was thinking this might not make much sense if you don't get on well
with maths. exponential means that if you increase Vbe by 0.1V and
this multiplies Ic by 10, say, then increasing Vbe by another 0.1V makes
another tenfold increase in Ic. And increasing by 0.3 V multiplies by
1000. So for pretty much any value of Ic or Ie, the corresponding value of
Vbe will be close to 0.7V.

put this together with the bit i said above about what happens when you
switch on, and maybe you can see how the circuit finds a stable state
where Ve is roughly 0.7V below Vb (which is set by the divider), and the
emitter current is roughly (Vb-0.7)/Re. Then the base current can be
worked out from this by dividing by Hfe.

14. ### JennyGuest

If I ignore Ib in calculating the drop across Re, and Vb=.7+Ve, and
use this Vb as basis for finding R2 bias resistor, then how can I
ensure that Ib will be Ic/hfe. I mean if its is not used in the
calculation, how can I be working backwards ?
Ok. let me assume that I add an R2 which is way greater than the Vb
required to forward bias the junction and have Ve to drop across Re,
then ok the transistor will heat up BECAUSE the current is really
high. Before all things get fried, wont Ic be REALLY huge for a few
instants when this Ib is large ?

15. ### JennyGuest

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. 16. ### John PopelishGuest

With this circuit, you decide what Ve you desire based on the values
of the emitter and collector resistors. Then assume that in order to
keep the transistor biased to produce those emitter and collector
voltages (based on nearly all the emitter current also passing through
the collector), Vb will have to be held .7 volts more positive than
the desired Ve. If there is no resistor in the emitter to allow a
predictable emitter current to pass at that voltage, this method falls
down and a completely different method is needed to determine the base
situation.

Here are some web tutorials that might help:
http://www.electronics-tutorials.com/amplifiers/small-signal-amplifiers.htm
http://www.rason.org/Projects/bipolamp/bipolamp.htm

17. ### Jonathan KirwanGuest

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
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

18. ### Jonathan KirwanGuest

Sorry, I mean constant 0.65V.

Jon

19. ### Jonathan KirwanGuest

I wasn't entirely consistent in my example, though the approach remains the
same. I substituted a 58k as one of the biasing resistors and that jacked up
the V(B) to some 1.76V and thus, V(E) to 1.11V (assuming the 0.65V V(BE)
figure.) That means the quiescent current is more like 1.1mA, not 1mA. And
this sets the quiescent V(C) to some 4.5V, not 5V. It changes a few of the
specific calculations I gave, but the approach hopefully is okay. Sorry about
that. Haste makes waste, I guess.

Jon

20. ### Kevin AylwardGuest

Yes, but this is a bit of reverse thinking.
Decide what your spec is *first*. *Then* *chose* a transistor that will
meet the spec. For example, if you are after low noise, you might
consider that the optimum noise (approx) is setting
re=Rsource/sqrt(hfe), to set Ic. Looking at the data sheet for the
transistor to *chose* the current in the first place makes little sense.
You are designing for the *application*, not the device.

You might want to be able to drive a large capacitive load at a given
frequency. You would then need to chose a highish current.

For high speed amplifier design you might want to select a transistor
with low Ccb.
But if *want* to use a high current, chose a different transistor.
Again, its the application that should be driving the choice of
operating current and device, not the other way round.
Often they are. Audio amplifies usually use feedback. The game plan here
is to have the amplifier as fast as possible so that one can apply
lashings of feedback in order to reduce distortion. To keep things
stable with lashings of feedback, one needs to minimise phase shift from
the transistors. This means very fast transisters.
Not really.
Well... yes!!!

It does vary a bit, but is often not a major issue.
Not usually a dominate distortion effect, but can be.
Probably the "best" reason for an emitter resister is to reduce
distortion. A simple transistor amplifier has around vi(mv)% distortion.
That is 1mv will give 1% distortion. This is a *huge* amount for such a
small signal. An emitter resister will reduce this by around (re/Re)^2.

http://www.anasoft.co.uk/EE/index.html
{snip tedious calculations}

Ok.... This can in fact be done automatically in SuperSpice http://www.anasoft.co.uk/DeviceDesigner.html

Simply chose node voltages and device currents, and press the button, it
will calculate out all the resister values for you.
You seem to have done pretty well, all things considered.

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.  