# Transistor biasing help

Discussion in 'Electrical Engineering' started by Peter, Nov 6, 2004.

1. ### PeterGuest

Hello.. I asked this question a while back and I did get some answers. But
I'm now taking an Analog Electronic course and the instructor is having us
go a little more into the circuit than I was taught.

To start off... I'm trying to understand a simple 4 resistor amplifier and
then add capacitors in there so I can amplify AC (I wont go into that now).
I'll use R1 and R2 for the voltage divider on the base of the transistor (R1
being the top resistor and R2 the bottom), R3 will be Rc and R4 will be Re.
Then we can assume a beta of 100, our input voltage will be 20 and a Vce of
approx 10 (half way from 20 so the signal can swing up and down).

Now the way I was originally taught... calculate your voltage divider and
ignore base current. So if R1 and R2 are both 1k, then your base voltage is
10volts, your Ve is 9.3 (assuming 0.7volt drop Vbe), you calculate your
emitter current and voltage, then you consider Ie and Ic equal and calculate

About a year ago I wanted to learn how to actualy "design" so someone
suggested the follow which is a GREAT way of doing it. A transistor data
sheet has beta for given values of Ic and Vce, so you first calculate your
R3 value to get your Vc to the value your looking for. In this case we can
consider Ic and Ie equal for purproses of discussion. (I know Ic is beta
divided by beta +1 times Ie higher). Then you set your R4 so your Vce will
be the 10volts that I'd like to get. Now at this point last year I was given
two pieces of advice. I could ignore base current if it's small enough or
(the way I used all the time) I calculate base current based on Ie/beta=Ib
(assuming worse case beta, for a 2n2222a I believe the lowest is 50, but
let's stick with 100) then I base my R1 on being 20 times higher than my
base current for a rule of thumb. Then R2 is selected based on the total
current calculated through R1 and subtract the base current to get your
current through R2.

This was the way I used and stuck with. But now that I'm in school, the
teacher is teaching us impedance and gains which is something I wasn't
using. He'll tell us to make Vce 10volts for max swing (understandable).
Then in order to get a gain of say 10, we make R3 10k and R4 1k, calculate
our currents and voltages. Now the totally confusing part for me is.... we
need to get our R1 and R2 values. The teacher uses Thev. equivelant by
saying our load looking into the base is Re*beta +1, our Vthev is what the
load sees looking the other way (into R1 and R2), and our Rb is R1 in
parallel with R2. This is easy to understand, but then the teacher says,
let's "assume" R1 and R2 is 1k, that would be a Vthev of 10volts and and
Rthev of 500 ohms and we do a loop equation based on that. I told him about
my 20 times higher rule and he said, that can be a way of doing it 'I
guess', but doing it this way you can come up with your own equation and
it's more accurate.

My confusion is what to make R1 and R2. My book (which is impossible to
understand, even the teacher doesn't understand why the school picked it)
goes into a few 'rules of thumb' as well which states something about make
this 1/3 of that and this 1/3 of this. But the strange thing is, if I'm
designing circuit after circuit, all my circuits are going to be exactly
alike because I'm making everything 1/3 of this and 1/3 of that.
Forinstance, my company pretty much uses 15volts for most of it's circuits,
5 volts for the logic and 24 for the higher powered stuff. All my amplifier
circuits would be based on 1/3 of 15volts and 1/3 of that, blah blah. I
could simply make an excel spread sheet and never have to calculate anything
ever again. Then how could I ever call myself an enginee? I'd never be able
to figure out where a problem is in my design if I only followed rules. but
anyway... I know I want my R1 and R2 high enough so that when base current
changes due to temp or the variations in beta that we don't have to worry
about it. But I'm trying to understand everything. The teacher gives us a
loop equation and distracts us with that, then plugs "assumptions" in for
values and bases his entire circuit on throwing whatever resistor value that
pops in his head in the circuit. I think it's becasue he has done if for so
many years, he knows what to use and a person that is trying to understand
why he does it doesn't follow his steps.

If someone can help me out with the R1 and R2 stuff, that would be great. I
don't think I need to waste anyone's time with R3 and R4, I know my formulas
and I know how to use the loop equation. We are really getting into the
details of a transistor and now I just found out that the emitter changes
25mv per ma I believe, so now does the transistor end up running away
because the emitter voltage increases, then the collector increases, changes
the beta then changes the base current, then changes the emitter voltage,
then changes the collector voltage, then the beta. I NEVER thought a 4
resistor and 1 transistor circuit can be so confusing. I thought I really
understood these circuits until I started this class.

The voltage gain is basically 40 times the supply voltage if there is no
external re./
This occurs near saturation of the transistor.

Nothing else matters.

I forgot to put my sig in the subject.

I once had to do a similer problem in class.

the gain of a transistor is simply 40 times Vcc and it occurs close to
saturation.

Nothing else really matters.

It there is large external resistors, the gain is Rc over Re.

4. ### Sylvia ElseGuest

I seem to remember this to be the the small signal voltage gain: 40 Ic
Rc, where R is the Ic is the collector current and Rc is the collector
load resistance. Ic Rc is voltage across the load resistor, which might
be said to approximate to Vcc when the emitter resistor is small and the
transistor near saturation. The value 40 is temperature dependent, and
is the approximate value for room temperature. The calculation only
makes sense if the emitter resistance is decoupled, since otherwise most
of the input signal voltage variation appears across the emitter
resistor, and is not amplified.

Sylvia.

Riight on sylvia.

If you write the equations for the gain of a transistor and take the
internal Re of the transistor as being 25ohms/Ire(in miliamps) then Re and
Rc drop out or the equation completely and the gain becomes 40 X Vcc,

Ass Re at a given current can vary with temperature if temperature is a
consideration then as you say the gain can also vary.

But Re does not really vary much with temp unless you go to extreems.
Of course "very much " is open to individual thoughts on the matter.

At close to saturation IcRc is equal to Vcc.
So you can drop them from the equation.

The example was given for a circuit with out an external emmitter resistor.

I got an A in the course for that one.

The Instructor just wanted to know the trick.

6. ### PeterGuest

Thanks for everyone's help. It is difficult to try and match all these
parameters. Right now I'm trying to work on a project for class.

Common emitter circuit, gain of -10, input impedance minimum 10k, output
impedance of 5k and the output has to be 5v rms.

I can get just about everything except matching the input impedance. If I
change a resistor to get that, then I change the gain, the I need to change
something else then I ruin the output impedance.

Thanks again