# Voltage -Divider Biased transistor circuit

Discussion in 'Electronic Design' started by [email protected], Mar 19, 2007.

1. ### Guest

hello
i am trying to find resistors values R1 and R2 in a voltage divider
biased transistor circuit.
the following info is given: Vcc=+24 volts, Rc=680
ohms(collector),Re=47ohms(emitter).

if anyone knows how to calculate this please let me know.
thanks

2. ### Jim ThompsonGuest

Yes. I do ;-)

...Jim Thompson

3. ### John PopelishGuest

There are many ways to calculate the resistor values,
depending on the quality of the approximations used.

Simplifying assumption #1: Assume the current gain of the
transistor is infinite (no base current).

Simplifying assumption #2: Assume the base to emitter drop
is a typical value, like .6 volts.

Simplifying assumption #3: Assume you want the collector
bias voltage to be half way between the positive supply, and
the voltage you would get if you replaces the transistor
with a short circuit. That short circuit voltage would be
24*47/(47+680)=1.55, s the half way point between that and
+24 would be 12.77 volts. Since we are doing
approximations, lets call that 13 volts.

First we calculate what the emitter voltage is when the
collector voltage is at this bias point of 13 volts. This
bias point puts (24-13)/680=.016 amps. Bases on S.A.#1,
this same current is passing through the 47 ohm emitter
resistor, so its drop is .016*47=.75 volts.

Adding the nominal base to emitter drop from S.A.#2, we find
the base voltage is .75+.6=1.35 volts.

So R1 and R2 have to divide the 24 volt supply to produce
1.35 volts base voltage. So (if R1 is the resistor to the
+24 volt supply), 24*R2/(R1+R2)=1.35 Solving for R1 in
terms of R2, that comes out R1=16.68*R2

Unfortunately, there are an infinite number of resistor
pairs that will divide 24 volts down to 1.35 volts, so we
need some additional piece of info to pick a pair.

To do that , we have to give up a simplifying assumption.
We have to estimate what the actual base current will be,
because S.A.#1 can't be right. So we can slightly improve
that assumption by guessing a current gain. So, I am going
to change that assumption to assuming that the current gain
is 100. If you have a transistor in mind, you can check the
data sheet for a better value.

So now we have to have a divider that produces about 1.35
volts while delivering .016A/100=.00016A to the base.

We can calculate the values with this exact current included
in the formula, but since it is an educated guess that might
be off by a considerable factor, either way, the
approximation often used is to just make the current passing
through the divider something like 10 times this estimated
base current, so that the base current distorts the divider
only a little. And when you get to picking actual values,
err on the side that produces slightly more base voltage to
compensate for the droop caused by the small base current.

So we have the above ratio that R1=16.68*R2 and the total
divider current 24/(R1+R2)=.0016A.

Solving these two equations, I get:
R1=14152
R2=848

Picking the nearest 5% values (the E24 set available at:
http://www.logwell.com/tech/components/resistor_values.html
that will produce a slightly higher voltage, I would use:
R1=13k
R2=910

A check of 24*R2/(R1+R2)=1.57 volts, without any base
will drop this slightly to 1.57-.00016/(1/R1+1/R2)=1.43
volts. So I may have over compensated by picking 910 for
R2, instead of 820. My choices would produce an emitter
voltage of about 1.43-.6=.83 volts. This represents an
emitter current of about .83/47=.0177ma. and based on our
guess of a current gain of 100, 99% of that will be
collector current, so the drop across the collector resistor
will be 680*.99*.0177=11.9 volts, down from 24 for a
collector bias point of 12.1, where we were shooting for 13.
Perhaps good enough, perhaps not.

When you get tired of such approximations, you move up to a
more accurate set of simplifying assumptions and
approximations and solve more complicated equations that
better model the details.

4. ### John PopelishGuest

Just a first fly over. Well, perhaps good enough for E100.

I was hoping you would do a more detailed version.

6. ### quandong nutGuest

Not likely. I don't believe they (students) learn when others do their
assignments, and I'm all for them actually learning their way through the
education system.

Mind you, I reckon you did a bloody fine job for him/her.

7. ### Guest

You still can learn a lot even if someone figures out the problem for
you because sooner or later that same type of problem can come up and
then you will know how to do it. Besides some teachers just dont know
how to explain things good.

8. ### Guest

I seen these formulas on a web site for calculating R1 and R2

R1 = (Vcc - Vb) / (10 x ib)
R2 = ( Vb) / (9 x ib)

9. ### Rich GriseGuest

How can you say this with a straight face? When someone else does your
homework for you, what you learn is to ask someone else to do your
homework for you.
Apparently your English teacher is an example of one.

Thanks,
Rich

10. ### Homer J SimpsonGuest

That will give you the voltage across the emitter resistor. Add about 0.1
volts to that and that's where the base voltage should be. That's your
starting point.

11. ### YDGuest

Uh, germanium transistor?

- YD.

12. ### Homer J SimpsonGuest

What part of "starting point" did you fail to understand?

13. ### John PopelishGuest

That shows the same generality that the bias divider should
pass about 10 times the base current that I used. I think
this pair of equations can be pulled out of the two I gave you.

Now, a question for you to think about. Since having the
divider pass so much more current than is actually needed to
bias the base, this really lowers the amplifier input
impedance and wastes some supply power. Why do you think a
current factor of 10 x ib is more common, than, say, 2 or 3?

14. ### John LarkinGuest

Um, 0.6 would be a better guess.

John

Maybe a Germanium Schottky ;-)

16. ### John LarkinGuest

Jfet near Idss?

Somebody recently introduced a sort of floating-gate mosfet where gate
threshold voltage can be factory programmed to be above zero
(enhancement region) or below (depletion) or very close to zero (which
I don't have a name for.)

John

17. ### Dr. Leonard H. McCoyGuest

There are many ways to calculate the resistor values,
Thank you, John, for a great explanation of a basic electronic concept that I
never did learn.

And to the critics: not all who ask such questions are teens who want help
with homework. Some are 50-something hobbyists who need to understand similar
things. Best to err on the side of "too much knowledge" (if there *is* such a
thing in America...)

Thanks again, John, for not doing the oh-so-common selfish "knowledge is
power" USENET thing...

Gratefully,
Doc

18. ### joseph2kGuest

Well, the reply is from a really nice guy. Just the same, the question
belongs in sci.electronics.basics. BTW i absolutely detest Google for
listing this ng as "popular"; the spam and OT posts went up 10 fold
afterwards.

19. ### Phil HobbsGuest

Used to be called "depletion-enhancement". Of course, that was before
you could buy depletion enhancers on the Internet.

Cheers,

Phil