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).

There are many ways to calculate the resistor values,

depending on the quality of the approximations used.

Lets start with the simplest.

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

current loading it down. The guessed .0016A of base current

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.