snipped-for-privacy@hotmail.com wrote:
>> 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:
>formatting link
>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.
Crikey, his electronics 100 class tutor WILL be pleased with your\\\\\\\\his very thorough assignment this week.