Voltage -Divider Biased transistor circuit

Yes. I do ;-)

...Jim Thompson

-- | James E.Thompson, P.E. | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona Voice:(480)460-2350 | | | E-mail Address at Website Fax:(480)460-2142 | Brass Rat | |

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

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

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:

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.

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