Re: Driving a transistor base from a voltage divider

BobW wrote:
"Jon Danniken" <jondanSPAMniken@xxxxxxxxxxxxx> wrote in message news:74c7gpF12rg7rU1@xxxxxxxxxxxxxxxxxxxxx
Hi, I am trying to determine the operating specs for operating a transistor base (as a switch) from a voltage divider.

What I am trying to figure out is the appropriate values of R1 and R2, given a desired Base current and source voltage/source current available (I know the source voltage, and how much current I can pull from it).

If I eliminate the "bottom" (R2) resistor in the voltage divider and only use one resistor, I can easily determine the current that flows through that resistor (E/R), and therefore the current that flows through the base ((E-0.7)/R), which is just computing the value of a Base resistor.

Right now I am getting hung up on what effect the "bottom" resistor has on the base current. I know that without the transistor, I can figure out the voltage at the center point as a ratio of the resistances, and the current through both resistors, but what current is available from the center point to drive a transistor base?

The closest I have come is by calculating a "resistance" for the transistor Base/Emitter junction by dividing 0.7V by the desired base current, and treating the transistor as a resistance in parallel with R1, but I'm not very confident with that solution.

BTW, here is an ascii of what I have:

---+
|
|
\
R1 / C/
\ | /
| |/
|-----|
| B|\
\ | \
R2 / E\
\
|
|
gnd

Thanks for any help,

Jon


Jon,

You can't really control the base current. It draws whatever is required to support the emitter current that is flowing.

Absolutely, positively, wrong.

The base voltage has a mild dependency on current, a moderate dependency on temperature (so pay attention if you want it to work outside of room temperature!) and a moderate to strong dependency on the transistor construction.

For a silicon transistor go with John Larkin's 0.7V.

In your resistor divider setup, R1 will provide current to the base of the transistor AND to R2. So, if you measure the current through R1 (volts across R1 divided by the value of R1) and subtract the current through R2 (volts across R2 divided by the value of R2) then you'll have the base current. The only problem with this technique is that you need to know the value of the voltages and resistances very accurately. An easier approach would be to add a small resistor in series with the base of the transistor and measure the current by reading the volts across this resistor and dividing it by the value of the resistor. Or, you could simply put a current meter in series with the base of the transistor.

Absolutely, positively, unnecessary (but at least it's not wrong).

Yes, the base voltage will vary somewhat, and you'll have to take this into account if you want the thing to work over temperature and manufacturing variations. But you just can't trust a transistor to maintain it's B-E voltage well enough that measuring the base current would do you any good at all. Instead, you need to design your circuit so that you always have sufficient base current (I'm assuming the OP wants it to be saturated, otherwise he needs to post again and ask for clarification).

The current that is available at the connection of R1 and R2 depends on how many volts (from gnd) you desire. At zero volts (from gnd), the current you'll get is simply V+/R1. To better visualize what you really have, first remove the transistor from your thoughts. Now, just look at the voltage divider. It can be reduced to (thought of as) a voltage source with a series resistor coming from it. The value of the voltage source is the open circuit voltage you would get at the R1-R2 connection. That is, you would have a voltage source equal to (V+) * (R2/(R1+R2)). The effective series resistor would be equal to the parallel combination of R1 and R2, and would be equal to 1/( (1/R1)+(1/R2) ).

For example, if V+ is 12V, and R1 is 1K ohm and R2 is 400 ohms, the "equivalent" circuit would "look like" a 3.43V battery with a single series resistor of 286 ohms. So, its open circuit voltage is 3.43V. Its short circuit current is 12mA (note that this 12mA value is the same as you get from the "real" circuit -- that is 12V/1K). Now, when you hook up your transistor, it's easier to see what the base voltage would be for a given amount of base current.

Well, at least _that_ part is correct.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details at http://www.wescottdesign.com/actfes/actfes.html
.

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