Re: Driving a transistor base from a voltage divider

On Sat, 11 Apr 2009 11:57:28 -0700, "Jon Danniken"
<jondanSPAMniken@xxxxxxxxxxxxx> wrote:

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


Assume the base is +0.7 when on. The current through R1 is
(Vin-0.7)/R1. The current through R2 is 0.7/R2. The difference current
is going into the base.

John


.

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