Re: Driving a transistor base from a voltage divider


"Jon Danniken" <jondanSPAMniken@xxxxxxxxxxxxx> wrote in message
news:74cjnlF11l510U1@xxxxxxxxxxxxxxxxxxxxx
"BobW"

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

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.

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.

Thanks, Bob, much appreciated. I think I was thinking of the voltage
source as I do with a linear power supply, ie so many volts *while*
delivering a certain current level.

In a linear power supply (or any type of power supply), its output voltage
and current are independant quantities. For a constant voltage power supply,
the output current is only determined by the load that's connected to that
power supply. If the load is a resistor then the output voltage and current
will be related (I=V/R), but it's not that simple for other types of loads
(like the base of a transistor).


Perhaps instead I should be thinking of voltage and current as opposites
given open or closed circuit behavior, as in when plotting a load line for
a tube? If that is the case, I'm guessing a minimum voltage being 0.7V
and the short-circuit current as whatever I need to get sufficient base
current to switch the transistor on?

I'm not sure it's helpful to think of voltage and current as being
opposites, in this case. Generally speaking, however, the output voltage
delivered will decrease as the output current increases. However, it's best
to try and understand each circuit on its own.

If the voltage at your emitter is fixed (e.g. at gnd) then it makes the
analysis easier because the base will be (about) 0.7V away from the emitter
(for any decent level of base current). So, if this is the case for your
circuit, then the base current will be easy to calculate using the previous
ideas that I presented.

Determining how much base current is sufficient to switch the transistor on
is a little more complicated. You need to know:

1 - The maximum collector current when the transistor is on.
2 - The minimum Beta (Ic/Ib) of your transistor when the collector is close
to being in saturation (e.g. when Vce = 1V)

If you know these two quantities then you can easily calculate the base
current required. When I do this, I usually double or triple the base drive
current to insure that the device is fully on (i.e. in full saturation).


Thanks again,


You're welcome.

Jon

Bob
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