Is the main reason for choosing a particular resistance value, the amount
of current you want from the divider?
Mike
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Mike Stramba Email: mi...@io.org
Toronto,Canada Internex Online - Toronto, Canada (416) 363-3783
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Precisely -- And the amount of energy you want to waste. As soon as you
put a load on it, it's no longer the same voltage divider.
If you want to supply Vcc/2 to an op-amp input, a couple of 100k resistors
will do fine; the op-amp input does not draw appreciable input.
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< Michael A. Covington, Assc Rsch Scientist, Artificial Intelligence Programs >
< The University of Georgia, Athens, GA 30602-7415 USA mcov...@ai.uga.edu >
< Unless specifically indicated, I am not speaking for the University. > <><
For information about any U.Ga. graduate program, email gra...@uga.cc.uga.edu.
: Precisely -- And the amount of energy you want to waste. As soon as you
: put a load on it, it's no longer the same voltage divider.
: If you want to supply Vcc/2 to an op-amp input, a couple of 100k resistors
: will do fine; the op-amp input does not draw appreciable input.
And if you want to draw some current and stay within, say, ten percent of
the supply voltage, then you would want at least 10 times the needed
current flowing thru the voltage divider. Or you could put a voltage
divider, as above, in front of an unused Op Amp with the - input tied to
the output, as a V follower. The op amp will then supply the variable
current without any voltage change.
>Is the main reason for choosing a particular resistance value, the amount
>of current you want from the divider?
>Mike
Are you taking the piss ?
: And if you want to draw some current and stay within, say, ten percent of
: the supply voltage, then you would want at least 10 times the needed
: current flowing thru the voltage divider. Or you could put a voltage
: divider, as above, in front of an unused Op Amp with the - input tied to
: the output, as a V follower. The op amp will then supply the variable
: current without any voltage change.
Good advice, but a small nit to pick. I think it's more like 5* the
load current to get 10% regulation: for example consider a 2 volt
source with 2 1-ohm resistors to give 1 volt. That's a 1 amp standing
current. Load the 1 volt output with 100mA, and you get 1.05 amps
thru the upper resistor and .95 amps thru the lower resistor, and .95
volts out. I think that's 5%. The divider's output resistance is the
parallel combination of the two resistors, assuming a zero-impedance
power supply on the input.
The op amp suggestion is a good one, and if you need more current than
than op amp can deliver, you can use the op amp output to drive an
emitter follower, and take the feedback from the emitter--but you do
have to be careful about stability then.