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I am looking for a circuit that will operate off of 10 to 20vdc, and will take a resistive input(pot) and convert that signal to a value ranging from one volt dc to five volts dc.......any ideas????

I'm a little lost, could you explain more about what it is and what it is for?

Sounds like a voltage divider to me.

How many amps do ya need?

What's the application? Like toejam said, how much current do you need?
The simplest thing that comes to mind is a 1k potentiometer with one leg tied to the +5 output of a 7805 regulator and the other leg tied to a 1V (or closest value) zener diode. Using an LED instead of the zener will give you about 1.2V at the bottom. The wiper will then give you a 1-5V output as you turn the pot. Note that the current provided will be pretty low (5mA max) and the output must be connected to a pretty high impedance, but it will work from a 10-20V input.<p>If you need more current, then you'll need to post more information

WHAT I HAVE IS A BOARD THAT IS ABLE TO TAKE AN ANALOG INPUT THAT RANGES FROM 1-5VDC AND COVERTS THAT TO A DIGITAL SIGNAL OR ADC. THE BOARD CAN BE USED TO JUST MONITOR THESE INPUTS OR CONTROL OFF OF THESE INPTS, BASICALLY A PLC UNIT.
THE INPUTS THAT I AM REFFERING TO ARE HIGH IMPENDANCE INPUTS SO CURRENT IS OF NO CONCERN, JUST AS LONG AS THERE IS ENOUGH TO DRIVE THE INPUT CURCUITS (VOLTAGE PROPORTIONAL TO CURRENT OR VISA VERSA).

hoping that no one took my caps lock on as being angry or mad

Since the inputs take very little current, you can do the whole thing with a voltage divider.<p>One or two fixed resistors and a potentiometer should do the trick.<p>Place the resistors and pot in series between the source (10V) and ground with the pot in the middle and the wiper connected to the input pin.<p>The fixed resistor at the top of the network will determine the value of your max voltage and the one on the bottom would be the min voltage<p>One solution:
for Vsupply=10V and Vout=1V to 5V
Assuming Iin is negligible (<20uA)
For a quiescent current of 1mA the three resistors must total R=10/0.001 = 10kohm<p>If you used only a 10K pot, you would have an adjustable range of 0V to 10V but you want 1 to 5V so you divide up the 10K among the three resistors so that when the wiper is at the max positions it sees 1V or 5V.<p>Therefore, the bottom resistor has 1mA and 1V so it has to be 1kohm. The top resistor has to drop 5V (10V-5V=5V) so its value needs to be 5Kohm. Since the whole network needs to be 10K the pot is the remainder, Rvar=4K<p>If the input to the ADC uses much more current you will need to account for that in the calculations. I figured 20uA (being 50x smaller than 1mA) would introduce a negligable error.<p>Using a voltage divider in this way to control larger loads (more current) not only introduces an error due to the additional IR drop but you risk burning up the POT when you dial the wiper to the lowest value. When you do this, lots of current flows through just a small portion of the resistive element near the end. This point will fuse open if sufficient curerrent is passed.<p>[ February 25, 2004: Message edited by: haklesup ]</p>

It just dawned on me, but would i be able to use some form of op amp, with your idea of a voltage divider for the input to the op amp. And yes, the output from the above is sent to the adc and from there to other circuits that would be able higher current loads such as relays or transistors

what it sounds like you have a dancer contoller board which takes a 1-5 0-10 volts input from a variable reluctance transducer ( detects arm position on take-up reel in winding operation ..tape ,duct ,wire ,ect controls take-up speed as the build on the spool drum increases in size slows down the reel speed also know as a tension controller) there are many stand alone modules on the market that will translate a pot rotation in to voltage 1-5 volts try looking in any process control catalog or omega on-line .
joe

sounds like you need a couple of millamps.I would take a a 5-10 k pot and hook the 12 v. source to the outside ends. Now, hook a wire to the middle terminal(wiper) and the other wire to the terminal attched to the negative side of the pot (adjustable resistor). Walla... an adjustable (low currant) voltage supply.If you more amps. go with the lm317. if you need more than an amp it can be done with a pass transistor, or a variety of other methods.
good luck
tj.