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-   -   Measuring resistance using analog breakout (http://www.chiefdelphi.com/forums/showthread.php?t=84106)

ajd 09-03-2010 23:45

Measuring resistance using analog breakout
 
Our team is interested in using sensors that change the resistance of a wire based on an input. We read the datasheet for the Analog Breakout and it included instructions for reading a potentiometer, which has three wires, so that there is a constant resistance between the +5V and GND and a variable resistance between both (+5V and SIG) and (SIG and GND). But what if a sensor only has two wires, and simply adjusts the resistance between them? Do we wire +5V to SIG through the resistor, or SIG to GND, or something more complicated?

Thanks.

dtengineering 10-03-2010 00:04

Re: Measuring resistance using analog breakout
 
I'd probably put a resistor in series with your sensor, hook one end up to +5 and one to GND, and measure the voltage (sig) between the two resistors. That voltage should change as the voltage of your sensor changes.

But there are a lot of folks around these forums who know more about electronics than I do... if you post the range of resistances you expect from your sensor, they might have other suggestions.

Jason

edit: Hey!! Anacortes, WA! We're darn near neighbours, in Chief Delphi terms. See you in Seattle?

Kevin Sevcik 10-03-2010 00:06

Re: Measuring resistance using analog breakout
 
Easiest to wire would be a voltage divider between a known resistance and your sensor. Think of it like your potentiometer, only with a variable resistance between +5 and GND.

Wiring would be so:
Sensor between SIG and GND
Known resistor between +5V and SIG

You'd get a varying voltage on your SIG equal to the following:
R - constant known resistance
r - variable sensor resistance

V = 5 * r / (r + R)

If you pick an R value about 5-10 times the maximum value of r, you'll end up with a moderately linear signal, if rather small.

Other options are various amplifier and/or transistor circuits which will all be a bit complicated.

Mike Betts 10-03-2010 03:35

Re: Measuring resistance using analog breakout
 
Quote:

Originally Posted by ajd (Post 934952)
Our team is interested in using sensors that change the resistance of a wire based on an input. We read the datasheet for the Analog Breakout and it included instructions for reading a potentiometer, which has three wires, so that there is a constant resistance between the +5V and GND and a variable resistance between both (+5V and SIG) and (SIG and GND). But what if a sensor only has two wires, and simply adjusts the resistance between them? Do we wire +5V to SIG through the resistor, or SIG to GND, or something more complicated?

Thanks.

There are several passive circuits such as resistance bridges and the aforementioned voltage dividers. There are also active circuits that can be used.

For resistance dividers, you have to be a little careful choosing resistance values based on the power rating of the sensor, input resistance of the analog module, et cetera.

Before you start implementing anything, why don't you post the part number you looking to use. This will allow the EEs out there to make specific recommendations.

I'll be tied up at the WPI regional for the next few days but I'm certain that someone will make a recommendation for you.

Mike

ajd 13-03-2010 00:27

Re: Measuring resistance using analog breakout
 
Thank you all very much for the information and help! I think now that we understand the basic concept of what needs to be done, I can work with a mentor on my time to figure out the math. (If anyone's curious, we're trying to implement a simple light sensor from a photocell for purely academic/learning purposes.)

Thanks! :)


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