Wheatstone Bridge Values Needed

I have a device that varies from 19 to 63 ohms and need to read it and determine its present value. It appears that a Wheatstone Bridge would allow a 5v input with the output used by an A to D converter.

My problem is that I am not up on the math needed to calculate the appropriate values of the other three resisters, and I don't actually understand what is going on.

I don't need references to Wheatstone Bridge theory, I have spent the last three hours reading numerous examples and explanations.

Can anyone who is versed in this give me some R values for the other three resistors, assuming a 5v input and the output going into an Arduino A to D?

Greatly appreciated. Thanks.

A Wheatstone bridge won't swing 0 to 5V if that's your objective.

Measuring low values of resistance directly is tricky. Using a bridge you need to amplify the output with a differential amplifier before applying it to the analogue inputs.
In essence all a bridge consists of is two potential dividers. One provides a fixed reference point and the other will change as the resistance of the one arm changes. So connecting a bridge directly to an Aduino is not going to happen. It needs amplifying first.
In essence you can use any value in the reference leg and in the other led pick a resistance equal to the mid range of your sensor. Drive it with 5V if that does not cause too much current for your sensor or power supply.

A technique you might like to consider is to use the resistance in a variable frequency oscillator and measure the frequency and change.

It doesn't have to swing from zero to 5 volts.

If someone familiar with this could give me some R values I would greatly appreciate it.

Thanks in advance.

I can use an op amp but first need to get a bridge working IMHO. Converting to a frequency doesn't seem to be the most often used way to do this based on my reading, but if there is a circuit I can look at I would like to see it.

Thanks.

If someone familiar with this could give me some R values I would greatly appreciate it.

I said:-

In essence you can use any value in the reference leg and in the other led pick a resistance equal to the mid range of your sensor.

Passive wheatstone bridge won't produce your desired 0-5 volts as previous responder stated. If you could provide details of your actual sensor device we may be able to offer suggestions

a) does it need power to energise or is it passive (not a 4-20mA device by any chance)
b) if answer to (a) is "yes", what is voltage and current demand
c) what is the actual application of sensor

If it really is a passive device and you want to avoid the complication of an op=amp at the expense of reduced resolution (0.5% as against 0.1%) then you could use a simple potential divider chain. Say 22ohm from ground to your device with other end of device tied to 5 volts. Common point of 22ohm and device connects to arduino input.

Input signal now varies from 2.683volts at 19ohms to 1.294volts at 63ohms a difference of 1.389volts

growerdick:
It doesn't have to swing from zero to 5 volts.

What range would work for you?

It is passive and a difference of 1.389 volts is enough.

I assume that taking the 5v from the board into the device, and connecting the device output to the 22 ohm resister, whose other end goes to ground, and connecting the device/resister connection to the A to D, will be okay for the analog input.

Thanks for the help.

No problem for the analogue input. Just can your device stand that much current, you know heating and such.

If my calculations are correct, around 40 ohms yields 125 ma .6 watts. I am sure it can handle that but I guess I shouldn't use a .25 watt resister.

Is there some way to knock things down more and retain at least a one volt range to measure?

What is this mystery device?

I would make the bridge work at a much higher impedenance level and use a instrumentation op-amp. So two top bridge resistor make 10,000 ohms, and these need to be closely matched, maybe make the one in the reference leg 9.1K ohms in series with a 1k trim pot for balancing the bridge. The lower reference leg can either be a 19 ohm precision resistor or a 100 ohm trimmer pot to better match the 'zero' starting value of your sensing resistor. You can easily do the math to see what the top and bottom millivolt output of the bridge will be for your sensor and then decide how much gain you will have to set up in your op-amp.

Lefty

It is possible to get a uni-polar voltage from a Wheatstone bridge. If the measurement side (variable side fixed resistor) is placed at one end of the measurement range and you use an isolated supply for bridge excitation, you can ground one side of the bridge output and thus have a 0 to whatever output voltage, center the fixed resistor and have a bi-polar output. If the output voltage caused by the excitation is too high at the extreme resistance difference lower the drive.
The really interesting is to use an AC voltage, two matched resistors and a capacitor for the "reference' component and measure reactance by balancing the bridge.
This is a simplified description. There are better ones here for both Wheatstone and Hay bridges and the math to go with it.

Bob

If my calculations are correct, around 40 ohms yields 125 ma .6 watts. I am sure it can handle that

No you need to check the data sheet. It sounds like a lot of current to me for a sensor.

Naaah, use small value medium wattage resistors but be aware that "Some" resistors will change by a few % in value when hot...

I forgot here is the reference... again, I thinkhttp://www.ustudy.in/node/6473. Good looking website with a LOT of high grade electronics lessons as far as I could see. I apologize for not making sense I slipped outside and hit my head on the curb...

Bob