Hi,
I'm trying to fix a problem with a load cell. The cell is rated 500 lbs full scale at 1.8901 mV/V. The issue is that when I excite the cell with 10V, I get 4.79V on the output with no load on the cell.

I know that wiring this load cell should be simple, so why am I getting such large output values?

 

Because you measure the voltage with respect to ground. You have to measure the voltage between the two output wires.

Which load cell is it?

 

It's a Sensotec model 53. I already have negative lead of the differential output grounded to the circuit ground.

I did another test exciting the cell with 5V. It's supposed to be excited with 10V but I figured why not try. The load cell behaved in a strangely ratiometric manner... I got somewhere around 2.25V (IIRC) which is about half the voltage I got when exciting with 10V.

 

The strain gauges in the load cell are arranged in a Wheatstone bridge. The signal + and - leads both ride on half the exciter voltage. You need an instrumentation amp to take the reading differentially.

You might like to go to www.omega.com and see their material on strain gauges and load cells.

 

Oops.

You need to use an instrumentation amplifier. The instrumentation amplifier measures the difference between the two outputs. It's more complex than that though, I'm afraid.

Intersil put out a great application note on instrumentation amplifiers for laypersons; AN1298.
Click here: http://www.intersil.com/data/an/an1298.pdf

 

It's a Sensotec model 53. I already have negative lead of the differential output grounded to the circuit ground.

I did another test exciting the cell with 5V. It's supposed to be excited with 10V but I figured why not try. The load cell behaved in a strangely ratiometric manner... I got somewhere around 2.25V (IIRC) which is about half the voltage I got when exciting with 10V.

The green wire shouldn't be connected to ground. Disconnect it and take the output from the green and white wires.

 

Ah ok, thanks for the links. I will check those out in a moment. What it sounds like you are saying is that I cannot directly ground the negative output of the load cell, and that I must use some sort of differential amplifier. Is that right?

I was planning on grounding the negative output and using a non-inverting amp on the positive output.

 

That will probably destroy one of the strain gauges through too much current. Plus, by using only 1/2 the signal developed, you will get only half the sensitivity of the load cell.

It is absolutely correct that you must use an instrumentation amplifier to handle the output of the load cell. Neither output lead can be tied to ground without causing problems.

 

That will probably destroy one of the strain gauges through too much current. Plus, by using only 1/2 the signal developed, you will get only half the sensitivity of the load cell.

It is absolutely correct that you must use an instrumentation amplifier to handle the output of the load cell. Neither output lead can be tied to ground without causing problems.

D'oh... thanks guys. I wish I didn't have to learn that bit of electronics the hard way.

I don't have any instrumentation amplifiers on hand. Would I be able to "get away with" using some op amps and 5% resistors following the schematic here (I'm guessing no)? http://en.wikipedia.org/wiki/Instrumentation_amplifier

 

D'oh... thanks guys. I wish I didn't have to learn that bit of electronics the hard way.

I just hope you didn't fry it. Those load cells likely aren't cheap.

I don't have any instrumentation amplifiers on hand. Would I be able to "get away with" using some op amps and 5% resistors following the schematic here (I'm guessing no)? http://en.wikipedia.org/wiki/Instrumentation_amplifier

"No" is the correct answer.

Please read the application note that I posted a link to. It is a lot to digest for a layperson, but explains in detail why you should not attempt the Wiki implementation of an instrumentation amplifier, and why CMRR is your worst enemy.

Then you could look at datasheets for such intrumentation amplifiers such as TI's INA114, INA118, INA128/INA129, just for starters.

 

I just hope you didn't fry it. Those load cells likely aren't cheap.


"No" is the correct answer.

Please read the application note that I posted a link to. It is a lot to digest for a layperson, but explains in detail why you should not attempt the Wiki implementation of an instrumentation amplifier, and why CMRR is your worst enemy.

Then you could look at datasheets for such intrumentation amplifiers such as TI's INA114, INA118, INA128/INA129, just for starters.

Thanks. Unfortunately I'm guessing it is toast. Shorting an output to ground doubles the current through the strain gauge, if I understand correctly. I wonder what kind of safety factor Sensotec used when they designed their strain gauges - maybe I'll get lucky ? And yes, I have seen the price tag on these sensors

Is there a test you could recommend to see if it is still functional? Since I only shorted the negative lead to ground, maybe I could compare the voltage of the negative lead to the voltage of the positive lead when the sensor is under no load. I would imagine that the strain gauge that got the extra current just burned out, so I should measure zero volts between the negative lead and ground, whereas the positive lead should measure approximately half of the excitation voltage. Ahhh it all makes sense why I was measuring 4.79 volts and 2.25 volts at the beginning of this thread (just resistor division with approximately equal resistances).


/long rant

 

You probably didn't fry it.

Read up on the documentation already supplied. Then come back and ask questions. Be patient.

 

You probably didn't fry it.

Read up on the documentation already supplied. Then come back and ask questions. Be patient.

You were right. The load cell is ok, but I had to cancel the project due to timelines.

Thanks a lot for all the useful information on common-mode voltage, differential amps, CMRR and so on. I'll be ready the next time a project like this comes around