I would like to build a load cell simulator to use with various scale indicators for software testing. I know a strain gauge is basically a resistor bridge, but what value resistors should I use and across which terminals is the potentiometer?
Thanks,
Dale

 

hi,
The range of bridge resistors I have used are from 100R thru 500R.
ie: the individual resistor elements are 100R thru 500R, giving a source impedance of 100R thru 500R.

The bridge excitation is either dc or ac, I use a Vx of 5Vdc, 10Vdc or 12Vdc.
The excitation voltage used should be as specified by the cell manufacturer.
[ its best if you use the cell indicator unit Vx, that normally excites the cell]

eg: 10kG cell, excitation 5Vdc, Vout 10mv/10kG

I would recommend that you do not use a variable resistor to set the simulated bridge offset, it will drive you nuts!!.

Get a good quality 12 pole rotary switch, buy some high precision resistors.
I use the 0.1% 249R version, use 4 to make a bridge [you may have to select a matching 4 from a batch],
leave the 'top' connection of the bridge open.
Thats where you connect 12, very low value precision resistors wired to the 12 pole switch.

The reason for the 12 poles is position 1 will give a -Vsig [-1mV] output and position 1 is Vsig [0mV],
the remaining positions have low value resistor giving +1mV signal increments upto +10mV.
The switch wiper is connected to the +Vx supply.

Use screened cables and a metal enclosure for the bridge and switch assembly, connect the case and screen to the analog ground,

Does this give you the general idea??

EDIT:
Attached a rough working drawing I used a couple of years ago.

The important point to remember is the accuracy of the DVM that is used to confirm the mVolt steps for each switch position.
Make a careful note of these values and they can be used to create a calibration graph/chart.

You will find it very difficult to get each switch position milli-voltage to within better then 0.1mV,
thats where the chart will be useful when the simulator is used to test an actual indicator.

Eric

 

Eric,
Thanks for the information. This not exactly what I was looking for...close, but not quite. I do not need specific load cell values, just a continuous variable control to simulate various load cell values. I will be able to view the resulting "weight" on the indicator. I will then use the serial output to develop averaging routines and to "lock" an appropriate value. I would be varying the value to simulate scale movement.

It appears, according to your drawing, that I would use a bridge with 251 ohm resistors and on the variable side, use 250 ohms plus a 1 ohm variable resistor. Unfortunately, the lowest value I can find is 100 ohms with 1K ohms being the most common lower value.
Thanks,
Dale

 

Hi Dale,
I thought you were looking for a 'stepped' input.
The low value variables are difficult to find. One solution, which isnt totally linear, is to use a 100R multiturn, with a 0.5R
from each end of the variable to the centre wiper.
Take off from the bridge at the junction of the two 0.5R's and the pot wiper.

Eric

 

Eric,
I understand about the variable resistor, but is 251 ohms correct for the other three sides of the bridge? I'm looking to simulate perhaps a 2 or 3 mV/V load cell.
Thanks,
Dale

 

Hi Dale,
Imagine a full 4 resistor bridge, made from 249R resistors.
The excitation is applied to the top and bottom of the bridge.

From the 'left' and 'right' sides of the bridge you woulld normally connect to a differential amp.
Now disconnect the two resistors on the right hand side of the bridge and insert the 100R and the 2 * 0.5R's, between the 249R's.

If you need to provide a bridge zero method, do the same on the left hand side of the bridge.

If this is not clear, let me know I'll do a sketch.

Eric

 

Eric,
I think a sketch would be most helpful.
Thanks,
Dale

 

Hi Dale,
Quick sketch, the adjustment will not be linear, you may have to select on test the 0.5R's

Use the leftside 100R to move the operating point of the rightside pot.

Eric

 

Eric,
Thanks, this is what I needed. If I do not need the zero trim (the indicator has an auto-zero function), could I simply use 250 ohm resistors?
Thanks,
Dale

 

Hi,
Yes, problem is, even with 0.1% resistors the fixed differential output could be well off ZERO mV.. Do a quick sum of the worst case 0.1% values to see what the fixed offset maybe.

As I say, you may also have to tweak the 0.5R's across the pot to give the mV range you require.

Eric

 

Eric,
I think I can take it from here.
Thanks so much for your help,
Dale