Hello,

I have set up a current-excited bridge (BGI) circuit for a DT800 to measure the strain across five quarter-bridges on a channel pair, as shown in the attachment. The problem I am having is that I do not know what scaling factor to use to convert the incoming ppm signal to microstrain for each measurement. In addition, I am not always running all five bridges, which I assume changes the scaling factor as well.

My question is: How do I calculate the scaling factor for this setup including when I have less bridges? particularly for five, four and two quarter-bridges on a channel pair.

I have been trying to figure this out but have been unable to do so, if you could also explain the theory behind calculating the scaling factor or what theory I should be using it would be much appreciated?

In addition, is there a more effective circuit and way of obtaining data for quarter bridges?

Thank you,

Matt Mysell

Hi Matt,

The scaling is the same as a single 1/4 bridge.

Cheers

Roger

Hi MattMysell,

To elaborate Roger's response, if you refer to our website for ppm to micro strain conversion, there is a number of active gauge needed for the conversion.

I know you are asking how to detect how many active gauge in the system, there is easy way.

1. You need to make offset reference by executing bridge channel before any schedule declaration (sometimes strain gauge reading is far off at 0 reading (i.e.: 80 ppm)).
2. Use this offset value as adder or substractor to your bridge channel inside a schedule and you will get zero reading ppm of your strain gauge.
3. You will find this zero reading will fluctuate a bit (normal if it is between -10...+10 ppm)
4. Use conditional statement for each ppm with normal zero value as test condition example: ppm1 -> if the reading fall between +/- 10 ppm set channel variable 1 (1CV) to 0 and if the reading outside +/- 10 ppm set 1CV to 1 IF(1CV><-10,10){1CV=0} IF(1CV<>-10,10){1CV=1}

Then add all the channel variables, in your case: 99CV=1CV+2CV+3CV+4CV+5CV and use 99CV as your divider.

Best regards,
Rudy Gunawan

Hi Roger and Rudy,

Thank you for the responses.

I did a simple experiment to test what the gage factor should be, using a beam with a mass on the end to cause bending which results in a known strain.

I then set up my circuit as shown before and measured the strain whilst varying the amount of 1/2 bridges (with one active gage each) on the circuit as well as changing between voltage excitation and current excitation. I have attached my results with my hand calculations. The gage factor is 2.08.

From the results it can be seen that the voltage excitation measurement is not affected by the added 1/2 bridges however the current excitation measurement is.

For the current excitation method the excitation voltage should decrease by a factor of [(the number of 1/2 bridges in the circuit, including the completion 1/2 bridge)/2] when adding additional 1/2 bridges to the circuit, shown in attached calculations.

Therefore, for one active 1/2 bridge the voltage reduction is 2/2, for two active 1/2 bridges it is 3/2, for three active 1/2 bridges it is 4/2 and so on. These factors are the same as the factors which the measured strain is out by for the current excitation method.

However, the output voltage is relative to the excitation voltage, so it should be reduced by the same factor. Therefore, the measured strain should not be affected as the reduced voltage factor would cancel out, as shown in calculations.

I have checked my circuit and done the test twice over, so I am not sure why the current excitation method is causing these results?

Thank you,

Matt

Good morning MattMysell,

It is best to use this configurations with Voltage excitation.
While you can use current excitation you must take care that your gauges do not fail during test.

Some theory:
When using current excitation the supply Voltage is derived from the circuit resistance.
From the formula Voltage = Current x resistance we can see that for a given precises current we can calculate the Voltage.
Also for resistors in parallel the circuit resistance can be calculated from 1/Rtotal=1/R1+1/r2+1/R3.....+1/Rn.
From this you can see that the over all bridge circuit reduces as you add more arms hence the drop in output.

Now I must get back to breaking things.

Cheers

Roger (Who is having way too much fun in his new job)