Dear Roger,

After reading a lot about scaling, can you please help me to find the solutions for our sensors with scaling factors for the DT800?

1. pressure sensor output: 0 to 100 mV excitation: 10 VDC gage factor: 1.0 kPa/mV config channel: 1BGV(4w) scaling value to DT800? Polynomial?
2. Strain gage (I think is full bridge) microstrain/mv: 283.39 at 2.5 Vdc excitation microstrain/mV/V: 708.475 config channel: 1BGV(4W) scaling value to DT800? Polynomial?
3. Resistive sensor resistance at full extension: 1471 Ohms factor: 1 inch/volt/volt config channel: 1R(4W) scaling value to DT800? Polynomial?

I saw that in DT800 we can measure bridges in current or voltage. Any bridges with mV output, can work with current too? Or load cells with mV output, must be excited with Vdc and load cells with mA output must be excited with current?

Best regards,
CPalha

Good morning CPalha,

To run through the maths..

All these sensors you mention are ratiometric. This means the output voltage is proportional to the input voltage.

When the dataTaker measures a load cell it measures the excitation Voltage and the output Voltage and then calculates the output.

PPM = (Vout / Vexc) x 10^6

Pressure sensor.
Your pressure sensor output 1kPa per mV at 10 VDC excitation. This is 0.1 kPa/mV/V therefore

0.1kPa = 1000 PPM so 1PPM = 0.0001 kPa

The second term of a polynomial is the slope so Y1=0,0.0001"kPa" for the pressure sensor.

Strain gauge.
What is the gauge factor? It usually some where just above 2 and will depend on the type of gauge and gauge material. The maths is described on page 146 of the DT800 user manual.

ue = (4/(GF x N)) x PPM

Where GF = gauge factor and n is the number of active gauges in the bridge. This can be pre-calculated, If your gauge factor is 2.11 and you have 1 active gauge per bridge the ue = 4/2.11 = 1.895734

ue= PPM x 1.895734

Potentiometer
What is the displacement of this sensor? If the resistance is 0 at 0 displacement and 1471 Ohms and the displacement is 1 inch then 1 Ohm = 1/1471 inch
So simply multiply the ohms by 6.798 x 10^-4

Once again I would use the second term of a polynomial.

If these devices are just plain Wheatstone bridges I would use current excitation in preference to voltage. As you are using 4 wire configurations on the bridge (pressure and strain gauge) there is no compensation for lead lengths and you will under estimate the input.

Cheers,
Roger

Dear Roger

Thanks for the prompt help.

About the PPM I understand the mathematics, and it a curious way to measure. I only have worked, until now, with mV.

About the pressure sensor, I calculate this calibration factor, but the output don't represent the really load, maybe I am applying light loads.

About the strain gage, this is a metallic bar with strain gage (s) (it's closed and I can't see). The GF = 2 and the resistance is 349 Ohms. Lead length 30 ft. In the other literature I saw output = ~2mV/V@1500 microstrains.

For the potentiometer: Length full extension = 6 in. and the travel range = 1 in.

My pressure sensor has 2 wires for remote sense, but I would prefer not use because I will need a lot of channels.

About the burst mode or fast mode, when I install the sensors under the road, I will need to saving data fast. I now it depends the number of channels connected. In your opinion you think I will get about 100 samples/sec in fast mode? Or I will need to use the burst mode?

We are planning to install or not the DT800 in the road. If yes, I must see how can I save the temperature of the asphalt each 30 minutes and saving some heavy trucks in burst mode or fast mode.

Can you give me some help to configure the pressure sensor pulling the trigger, and if it is possible, measure the load, if it is bellow some value, discard data, if not active the trigger and save own data and the other sensors?

If not I will do it manually with a wire connected to the D1 and pull the trigger when touch the GND.

Sorry for the long message

Best regards,
Carlos Palha

Good morning Carlos,

1. PPM
   Not that curious. If you think about it 1ue is 1 PPM so there is a direct relationship between the two. Measuring the supply Voltage and the output and dividing the two ensures a stable reading.

2. Lead wire compensation.
   You should always use lead wire compensation with especially with Voltage excitation and even more so with long lead lengths. The Voltage drop in the leads will mean the excitation Voltage at the load cell will be lower than the Voltage measured at the terminals.

   Because load cells are ratiometric the Voltage drop will mean the output will be lower than expected.

3. Strain gauges.
   Have a look at the packet the strain gauge came in. The stain gauge manufacturers provide the gauge factor on every packet.

   You have 350 Ohm strain gauges so you can do a shunt calibration.
   For example if your gauge factor is 2 then a 174650 Ohm resistor shunted across the gauge is 1000 ue. From there you can calculate the GF. Remember to use a three wire configuration on the strain gauge.

4. You will have to use burst mode. While you might just be able to get 100 Hz if you are using a memory card then the maximum speed will be around 50 Hz due to the card write speeds.

Cheers,
Roger

Dear Roger

I am sorry to insist. The calculation below it seems give reasonable results for the expected microstrains.

Please comment the calculation below:
Asphalt Strain gage:

Excitation: 2.5 V DC
Calibration factor: 283.39 ue/mV

Output from data acquisition: PPM 
PPM=(Vout x 10^6)/(Vexcit)
PPM = -530 (from datataker)

Calculating:

PPM=(Vout x 10^6)/2500mV <=> -530= Vout x 400 <=> Vout=-1.325 mV
ue = -1.325 mV x 283.39 ue/mV
ue= -375.49

Or using strain gage manufactur calculation

GF=Vexcit x ue/mV
GF=2500 x 283.39 eu/mV = 5.64593

Calculating:

ue = (4/(GF x N) x PPM ; No of active strains (1)
ue = (4/(5.64593)) x PPM
ue = 0.708475 x PPM

ue = = 0.708475 x -530 
ue = -375.49

Best regards
Carlos Palha

Good morning cpalha,

What is your question? Where does the 2.5 VDC excitation come from?
What is the gauge factor of the gauges?

Cheers,
Roger

Dear Roger

I am sorry for the confusion. The last message was only to show my calculation. The sensor bulletin is below: http://www.ctlgroup.com/FileUploads/1

With sensor came a calibration certificate that has the additional information: ue/mV/V: 283.39 at 2.5 V excitation

In the bulletin You can see the maximum strain is 1500. If I use your value 1.89 x PPM (without forces) => -530 x 1.89=1000 ue is almost the full scale.

Sorry once again for the confusion.

Best regards,
CPalha

Good morning cpalha,

Ok, now I understand.

This is not a single strain gauge but a full bridge configuration. So according to the document has a full bridge the output is 1500 ue @ 2mV/V

2mV/V is 2000 PPM so 1500 ue = 2000 PPM

Cheers,
Roger