INA125 Reference Voltage lower than expected

Question

I am using a INA125 to amplify the signal from a load cell. When I connect everything according to the datasheet as seen below, the reference voltage drops when I connect the load cell. I am supplying the IC with 24 V and 0 V.

If I measure the output voltage when nothing is connected it reads 10 V. When I connect the load cell it drops to 6.8 V. the two signal leads are 3.4 V as expected.

I am thinking that the INA125 can not supply enough current but I cant find the limits in the datasheet.

Can somebody give me any advice on how to solve this? Maybe adding the transistor will help?

Thank you very much!

Datasheet: https://www.ti.com/lit/ds/symlink/ina125.pdf)

Answer 1

The reference is only characterized for 0-5mA output current. "IL = 0 to 5mA"- anything more than that is not guaranteed.

If you add the transistor there will be a few effects:

1. The transistor needs about 0.7V to work, so the chip has to output a bit higher voltage. You need about 2V higher supply than the reference (1.25V from the datasheet plus 0.7V). A bit more if it has to operate in extreme cold conditions. 3V would be better. If you have a higher clean supply, you can connect that to the transistor collector. You have 24V, so no problem.

2. The power dissipation associated with supplying the reference voltage to the bridge will mostly be moved to the transistor. So you may have to use a power transistor such as the TIP series rather than something like an SOT-23 part. Higher supply voltage relative to the output reference voltage means more dissipation, like any linear regulator. Eg. 50mA to the 10V bridge with a 24V supply would mean the transistor has to dissipate about 700mW.

3. The current the chip has to be supplied will be reduced by the current gain of the transistor. The gain of the TIP29C will be at least 40 at 25°C, but could be 20 at -40°C. Even so you could get 100mA out of it.

4. You forfeit any thermal protection or short-circuit current limiting that is built into the chip. If the user shorts the bridge the transistor power dissipation could be very high. You should consider adding a 'PTC' resettable fuse and calculating whether that will actually protect the transistor (it may need a much larger heat sink than that otherwise would be indicated).

The TIP31C has beta bins and the lowest one 'R' is pretty miserable (guaranteed gain of only 10 @ 25°C 1A and Vce = 4V, so it might only have a gain of 5 worst-case.. ugh). If you don't have a really horrible one, and don't store it in the freezer, it should work, however.

You need about 30mA with a 330Ω bridge. The transistor will be dissipating about 420mW so it will need to be something substantial. I would probably use something more modern like this TO-252 part with minimum gain of 500 @ 100mA.

It's good to physically separate the high power dissipation parts from the precision analog anyways to get better accuracy.