0
\$\begingroup\$

I want to get the current through a conductor by measuring the magnetic field of a conductor by using a DRV425. I use the DRV425 Evaluation Board with the original settings (R1 = 100 Ohms) for my research purposes.

I use a 12 Ohms resistor and connect the resistor with the output of my signal generator. The signal generator generates a square signal with the following signal parameter:

  • High level: 5 V
  • Low level: 0 V
  • Frequency: 1 Hz

I use some tape to fix the wire on top of the DRV425. The thickness of the wire is around 2 mm, so the distance between the center of the wire and the chip is around 1 mm.

enter image description here

The green arrow indicates the flow direction of the current and the black line indicates the sensitivity axis of the sensor. Then angle of the wire and the sensitivity axis is around 90° (can´t rotate the green arrow that fine, so the angle of the arrow isn´t the same as the angle of the wire).

I also measure the output signal of the signal generator and the output voltage of the DRV425 with an oscilloscope.

enter image description here

As you can see from the screenshot I measure the following voltages with the sensor:

  • Low: 1.461V
  • High: 1.358 V

Now I calculate the H and B of the conductor with r = 1 mm during the high phase of the signal from the signal generator. For this I measure the current with a multi meter

equation

equation

equation

equation

equation

Next I calculated the B with the equation from the DRV425 evaluation board manual:

equation

I have to use the difference between the high and the low voltage of the signal for VOut, because I want to calculate the resulting field based on the current difference when I turn on the signal. So my difference is

equation

Which results in a B field of

equation

So I have a difference of 10µT which I can not explain. Where is my mistake?.


Update:

Based on the answer of Andy aka I replace the calculation of the current with a direct current measurement with a multi meter.

I also check the distance tolerance and it seems that the results are equal when I increase the distance by 0,5 mm. So a possible answer is that my distance measurement wasn´t exact enough and I haven´t consider the tolerances enough. So the next idea would be to use a thinner conductor to replace the thick (and flexible) isolation of the current silicone wire.

\$\endgroup\$
  • 1
    \$\begingroup\$ Please see my updated question. I have added the sensitivity axis of the sensor. \$\endgroup\$ – Kampi 2 days ago
2
\$\begingroup\$

Most signal generators have a default output impedance of 50 ohms so, if you recalculate current through the load resistor based on it actually being 62 ohms, you are a lot closer to measuring the proper B theoretical value implied by the H field.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Thank you for this hint! I have updated my question. Now I measure the current directly, because the signal voltage of the signal generator drops with my current load. \$\endgroup\$ – Kampi 2 days ago
  • \$\begingroup\$ Unfortunately, your edited question makes my answer look misguided. You should edit your question to ensure that anyone reading it recognizes that you have made corrections based on the information in my answer. \$\endgroup\$ – Andy aka 2 days ago
  • \$\begingroup\$ Thanks are normally recognized by upvotes by the way. I see the readings are much closer now. Have you calculated what distance change might bring the two readings to be equal? Have you looked into the error systems of the TI chip to see if errors could produce this measurement difference @Kampi \$\endgroup\$ – Andy aka 2 days ago
  • \$\begingroup\$ Offset: ±8 µT (Max) for the DRV425. \$\endgroup\$ – Andy aka 2 days ago

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.