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I am using a current transducer (LPSR 15-NP) to measure the current that is flowing through my AC servo motor powered by my SVPWM PCB. I have currently implemented an open loop control to the motor and it is working perfectly. I need to now design a current control closed loop. For that, I am measuring the current going through the U-V-W phases of my PMSM motor. On the output I am getting a variation of 2.407V to 3.047V. I need a constant voltage so that I can implement my current loop.

How do I do this? Are the connections for my LPSR correct or do I need to add some capacitors for signal conditioning? I do not understand why the LPSR is not simply giving me an output voltage in the range 0-5V as the datasheet suggests.

Current transducer wiring in the SVPWM board

Do I need to add a True RMS to DC converter on the output of my LPSR 15-NP to get the DC voltage I require? Do I need to add some rectification or some capacitive filtering in my output? Or am I missing something here?

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    \$\begingroup\$ This sensor output directly follows the high frequency AC current created by the SVPWM driver. Yes, you need filtering and a rectifier. \$\endgroup\$
    – Jens
    Commented Sep 11, 2023 at 20:25
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    \$\begingroup\$ You need a high sampling rate if you want to filter this in software, but, yes, I did so myself. Have a look at the minimum load requirement in the datasheet, I am not sure what impact this might have. \$\endgroup\$
    – Jens
    Commented Sep 12, 2023 at 12:35
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    \$\begingroup\$ I would solder an R/C low pass and connect this to a scope. Then compare this with the calculated values. Is there always enough CPU power to do the math or is this lagging under some conditions, e.g. other CPU tasks? \$\endgroup\$
    – Jens
    Commented Sep 12, 2023 at 12:56
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    \$\begingroup\$ But the sensor delivers a signed value relative to the center voltage of 2.5 V. This includes the current direction. \$\endgroup\$
    – Jens
    Commented Sep 13, 2023 at 14:28
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    \$\begingroup\$ No, the LPSR series are true bipolar devices. The signal is a signed value Vout - Vref, where Vref is 2.5 V. The switching transients of PWM current reversal may have such a high amplitute, that you don't see the effective current in this noise. There may be capacitive coupling from the phase line into your ADC. Apply a low frequency sine wave current to verify the bipolar nature of the output. \$\endgroup\$
    – Jens
    Commented Sep 14, 2023 at 12:48

1 Answer 1

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So, what I basically did was I made a comparator circuit to remove the 2.5V DC offset from the response. Then I received the values in my ADC. I calculated the RMS values for Ia and Ib. Ic can be calculated by this formula:

Ia + Ib + Ic = 0

Then I did the clark transform to get Ialpha and Ibeta.

Then I applied the park transform to get the Id and Iq values. I calculated the RMS value of Id and Iq once again. This seems to give me a steady state response. Id needs to be close to zero and Iq needs to increase as the current across the motor increase. I am not sure why this works but this is what I did. If there is a problem in my methodology please let me know.

I am going to try and implement the current PI control on Id_ref = 0 and Iq_ref = User defined, and then see how it goes.

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