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I'm working on part of a project and I need to build the circuit that reads the current through a phase of a motor using this IC from Allegro.

Assuming there will be no more than 5A, and using a VCC=5V, the IC will output between [0.5;4.5]V.

We need this to be turned into 3.3V to enter some microcontroller.

I thought a simple voltage divider would suffice but I was suggested opamps. This is what I came up with. I picked this OpAmp so I could use the same 5V rail that powers the Allegro IC.

The first OpAmp turns [0.5,4.5] into [-3.3,-0.3667] and the second one turns it into [0.3667,3.3].

I understand a buck converter would not be ideal here because it takes a bit to get into steady-state again and the purposes for which this will be used may need a faster transition.

But why not a voltage divider? Surely with the limited amount of current (aprox 10mA, max 14mA) coming out of the sensor, it wouldn't be wasting any considerable amount of power. Each phase of the motor has ~125W of nominal power.

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    \$\begingroup\$ I would recommend double checking your μC's ADC input ranges; a lot of 3.3V micros can only read up to 3V on the ADCs. In any case, a voltage divider would probably work fine, or a voltage divider followed by an op amp buffer. A buck converter is not at all what you want; those are really only meant for power supplies, not for analog voltage conversion. \$\endgroup\$
    – Hearth
    Jul 9, 2018 at 14:54
  • \$\begingroup\$ What you use will depend on how much of the range of the ADC you want (8,12,16 bit) to use. If you current resolution is not too challenging, then a resistive divider would be ideal. If you need to use the limits of the ADC then an Op-amp is a better choice. \$\endgroup\$
    – Jack Creasey
    Jul 9, 2018 at 15:12
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    \$\begingroup\$ Apart from all of that ^^, I'd be very impressed if you managed to get a negative output from the op-amp if you do not actually supply it with a sufficiently negative supply voltage. \$\endgroup\$
    – Asmyldof
    Jul 9, 2018 at 16:47
  • \$\begingroup\$ I will ask about the input range, thank you! No negative input needed. Not a sine wave. PWM and the current will only flow in one direction. \$\endgroup\$
    – Marcelo Queirós
    Jul 9, 2018 at 22:49

3 Answers 3

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Assuming there will be no more than 5A

If the peak current is 5 amps then the RMS current (sine waves assumed) will be 3.536 amps. I mention this because it does seem a little light in value for what appears to be a multiphase motor on AC. Starting currents and full load current possibly needs to be accounted for. Just an observation.

The ADC will have a range of voltages that it can be relied upon to work with. The loswet voltage is usually regarded as 0 volts but, due to variations in manufacture and temperature might be only as low as 50 mV. The upper limit of the range is unclear so you need to take that into account and, for safe measure assume it is 50 mV less (or just do a lot of reading in the data sheet to ascertain precisely what can be relied on.

And yes, a voltage divider will work just fine. Choose resistaors to do the division accurately.

For worst case considerations make sure that the voltage division ratio used doesn't exceed the absolute limit value for the ADC when 5 volts is produced from the Allegro chip. Also, choose resistor values that are low enough in value not to exceed the limit imposed by specifications relating to the ADC used (typically less than 10 kohm but will vary with different ADC manufacturer).

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  • \$\begingroup\$ Not sine wave actually. PWM. The 5A I was referring to are already RMS. I'll ask again if a voltage divider isn't best in this situation. Thanks! \$\endgroup\$
    – Marcelo Queirós
    Jul 9, 2018 at 22:52
  • \$\begingroup\$ Well you have to ascertain what the peak current signal is for your RMS of 5 amps and ensure the potential divider accommodates it. I’d still use a pot divider though. \$\endgroup\$
    – Andy aka
    Jul 9, 2018 at 23:34
  • \$\begingroup\$ Peak will maybe surpass it, I'm not sure. However start-up current will surely surpass it.. \$\endgroup\$
    – Marcelo Queirós
    Jul 10, 2018 at 10:03
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OpAmp VS Voltage Divider VS Buck Converter

How about "none of the above" approach?

The IC in your link is available in four different sensitivity ranges. For example ACS723LLCTR-40AU is 100 mV/A device. With 5A maximum it outputs only 0.5 V, which can be sent directly to MCU ADC.

For AC current ACS723LLCTR-40AB has 50 mV/A sensitivity and 0.5 Vcc offset, resulting in 2.25-2.75 V output also compatible with ADC.

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  • \$\begingroup\$ The professor wanted the one with the highest sensitivity. We'll be using the 20A uni-direccional one. \$\endgroup\$
    – Marcelo Queirós
    Jul 9, 2018 at 22:48
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    \$\begingroup\$ Ah, I see. The point is not to make simple working circuit but to make you suffer. Interesting approach to education. Just not make it a habit when you start designing your own devices. Simpler is always better. \$\endgroup\$
    – Maple
    Jul 9, 2018 at 23:40
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I'm also who think, a voltage divider be enough and an op.amp as follower for impedance transformation and try to find a uC which has a built in reference (FVR) what usually make easy to calculate with the converted value and for accuracy. Furthermore, I think make no harm to attach a clamp diode at the input of the follower.

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