arduino current sensor


There are, I think, three types. This is the 5 amp there’s, also a 20 amp and a 30 amp type. And now, if you want to buy this chip on eBay, just the chip you’re looking at about four or five pounds, but due to the wonders of Chinese economics, if you want it mounted on this board, with all the connectors in place, you’re looking at about 2 Pound 50. Now one gripe I do have with this type of terminal block is that when you try and turn the screws, the whole block twists on the board and that’s, because there are no locating plastic lugs and it just turns on the the metal work that’s. Underneath but fortunately, the supplier of this Court has very kindly not soldered the connector in so I think I’m going to use it like this, where I’ll just put ring terminals directly onto the holes in the circuit board. Now my plan is to mount the current sensor. Sort of here on my solar panel voltage and current test rig, the voltage sensor is already mounted there. The microcontroller is here with the display and I’m going to use it to measure solar panel voltage and current now. The conventional way to measure current and it’s used in this cheap nasty multimeter on the tenant range is to measure the voltage across a low value resistor and the resistors here it’s. This thick piece of copper wire that spans the common connection and the 10 amp socket, and the voltage across a thick piece of wire, like that with an extreme Nilo resistance, is going to be very low.

So the chip up here will have an amplifier to increase that voltage and make it measurable, and the same technique is used with this. On this DC power supply. You can see there is a low value, resistor it’s, a 0.01 ohms and the current will flow through it. The voltage across it will be measured, but of course, that voltage is very small, so here down there there’s an op amp and that scales the voltage up, multiplies it up, but of course, that’s, subject to interference by noise temperature drift and other problems like that. So I now have the Hall effect current sensor attached to my circuit. Here it is, and this module will have solar panel on the left hand, side and a battery on the right hand, side and the current needs to be measured on the high side of the circuit. So I don’t want to be measuring current in the ground line. Now this chip works a little bit like an opto isolator, in the sense that the current sense element on the left hand, side is electrically isolated completely from the measuring device. On this right hand, side – and you can have something like 1.2 kilovolts between the two sides and the chip will still function completely. Normally it isn’t opto isolation in this case it’s magnetic isolation, there’s a magnetic field surrounding the piece of wire when current flows, through here and there’s, a Hall effect sensor on this side of the chip which detects that current and through a little bit of integral electronics, Provides a signal on these pins, that is between 0 and 5 volts, and you can ground or have to ground this ground, pin here to system, ground and that’s perfectly OK, it doesn’t interfere with the high side measurement of current right, so I’ve put 5 volts onto The VCC and ground pins on the module there and I’ve got the voltmeter connected to the center pin measuring volts coming out of the sensor now.

The way this works is that it’s bi directional so currently there’s no current flowing through the yellow and red wires. So the voltage coming out sits at the midpoint between VCC at 5, volts and ground at 0. So you can see that it’s 2 and a half volts now without putting a current through it I’m just going to disturb the magnetic field using a magnet attached to a screwdriver, and you can see that if I get the magnet really near the chip, the voltage Swings right down almost to zero. I spin the magnet round and bring it in again. Then the voltage shoots up so you do have to be a little bit careful of magnetic fields because they will disturb the operation of this device. And now I’ve got some current flowing through the current sensor: it’s a bit of a lash up it’s, just a 12 volt bulb and I’ve got 12 volts coming round the circuit and it’s reading 2.1 9 volts. Now that doesn’t tell you anything, but you have to know that this current sensor has a scale of 185 million bucks per amp. So all you need to do is take the original 2.5 volts subtract. The 2 point 1 9 volts that is reading now and then divide by point 1, 8 5, which will finally give us amps. So there are 1 point 6 7 amps flowing through this circuit. Now. That might seem an extremely difficult way to read current and, of course, it is by just putting a DVM on the output of the current sensor module.

But of course, this sense module is designed to work with a micro computer microcontroller like the Arduino and so in part 2 of this video.


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Arduino monitor tegangan dan arus (voltage and current) AC KWH Daya FREQUENCY sensor PZEM-004T V3.0
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#arduino Looking to measure current as part of a personal project but not sure what current sensor to buy










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Comment (28)

  1. hello sir !
    i am measuring power factor through ATmega8 micro controller..
    i used “zero cross detection” technique to find power factor..
    but the output wave of this “ACS712” sensor does not touch zero (ground level) ,
    it give me output wave form from “2.5” to “5”.
    i want to get its output symetrical across zero… what i need to do… ?
    plz give me sugeestion to solve this problem..?

    1. Delving into the world of split rail (or dual rail) power supplies is very tricky. You could make a power supply that outputs negative voltage and make amplifiers that operate from the positive to the negative rail to remove the DC offset of this device. And then make protection circuits because the ATmega8 can’t withstand negative voltages. That would give you your plus and minus signal centered about ground. Or you could make a rail to rail amplifier circuit that removes the DC offset, in which case the negative half of the signal would simply be lost. 

      My suggestion is to do it in software. There are two possible ways to do this. If you know that your signal is symmetric (i.e. +5 to -5) then you can simply record the max and min values in software and use the point halfway between them as your reference (3.75 in your example). If the signal you expect is not symmetrical, then you can calibrate your software by measuring the signal with no current flowing, and using that value as your reference.  All of these methods above will require physical or software calibration anyway, so I personally would’t bother making any additional circuit if you’re going to have to write the software anyway. Just do it all in software.  Let me know if that helps.

  2. Why do you want to only measure current on the high side? The amount of current flowing directly out of the positive end should be the same flowing into the negative end (reverse if you’re talking about electrons), so measuring current at both ends should work the same, no?

    1. The reason is not because the amount of current will change, it’s because of the use of common grounds. This chip has a nominal 1.2 mOhm internal resistance. If you put 20 amps through it, it will drop 24 mV. There are some situations where you need the ground potential of your device being powered to be exactly equal to the ground potential of your power supply. If you put this device on the high side, then the ground of your power supply and the ground of  your load stay the same, and it makes everything much easier as you no longer have to account for that 24 mV difference.  For example if you also need to measure the voltage across your load, putting this device on the low side would required taking voltage measurements on both sides of the load, and subtracting. If you put this device on the high side, you only have to measure the voltage at the positive rail of your load, because you know that the ground of your voltage measurement is the same potential as the ground of your load.

    2. gizmoguyar Awesome, thank you! If you have no need for a common ground between components (such as no serial communication between them) then there should be no issue with placing this on the low side, right? Although I suppose placing this on the high side would be a best practice regardless.

    3. Yes, that’s absolutely right. It would be fine to place it on the low side if you want. Although since it doesn’t change anything, I would place it on the high side anyway as you said.

    1. Yes, an 11 watt bulb should draw about 50 mA, so you won’t get very high resolution. It should output an AC voltage of 26 mV peak to peak, centered around 2.5 volts. That’s going to be quite hard to read, so you probably would be better with something else but it shouldn’t be dangerous. Please do be super careful when messing with high voltage! That Sh!t will kill you faster than a rogue US navy drone.

  3. Hi, may I know is there any current transformer/sensor to measure 240V AC and able to interface with arduino? This will helps a lot….thanks

    1. Hey..Jeffrey Ho did u get the solution for this..i am also facing same problem..tell me if u got d ans..ty in advance

  4. I think with all those ring terminals you’d be better with a soldered in low value resistor! But I can see that the purpose is to use it as a test rig and or demo. Just out of interest stick your Dmm set to mV across the 2 terminal sockets that you fitted to the end of the wooden test board when you have a high current flowing. Be interesting to see what drop you get.
    Great demo and thanks for sharing!

  5. Please suggest me to measure th DC voltage upto 500V and Current of 20A I want to make power meter for my solar panals whcih are capable of giving volage upto 300V and current upto 5A please suggest me the sensors

  6. Julian, you stated the isolationvoltage of the acs712 is 1200V. According to the specs of Allegro it is 2.1 kV RMS (60Hz) for the duration of max. 1 minute. Did you test the 1200V as a continuous DC voltage during, lets say, 1 day?


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