Sorry, I guess I mean the connector!

this was done

The 2.1kVrms insulation is for the ACS712 chip only. Since the manufacturer of this module has added some conductors between the inputs and the outputs of the chip, the actual insulation is less than that. In fact, this module does not meet the minimum security requirements in several countries, and should not be used in a 230Vac device.

thanks and noted

@@HoBrian Old Post but pertinent to some. This circuit would not function on AC current. The magnetic Phase Change would end up as ZERO. You would have nothing to measure. You would need to convert your voltage to DC and that would nullify your true values.

@@pulesjet ACS712 provides economical and precise solutions for AC or DC current sensing

Yes without any problem

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

It is the same ground for both, the 5V input that powers up the chip and the 5V output that gives the reading

For those who accustomed to the convenience of libraries for arduino: github.com/rkoptev/ACS712-arduino

+lajolo tenes que acondicionar tu señal para terner mayor sensiblidad el datasheet te da algunas opciones para eso

+lajolo If a lower bandwidth suits your application, you can use the WCS2702.

+Rachit Pradhan Thanks. I had a look at the datasheet and it looks interesting. Much more accurate (1 mV/mA) and low operating current (3 mA).

Try increasing the value of the filter capacitor

+Julian Ilett thanks for the quick response, I'm using a 474nf filter cap as per your suggestion in video 3

your signal could be dirty

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.

I'm guessing "voltage divider" ;)

kattejuice ..search on ZMPT 101B

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.

gizmoguyar thank you sir...

found it! thanks

so, can you tell the name of that voltage sensor plz?

Georgios Diamantis acs712. its in the video description

Neutral or phase no issues.

Try this library for RMS measurements: github.com/rkoptev/ACS712-arduino

Habib Ishtiaq Use a simple potential divider, but I can't help you with the calculations for AC power

Kishan Prudhvi Guddanti you can use this one www.ebay.com/itm/390902157627Voltage Transformer Module Active Single Phase Voltage Sensor Module AC output

The voltage depends on where you live, typically 110, 120 or 230 volts. You dont need a second chip as the voltage should stay the same.

+Habib Ishtiaq For power measurement, you need to integrate over the waveform because of phase shift between voltage and current waveforms. It is not a simple multiplication of v x I. You will need to do some study of power electronics to get the necessary knowledge of how to do it. For example, the current waveform my be in the form of distorted pulses whilst the voltage waveform is more of a sine wave.

yes

+Cesar Coronel it will do both :)

there is a type of AC/DC current transducer. You input the AC current, you get AC signal output.,and you input the DC current, get the DC signals. see g-sensor.net/signal-conditioner/Synchro-Tracking-Transducer//2018/0307/151.html

Habib Ishtiaq God, Indians and Arabs are always the ones demanding videos and schematics and code. Do your final project yourself.

i have done it by my self thanx for your very late reply...

+Habib Ishtiaq Did you try out this module with ac loads?When a current going thru the connector there is a voltage and no current no voltage at out??

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.

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.

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.

learn C

does not really answer the question tho I asked not just how I can do it but if I can do it.

Yes, it'll work for what you want. There's code in the video description. Add a simple if/then statement and you're done.

cheers thank you should be able to go fine from here then

Try this library: github.com/rkoptev/ACS712-arduino

problem is your voltage has to be under 8v... what if you want to monitor amps for a brushless dv 48v motor at 30a...?

@@artbyrobot1 I bought some of these and put them in a box years ago :) Max input internal power 5v- 8v MAX. current being measured has no potential, just current. 48v 30amps is redline. Use 2 or redo the math and use a shunt :) To anyone who might find this useful.

Judging by all none of the videos you've contributed to the community, you clearly have all the reason in the world to bitch...

He's measuring apples to oranges though. He should have compared the price of the chip in china too.