Is this correct?

Hi,

I am trying to measure the current from my electrical circuit using Arduino Mega.I made up this schematic,but can someone please tell me if this is correct?
(I know that initally I’ll measure the voltage.I’ll get the value of the voltage and apply Ohm’s law to calculate the current. I = U/R)

I made it in paint,since I’m not home and didn’t had any other software products on my hand.I do apologise for my drawing skills.

Thank you!

Some details:

My electrical circuit is powered from a 230 Volts AC,it has multiple consumers.I have a digital multimeter/Voltmeter,a shunt resistor or instead of it I could use multiple resistors conected in series.
One wire goes to Arduino gnd and the other one goes to one of it’s analog input pin.

schem.jpg

Wow, that doesn't make a lot of sense. You are showing a connection across a voltmeter but voltmeters are not set up that way, they take a measurement across a voltage drop, it isn't like one probe is an input and the other is the output. Is that actually an ammeter? Why is there a voltmeter in this schematic as if it is a drop-in component? A voltmeter isn't a component in this way. In fact, the idea is to use the Arduino itself as the voltmeter, which it is fairly good at (1 part per 1024). What's up with the shunt? If you plan to measure current by measuring voltage first and applying Ohm's Law you place the shunt in parallel with the load and measure the voltage drop across the shunt. That shunt is in series and it won't change the voltage in any meaningful way, it's still 230V on the other side of the shunt. And in the end, have you noticed that you are actually connecting 230V directly to your Arduino, only through a low impedance ammeter (I guess) and a low impedance shunt resistor? I assume you have not actually done this yet because the results will be hilarious if you try that. I don't recommend it.

The big problem here is even when you get this all right, you will be measuring AC on your Arduino so you will have to take a lot of fast measurements and average it out.

Want to spend ten bucks and save yourself a laboratory fire or worse? This is a safer solution and looks pretty professional in the end:

http://openenergymonitor.org/emon/node/58

Here is another interesting product, though it is hard for AC because the voltage it throws off is going to be a sinusodal wave:

Works great for DC though, personal experience.

I think I'll be using this current sensor Non-Invasive Current Sensor - 30A - SEN-11005 - SparkFun Electronics
But how do I wire it and connect it to arduino.A schematic would be fantastic.
And can I connect it dirrectly to my circuit(which is powered from a 230 V source)
Thank you

I think I'll be using this current sensor Non-Invasive Current Sensor - 30A - SEN-11005 - SparkFun Electronics

What size of current will you be measuring, while it measures up to 30A it is not very good at lower currents.

But how do I wire it and connect it to arduino

Look at the data sheet. Connect a 10R resistor across the output.
You connect one end say black to ground and the other end to the anode of a diode. Connect the cathode to the analogue input. Then connect a capacitor from the analogue input to ground.
This will measure the peak voltage which is 1.414 times the RMS voltage.

What current do you expect.
Look at the graph. The output of that current transformer is ~50mV@10A.
10A is already quite a lot for a 230v device. That's e.g. a 2300W heater...
A diode rectifier will be useless. Diodes need ~500mV to start conducting.

I'm affraid this small signal has to be amplified first with an opamp, to give a usefull resolution.

Another way to increase the output voltage is to have more turns through the hole, but is looks quite small.
Leo..

So what do you guys think about this schematic.Can I measure my current with it?

So what do you guys think about this schematic

Not a lot.

The 741 is a very old amplifier and you can't run it from 5V. You need a split supply and this will take the output higher than the Arduino's input can stand. That output capacitor is a stupid value.
Why the diodes in the Op-amp's output.

Hi,
The cap means that the output is Peak Current, the diodes and op-amp approximate a perfect diode for the rectification.

Tom...... :slight_smile:

A cheap LM324 would do better there. They even run on 3volt, and have a common mode that includes ground.
Output swing goes not high enough (3.5v), but that could be fixed by using A-in with 1.1v Aref.
Leo..

Hi Grumpy_Mike,

I made up this schematic,and in theory I should be able to measure the current using this sensor (Non-Invasive Current Sensor - 30A - SEN-11005 - SparkFun Electronics) connected to my arduino board.Can you please check it out and tell me if it is okay?I made it up acording to your indications.

Grumpy_Mike:
Not a lot.

The 741 is a very old amplifier and you can’t run it from 5V. You need a split supply and this will take the output higher than the Arduino’s input can stand. That output capacitor is a stupid value.
Why the diodes in the Op-amp’s output.

Well it will give something but it will not be very sensitive. Have you not read or understood the replies you have been getting?

Grumpy_Mike:
Well it will give something but it will not be very sensitive. Have you not read or understood the replies you have been getting?

To be honest I haven't fully understand them.How the circuit should look?

I have a 741 in my ‘museum’ with a 7024 datecode!

The arduino can only detect voltages in steps of 5 / 1024 Volts, that is 4.88mV per step.
So for a mains load of 30A you get 15mA of current, that is in the data sheet.

If you put 15mA of current through 10R you get a voltage of 0.15V. That means an analogue reading of about 30 on the Arduino. So it is not very sensitive. You get an increase of 1 in the reading per amp.

So to get any meaningful measurements you have to amplify that 0.15V to 5V. That means a gain of 5 / 0.15 which is a gain of about 33. You can do this with one op amp but not a 741. You need to get a rail to rail op amp.

Would this be good for my project? http://www.adafruit.com/products/808
I’m a programmer actually…

geniulmalefic:
Would this be good for my project? http://www.adafruit.com/products/808

Yes that looks like it will do.

If you plan to measure current by measuring voltage first and applying Ohm's Law you place the shunt in parallel with the load and measure the voltage drop across the shunt.

No, you put the shunt in series with the load. The load current has to flow through the shunt, for that method to work.