AC mains low current measurement

[caffeinatedbard]

I am trying to measure the current draw of AC solenoid valves.  They are only 14W rated for 120VAC.  The current draw is small so my question is what would be the best/simplest/best bang for the buck way to do this with?  There are beautiful ways to this for DC but for mains AC at low currents I have found very few ideas...

Here is what I have already tried...
-Hall effect sensor 0-5A from Allegro systems (Vnoise floor of the part is 20mV peak to peak) and response is about 180mV/A, so a 80mA is just about 15-20mV range making it impossible to get reliable readings.
-Current transformer, again same issue -- at 80mA of current, cheaper current tranformers don't give a big enough signal.  The one I tried was from CR Magnetics. 

I eagerly look forward to your suggestions!
bard

[Psi]

Best way to measure the current is to measure Vrms drop across a resistor. You can spec the resistor to produce a voltage drop which you can measure accurately.
However you have to be careful, since its mains voltage.

Since it's 120VAC at 116mA you can afford to lose say 1-2V, which is enough to run the IR LED inside an optoisolator.
So, if you used an optoisolator with a linear response you could make up a circuit were the LED light was proportional to current through the resistor.

It would get you a nice isolated way to measure the current.

[caffeinatedbard]

Psi, thanks for the very useful suggestion. I will definitely look into implementing this.

Do you have any parts you can recommend given that the quality of these parts could be affect the accuracy of the measurement?  Price is not a concern.

Thanks,
bard

[IanB]

I am trying to measure the current draw of AC solenoid valves.  They are only 14W rated for 120VAC.  The current draw is small so my question is what would be the best/simplest/best bang for the buck way to do this with?  There are beautiful ways to this for DC but for mains AC at low currents I have found very few ideas...

What is the significance of the "mains" AC in your question that you keep mentioning it? As far as I can see you are trying to measure ~100 mA AC, and that would seem to be within the normal operating range of an AC milliammeter (or the AC mA range of a multimeter).

So I'm missing something. Are you trying to measure the current in-situ where the devices are inaccessible, rather than on a test bench? Because otherwise you would just put a DMM in the circuit on AC mA and measure the current. What is the unstated part of your requirements?

[caffeinatedbard]

@IanB-  Significance of mains AC is just to state it's 120VAC as opposed to isolated through a transformer low voltage AC.
The second aspect is I want to use it with a microcontroller so isolation is important because I have other hardware already in place that I want to be careful to protect.  I understand what you are trying to say and to answer your question, the unstated requirement is that it's not going to be a quick measurement on a test bench with a DMM.

[SeanB]

Use a current transformer to do the isolation, and then either measure the secondary voltage across the load resistor directly or use a simple diode capacitor to rectify it. If you want dc with better accuracy then replace the diode with a rms converter chip.

[JVR]

100mA?

More than enough to measure via a CT.

1000 turn solid ferrite CT + ~220R will work wonders.

I've used the above with a 22R resistor, and a bit of code to measure from 5W -> 13.2Kw on the same CT with <1% accuracy.

[caffeinatedbard]

@JVR- I might have too cheap of a CT.  Can you tell me which one you used or can you recommend one?

Thanks,
bard

[SeanB]

1 old PC power supply and whip out the powdered iron toroid inside, and wind the secondary you need on it. A single turn for the primary and you are good.

I used an old burnt out lighting toroid ( cable short blew it up) as the core to make a simple CT, using a 12 turn 2.5mm wire secondary leading to a 5A industrial moving coil AC ammeter and a single wire from the mains supply as primary. Another has a 10 turn primary and a 1 turn secondary ( I needed a 6:60 ratio) and is connected to a kilowatthour meter just to see power use of the PC, currently 503.7 kWh used since installed 2 years ago. That uses the core out of a broken RCB unit. Nice permalloy tape winding in a pair of plastic cups.

[Psi]

Just keep in mind that CT's need a load resistor.

[caffeinatedbard]

@Psi--Thanks for the tip.  I did know that not having a burden resistor will cause unpleasant results.

[jeroen74]

If you decide to use an optocoupler, that in account that these are very temperature sensitive. They suck for linear applications (and sometimes digital applications too).

If it's only for a digital status indication (e.g. solenoid really draws current), it should work.

[jeroen74]

There are a few optocouplers like the IL300 that have an extra phototransistor on the LED side. With a simple opamp circuit you can compensate for the imperfections. But even just another optocoupler of the same type does a reasonable good job.

It's nice to see the circuit compensate the LED current when temperature changes with the net-effect the output stays rock-solid at the same output level.

[JVR]

@JVR- I might have too cheap of a CT.  Can you tell me which one you used or can you recommend one?

Thanks,
bard

Unpossible, the ones I used are ~$2ea, new from the factory. They are ZA made though, so somewhat useless to you.

Just get one that has the smallest possible aperture, and a hell of a lot of windings.

[JVR]

Just to add, I fed the output from the CT to its burger resistor, and that was connected to a resistor mid-supply on the board, with the other end directly to the MCU, bit of RMS calculations and some averaging and your golden