Can I measure AC vampire current this way?

Measuring the drop across a series resistor is the standard way to measure current with a digital meter. To get accuracy, you need a precision resistor. One ohm is a good starting value for low currents; it gives one millivolt per milliamp. If the device draws an amp, the resistor will have to dissipate one watt. It would be a *really* good idea to put the resistor in an insulated box with a fuse and insulated jacks for your meter probes. The other (Safer!) way is a clamp-around current probe and a line splitter, but a 1-ohm, 1-watt, 1-percent resistor, plus the fuse, jacks, and a little plastic box, will cost less.

There is one disadvantage though. That AC-voltage meters are calibrated for sinusoidal waveforms. Which is about right for most commmen voltage measurements. Currents however tend to be far from sinusoidal, especially when the load is equipped with rectifiers. So you can get an idea of the current involved but the measured value may be far from accurate. Accurate measuments would require a true RMS meter. Which is an expensive piece of equipment.

petrus bitbyter

Drop $20-30 on a Kill-A-Watt. It's inexpensive, and very good at what it does, which is what you are trying to do (and a few more things).

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Cats, coffee, chocolate...vices to live by

:I'd like to measure how much current various plugged-in devices draw :when they are "off". I have a digital multimeter that will measure :AC voltage. So I thought maybe I could place a small-value resistor :in series in one of the AC lines, and measure the voltage drop :across it. And then just calculate the current. But I don't know :if the results would be at all accurate. : :Has anyone tried that? Is there a better way? :

There certainly is a better way, although it will cost you. The Cent-A-Meter was designed and developed in Australia to do exactly what you require. It is suitable for all AC supply voltages and both 50/60Hz frequency. Cost calculations can be set to $, UKP and Euro's.

Thanks very much. Since I'm only going to be measuring phantom current, which I assume would be a few milliamps of AC, I think it should work ok so long as the readings are valid.

They are valid in the sense that they provide an indication, but not an accurate measurement of power being consumed. If the current is passed in narrow spikes (as it is with rectified supplies) or is phase shifted , as it is with reactive loads, then it does not indicate exactly what power is being consumed. That takes a true watt meter that averages the instantaneous product of voltage and current.

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Regards,

John Popelish

Well, I think in most cases the load will be a switching power supply, which would involve rectifiers. Probably not very sinusoidal.

If it was a resistive load, I could calibrate it with another precision resistor (a high-valoue one) as the load. But I don't quite know how to calibrate it using rectifiers.

That would violate the Prime Directive, and take all the fun out of it.

But I wonder how that device works.

"Ross Herbert"

** In no way shape or form will it do that. ** Does not connect to INDIVIDUAL appliances.

Says 50 Watt minimum - so it is TOTALLY useless.

Also says it MUST be installed by a licensed electrician.

...... Phil

: :"Ross Herbert" :>

:> There certainly is a better way, although it will cost you. The :> Cent-A-Meter was :> designed and developed in Australia to do exactly what you require. : : :** In no way shape or form will it do that. : : :> It is :> suitable for all AC supply voltages and both 50/60Hz frequency. Cost :> calculations can be set to $, UKP and Euro's. :>

: : :** Does not connect to INDIVIDUAL appliances. : :Says 50 Watt minimum - so it is TOTALLY useless. : :Also says it MUST be installed by a licensed electrician. : : : : :...... Phil :

Ok Phil... you are correct. I did err in regard to the very much lower currents involved in the OP's request and the Cent-A-Meter's minimum power measuring spec.

It's still a great device for people interested in finding out more info on their power consumption though. Both you and I know that the average electronics enthusiast who can read circuits and build stuff knows how to clip a current sensor around the phase wires in the meter box and wouldn't bother getting a licensed sparky in. Would you get a sparky in to install it for you?

"Ross Herbert" "Phil Allison" :>

** Of course.

** Makes it 10% useless.

Has no way to monitor an individual appliance.

** The bill tells you that.

** You presume far too much.

....... Phil

: : : :> It's still a great device for people interested in finding out more info :> on :> their power consumption though. : : :** The bill tells you that.

Yes, but well and truly after the event... A paper bill doesn't allow real-time monitoring of consumption.

: : :> Both you and I know that the average electronics :> enthusiast who can read circuits and build stuff knows how to clip a :> current :> sensor around the phase wires in the meter box and wouldn't bother getting :> a :> licensed sparky in. : : :** You presume far too much. :

And you didn't answer the question...

"Ross Herbert" "Phil Allison"

** Yawn.....

** I see only a series of dubious presumptions.

Hiding a giant red herring.

...... Phil

Yeah, but it is the right tool for the job. Cheap, accurate, safe, and feature rich. Get one, seriously.

If you really want to DIY then you can get an energy meter kit:

See the above silicon chip article.

Dave.

Unless you use an RMS reading meter the result will very likely be somewahat inaccurate due to the non-sinusoidal current likely being drawn.

However there is another point. It will NOT tell you the WATTS being used which is surely what you want to know ? For non-sinusoidal or phase-shifted waveforms the watts are not equal to volts times amps.

Graham

You can't as the waveforms may vary.You need a true RMS meter for this.

Graham

And does the electric company's meter on the side of my house measure things correctly? Aside from various "green" considerations, I think my meter should read the same as theirs, whether it's technically right or wrong.

Yes.

Well, the problem is that the cost of the kit is so high that it would take a lifetime to recover the cost from whatever savings I could get from it. Furthermore, the cost to the planet in terms of raw materials and energy to manufacture the device also exceeds any related benefit I might get from it in measuring low-power devices. It also looks like I have to pay $8 even to read the rest of the Siliconchip article. :-)

The Kill-A-Watt is certainly more reasonable. I just wondered if there's a home-brew way to get similar information about power consumption for esentially no money.

Let me try to get back to the original idea. Using my multimeter and other things I already have, which are free, is there any measurement I can make that would give me *any* useful information about the power consumed by electronic devices which are turned "off"?

If the measurements wouldn't be correct in absolute terms, would they at least be relatively correct in comparing several devices?