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?

On Fri, 15 Feb 2008 19:43:17 -0600, George wrote:

> 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?

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.

"Stephen J. Rush" <sjrush@comcast.net> schreef in bericht
news:a8qdnebOY5Q72yvanZ2dnUVZ_vHinZ2d@comcast.com...
> On Fri, 15 Feb 2008 19:43:17 -0600, George wrote:
>
>> 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?
>
> 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

In article <Vcrtj.801$QC.294@newsfe20.lga>,
George <gh424NO824SPAM@cox.net> wrote:

> 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?

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).

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

On Fri, 15 Feb 2008 19:43:17 -0600, George <gh424NO824SPAM@cox.net> wrote:

: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.
http://www.cispl.com.au/centameter/centameter.cfm

Stephen J. Rush says...

> 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.

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.

George wrote:

> 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.

--
Regards,

John Popelish

petrus bitbyter says...

> 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.

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.

Ecnerwal says...

> 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).

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

But I wonder how that device works.

"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.
> http://www.cispl.com.au/centameter/centameter.cfm


** 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

On Sat, 16 Feb 2008 15:41:19 +1100, "Phil Allison" <philallison@tpg.com.au>
wrote:

:
:"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.
:> http://www.cispl.com.au/centameter/centameter.cfm
:
:
:** 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"
:>
> :> 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.
> :> http://www.cispl.com.au/centameter/centameter.cfm
> :
> :
> :** 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.
> :
> :
>
> Ok Phil... you are correct.


** Of course.

> 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.


** Makes it 10% useless.

Has no way to monitor an individual appliance.


> It's still a great device for people interested in finding out more info
> on
> their power consumption though.


** The bill tells you that.


> 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.




........ Phil

On Sat, 16 Feb 2008 15:56:10 +1100, "Phil Allison" <philallison@tpg.com.au>
wrote:

:
:
:
:> 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"

> :
> :> 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...


** Yawn.....





> :> 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...


** I see only a series of dubious presumptions.

Hiding a giant red herring.



....... Phil

On Feb 16, 3:30 pm, George <gh424NO824S...@cox.net> wrote:
> Ecnerwal says...
>
> > 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).
>
> That would violate the Prime Directive, and take all the fun
> out of it.

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:
http://www.altronics.com.au/index.asp?a ... m&id=K4600
http://www.siliconchip.com.au/cms/A_102045/article.html

> But I wonder how that device works.

See the above silicon chip article.

Dave.

George wrote:

> 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.

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

George wrote:

> But I don't quite know how to calibrate it using rectifiers.

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

Graham

John Popelish says...

>> 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.

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.

George wrote:

> And does the electric company's meter on the side of my
> house measure things correctly?

Yes.

David L. Jones says...

>> > 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).

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

> 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:

> http://www.altronics.com.au/index.asp?a ... m&id=K4600
> http://www.siliconchip.com.au/cms/A_102045/article.html

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?

> =A0Phil Allison > =A0 =A0 Hiding a giant red herring.


In his pants.

George wrote:
> John Popelish says...
(snip)

>>... That
>> takes a true watt meter that averages the instantaneous
>> product of voltage and current.
>
> 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, the watt hour meter on your house does a quite accurate
job of totaling the actual energy you use from your power
service. The old ones have a motor whose drove torque is
produced by the instantaneous multiplication of voltage and
current, while it is braked in proportion to its rotational
speed. It may not handle microsecond pulses, but it does
take phase shift and non sinusoidal waveforms into account.

More modern ones have microprocessors that total the product
of high speed samples of voltage and current.

Neither will charge you for inductive or capacitive current,
nor miss the energy a rectifier and capacitive filter draw
from the service at the peaks of the voltage waveform.

--
Regards,

John Popelish

On Feb 17, 2:58 am, George <gh424NO824S...@cox.net> wrote:
> David L. Jones says...
>
> >> > 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).
>
> >> That would violate the Prime Directive, and take all
> >> the fun out of it.
>
> > 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:
>
> >http://www.altronics.com.au/index.asp?area=item&id=K4600
> >http://www.siliconchip.com.au/cms/A_102045/article.html
>
> 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.

You've missed several important points:
- Doing it manually can be dangerous.
- Doing it manually takes a lot more time and effort.
- Doing it manually has many traps
- The Kill-A-Watt gives you a LOT more useful info that is hard to get
with just manual measurements.

How much energy have you wasted discussing this on this forum?

> 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.

Geeze, just buy the Kill-A-Watt meter, they are under $20:
http://www.google.com/products?q=kill-a ... &scoring=p

> 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"?

I believe some others have discussed that already.

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

Possibly not. Manual measurements may have a high degree of both
absolute and relative error. Unless you know all the factors involved
and understand exactly what's happening, you'll probably get caught
out. For under $20 you can do it correctly and get a LOT more useful
info safely.

Dave.

"George" <gh424NO824SPAM@cox.net> wrote in message
news:Vcrtj.801$QC.294@newsfe20.lga...
> 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?
>
>


Bob Pease has a good article about how a watt meter works, and a working
schematic.

http://electronicdesign.com/Articles/In ... cleID=2190

OTOH, I concur with the folks who say a Kill-A-Watt is the way to go. Much
more reliable, less dangerous, and quite fun.

Regarding how it works, all you need to do is to get a very low value shunt
resistor, and then multiply the voltage across the resistor with the voltage
across the device in series (in real time). That gives you instantaneous
power. You then integrate that to get the result over time, which is of
course energy. You need to scale the result to get it in kW hours. His meter
given in the link above does it with an analog circuit, which is pretty
cool. One can also do the math with a microcontroller, which is probably how
the Kill-A-Watt does it.

Electricity is dangerous, of course, so if you do this, keep one hand in
your back pocket at all times when you are near any live wires. The thing
that kills is AC across your heart, which can induce fibrillation.

Regards,
Bob Monsen

On Sat, 16 Feb 2008 09:58:59 -0600, George <gh424NO824SPAM@cox.net>
wrote:

>David L. Jones says...
>
> >> > 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).
>
> >> That would violate the Prime Directive, and take all
> >> the fun out of it.
>
> > 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:
>
> > http://www.altronics.com.au/index.asp?a ... m&id=K4600
> > http://www.siliconchip.com.au/cms/A_102045/article.html
>
>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.

---
Try one of these and your multimeter:

http://www.analog.com/en/subCat/0,2879, ... 5F,00.html

First, use one to measure the RMS current into the DUT, then the RMS
voltage across it, then multiply the two readings and you'll have
the power your DUT is dissipating.
---

>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 you don't know the impedance the device presents to the mains,
then the information you get from your measurements will be largely
irrelevant.
---

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

---
Not unless the devices' inputs were identical in terms of the
impedance they presented to the mains.

--
JF