Can I measure AC vampire current this way?

[Ross Herbert]

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...

 

[Phil Allison]

"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

 

[Ross Herbert]

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?

 

[Phil Allison]

"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

 

[George]

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.

 

[George]

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.

 

[John Popelish]

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

 

[George]

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.

 

[Ross Herbert]

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

 

[Ecnerwal]

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

 

[petrus bitbyter]

"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

 

[Stephen J. Rush]

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.

 

[George]

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?