Let's build a power meter for five bucks. It's too bad Tech

[AC/DCdude17]

Haha why spend $100 on a true wattmeter when a $5 resistor shunt and a
cheapo DMM will do the samething?

http://www.techtv.com/screensavers/howto/story/0,24330,3403553,00.html

article mentions
http://www.silentpcreview.com/modules.php?op=modload&name=Sections&file=i
ndex&req=viewarticle&artid=19&page=1

It only cost $5, because it doesn't give out any meaningful result when
you're measuring anything more complex than Christmas lights and table
lamps.

Measure computer or refrigerator's power consumption? yeah right. It's
sad TechTV is supporting this non-sense.

Some people just don't get that things aren't as simple as basic P=I*V

First of all, they're totally ignoring power factor and when you're
dealing with a load with a PF of 0.5 to 0.6 and decide to ignore it, you
might as well not measure it.

You can't measure the current draw of most computer power supplies with a
cheap DMM either in A mode or V mode with a shunt. Except with a true
DMM multimeter, something cheapskate is unlikely to have, the readings
from measuring a peculiar current waveform of a computer power supply is
incorrect.


Test result for my CRT(which has a switch mode power supply just like a
PC and I'm not going to unplug my PC right now):

Line voltage is 123.3V

True power meter reports:
1.29A
PF 0.63
100W

Regular DMM reports:
0.67A
123.3*0.67A= 83W

Damn talk about way off.

 

In alt.engineering.electrical AC/DCdude17 <dude17@sactakethisoutbeemail.com> wrote:

| Test result for my CRT(which has a switch mode power supply just like a
| PC and I'm not going to unplug my PC right now):
|
| Line voltage is 123.3V
|
| True power meter reports:
| 1.29A
| PF 0.63
| 100W
|
| Regular DMM reports:
| 0.67A
| 123.3*0.67A= 83W
|
| Damn talk about way off.

Now calculate what the heat loss is in the wiring due to the current spikes.

--
-----------------------------------------------------------------------------
| Phil Howard KA9WGN | http://linuxhomepage.com/ http://ham.org/ |
| (first name) at ipal.net | http://phil.ipal.org/ http://ka9wgn.ham.org/ |
-----------------------------------------------------------------------------

 

AC/DCdude17 wrote:

Haha why spend $100 on a true wattmeter when a $5 resistor shunt and a
cheapo DMM will do the samething?

http://www.techtv.com/screensavers/howto/story/0,24330,3403553,00.html

article mentions
http://www.silentpcreview.com/modules.php?op=modload&name=Sections&file=i
ndex&req=viewarticle&artid=19&page=1

It only cost $5, because it doesn't give out any meaningful result when
you're measuring anything more complex than Christmas lights and table
lamps.

Measure computer or refrigerator's power consumption? yeah right. It's
sad TechTV is supporting this non-sense.

Some people just don't get that things aren't as simple as basic P=I*V

First of all, they're totally ignoring power factor and when you're
dealing with a load with a PF of 0.5 to 0.6 and decide to ignore it, you
might as well not measure it.

You can't measure the current draw of most computer power supplies with a
cheap DMM either in A mode or V mode with a shunt. Except with a true
DMM multimeter, something cheapskate is unlikely to have, the readings
from measuring a peculiar current waveform of a computer power supply is
incorrect.

Test result for my CRT(which has a switch mode power supply just like a
PC and I'm not going to unplug my PC right now):

Line voltage is 123.3V

True power meter reports:
1.29A
PF 0.63
100W

Regular DMM reports:
0.67A
123.3*0.67A= 83W

Damn talk about way off.

Never mind the inaccuracy - how about he safety
hazard? The jerk put the terminals in the hot lead.
They will both be about 120v with respect to ground.
He should have put them in the neutral side where
they would be only a few volts above ground,
worst case.

Next, he specified a 10 watt resistor. That's
probably fine if he uses it only on a PC. But
there's no disclaimer - someone who blindly
follows the article might plug his laser printer
in. Ooops! At 5 amps (or better) the resistor
will need to dissipate 25 watts. Or he may decide
to measure anything - a hair dryer for example,
drawing in excess of 10 amps. Ooops!! That poor
resistor may be trying to dissipate over 100
watts.

The right way to do this is to use a .1 ohm,
50 watt aluminum housed resistor, mounted to the
inside of a deep 4" metal junction box with a single
duplex receptacle. The line cord is not broken.
The neutral is soldered to the resistor inside the
J box, and the other side of the resistor has a
jumper soldered to it, which connects to the
neutral side of the receptacle. Add a couple of
pin jacks wired across the resistor, read the
voltage at the jacks, and move the decimal point
one place to the right to see the amperage. You
add only three things to the cost - a j-box and
cover, a receptacle and a fitting for the line cord
to enter the j box. (The pin jacks replace the screw
terminals.) Safe, simple, cheap, and effective.
Even a 1500 watt toaster won't be a problem - the
resistor will need to dissipate only about 16 watts
into the heatsink, which is the metal J box.

You still have the other limitations with a cheap
DMM - no true rms reading so it's not accurate
with switching supplies, won't measure Pf, but
it's one hell of a lot safer than the referenced
article. And for a few bucks you can't expect
true RMS or PF readings.

 

[bushbadee]

Some years ago I had an occaision to build a true power meter to measure RF
power
I invented a very simple device to do this.
I took a vsr meter and stripped the guts.
I took a photo sensitive resistor and placed it into a tight tube.
With in the same tube I placed a very small low power light bulb.
Now the resistance of the resistor controlled the current that flowed as
its resistance was quite high compared to the bulb.
In series with the resistor light bulb combination, I placed a 1.5 volt D
cell.
No on-off switch. The current drain with out power input was in the micro
amps..
Across this combination I placed a 50 ohm resistor to act as the RF load.
I then used a know DC to calibrate the power that the meter read.

The meter was calibrated to the know DC power in the power resistor with a
carefully measured DC input.
I used this combination to measure the RF power out of transmitters and it
tracked much more expensive power meters quite well.
The battery, even with out a switch lasted 2 to 3 years.




"AC/DCdude17" <dude17@sacTAKETHISOUTbeemail.com> wrote in message
news:MPG.1ad033e44189c0d69896ac@news.verizon.net...

Haha why spend $100 on a true wattmeter when a $5 resistor shunt and a
cheapo DMM will do the samething?

http://www.techtv.com/screensavers/howto/story/0,24330,3403553,00.html

article mentions
http://www.silentpcreview.com/modules.php?op=modload&name=Sections&file=i
ndex&req=viewarticle&artid=19&page=1

It only cost $5, because it doesn't give out any meaningful result when
you're measuring anything more complex than Christmas lights and table
lamps.

Measure computer or refrigerator's power consumption? yeah right. It's
sad TechTV is supporting this non-sense.

Some people just don't get that things aren't as simple as basic P=I*V

First of all, they're totally ignoring power factor and when you're
dealing with a load with a PF of 0.5 to 0.6 and decide to ignore it, you
might as well not measure it.

You can't measure the current draw of most computer power supplies with a
cheap DMM either in A mode or V mode with a shunt. Except with a true
DMM multimeter, something cheapskate is unlikely to have, the readings
from measuring a peculiar current waveform of a computer power supply is
incorrect.


Test result for my CRT(which has a switch mode power supply just like a
PC and I'm not going to unplug my PC right now):

Line voltage is 123.3V

True power meter reports:
1.29A
PF 0.63
100W

Regular DMM reports:
0.67A
123.3*0.67A= 83W

Damn talk about way off.

 

[Watson A.Name - \"Watt Su]

"AC/DCdude17" <dude17@sacTAKETHISOUTbeemail.com> wrote in message
news:MPG.1ad033e44189c0d69896ac@news.verizon.net...

Haha why spend $100 on a true wattmeter when a $5 resistor shunt and a
cheapo DMM will do the samething?

http://www.techtv.com/screensavers/howto/story/0,24330,3403553,00.html

article mentions

http://www.silentpcreview.com/modules.php?op=modload&name=Sections&file=

i

ndex&req=viewarticle&artid=19&page=1

It only cost $5, because it doesn't give out any meaningful result
when
you're measuring anything more complex than Christmas lights and table
lamps.

Measure computer or refrigerator's power consumption? yeah right.
It's
sad TechTV is supporting this non-sense.

Some people just don't get that things aren't as simple as basic P=I*V

First of all, they're totally ignoring power factor and when you're
dealing with a load with a PF of 0.5 to 0.6 and decide to ignore it,
you
might as well not measure it.

You can't measure the current draw of most computer power supplies
with a
cheap DMM either in A mode or V mode with a shunt. Except with a true
DMM multimeter, something cheapskate is unlikely to have, the readings
from measuring a peculiar current waveform of a computer power supply
is
incorrect.


Test result for my CRT(which has a switch mode power supply just like
a
PC and I'm not going to unplug my PC right now):

Line voltage is 123.3V

True power meter reports:
1.29A
PF 0.63
100W

Regular DMM reports:
0.67A
123.3*0.67A= 83W

Damn talk about way off.

I had the same problem when I tried to measure the current and power of
a $5 Intermatic timer, the one with the dial with the pegs on it on the
front. It measured more than 2.5W with the DMM, but it was probably
less than 1.8W when the PF was taken into consideration. The motor has
a lot of reactance.

 

[bushbadee]

My device does not care about power factor.
It uses a resistor which has a powerfactor of 1 as the load.
All I really was interested in was determining the true ouput of CB and
other radios.


"Watson A.Name - "Watt Sun, the Dark Remover"" <NOSPAM@dslextreme.com> wrote
in message news:106g8a5aqjbh79e@corp.supernews.com...

"AC/DCdude17" <dude17@sacTAKETHISOUTbeemail.com> wrote in message
news:MPG.1ad033e44189c0d69896ac@news.verizon.net...
Haha why spend $100 on a true wattmeter when a $5 resistor shunt and a
cheapo DMM will do the samething?

http://www.techtv.com/screensavers/howto/story/0,24330,3403553,00.html

article mentions

http://www.silentpcreview.com/modules.php?op=modload&name=Sections&file=
i
ndex&req=viewarticle&artid=19&page=1

It only cost $5, because it doesn't give out any meaningful result
when
you're measuring anything more complex than Christmas lights and table
lamps.

Measure computer or refrigerator's power consumption? yeah right.
It's
sad TechTV is supporting this non-sense.

Some people just don't get that things aren't as simple as basic P=I*V

First of all, they're totally ignoring power factor and when you're
dealing with a load with a PF of 0.5 to 0.6 and decide to ignore it,
you
might as well not measure it.

You can't measure the current draw of most computer power supplies
with a
cheap DMM either in A mode or V mode with a shunt. Except with a true
DMM multimeter, something cheapskate is unlikely to have, the readings
from measuring a peculiar current waveform of a computer power supply
is
incorrect.


Test result for my CRT(which has a switch mode power supply just like
a
PC and I'm not going to unplug my PC right now):

Line voltage is 123.3V

True power meter reports:
1.29A
PF 0.63
100W

Regular DMM reports:
0.67A
123.3*0.67A= 83W

Damn talk about way off.

I had the same problem when I tried to measure the current and power of
a $5 Intermatic timer, the one with the dial with the pegs on it on the
front. It measured more than 2.5W with the DMM, but it was probably
less than 1.8W when the PF was taken into consideration. The motor has
a lot of reactance.

 

[~^Johnny^~]

On Mon, 29 Mar 2004 05:15:30 -0800, "Watson A.Name - \"Watt Sun, the Dark
Remover\"" <NOSPAM@dslextreme.com> wrote:

"AC/DCdude17" <dude17@sacTAKETHISOUTbeemail.com> wrote in message
news:MPG.1ad033e44189c0d69896ac@news.verizon.net...
Haha why spend $100 on a true wattmeter when a $5 resistor shunt and a
cheapo DMM will do the samething?

http://www.techtv.com/screensavers/howto/story/0,24330,3403553,00.html

article mentions

http://www.silentpcreview.com/modules.php?op=modload&name=Sections&file=
i
ndex&req=viewarticle&artid=19&page=1

It only cost $5, because it doesn't give out any meaningful result
when
you're measuring anything more complex than Christmas lights and table
lamps.

Measure computer or refrigerator's power consumption? yeah right.
It's
sad TechTV is supporting this non-sense.

Some people just don't get that things aren't as simple as basic P=I*V

First of all, they're totally ignoring power factor and when you're
dealing with a load with a PF of 0.5 to 0.6 and decide to ignore it,
you
might as well not measure it.

You can't measure the current draw of most computer power supplies
with a
cheap DMM either in A mode or V mode with a shunt. Except with a true
DMM multimeter, something cheapskate is unlikely to have, the readings
from measuring a peculiar current waveform of a computer power supply
is
incorrect.


Test result for my CRT(which has a switch mode power supply just like
a
PC and I'm not going to unplug my PC right now):

Line voltage is 123.3V

True power meter reports:
1.29A
PF 0.63
100W

Regular DMM reports:
0.67A
123.3*0.67A= 83W

Damn talk about way off.

I had the same problem when I tried to measure the current and power of
a $5 Intermatic timer, the one with the dial with the pegs on it on the
front. It measured more than 2.5W with the DMM, but it was probably
less than 1.8W when the PF was taken into consideration. The motor has
a lot of reactance.

You could take a simultaneous voltage [Not the voltage drop across the series
(shunt) resistor, but the SOURCE voltage] reading, integrate the two
instantaneous values over a period of 2*pi radians, and do a vector sum.
That would give you true RMS power, no?


--
-john
wide-open at throttle dot info

~~~~~~~~
"The first step in intelligent tinkering is to
save all the parts." - Aldo Leopold
~~~~~~~~

 

~^Johnny^~ wrote:

On Mon, 29 Mar 2004 05:15:30 -0800, "Watson A.Name - \"Watt Sun, the Dark
Remover\"" <NOSPAM@dslextreme.com> wrote:


"AC/DCdude17" <dude17@sacTAKETHISOUTbeemail.com> wrote in message
news:MPG.1ad033e44189c0d69896ac@news.verizon.net...
Haha why spend $100 on a true wattmeter when a $5 resistor shunt and a
cheapo DMM will do the samething?

http://www.techtv.com/screensavers/howto/story/0,24330,3403553,00.html

article mentions

http://www.silentpcreview.com/modules.php?op=modload&name=Sections&file=
i
ndex&req=viewarticle&artid=19&page=1

It only cost $5, because it doesn't give out any meaningful result
when
you're measuring anything more complex than Christmas lights and table
lamps.

Measure computer or refrigerator's power consumption? yeah right.
It's
sad TechTV is supporting this non-sense.

Some people just don't get that things aren't as simple as basic P=I*V

First of all, they're totally ignoring power factor and when you're
dealing with a load with a PF of 0.5 to 0.6 and decide to ignore it,
you
might as well not measure it.

You can't measure the current draw of most computer power supplies
with a
cheap DMM either in A mode or V mode with a shunt. Except with a true
DMM multimeter, something cheapskate is unlikely to have, the readings
from measuring a peculiar current waveform of a computer power supply
is
incorrect.


Test result for my CRT(which has a switch mode power supply just like
a
PC and I'm not going to unplug my PC right now):

Line voltage is 123.3V

True power meter reports:
1.29A
PF 0.63
100W

Regular DMM reports:
0.67A
123.3*0.67A= 83W

Damn talk about way off.

I had the same problem when I tried to measure the current and power of
a $5 Intermatic timer, the one with the dial with the pegs on it on the
front. It measured more than 2.5W with the DMM, but it was probably
less than 1.8W when the PF was taken into consideration. The motor has
a lot of reactance.


You could take a simultaneous voltage [Not the voltage drop across the series
(shunt) resistor, but the SOURCE voltage] reading, integrate the two
instantaneous values over a period of 2*pi radians, and do a vector sum.
That would give you true RMS power, no?

--
-john
wide-open at throttle dot info

I'm guessing that you meant a simultaneous voltage
and current measurement. You seem to be proposing
an incorrect mathematical approach to a real world
issue. You can't integrate two instantaneous values
of two different things. Could you please re-state
this to make it clear.

My guess is you really mean integrate all V readings
from 0 to 2 pi radians, and also integrate the I readings
for the same 360 degrees. But that mathematical
solution is still incomplete - you then need to multiply.
Is that what you have in mind?

Now, if you mean this to be a practical solution,
what instrumentation do you buy to sample the V and
I, how many samples does it take during the cycle,
how are the measurements stored, and how do you
integrate them?

It's much be cheaper and easier to buy a true RMS DMM
and measure the V across a shunt resistor, then compute
the power. Hell, I've read even have a power measurement
device for about 40 bucks, which is even cheaper than a
true RMS DMM, and involves no computation. I don't
know if it's any good.

 

[KR Williams]

In article <9sp390dfki2tjffbh7bv3ln913gbqde62p@4ax.com>,
NOSPAM.Long.Live.Linda@MUNGED.OkBy.UsINVALID says...

On Mon, 29 Mar 2004 05:15:30 -0800, "Watson A.Name - \"Watt Sun, the Dark
Remover\"" <NOSPAM@dslextreme.com> wrote:


"AC/DCdude17" <dude17@sacTAKETHISOUTbeemail.com> wrote in message
news:MPG.1ad033e44189c0d69896ac@news.verizon.net...
Haha why spend $100 on a true wattmeter when a $5 resistor shunt and a
cheapo DMM will do the samething?

http://www.techtv.com/screensavers/howto/story/0,24330,3403553,00.html

article mentions

http://www.silentpcreview.com/modules.php?op=modload&name=Sections&file=
i
ndex&req=viewarticle&artid=19&page=1

It only cost $5, because it doesn't give out any meaningful result
when
you're measuring anything more complex than Christmas lights and table
lamps.

Measure computer or refrigerator's power consumption? yeah right.
It's
sad TechTV is supporting this non-sense.

Some people just don't get that things aren't as simple as basic P=I*V

First of all, they're totally ignoring power factor and when you're
dealing with a load with a PF of 0.5 to 0.6 and decide to ignore it,
you
might as well not measure it.

You can't measure the current draw of most computer power supplies
with a
cheap DMM either in A mode or V mode with a shunt. Except with a true
DMM multimeter, something cheapskate is unlikely to have, the readings
from measuring a peculiar current waveform of a computer power supply
is
incorrect.


Test result for my CRT(which has a switch mode power supply just like
a
PC and I'm not going to unplug my PC right now):

Line voltage is 123.3V

True power meter reports:
1.29A
PF 0.63
100W

Regular DMM reports:
0.67A
123.3*0.67A= 83W

Damn talk about way off.

I had the same problem when I tried to measure the current and power of
a $5 Intermatic timer, the one with the dial with the pegs on it on the
front. It measured more than 2.5W with the DMM, but it was probably
less than 1.8W when the PF was taken into consideration. The motor has
a lot of reactance.


You could take a simultaneous voltage [Not the voltage drop across the series
(shunt) resistor, but the SOURCE voltage] reading, integrate the two
instantaneous values over a period of 2*pi radians, and do a vector sum.
That would give you true RMS power, no?

Not vector sum, rather multiply the two, point by point and
average over the cycle. BTW, it's average power (RMS power makes
no sense). The vector sum would give you the power factor (or
more precisely lead/lag angle). Of course the power factor is
also the RMS(V) * RMS(I) / average power. Yes, this all is fairly
easy to do with a microcontroller and a couple of ADCs.

--
Keith