"Tweetldee" wrote in message
...
"Robert Monsen" wrote in message
newsEkgb.508681$Oz4.358896@rwcrnsc54...
I have a manutech current sense transformer I picked up from a junk box
buy,
and I've been trying to build a simple kill-a-watt meter. Here is my
schematic:

3 to 2000
| |
| | -----+----------------
+----| | | | | ^
| C|| | C| .-. |
AC C|| | C| | | 100 ohms V Sense
| C|| | C| | | |
+----| | | | '-' |
| | | | v
| | ------+----------------
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

Now, this seems to approximately work for the three devices I've tried
it
on:

100W light bulb, Vsense = 0.120V, computed power = 94W
25W light bulb, Vsense = 0.028V, computed power = 22W
AC Motor , Vsense = 0.220V, computed power = 167W

However, the question arises whether this will work for appliances.

1) I have a simple A/C meter. Does that do a passable job in this
circumstance?

2) What about power factor? Since I'm residential, I know I'm paying for
'real power'. Does this setup measure what I'm paying for?

3) This is a pretty small transformer. I have no specs for it. It seems
like
the power through the thing is miniscule. However, I'm slightly worried
about subjecting my refrigerator to this device. Any thoughts on that?

4) I could do some fancy software on a PIC to try and integrate the
waveform
in order to get a better approximation of real power in the cases where
the
waveform has been mashed by appliances. Is it worth it? Will the shape
of
the waveform be preserved through the transformer? Do you know the
percentage difference between this and just using power = Vpeak^2 / 2R?
(which is what I'm guessing my A/C meter does.)

with thanks,
Bob Monsen


Bob,
What you have there is just a current sensor.. only the beginnings of a
wattmeter. You need a bit more circuitry to get a wattmeter. Consider
that
Watts = Volts * Amps. You only have amps (actually, a scaled
representation
of amps) from your circuit.

Thanks for the help. I guess I was assuming the voltage was line voltage,
ie, around 115V. That makes it possible to determine power using only a
current sensor. An essential part of the posted 'wattmeter' was my
calculator...

However, you are right that a better way to determine the voltage across the
device makes sense. Another voltage sensor would be one way, say a small
power transformer. Using that, I can determine the voltage across the line
directly, and use these to generate the display.

Regards,
Bob Monsen