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Topic: Using CST-1020 (Read 3834 times) previous topic - next topic

Nunov

I found the error.
The datasheet is right. I have to divide the voltage by 0.98 for a burden resistor of 100 ohm.
My mistake was calculating the current for the load. I was adding a sqrt(3) on a monophasic load..  :smiley-red:


mauried

What sort of accuracy are you expecting with this energy meter?
Current only sensing energy meters can be highly inaccurate as they dont measure the voltage.
So with loads that have a power factor of less than 1 , the meter will read incorrectly.
Loads that have switch mode power supplies in them , like computers or plug packs will read even more
innacurately, due to the current waveform not being sinusoidal.

Nunov


What sort of accuracy are you expecting with this energy meter?
Current only sensing energy meters can be highly inaccurate as they dont measure the voltage.
So with loads that have a power factor of less than 1 , the meter will read incorrectly.
Loads that have switch mode power supplies in them , like computers or plug packs will read even more
innacurately, due to the current waveform not being sinusoidal.


I'm starting by the current. The next step is to measure the voltage. How do you recomend to measure the voltage?

mauried

If you really want to measure energy accurately, you are far better off using a device specifically
for this purpose like a ADE5569.
http://www.analog.com/en/analog-to-digital-converters/energy-measurement/ade5569/products/product.html

It does all the hard work for you.

RIDDICK

for the voltage i saw a transformer here, i think...
http://arduino.cc/forum/index.php/topic,93779.0.html
http://openenergymonitor.org/emon/
-Arne

Nunov


If you really want to measure energy accurately, you are far better off using a device specifically
for this purpose like a ADE5569.
http://www.analog.com/en/analog-to-digital-converters/energy-measurement/ade5569/products/product.html

It does all the hard work for you.


Hi.
Thanks for your suggestion.
I got and ADE7763 ( http://www.analog.com/static/imported-files/data_sheets/ADE7763.pdf ) to help me out to compute the power and energy, but this chip is supposed to receive a voltage between 0.5v and -0.5v.
How am I supposed to reduced a 230V RMS to 0.5 V? Using a voltage divisor? Am I getting any good resolution with such low voltage?

RIDDICK

#22
Oct 21, 2012, 09:37 pm Last Edit: Oct 23, 2012, 08:40 am by RIDDICK Reason: 1
welcome back Nuno!  :)

According to the datasheet (page 12 of 56, figure 22) a voltage divider is a good idea...
but use sufficiently big resistors, so that u dont waste more energy than necessary...
they say 600kR+1kR is good for 110V...
for 240V u can use 1MR+1kR, i guess...
maybe u need to adapt the capacitor?
maybe u can just omit the capacitor (just in case ur load distorts the voltage...)?  ;)

i dont know why they use transformers so often in this forum...
a voltage divider seems to more natural and easier...
maybe it is about isolation... if u use a voltage divider the meter has to share the ground/neutral line...
-Arne

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