They are rated at 3v AC for a 200a load. I measured the signal, and it is 3v RMS, which means it is about 4.2v AC VPP. Now I need to feed this signal into the analog pin. I was reading this post here: http://arduino.cc/forum/index.php/topic,15362.0.html
and am trying to understand the circuit. Would I need to bias with 4.2V DC, or 2.1V DC? What net effect does the DC bias have on the AC signal - I mean, what would it look like on an o-scope after the bias is added? Also, what purpose do the caps serve - are they just there for smoothing out the DC signal?
I have tried to find info on DC biasing of AC signals, but so far, my searches have been rather fruitless. :~
Generally one doesn't apply the AC voltage (even offset to fit in 0-5vdc range) into a analog input pin because one already knows it's a constant frequency, variable amplitude proportional to current flowing through the CT. If you just want to measure the frequency of the AC power line being monitored then there are easier ways to perform zero crossing detection and then measure the time period between zero crossings.
One usually rectifies and filters the AC output voltage of the CT to gain a DC voltage that is proportional to the current passing through the CT. This DC voltage may then need to be reduced through a voltage divider to keep it in the arduino's 0-5vdc input measurement range.
"One set of Two Current Transformers (CTs) to measure individual breakers or
addtional panels.
I didn't get it, "individual breakers", what does it mean?
And 47 $?
If I understand, all you need a current sensor, with output 0-5V. and input?
100A, 1000A?.
Should be a lot of them on a market, just google "current sensors",
retrolefty:
I am really trying to measure power for the mains coming into my house. As I've read elsewhere, rectifying the AC signal with a bridge rectifier would lop off 0.7v, and I am trying to not do that - hence feeding it a DC bias. I guess I'm just confused on what the net effect is on the AC signal after applying DC bias.
Magician:
I don't know what they are thinking about measuring indiividual breakers, being as it's for a 200a load. Perhaps they meant measuring loads in subpanels? I have found other current sensors online, in fact there are ones that produce DC output directly. However, they are quite pricey. Eventually, I'll be adding CT's for all the circuits, but I have to get past this hurdle first.
"rectifier will lop off 0.6 volts" - not true all the time
It is all depends on design. If you supply initial current to the diod, it will be conductive
from the beginning.
If we are discussing "envelope detector", there are at least two variety.
I build one, like on drowning below to modulate HF transmission,
as low as 10 mV.
Ooopsss,
can't post a picture,
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Anyone knows where is upfolder, and how to clean it up?
I am really trying to measure power for the mains coming into my house. As I've read elsewhere, rectifying the AC signal with a bridge rectifier would lop off 0.7v, and I am trying to not do that - hence feeding it a DC bias. I guess I'm just confused on what the net effect is on the AC signal after applying DC bias.
I really don't think you have a choice. Measuring the AC voltage proportional to current via a CT is a lot more complex then you might imagine. Your house AC power voltage is a sine wave, but the current draw is anything but a pure sinewave, as many loads are non-linear. Your PC is a good example as it's switcher power supply is anything but a linear load. Anything with a motor will have a 'power factor' effect as it causes voltage and current to become out of phase with each other by some degrees and thus need special measurement and correction factor to convert to true power consumed and paid for.
Anyway rectify/filtering/scaling. and coming up with magic calibration numbers are what I think you will end up having to do. Or buy a 'smarter' AC current or power sensor/monitor that will do all the messy part and just give you a pure DC measurement value to read with your Arduino.
I'm afraid I can't give you a detailed circuit or method to use, the task is more complex then one might imagine. AC is not as easy to measure and analyze as DC.
Your best bet might be to search out a already proven project on the web that has proven results and accuracy. I know home power monitors have become more popular and available, but I don't have any specific recommendation or knowledge of what's actually out there.
I realize that non-resistive loads will become an issue, and will have to play with the calibration a bit I'm sure. I am trying to base my design on openenergymonitor: http://openenergymonitor.org/emon/node/58
Here, 2.5DC is fed/mixed into the AC signal, to "bias" it. Here is the source of my confusion. How is the 2.5v arrived at? Is it because the AC signal is 2.5vpp, or 5vpp? What would the resulting signal look like after biasing? In the arduino.cc post I mentioned in my original post, pluggy posted a circuit that is close to the one on openenergymonitor.
Well to put ac rms Vs peak Vs peak to peak is pretty simple:
AC RMS voltage is the heating equivelent value to DC voltage of the same voltage value. That is 120vac (rms) at 1 amp can generate 120 watts of energy just like 120vdc at 1 amp of DC power.
120 vac rms is equal to (X 1.414) =169.68 volts Peak. Peak to Peak AC value (which measures the complete AC wave form) is 2 X Peak or = 339.36 volts.
To retrolefty:
you are right regarding amplitude of the AC, it multiply by square root of 2, that is 1.414213562
So 120 AC will be detected as 120*1.414213562=169.705627485 DC voltage.
But you are wrong when you try to multiply it by 2 .
"form) is 2 X Peak or = 339.36 volts" - is not correct.
voltage will be exactly 169.71 V after rectifying by envelope detector or bridge.
OK - so that answers the first part of my question. Would I be correct in stating that given 3v AC RMS, that would be ~4.2V Peak, or 8.4vpp? So it sounds like I would bias it with 4.2V DC?
So given an AC signal of 8.4vpp, and mixing in or biasing it with a 4.2V DC source, what would the resulting signal even be? Would it be 0-12.6vpp AC? If so, I believe I could then use a voltage divider to get that down to the 0-5v that the arduino needs.
is not informative, there is no link to data sheet of sensor.
Or I 've missed one?
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Magician:
Thanks for the clarification on the voltage. I guess the 3v RMS would produce 4.2vpp, which means I would bias by 2.1v DC to raise the waveform to zero. I'm still not sure what the net effect of doing this would be though...
If I visualize the waveform of a 4.2vpp AV signal, we have 2.1v above the zero volt "line", and 2.1v below the zero volt "line". Isn't the point in biasing the signal to push it up by 2.1v, so the bottom of the waveform at its lowest point is at the zero volt "line"? Wouldn't this produce 0-4.2v?
It seems to be that if I bias said signal by 4.2v DC, that would result in a waveform that was +2.1 to +6.3V. I am very confused as to how adding 4.2V DC to a 4.2vpp AC signal would result in 8.4v. If you could explain this to me, I would be ever so grateful. Also, if anyone has any links that explains this, it would also be most helpful.
read , please, than ask a specific question.
don't want to be a impolite, but there is public forum,
I will give an answer on private message, if you have a question.
At least as typed, some of the posters here are mixing up peak values with peak-to-peak values, they are two different measurements of the same waveform. Again 3 vac (rms) has a peak value of 4.24 volts but indeed has a peak to peak value of 8.48 volts. So if you want to 'center' the complete ac waveform onto a single polarity DC measurement range you must have a DC offset equal to one half of the peak to peak value, or 4.24vdc. The complete 3vac (rms) wave form will then totally fill the a 0-8.48vdc measurement range. That would then require a voltage divider to cut down to a arduino compatible 0-5vdc measurement range.
So lets try and keep our peak and peak-to-peak values separated. Peak (p) is the value spanning from zero crossing to either the negative or positive peak. Peak-to-peak (pp) is the value spanned from the maximum negative peak to the maximum positive peak.
Yes, I agree,
But if someone want to measure AC level, there is no point to "shift ,or bias"
a level.
As negative self period of sine wave is equiale to positive half wave,
ordinary "envelope detector" will produce an output correctly.
There is only one reason to shift or bias if you try to interpreter input,
perform FFT or other sound analisis function.
There would be a trade off between frequency of input signal an frequency of scaning input.
In other words, if software check input with 40 Hz, it's cannot see anythyng below that.
For regular AC measurements, 1 Hz would be o'k, and there is no needs to measure both half
periods of the sine wave. One would be enough.
.
'Peak value' for a sinusoidal signal is usually called 'amplitude'. Of course people have noted that the output won't be sinusoidal in general due to non-linear loading, so 3V RMS could mean peaks in excess of 4.2V - allowing some headroom call it 5 or 6V.
Retrolefty's strategy (rectify/filter/scale) would be the best, in my opinion. If the maximum range will truly be significantly less than +/-12v, then you can use opamps to do the rectification so there will be no .7v rectifier losses. (That's how ac voltmeters do it.) Another advantage of using opamps to do the rectification, is that you may also be able to do the scaling using the same opamps. Also, the opamps won't load the output of the CT, (in case it has a higher output impedance, in which case other types of scaling would affect the voltage levels). Once it's scaled and rectified with opamps, you can use a simple capacitor for filtering high frequency noise.
I think I read the purpose of this is to measure the power usage at the mains of the house. In that case you will need more info than the simple current to time function. You will need to correlate the current to the voltage of the mains. As was mentioned, this is because current and voltage don't always line up due to inductive, capacitive, or active loads. You may have to research how the power meter from your utility company compensates for power factor correction. At larger industrial facilities, at least here in Illinois, we are required to install large capacitor banks in order to compensate for the large motor loads we have. This is simply so the power factor is correct for the usage meters, so we get billed correctly.
If all you have around the house is lighting, then current and voltage will be in phase, but if you have a large refrigerator, airconditioning, etc, then the current and voltage will be out of phase slightly, and this will affect the wattage measurement.
A quick google search for opamp rectification found this, it was the first one I clicked on: http://sound.westhost.com/appnotes/an001.htm
Be sure to take note of when to use buffer blocks in case the CT is a high impedance output.
