I am wanting to monitor my aquarium power usage, this will be on a New Zealand 230v, 50 hz system. It all runs from a single power point.
In my setup there is the following:
2 Filters - about 10 watts
2 heaters - potentially 600w if both are on
Lights
Air
So I am wanting to use the CT Clamp with produces via a burden resister an AC voltage.
In my testing with a 1000w/2000w fan heater which would draw close to 5 Amp at 1000w
Now running the heater at 1000w I see an AC voltage of 0.023, if I switch the heat to 2000w it jumps to 0.46 volts AC
I have been looking around and playing with some circuits to try to convert this to DC. At the moment I am running this on a multimeter but I want to use an Arduino to save this data.
Since the voltages are so low what are induced from the CT Clamp I am thinking of a precision rectifier circuit to convert to DC. I would like to make 10amp = 5v.
So 5 amp (heat running at 1000w) = approx 2.5 volts.
ok, so you clamp your cable and now you're monitoring the current draw.
um, is there any reason why you're not controlling each of the devices via relays from your arduino? then there's no guessing involved you have complete control over everything!
How accurate do you want the result to be?
CT type power measurements arnt very accurate because they assume the supply voltage never varies
which is almost never the case.
I guess accuracy as good as a plugin power monitor like you can buy for $20 would be good.
Form my simple test with the clamp as I said in the first email I received:
0.023v from 1000w load
0.046v from 2000w load
All I really want to do is convert the small AC to DC and then after that conversion amplify the signal so it is between 0 and 5v - so something like a 100X Multipler.
I am just not sure on the final two components a resister and capacitor I assume to filter the signal, but I do not know how to calculate this.
The pdf has this text
The CT secondary current is applied to the resistor R1, which generates a voltage equal to the primary current
divided by the turns ratio and multiplied by the value of R1. This AC voltage is rectified by the first op amp,
and then amplified by the second op amp.
The gain of the first stage is always kept at 1 or unity (R2 = R3) to guarantee symmetry of the rectified
waveform. R2 should be chosen at least 10 times greater than R1 for proper accuracy.
The gain of the second stage is R5/R4 + 1. This gain is chosen to get the desired output DC voltage for the
designed input voltage.
The main advantage of this circuit is the removal of the diode drop as a variance in the signal. DC can be
generated directly from R1 by applying the AC voltage to a diode bridge. However, the AC voltage required to
do this must be greater than 2 diode voltage drops, or over 2 VAC. This limits the designer to use a silicon steel
core that can generate enough voltage before going into saturation. By using the above circuit, the AC voltage
input can be very low (10-100 mV) and then amplified to the level desired. This then allows the designer to
choose smaller core devices and nickel core devices which saturate at low voltages. Accuracy and cost are both
improved.
Use standard op amp design guidelines when setting up this circuit. Keep resistors at 1 M ohm or less, and
keep gains to 100 or less.
The output must be filtered for pure DC. The RC output network shown should be designed with a time constant
at least 10 times greater than the period of the waveform sensed. For 60 Hertz, use a time constant of 1/6.
For 400 Hz, use a time constant of 1/40, minimum.
So for me I am thinking of using
Z1, Z2 = 1N4735A - not sure if this Zener is OK, 6.2v 1W but I assume it is.
R1 = 10 Ohm
R2 = 100 Ohm
R3 = 100 Ohm
R4 = 1k Ohm
R5 = 100K Ohm (to give a 100 x multipler)
R6 = ?
C1 = ?
Technically with my 100 times multiplier I believe:
0.023v * 100 = 2.3v
0.046v * 100 = 4.6v
So I should be able to read this on an analog in on the arduino?
A colleague at work suggested that maybe a Toroidal current transformer may be a better thing instead of a current clamp since my current is quite small - generally 1000w would be the maximum.
I am going to look into this and see how I get on.
A toroidal current transformer would give you a greater output that the current clamp. You would then need less amplification; or you could read the output without amplification and then do the AC-to-DC conversion in software (this is what energy monitors usually do).
If you want to use the current clamp, then I find the circuit you linked to over complicated. In particular I don't believe that the first (unity gain) stage does "guarantee symmetry of the rectified waveform". I would use the attached schematic.
The clue to the output R and C is in the text
The RC output network shown should be designed with a time constant at least 10 times greater than the period of the waveform sensed. For 60 Hertz, use a time constant of 1/6.
But you don't need the output R because R5 + R4 performs that function already. So I've scrubbed it out in my version. I've also simplified the input protection, because your input voltages are very small (so you can use a normal diode instead of a zener) and the LM358 can stand up to 50mA of negative input current. I've also reduced the gain from 100 to 48, because if you power the LM358 op amp from the Arduino +5V supply the then output will only go up to about 2.5V because the LM358 is not rail-to-rail. BTW the LM358 and LM324 are the same, except that you get 2 op amps per package in the 358 and 4 in the 324.
[EDIT: I forgot to allow for the voltage drop in the output diode when considering the output voltage swing. You might want to power the LM358 from more than +5V to avoid this issue. Alternatively, use a Schottky diode such as BAT85 for the output diode, and reduce the 470K resistor to 220K or 330K to further reduce the gain.]
iisfaq:
Are you sure that the voltage is enough to drive the first diode?
My ac voltage from the CT Clamp is only 0.023v for a 1000w load on the CT clamp at 240v.
The first diode is only there to protect the op amp from positive transients (like the zener diodes in your original). I am relying on the fact that the signal from your CT is too small for it to conduct during normal operation. The second diode does the rectification.
