Hi, I'm working on a PCB and need to monitor the current so I've been searching for an appropriate current sense transformer. I came across one that looks promising but I'm disappointed in the lack of info in the datasheet so have some questions. Here's the datasheet for it.
robsworld78:
Which pin is the AC source side, 1A or 2A?
That must be 1A AND 2A.
Load current flows THROUGH the thick 30A wire of the transformer.
5 and 8 is the secondary winding.
The value of the burden resistor sets output voltage (mV/Amp) of the CT.
Your project could be much safer if you choose a clip-on CT, or one with a hole for the load wire to pass through.
So there is no mains voltage on your circuit board.
Search for "current sense transformer" or "current transformer" on ebay.
Leo..
I came across one that looks promising but I'm disappointed in the lack of info in the datasheet
I agree with that!!! If you have a multimeter, the primary (AC side) should measure virtually "shorted". "Traditionally" on a schematic the input is on the left and output on the right so 1A/2A is probably the primary.
2.Which pin is GND and "data", 5 or 8?
Either of the secondary terminals can be grounded (to the Arduino ground.)
3.Is the data output from 5 and 8 in a range of 0-5v DC? I want to use an analog pin to read.
No, it's not DC and its not "data". __It's an AC voltage. Transformers are AC.* __
The Arduino can't read the negative-half of an AC voltage... In fact it can be damaged by negative voltages or voltages greater than +5V.
There are a couple of ways to deal with the AC. You can bias the input (see attached schematic... It's the same circuit used for AC audio signals). With the bias, zero AC voltage will read 512 and you can subtract that out.
Or, you can use a diode/resistor [u]protection circuit[/u]. You can make a protection circuit that protects against negative or excessive voltage so that could be an advantage if the current you're measuring is unpredictable. With the protection circuit, you'd ignore the negative half of the cycle and just assume it's equal to the positive half.
In either case the readings will appear random because you are "sampling" an AC waveform that crosses through zero twice per cycle. So, typically you sample in a "fast loop" and you find the peak. Assuming a sine wave, the peak is about 1.4 times the RMS value and the peak-to-peak is about 2.8 times RMS. (i.e. If the transformer is putting-out 5V RMS, that's too much and you'll need a voltage divider.)
The specs for the transformer will be RMS so you'll have to scale the readings. It's also possible to calculate RMS, or calculate an average** and apply a correction factor to get RMS (again assuming a sine wave where these factors are known).
** The true average of a sine wave is zero (it's positive half the time and negative half the time) so you have to calculate the average of the absolute values, or just the average of the positive values (ignoring the negative readings).
Wawa:
That must be 1A AND 2A.
Load current flows THROUGH the thick 30A wire of the transformer.5 and 8 is the secondary winding.
The value of the burden resistor sets output voltage (mV/Amp) of the CT.Your project could be much safer if you choose a clip-on CT, or one with a hole for the load wire to pass through.
So there is no mains voltage on your circuit board.
Search for "current sense transformer" or "current transformer" on ebay.
Leo..
Thanks Leo, I guess my main confusion is the output, I read that not all CTs output the same and confused which direction the power should flow but that I can all test but didn't want to ruin an Arduino. I can't use a clip on and don't want a through hole CT as I don't want to mess with wires. The PCB already has AC voltage so I'm not worried about that, I just don't want to cut, trim, tin and solder wires. 
DVDdoug:
I agree with that!!! If you have a multimeter, the primary (AC side) should measure virtually "shorted". "Traditionally" on a schematic the input is on the left and output on the right so 1A/2A is probably the primary.
Either of the secondary terminals can be grounded (to the Arduino ground.)
No, it's not DC and its not "data". __It's an AC voltage. Transformers are AC.* __
The Arduino can't read the negative-half of an AC voltage... In fact it can be damaged by negative voltages or voltages greater than +5V.There are a couple of ways to deal with the AC. You can bias the input (see attached schematic... It's the same circuit used for AC audio signals). With the bias, zero AC voltage will read 512 and you can subtract that out.
Or, you can use a diode/resistor [u]protection circuit[/u]. You can make a protection circuit that protects against negative or excessive voltage so that could be an advantage if the current you're measuring is unpredictable. With the protection circuit, you'd ignore the negative half of the cycle and just assume it's equal to the positive half.
In either case the readings will appear random because you are "sampling" an AC waveform that crosses through zero twice per cycle. So, typically you sample in a "fast loop" and you find the peak. Assuming a sine wave, the peak is about 1.4 times the RMS value and the peak-to-peak is about 2.8 times RMS. (i.e. If the transformer is putting-out 5V RMS, that's too much and you'll need a voltage divider.)
The specs for the transformer will be RMS so you'll have to scale the readings. It's also possible to calculate RMS, or calculate an average** and apply a correction factor to get RMS (again assuming a sine wave where these factors are known).
- Some people call "wall wart" power supplies "transformers" but if they put-out DC they are actually power supplies built with a transformer, rectifier, capacitor filter, and perhaps a voltage regulator circuit and perhaps with a high-frequency switching circuit if it's a switchmode power supply.
** The true average of a sine wave is zero (it's positive half the time and negative half the time) so you have to calculate the average of the absolute values, or just the average of the positive values (ignoring the negative readings).
Thanks for the detail Doug, really appreciate it. Had a feeling it would be an issue connecting directly to Arduino and yes now you spell it out it makes sense the output is AC voltage, guess I was dreaming when I thought it could possibly be DC haha. I know they don't technically put out data that's why I put it in air quotes.
Unfortunately you use a lot of terms I'm not very familiar with so I'll have to read this many times but bottom line is this will work using one of the two methods you mention? I haven't ordered any yet but if I know it will I'll get some and can figure out the flow. Nice to know the GND from Arduino can go to any side of secondary winding, that will definitely help with traces.
I'm using a SSR and want to monitor the load on it. Anything could be plugged in drawing very little current up to 5amps. I don't need absolute precision, hoping to match a cheap kill-a-watt meter. I assume I would put the CT on the output side of the SSR?
Thanks
DVDdoug:
Either of the secondary terminals can be grounded (to the Arduino ground.
Can be, but in this case shouldn't be.
It is common to "ground" one pin of a CT to a mid-voltage point.
Google "Arduino CT schematic" (images).
One of the first image(s) (OpenEnergyMonitor) shows you how to connect it.
Leo..
Wawa:
Can be, but in this case shouldn't be.
It is common to "ground" one pin of a CT to a mid-voltage point.Google "Arduino CT schematic" (images).
One of the first image(s) (OpenEnergyMonitor) shows you how to connect it.
Leo..
Thanks Leo, that's very helpful. I just redrew the schematic to make it a little clearer, at least for me, I attached it to the post. A few questions.
Sorry for all the questions.
Pins 1A and 2A are completely equivalent, this is an AC device, its doesn't matter which way they
are connected.
Similarly pins 5 and 8 on the low voltage side are completely equivalent (unless you care about
phasing).
From the Arduino's point of view you just have an AC voltage source that you refer to a mid-rail
reference voltage from the divider. Thus the voltage to the Arduino analog pin is centred around
2.5V allowing the whole AC waveform to be samplde.
I'd add a 10k resistor in series with the Arduino analog pin to protect it from any over-voltage event and
for when the Arduino is powered down.
The capacitor in the divider can be anything, completely non-critical, a few uF is plenty, just reduces
noise pickup.
R1? There is no R1 in that diagram.
It is absolutely essential that the secondary of a current transformer has its burden resistor connected
securely with no possibility of coming lose - if it comes disconnected you'll get extremely high voltages
generated (and even flashover and fire). It you get the chance always chose a model with a built-in burden
resistor, its inherently much safer. (A current transformer needs an effectively shorted secondary, otherwise it will behave like a voltage step up transformer).
The output voltage form a CT is input current / turns ratio x burden resistor. Thus
I / 500 * 33 = I * 0.066.
Or put another way its like a 66 milliohm shunt (but isolated and AC only).
You need to ensure the peak-to-peak voltage is <= 5V so that it can be read by the 5V Ardiuno's ADC.
With that 33 ohm burden resistor and 500:1 ratio that means the max peak-to-peak input current is limited
to 75A, which is 26A rms.
Note that the datasheet for that transformer has incorrect calculation for the sensitivity, as
60 ohms / 500 = 0.12 ohms, giving 120mV/A, not 110mV/A - makes me wonder what else
is wrong in the data...
MarkT:
Pins 1A and 2A are completely equivalent, this is an AC device, its doesn't matter which way they
are connected.
Hi Mark, thanks for all the info. I know it generally doesn't matter to much however I don't think that's entirely true either as most plugs now have one spade slightly wider so they can only be plugged in one direction but for the CT everything I've read says it must be installed in the correct direction. I'm sure you know much more than me but I have to question you on this one.
MarkT:
From the Arduino's point of view you just have an AC voltage source that you refer to a mid-rail
reference voltage from the divider. Thus the voltage to the Arduino analog pin is centred around
2.5V allowing the whole AC waveform to be samplde.
One big surprise to me is AC voltage can be connected directly to an Arduino pin if it stays under the 5v or on some boards 3.3v.
MarkT:
R1? There is no R1 in that diagram.
Ooops, sorry, I updated the schematic.
MarkT:
Note that the datasheet for that transformer has incorrect calculation for the sensitivity, as
60 ohms / 500 = 0.12 ohms, giving 120mV/A, not 110mV/A - makes me wonder what else is wrong in the data.
If that's the formula then yeah definite problem there, good eye.
MarkT:
The output voltage form a CT is input current / turns ratio x burden resistor. Thus
I / 500 * 33 = I * 0.066.Or put another way its like a 66 milliohm shunt (but isolated and AC only).
You need to ensure the peak-to-peak voltage is <= 5V so that it can be read by the 5V Ardiuno's ADC.
With that 33 ohm burden resistor and 500:1 ratio that means the max peak-to-peak input current is limited
to 75A, which is 26A rms.
Ok I think I understand most of this.
I / 500 * 33=I * 0.066
30a / 500 * 33 = 1.98v output
So if a max of 30amps was put through the CT the output would only be 1.98v max using a 33R burden resistor?
I see 75a / 500 * 33 = 4.95v the limit for the Arduino but 75amps would never be achieved.
I will never be putting 30amps through the CT, I'm limiting the circuit to only 5 amps so lets use 10amps as a max ceiling with a 100R burden resistor.
10a / 500 * 100R = 2v output.
Doing it that way the max peak-to-peak input current is limited to 25A which is 8.8A rms?
Maybe I'm wrong but I also see this giving better accuracy on the low current going through the CT?
robsworld78:
One big surprise to me is AC voltage can be connected directly to an Arduino pin if it stays under the 5v or on some boards 3.3v.
AC voltage is just a slowly changing DC. That is, slow for an Arduino
If you sample that changing DC voltage fast and long enough (at least one AC wave length), then you can work out the two peak voltages by subtracting min and max A/D values.
75Amp?
1.98volt AC is 1.981.42= 5.5volt peak/peak...
Leo..
Wawa:
AC voltage is just a slowly changing DC voltage. That is, slow for an Arduino75Amp?
1.98volt AC is 1.981.42= 5.5volt peak/peak...Leo..
I'm a DC guy so don't really understand AC, just know not to mix the two.
Thanks for clarifying, I thought I could be off on the 75amp but wasn't sure.
So if 30amps was the goal the burden should actually be 30R, as 30A / 500 * 30R = 1.8v1.42 = 5.04v?
But as I only need 5A is it ok to base the math off 10A? 10A / 500 * 90R = 1.8v1.42 = 5.04? In reality because it'll never be over 5A the Arduino should never see more than 2.52v. That seems very safe to me, maybe too safe. Maybe 110R is better for my application?
robsworld78:
Hi Mark, thanks for all the info. I know it generally doesn't matter to much however I don't think that's entirely true either as most plugs now have one spade slightly wider so they can only be plugged in one direction but for the CT everything I've read says it must be installed in the correct direction. I'm sure you know much more than me but I have to question you on this one.
Whatever makes you think your present concern regarding the current transformer has any relationship whatsoever with a mains socket?
robsworld78:
I'm a DC guy so don't really understand AC, just know not to mix the two.
So it seems. 
I'm a DC guy so don't really understand AC, just know not to mix the two.
Absolutely fine to "mix" ac with dc when you need to, easy to separate again, this is just
called voltage offsets and coupling capacitors.
The constraint in the datasheet is that the burden resistor needs to be small enough to limit the
rms output voltage to 4.0V, which it will be with 5A input and 110 ohms.
MarkT:
Absolutely fine to "mix" ac with dc when you need to, easy to separate again, this is just
called voltage offsets and coupling capacitors.The constraint in the datasheet is that the burden resistor needs to be small enough to limit the
rms output voltage to 4.0V, which it will be with 5A input and 110 ohms.
Great, thanks for all your help.
