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Current Coil interfacing with controller

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venkates2218

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In my project there is an cooling fan with 230VAC and 0.10A current.
I need to measure the current taken by the fan.

So I used CT coil with 10A rating and 100 turns to read the current.By using 15 OHM resistor across the two terminals and connected the multimeter(Multimeter parameter set to mA) to measure the voltage,which produced by CT coil.But there is no output measured in multimeter.

So I tried an circuit by keeping the reference as above circuit to connect with PIC controller.

But there is no response in pic controller with above circuit..

How to solve this issue.?
 

But there is no output measured in multimeter.
The multimeter and PIC should show the same measurement. How are you measuring the voltage, on the DC range or AC range? You should get about 2.5V DC (if the resistors are both 10K) if nothing else. To measure AC superimposed on the DC you may need a blocking capacitor on the meter but the PIC should be OK.
I presume you are using suitable software in the PIC that can sample at a fast enough rate and do the necessary math.

Brian.
 

Measure on both DC and AC range..

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The multimeter and PIC should show the same measurement. How are you measuring the voltage, on the DC range or AC range? You should get about 2.5V DC (if the resistors are both 10K) if nothing else. To measure AC superimposed on the DC you may need a blocking capacitor on the meter but the PIC should be OK.
I presume you are using suitable software in the PIC that can sample at a fast enough rate and do the necessary math.

Brian.

Is it possible to measure the voltage in multimeter by connecting an resistor across the CT coil and by connecting the junction to multimeter's probe..?
 

The calculation in post #1 is wrong, you forgot the transformer turns ratio. With 1:100 ratio, the output voltage is only 1/100 of the calculated. As a first step, I would check if the primary current is actually 50 mA rms. You should also check what's the maximal burden resistor value for the CT and if you get sufficient sensitivity with this value. Otherwise an amplifier between secondary and ADC may be necessary.
 

Hi,

Your CT is a 1:100 ratio one.
50mA RMS of load current become 500uA RMS on secondary side.

--> You need 3300Ohms burden

Klaus
 

Hi,

Your CT is a 1:100 ratio one.
50mA RMS of load current become 500uA RMS on secondary side.

--> You need 3300Ohms burden

Klaus

Can you explain please...?
how it is 1:100 ratio coil and how did you get 3300Ohms burden value..?
 

Connected 3300 ohms across the CT coil and measured with multimeter in mA position and got the value of 10.2mV in AC..
How to convert it to DC and have to interface with ADC in PIC controller..

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Shall we use gain amplifier to convert AC output to DC to interface with ADC pin in PIC controller..?
 

Connected 3300 ohms across the CT coil and measured with multimeter in mA position and got the value of 10.2mV in AC..
Brings up two questions:
1. Is 3300 ohm burden in the permitted range according to CT datasheet?
2. How does your multimeter measure voltage in the mA position? Mine doesn't.

You can either process AC voltage in software or use a precision rectifier circuit to read the measurement as DC value.
 

Hi,

Can you explain please...?
how it is 1:100 ratio coil and how did you get 3300Ohms burden value..?
I don't know how many turns you wind the load current wire through the CT...thus I assumed you did it the usual way: only once
But you wrote:
and 100 turns
Thus I assumed 1:100 winding ratio, that gives 100:1 current ratio.

How do I get 3300 Ohms?
I saw you calculated with 1:1 ratio...and your result was 33 Ohms.
This value I multiplied with the estimated "100" of 1:100" ratio.

Klaus
 

Brings up two questions:
1. Is 3300 ohm burden in the permitted range according to CT datasheet?
2. How does your multimeter measure voltage in the mA position? Mine doesn't.

You can either process AC voltage in software or use a precision rectifier circuit to read the measurement as DC value.
img.PNG
Sorry its mV position.
i'm using 56100C model
They didn't give any details about burden resistor...

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Hi,


I don't know how many turns you wind the load current wire through the CT...thus I assumed you did it the usual way: only once
But you wrote:

Thus I assumed 1:100 winding ratio, that gives 100:1 current ratio.

How do I get 3300 Ohms?
I saw you calculated with 1:1 ratio...and your result was 33 Ohms.
This value I multiplied with the estimated "100" of 1:100" ratio.

Klaus

I didn't wind the load current wire on CT.Simply put the load wire into CT's hole.
In datasheet they given the current ratio as 100:1 and current rating as 10A
 

The datasheet gives many details. The CT is intended for 20 to 200 kHz frequency range. Primary inductance is relative low, thus you don't get specified sensitivity at 50 Hz, even with very low burden.
 
Hi,

I didn't wind the load current wire on CT.Simply put the load wire into CT's hole.
This means:
* primary winding count: 1(you put the wire 1 times through the hole)
* secondary winding count: 100 (given from datasheet)

....but as FvM already noted, this CT is not suitable for mains frequency applications.

Klaus
 
The datasheet gives many details. The CT is intended for 20 to 200 kHz frequency range. Primary inductance is relative low, thus you don't get specified sensitivity at 50 Hz, even with very low burden.

Inductance value should be high..?
How to choose the CT for 50Hz with correct inductance value..?

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Hi,


This means:
* primary winding count: 1(you put the wire 1 times through the hole)
* secondary winding count: 100 (given from datasheet)

....but as FvM already noted, this CT is not suitable for mains frequency applications.

Klaus

These are equation used to calculate the burden resistor for 1:100 ratio
1.Primary peak-current = RMS current × √2
2.Secondary peak-current = Primary peak-current / no. of turns in secondary
3.Ideal burden resistance = (AREF/2) / Secondary peak-current


Suppose if we having no.of turns in primary(Load current wire wind over CT) means,is any other method to calculate burden resistor or shall we use this equations itself..?
 
Last edited:

Hi,

Use a selection guide to choose the correct CT.
The selection guides should be avaliable at CT manufacturers as well as distributors.
If you want it to operate down to 50Hz, then look for one that is able to...

Current calculations are similar to every transformer.
Also every manufacturer usually provides free application notes.

1.Primary peak-current = RMS current × √2
Is true only for perfect sine waveform... mains waveforms are not perfect.

Suppose if we having no.of turns in primary(Load current wire wind over CT) means,is any other method to calculate burden resistor or shall we use this equations itself..?
As already said: It's like every other transformer..
Obviously when you wind the current twice through the hole the output signal amplitude will be twice, too.

Klaus
 

You won't find an of-the-shelf CT allowing higher output swing than a few 10 or maximal 100 mV for 50 or 100 mA primary current with passive burden.

Possible solutions are:
1. Multiple primary turns
2. An AC amplifier between burden and ADC
3. An active burden circuit, e.g. transconductance amplifier

See appended the datasheet of a popular 5A 50 Hz CT
 

Attachments

  • AC1005.pdf
    46.3 KB · Views: 87
How to design an amplifier circuit to interface with controller with CT..?
Please give some guides..
 

Hi,

It has been discussed many times here and even more often in the internet:

use the forum "advanced search", Keywords: "microcontroller current" and choose "search Titles only".

Klaus
 

Can you give some guide to design amplifier to interface CT coil with PIC controller please.
Lot of circuits are there,I don't where to start...
 

Hi,

Can you give some guide to design amplifier to interface CT coil with PIC controller please.
My recommendation: requirements --> specification --> part selection --> circuit parts calculation

Lot of circuits are there,I don't where to start...
Just pick one that suits your needs.

It´s like:
If you don´t want to starve...just eat. It´s not that important what you eat.
Later, when you have tired out several things, you will learn what you like and dislike

Klaus
 

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