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.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.
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
Brings up two questions:Connected 3300 ohms across the CT coil and measured with multimeter in mA position and got the value of 10.2mV in AC..
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 onceCan you explain please...?
how it is 1:100 ratio coil and how did you get 3300Ohms burden value..?
Thus I assumed 1:100 winding ratio, that gives 100:1 current ratio.and 100 turns
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,
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
This means:I didn't wind the load current wire on CT.Simply put the load wire into CT's hole.
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,
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
Is true only for perfect sine waveform... mains waveforms are not perfect.1.Primary peak-current = RMS current × √2
As already said: It's like every other transformer..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..?
My recommendation: requirements --> specification --> part selection --> circuit parts calculationCan you give some guide to design amplifier to interface CT coil with PIC controller please.
Just pick one that suits your needs.Lot of circuits are there,I don't where to start...
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