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how to interface a current transformer to pic16f877a to measure the current ?

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Prabhu Siva

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Actually i need to interface the pic16f877a with current transformer to measure the current flow to the compressor. when the current consumption is more i need to turn it off so i need to measure the current . kindly help me in program in pic16f877a



thanks in advance
 

KlausST

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Hi,

You don't say anything about current and voltage...so we have to guess...
There are many current transformers.
If you use a passive one it has a current transfer ratio.
Maybe 1000:1. It means if you have 10A input current, then it has 10mA output current.
Let's say you want an input range of 10A AC, then you have +/- 14A voltage range --> +/- 14mA output current.
Now if your PIC ADC input voltage range is 3.0V you need a resistor of 3V/28mA = about 100 Ohms.
Additionally you need a voltage divider to lift the transformer output to VRef/2.
I recommend to use low pass filter between transformer and ADC.

Klaus
 

Prabhu Siva

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Hi,

You don't say anything about current and voltage...so we have to guess...
There are many current transformers.
If you use a passive one it has a current transfer ratio.
Maybe 1000:1. It means if you have 10A input current, then it has 10mA output current.
Let's say you want an input range of 10A AC, then you have +/- 14A voltage range --> +/- 14mA output current.
Now if your PIC ADC input voltage range is 3.0V you need a resistor of 3V/28mA = about 100 Ohms.
Additionally you need a voltage divider to lift the transformer output to VRef/2.
I recommend to use low pass filter between transformer and ADC.

Klaus



I need to measure the current flow to the compressor if the compressor is consuming more amount of current i need to switch off the compressor. the compressor is consuming 28amps right now if it starts to consume above 30amps i need to switch of it.
 

Ranbeer Singh

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You would convert current to voltage for measurement.

v=ir

for a example

CT = 50/1A. 50/1A means that if 50 amps currnt will flow in primary then 1 amps current will flow in secondary. Actually it is the transformation ratio of the ct Ip/Is. Second you will use a voltage divider.

r = v/i
r = 5/1 = 5E
 

Prabhu Siva

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You would convert current to voltage for measurement.

v=ir

for a example

CT = 50/1A. 50/1A means that if 50 amps currnt will flow in primary then 1 amps current will flow in secondary. Actually it is the transformation ratio of the ct Ip/Is. Second you will use a voltage divider.

r = v/i
r = 5/1 = 5E

i am using pic16f877a controller so it required current only in milli amps, from your input i decided that i need to choose the current transformer with a ratio of 30:5milliamps. but i need to how to configure with pic16f877a USING voltage divider bias and the concept of calculating rms current rms can u help in designing the schematic diagram to interace the CT
 

KlausST

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Hi,

I doubt the circuit shown in the link above gives (good) results...

***
Better use this:
curMeas.jpg
* CT/L2 is the current transformer. You need to know transfer ratio
* R2 is for setting output voltage
* R3 & R4 & C2 are for biasing output signal to V_Ref/2. Center of ADC value is zero. Below center is negative current, above is positve current.
* R1 & C1 form a low pass filter (used for attenuate high frequency, to lower noise level, as anti aliasing filter)

Klaus
 
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