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General Category => General Discussion => Topic started by: fromdawest on December 25, 2012, 11:49:05 AM

Title: How to interface Current Transformer ( 30:5 ) into 16F877 PIC
Post by: fromdawest on December 25, 2012, 11:49:05 AM
I want to make a simple ammeter circuit using PIC16F877 . To measure the current I'm using a simple CURRENT TRANSFORMER ( 30:5 A ) . So what i want is to, get this current transformer output and make it the input to the PIC IC ( to it's ADC ), and then to drive a small LCD to show the current value .

So I want to know how to interface a CURRENT TRANSFORMER with the pic 16f877 in proteus . I tried but even i cant find the current transformer element in proteus 7.x .

Please help .
Title: Re: How to interface Current Transformer ( 30:5 ) into 16F877 PIC
Post by: kam on January 02, 2013, 10:53:37 AM
you will use a shunt resistor. Connect the resistor to the output of the transformer. Then you measure the voltage across the resistor. Ohms law will give you the current.
Title: Re: How to interface Current Transformer ( 30:5 ) into 16F877 PIC
Post by: cheerio on January 02, 2013, 19:54:48 PM
i bet you want to measure AC... so you have to rectify the shunt signal without losses. check this out: http://sound.westhost.com/appnotes/an001.htm (http://sound.westhost.com/appnotes/an001.htm)
Title: Re: How to interface Current Transformer ( 30:5 ) into 16F877 PIC
Post by: kam on January 14, 2013, 09:00:11 AM
hmmmm i've never thought of this diode trick... But why the diode is at the output instead of the input? The result would be the same, right? Am i missing something here?
Title: Re: How to interface Current Transformer ( 30:5 ) into 16F877 PIC
Post by: cheerio on January 14, 2013, 15:51:24 PM
let's assume that we have a signal of 100mv and you put the diode at the input... problem isn't it?
Title: Re: How to interface Current Transformer ( 30:5 ) into 16F877 PIC
Post by: kam on January 14, 2013, 15:58:48 PM
let's assume that we have a signal of 100mv and you put the diode at the input... problem isn't it?
Sure, but the op-amp is a buffer, not amplifier, so the problem stands for both situations (diode at input or output).

hmmmm i wonder....

Now that i think of it again, the 0.7V of the output diode will be compensated by the op-amp... 1V input will cause the opamp to output 1.7 in order to have same + and -...
I knew that it was a good idea....