500v to 5v?

Thread Starter

picboy

Joined Dec 17, 2011
2
I have to measure voltage (0- 500v) and current (0 to 5a) using a PIC microcontroller.
Since it takes 0 to 5v, how do i bring the voltage and current down linearly such that.. if i measure 5v, it is equal to 500v , 4=400v, 3=300v etc..
n the same goes for the current.
plz help :)

thanks in advance..
 

Georacer

Joined Nov 25, 2009
5,182
Please post your question only once. Posting the same question over several forums results in confusion and mixed answers.

All your other identical threads have been deleted.
 

t06afre

Joined May 11, 2009
5,934
And also good questions draw good answers. Try to be as descriptive as you can in your first post.
As a start I will give you some pointers to elaborate on
Is it 500 Volt AC, or DC?
If it is AC, what is the frequency? Why not post some picture of this voltage. And to avoid any confusing. Please specify RMS or peak value.
 

Thread Starter

picboy

Joined Dec 17, 2011
2
its a 400V AC Supply at 50 hz from a 150kva/120kw generator with a current of 216A.
Since the voltage can go to even 415V , I was looking for something in 500V.
 

evilclem

Joined Dec 20, 2011
118
I would use a potential divider for most cases where I need to measure voltage on an ADC but would also be looking into some sort of isolation for this one as well.

Are you wanting to measure the instantaneous or RMS voltage?
 

t06afre

Joined May 11, 2009
5,934
its a 400V AC Supply at 50 hz from a 150kva/120kw generator with a current of 216A.
Since the voltage can go to even 415V , I was looking for something in 500V.
The 400 AC is probably RMS voltage. So the peak value will be (square root(2)) higher. The only safe method of doing this is by using a transformer. Discussing any other method will close this thread. For current measurement Google current transformer
 

tom66

Joined May 9, 2009
2,595
Many small transformers will distort the waveform considerably, because they run close to saturation. Make sure to compensate for this; either as a calibration factor or by using a bigger transformer.
 
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