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Hi thank you for reading.
I have the enclosed circuit using a 100Amp / 5Amp CT ... Bridge Rectified and smoothed.
After searching internet on how to calculate the Burden Resistor I would appreciate some help.
I am looking for 0-10V IP from 0-100Amp
Attachment 7674
Anybody .....????
Thank You.
I never used a current transformer but I suggest you place the burden resistor before the bridge.
Then you CT is 100:50 = 20:1
@100A on the primary you'll have 5 amps on the secondary, so 2ohms will give you 10VAC rms.
The bridge has a 1.4V drop you must take in account. Also, AC voltage fully rectified with a capacitor and no load will give you a peak voltage of (v * 1.4142) (1.4142 = square root 2).
So for a 10VDC max:
(10/1.4142) + 1.4 = 8.47VAC on the secondary.
8.47VAC / 5A = 1.694 ohm. Good luck to find one.
My suggestion: pick a standard value < 1.694ohm and then compute the results. Finally, use a good multimeter to caliber your system.
HTH
Thank you for that…
2R is the answer I got to…. I also worked out the Wattage of the resistor needs to be > 50Watt (Large and Hot)
Possible temperature drift !!!!
Yep! Lot of power! What about shorting the secondary and measuring the current with a current sensor like this:
http://www.allegromicro.com/en/Produ...Cs/ACS712.aspx
on a breakout board:
http://www.dx.com/p/acs712-5a-curren...7#.VLRjedKG_CY
[Board Admin: I'm sorry for the ad :frown:, I'm sorry! I'm sorry!]
if you have way to calibrate and check linearity......
Attachment 7676
I like this.....Quote:
Originally Posted by MichelJasmin
Also a 30Amp Version.... Might not need a CT !!!
Thank you
Thank you Amgen
Nice Idea.... but pot Wattage needs to be decent size. (More power) ?
ok check these..... http://amploc.com/
Thank you
The sensors all ready exist.. I am attempting to use the output for my purpose.
But will prob look at those in future.
Thank you again
From my experience working with CT's, putting a higher value resister across ct secondary will lower the secondary current by some factor but the voltage still has some relation to the primary current...... that's why a 100 ohm or 1000 ohm resistor will not create that much heat. My guess is that a 100 ohm load resister will probably drop the secondary current to less than 1 amp with 100 amp thru CT.
don