Measuring AC current below 1A

Status
Not open for further replies.
T

tictac

Full Member level 5
Joined
Oct 22, 2006
Messages
297
Helped
3
Reputation
6
Reaction score
3
Trophy points
1,298
Location
Polland
Activity points
3,572
Hi
I want to measuring the current of one 220v fan(Imax=500 mA). I want to read the current value with the ADC of mega16 and show it at the LCD. How can I do this with isolation measuring(by using of opto-coupler or CT)?

Regards
 

Dear tictac
Hi
It is simple . try to use linear optocouplers , such as IL300 . or use a CT in your AC line . then a resistor and a simple instrumentation amplifier , or an amplified CT ( you can make it with your hands . )
Best Wishes
Goldsmith
 
Reactions: tictac

    T

    tictac

    Points: 2
    Helpful Answer Positive Rating
Hi Dear Goldsmith
Where can I found IL300 in Iran? How can I make one CT that can measuring AC current below than 1 ampere? which is cheaper and simpler?

Regards
 

Hi again
I'm confused , because the location that you wrote , is ........... and you are asking about which place you can find ....... !!!!
Anywhere . this issue that which one is better , depends on the precision that you need .
 

Hi,I live in Iran.
I cant find IL300 in electronic shops. The precision is about 5% .

Regards
 

Which city you are in ? if you are in Tehran , you can find it easily . and a thing that you should think about it : because your current is simultaneous , you can neglect from a half cycle and then consider that it is a DC signal , my mean is , that you can rectify it with a precise rectifier .
Let come back to CT : it is a transformer simply , ! it has a primary winding and a secondary . it's turn ratio can decrease the secondary current or increase . it depends on your abilities in design .
you can use usual current transformers or wind your one !
Best Events
Goldsmith
 

Im in Tehran. Can you tell me the name of the electronic store in Tehran,please? which one is cheaper? IL300 or CT?
I read about making CT. I think CT is cheaper. what is your idea? Can you help me,please?

Regards
 

If you're in Tehran , go tho the jomhouri street , and after hafez bridge , go into the abbasian , passage , and then go to the javan electronics , you'll find it ( 2 years ago i bought it from there ) .
And i think if you use a precise rectifier and an IL300 you can achieve good precision , but if you like use CT , no problem . it will be good , too . ( i used both of those methods , both of them are good )
Respect
 

I search javan's website. "http://www.javanelec.com/Search.aspx" but they dont have IL300. so I want to use one toroid core.
one of the fan's wire passes through the core, then I want to wind 10 turn as a secondary winding.
If I(fan)=1 Amp then I(out)=0.1 Amp. I want to use one resistor in parallel with the secondary winding at about 20 ohm. so Vout=2 Volts and rectify it and measure it with micro controller. Is it true?
Can you improve my idea,please?

Regards
 

Are you ok ? how it is possible that vout=2 volts ??!! did you think that you have a current source ? surely not ! if you short circuit the secondary , you'll have 0.1 amperes , ( maximum ) but if you choose a high , load ( 20 ohms ) your current will decrease and your voltage can't increase . you should use a series ammeter ( sort circuit the secondary with an ammeter ) or use a pretty low value resistor ( such as 0.1 or 0.27 ohms ) then measure the voltage across it , or give it to the ADC and ........... .
And about IL300 , it isn't in their site , but they have it .
 
Reactions: tictac

    T

    tictac

    Points: 2
    Helpful Answer Positive Rating
Thanks you. Resistor with the value equal 0.1 ohms generate 10 mV for 1 ampere(maximum current ). The noise and ripple of output voltage of my power supply is at about 100 mV, and I think with this very little output cant be measured correctly! what can I do?
I can amplify the secondary winding output with an opamp such as op07. but I think the ripple and noise of the power supply can effect on the amplified voltage.

Regards
 

Hi again
If you have that value of noise in your out put , it will divide by that value too . BTW: if the noise can suffer you , you can create a loop for noise and remove it , simply ( from your measurement ! )
 

Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…