Still Struggling with the formula!

[Ramius]

Hi Ioannis!
Thank you for the suggestion and I still seem to be missing something fundamental. At 0.0 amp (no load) the A/D gives a digital reading of 485 and after the >>3 this becomes the number 60. Therefore 60 I think becomes the offset value. At .461 amps the A/D gives 506 and after the >>3 this becomes 63. Last at .869 amps the A/D gives 516 and after the >>3 you now have 64! Just as an FYI, the spec sheet shows the offset to be 0.5 volts so the number 60 makes sense.

[mackrackit]

Who knows, maybe this will help?
http://www.picbasic.co.uk/forum/cont...Sensor-Scaling

[Ioannis]

On page 4 of the data sheet in Quiescent Output Voltage it clearly indicates that the output voltage will be the half of the power supply. So if the Vcc is 5 volts then the output will stay at 2.5 volts.

Ioannis

[Ramius]

Thanks Ioannis and Dave!
Ioannis you are correct and the actual device is the 50U. I am not sure we are looking at the same spec sheet and if so then on page 6 it shows that with this sensor you get 60mv/A. Since the device is a 50 amp max. device then 60 x 50 would give 3000 or 3.0 volts. Subtract the 0.5 volt offset and then you get the 2.5 volts at 50 amps. It could very well be that there is an error in my ampmeter's scale as this is what I have been using to compare to the program results.

Best, Ed

[Ioannis]

No you are not seeing the specs correct.

If there is no current the device just sits inthe middle of the Vcc. Ifyour Vcc is 3.3 volt then the Vcc/2 is 1.65volts. If it is 5 then the output sits at 2.5 volts.

It is made in such way because it measures AC and DC currents.

Ioannis

[HenrikOlsson]

Hi,
If it indeed is the 050U device then, if I read the datasheet correct it's an Unidirectional version and its output voltage at 0A is 0.6V - not Vcc/2.

So if VRef for your AD-converter is 5.000V and you're using 12 bits then you should get an ADC reading of 4096/5*0.6=492 at 0A current. In an earlier post you said that you got 485 - that's pretty close. You'll then get 60mV per A.

So at 1A current you should get a reading of offset+4096/5*0.06=541. To convert this to a readable value, if that's what you want, you first subtract the offset (541-492=49) and then multiply it by say 204 and then divide by 10. 49*204=9996, then divide by 10 = 999mA

If you have 3.75A then you'll get an ADC reading of offset+4096/5*(0.06*3.75)=492+184=676. Start by subtracting the offset to get the "real" value (184) multiply by 204 and divide by 10 and you get 3753mA.

If doing it this way watch out so you don't overflow you variables when multiplying. Another aproch would be to use the ** or */ operators as I showed you in another thread.

/Henrik.

[Ioannis]

I had the old Data sheet where there was no such device.

Please accpet my appologies. Now downloaded the new pdf and see that indeed there is a variation of bidirectional device.

050U is unidirectional and has 0.6 volts offset.

So sorry.
Ioannis