Still Struggling with the formula!

[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

[Ramius]

No problem Ioannis it happens to the best of us. Lol It is actually my fault if I had been thinking I would have included the spec sheet rather than just referencing it. It turns out from trial and error that the A/D readings includes the offset so in the calculations you have to subtract it.

Best, Ed

[Ioannis]

Yes, thats right. It is a constant value that has to be subtracted. The only little matter is that this constant is not so constant and may change with temp or other factors (aging etc.)


So a complete zero after calculation cannot be guaranteed. Even worse,the calculation can result in a negative number, that in absolute valu can be very large positive.


So a regular update of the offset value could be a good idea.

Ioannis

[Ramius]

Hi Ioannis!
Thank you and you are very correct. What will probably be necessary is to add an external 3 second delay to measure the zero point each time time circuit is powered up. That is, power up the PIC, wait 3 seconds, then add the load. Where I have become very stuck is in what I call 4994 land. At 50 amps the A/D will be giving you value of 2987. Subtract a zero point of 490 and you get 2497. Times 2 and you get 4994. It seems no matter what I do this number keeps showing up!