Still Struggling with the formula!

[Ioannis]

Taking into account that your power supply is 5Volts and the max output of the device is 3.6 volts in respect with the flowing current, the reading of the ADC I think is correct.

I don't understand why you double it.

Ioannis

[Ramius]

When things go wrong is that yes (2987 divided by 10 gives 298 -49 = 249. 249 x 200= 49800. 49800 /4096 = 12. 12 x 4167 = 50004 close enough to 50.00 amps. Now do the same with an A/D OF 700 which is very close to 4.24 amps measured and 700 / 10 = 70. 70 - 49 = 21. 21 x 200 = 4200. 4200 /4096 = 1 (in a PIC) 1 x 4167 = 4167 or .4167 amps not the 4.24. What I am trying to say is that if I get the high end of the scale to work the low end does not! Even at a change to 800 the results will not change. Other approaches such as ((700-490) ** 8000) >>4 = 1 so it does exactly the same!

[HenrikOlsson]

At 50A: ADC=2987, offset=485. (2987-485) * 20 = 50040mA (50.04A)
At 4.24A: ADC=700, offset=485. (700-485) * 20 = 4300mA (4.3A)
At 1A: ADC=534, offset=485. (534-485) * 20 = 980mA (0.98A)

Not perfect. However, there are some things that doesn't quite add up here... You should be getting 60mV per A so at 50A you should get 3V or a count of 2457+offset. Now, you say that the offset is 485 which should give you a total of 2942 at 50A, yet you say that you get 2987 but never mind.

In theory, provided that your VRef is exactly 5.000V you will get 49.152 'counts' per A. If you get yoursefl a voltage reference of 4.096V then your ADC will report the measured voltage in units of 1mV instead which means that you'll get 60 counts per A - might make it easier for you.

/Henrik.

[Ramius]

Hi Henrik!
Thank you, thank you, thank you! Yes you are correct and it ended up being a low battery which gave the 485 value. After changing the 9 volts battery which was feeding a 5.0 volt regulator the 0.0 amp reading became 490. Before the battery failure, I placed an external ampmeter is series with the load which was two automobile brake lights. This external ampmeter was showing 4.22-4.24 amps. The other test load was two resistors which gave a load of .869 amps. To estimate what the A/D reading would be (without a known 50.0 A load) I have been testing the formula in Excel. Now that I plug in your formula Excel shows the A/D would be 2989. It is very clear after working on this for about a month I completely over complicated the math! The other part is that you tend to believe your PIC information as being more accurate than an external meter, amp or volt. What also did not help was calling Allegro and being told the 0.0 amp voltage to the A/D would be 0.5 volts and not 0.613 volts measured! Anyone need a approximate 1/4 bottle of 350 count bottle of Excedrin?

[mackrackit]

Anyone need a approximate 1/4 bottle of 350 count bottle of Excedrin?

Why? Are you giving up programming?

[Ioannis]

I do not really know why, but such prescriptions are very rare here. Maybe it is the sun, the sea, the frappe or the good heart most we have. Not for me, thanks!

Ioannis