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46  Using Arduino / Project Guidance / Re: Is it possible to use Arduino as Ampmeter for 100mA only ? on: April 20, 2012, 05:22:59 pm
So I've changed the resistor and put it in between the + of the power supply and the 5V of the Arduino.

Voltage power supply = 4.98V (before the shunt)
5V of the Arduino = 4.95V (after the shunt)

Both A2 and A3 give me an analogRead = 1023 !!!!

I think there is a 5V reference problem.
The Arduino must think that the 4.95V supplied is 5V  => full scale of the ADC, so it gives 1023 for the after shunt voltage (4.95V) instead of 1013 !

And for the before shunt voltage (4.98V), of course, it can't give more than 1023 !

It must have something to do with AREF.
I guess, I should give AREF a proper external 5V reference, but where to get it from.

In fact, as I want to use a difference voltage, A2-A3, I need 4.98V<=AREF<=5V.

But If I would need to read a proper voltage, I must give the Arduino a proper and stable voltage reference, otherwise, the reading could fluctuate as the reference is not stable.
Am I right ?

Now,  have to figure out how to use the analogReference function.
47  Using Arduino / Project Guidance / Re: Is it possible to use Arduino as Ampmeter for 100mA only ? on: April 20, 2012, 04:05:50 pm
To michael_x :
Well, many thanks, I made a big mistake !
I put the shunt in between Arduino GND and power source (-)  after regulator.

And yes , I didn't notice but I find negative voltage on resistor ends !!!

I have to reconsider my schematics ! smiley-red

To CrossRoads :

I'm trying this.
Thanks a lot.
48  Using Arduino / Project Guidance / Is it possible to use Arduino as Ampmeter for 100mA only ? on: April 20, 2012, 02:24:08 pm
Hi.

I was wondering whether it is possible to use Arduino as Ampmeter for  about 100mA only .

I've seen some example with a shunt resistor so I wanted to play with mine and print on my LCD the current drain of the system (Arduino, lcd, and other stuff around.

I know it takes about 120-150mA but I can't get my project of printing it on the lcd (neither on the serial monitor). It always says 00.00mA !
I'm using a 0.48 Ohm resistor in series in the power line (two 1/4W ,  1 Ohm resistor in parallel).

Code:
mesureUshunt = analogRead(A2);                         //prend l'intensité par résistance SHUNT 0.48 Ohm
  mesureUshunt1 = map(mesureUshunt, 0, 1023, 0, 1000);   
  Icons = mesureUshunt1 / 475.953;                      //résistance de 0.47953565 Ohm

//print different data on serial monitor for control
Serial.print(" mesureUshunt ");
Serial.print(mesureUshunt);
Serial.print(" Ushunt ");
Serial.print(mesureUshunt1);
Serial.print("  Icons ");
Serial.println(Icons);

//print current drain (Icons) in mA on LCD
lcd.setCursor(8,2);
lcd.print(Icons);
lcd.setCursor(13,2);
lcd.print("mA");


Is my resistor to small , so it is the voltage I'm trying to measure with the analogpin ?

I'd like to be able to measure from 0 to 500mA.

Any advice welcome.
I'm a beginner.
49  Using Arduino / Installation & Troubleshooting / Re: How to switch off Keypad LCD Shield V2.0 -Arduino Compatible on: April 17, 2012, 03:21:37 pm
Yes, I've just done this and it switched off completely !

But how is it possible ?
How much current takes the LCD and its backligth on the Arduino pin when the VDD 5V is off ?

Can it destroy the Arduino taking too much current on its pin. I guess it could because the voltage was falling down 3.5V in that case and even more when I unplugged some of D4-D7  or RS, EN to make some test !
Quite scarry !

Is it normal ?
50  Using Arduino / Installation & Troubleshooting / How to switch off Keypad LCD Shield V2.0 -Arduino Compatible on: April 17, 2012, 02:31:12 pm
Hi.

I get this "Keypad LCD Shield V2.0 -Arduino Compatible" from emartee :
http://emartee.com/product/42054/Arduino%20Keypad%20LCD%20Shield%20V2.0

I'd like to switch it off to save battery  but when I cut off the 5V power attached to VDD of the LCD, the written data on the LCD go off, but the backlight stays on.

It just dim a little. If I unplug RS or EN, it also dim a little more.

Any solution ?

I've been reading on how to switch off LCD when it's alone, and it is said that there is different solution, but to save battery, the best is to cut off main power on VDD.

On my Shield, when I cut off the 5V, I still have 3.50V on VDD
It must get some "back current" from other pin (RS, EN, and D4, D5, D6, D7). How is that possible ?

Here is a schematic I found for the Shield :
http://www.geeetech.com/Documents/LCD%20Kepad%20Shield%20Schematic.pdf.

I'm full beginner.
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