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[on:]

I'm trying to use arduino with LCD as car voltmeter. I've got R1=33kohm and R2=15kohm resistors as voltage divider. So if voltage stays below 15V from alternator it wont break arduino by going over 5V(it is the Vmax for inputs?)

But i cant figure out the how to make it show voltage correctly. I've got R1 and R2 in A0.

I tried to do it with this http://www.arduino.cc/cgi-bin/yabb2/YaBB.pl?num=1292921201 but i dont know the right formula for me.


I've got now

Code:

value = analogRead(A0);
vout = (value * 3.4)/1024.0;  //voltage coming out of the voltage divider
vin = vout / (R2/(R1+R2));  //voltage to display

But i think i need the right value to replace 3.4?

[Reply #1 on:]

But i think i need the right value to replace 3.4?

Whatever value is connected to Vref.

[Reply #2 on:]

You mean Ref-pin? Nothing?

[Reply #3 on:]

How about 5 then?

[Reply #4 on:]

I've tried that. But when fluke shows 6.9V at resistors, arduino shows 6.6V. Is it possible to make it accurate or do i just need to finetune that value? Or are 1% resistors just not right for it?

[Reply #5 on:]

I'd leave the 5V as it is (assuming the supply voltage really is 5V) , and check that the values of resistors plugged into the potential divider equation are correct.

[Reply #6 on:]

Arduino adc is max rated for 5.5v, so 6.9v is good to kill it.

[Reply #7 on:]

He's right - I assumed when you wrote "at the resistors", you meant across the potential divider.

[Reply #8 on:]

15Volt  over 33K and 15K gives  max 4.7 volt over the 15K resistor.
As Vref is not connected it uses 5Volt from the Arduino Power supply - you need to measure this as between 4.9 and 5.1 reference voltage = 4% difference!

raw = analogRead(A0);

// raw = 0..1024    representing 0..Vref  volt

vout = 5.00 * raw / 1024.0;  // as Vref = 5.0 Volt (to be measured)

convert to car voltage;

voltageCar = (vout  / 15.000) * (15.000 + 33.000);  // or in short voltageCar = vout * 3.2:

Please fill in the resistors measured values .

The final code could become

Code:

// these defines have to be measured
#define VREF  5.10
#define R1 32767.0
#define R2 15140.0

float raw = analogRead(A0);
float vout = VREF  * raw / 1024.0;
float carVoltage = vout / R2 * (R1 + R2);  

// or in a oneliner
// float carVoltage = analogRead(A0) * VREF / 1024.0 / R2 * (R1 + R2);
// as VREF / 1024.0 / R2 * (R1 + R2); are all constants one could compact this to

carVoltage = analogRead(A0) * 0.01576; // used sample constants above.  

-- updated layout --

[Reply #9 on:]

I tried to add measured values from resistors but its still off. Can this even be very accurate? I've changed a bit of R1 and R2 values and now im tenth off comparing to fluke. I think im satisfied with this

Thanks guys!

[Reply #10 on:]

The precision of tha ADC of the Arduino is 10 bit, in practice the last bit may be inaccurate. That means you have approx 500 steps for 0..15 volt ==> at best 1/30 volt precission = 0,2% . Assume the resistors have an accuracy of 1% . suppose VREF has an accuracy of 2%.

Then - float carVoltage = analogRead(A0) * VREF / 1024.0 / R2 * (R1 + R2); - would give an max accumalate error of
                                                    0.2% +     2%  +    0%     + 1% + 1% + 1%  = 5.2%

=> 15V * 5.2% = 0.8 Volt max error => so one tenth of is indeed very good.

Disclaimer not official error math, just an first order approximation.

One can improve the measurement by doing the AnalogRead multiple times:

Code:

float readVoltage()
{
  #define SAMPLES 3.0
  float rv = 0;
  for (int i=0; i< SAMPLES ; i++) rv += analogRead(A0);
  rv /= SAMPLES;
  return rv * 0.01576;   // average of 3 reads times the calculated constant.
}

If time is not an issue one can take more samples ...