Convert voltage to dB

[louisbourdon]

Hi
I'm looking at this link (http://www.sengpielaudio.com/calculator-gainloss.htm) to get the formula for converting voltage to dB. I use a sound sensor connected to Arduino (0-1024) and I want to convert this to acoustic energy (dB) (for display, etc).

As I understand this, 1024 values (or 0 to 5V) would equal to 60 decibel units, so if I use the formula -> (20*log10)*V/V0, I will get from 0 to 60 - and then I can offset by an amount if I want to match proper settings (i.e. 0 would then equal approximately 20-30 dB).

Does this sound alright?

[robtillaart]

yep except that Arduino goes from 0..1023

[Magician]

As I understand this, 1024 values (or 0 to 5V) would equal to 60 decibel units, so if I use the formula -> (20*log10)*V/V0, I will get from 0 to 60

No. Voltage is RMS value in the formula, your range 0-1023 is peak-to-peak. You have to divide by 2 in the first, and than by sqrt(2). 1024 / 2 * sqrt(2) = 362.  Roughly, you can get about 53-54 dB at the best - perfectly centered steady sine wave.  Less than 40 dB for normal audio content.
  Options: use ADC with more bits, or averaging.
http://coolarduino.wordpress.com/2013/01/09/audio-vu-meter/
 

[louisbourdon]

I try this code, but it doesn't give correct results. Someone can help to fix the formula?

Code: [Select]

const int analogInPin = A0;

int sensorValue = 0;

void setup() {
  Serial.begin(9600);
}

void loop() {
  sensorValue = analogRead(analogInPin);           

  Serial.print("sensor = " );                       
  Serial.println(sensorValue);     

  double db = (20. * log(10)) * (sensorValue / 5.);
 
  Serial.print("db = ");
  Serial.println(db);
   
  delay(500);                     
}

[Magician]

Try:

Code: [Select]

double db =  20.0  * log10 (sensorValue  +1.);

[MarkT]

As I understand this, 1024 values (or 0 to 5V) would equal to 60 decibel units, so if I use the formula -> (20*log10)*V/V0, I will get from 0 to 60

No. Voltage is RMS value in the formula, your range 0-1023 is peak-to-peak. You have to divide by 2 in the first, and than by sqrt(2). 1024 / 2 * sqrt(2) = 362.  Roughly, you can get about 53-54 dB at the best - perfectly centered steady sine wave.  Less than 40 dB for normal audio content.
 Options: use ADC with more bits, or averaging.
http://coolarduino.wordpress.com/2013/01/09/audio-vu-meter/
 

But the smallest signal you can see is 1 count peak-to-peak, not 1 count RMS,
so its 1023 times smaller than the max, ie 60.2dB less.

[Magician]

1 count peak-to-peak, not 1 count RMS,
so its 1023 times smaller than the max, ie 60.2dB less.

I never heard someone would measure DC voltage in dB. For AC there is an offset,  and full span is only 512 , it gets to 6.02 * 9 + 1.76 = 55.94
I agree, that 512 is higher than 362, which I counted in integer math term, and basically isn't always the case

[louisbourdon]

So, is there any formula that works to convert the voltage in dB SPL? I have tried many different ways, but I always get false results. There has to be something out there - or if could someone direct me to resources?

Again, I want to convert the voltage in db SPL (sound pressure level), with a minimum value of around 30dB, and a maximum at around 90dB.
Thanks!

[Magician]

So, is there any formula that works to convert the voltage in dB SPL?

No, otherwise it has to include mic sensitivity , amplification coef. and arduino reference voltage.  You have to calibrate with a standard device, in special soundproof chamber.

[juanernesto40]

maybe can try with this:
double db = 20. * (log10 (soundSensor));

[KeithRB]

Is your sensor just a microphone? If so, you cannot just convert any single value to dB, you need to know the amplitude of your signal, and possibly filter it for A-weighting if this is for an SPL meter.

[louisbourdon]

After some time I returned to finish this project. So far, still, no answer to make it work as closely as possible to as a dB meter.

I'm reading this article (http://www.inmotion.pt/store/phidgets-sound-sensor) saying that there is a formula that can be used to get the dB conversion (for 1kHz sound for example).

SPL at 1kHz tone (dB) = 16.801 x ln(sensorValue/1023) + 9.872

I tried the formula but I got nothing. I don't know what the ln stands for 

Maybe someone could help?

[KeithRB]

The *second* hit if you google "ln" tells you it is the natural logarithm. (i.e., base e, not base 10)

[louisbourdon]

Sure thing, but how this is implemented in code? Cause this is not working...

double db = 16.801 * log (sensorValue/1023) + 9.872;

[KeithRB]

You probably need to cast sensor value to a float so that the division is done in floating point.