How to programme the waspmote to get CO value?

AmroQuandour · **Posted:** Sat May 07, 2011 12:03 pm

The humidity sensor graph is a linear function: -

That is: -

y = mx + b

on the humidity sensor datasheet

y = Voltage out >> ie.- SensorGas.readValue(SENS_HUMDITY) << that value is your voltage out.

x = Relative humidity(%).

From the graph, just pick 2 points on the graph >> I picked the following: -

Y1 = 1.75 volts , X1 = 30

Y2 = 3 Volts , X2 = 70

the gradient m = ( Y2 - Y1 ) / ( X2 - X1) = ( 3 - 1.75 ) / (70 - 30) = 1.25 / 50 = 0.025

to find the intercept ie. b : -

b = y - mx

for y = 3
m = 0.025
x = 70

b = 3 - (0.025*70) = 1.25

Hence Humidity x is : -

x = (y - b)/m

Humidity(%) = ((SensorGas.readValue(SENS_HUMIDITY) - 1.25) / 0.025

That equation will be sufficient for your needs.

The reason for this equation Humidity(%RH) = ((Voltage_Read*5000/3.3 )- 800)/31; is calculated with the maximum resolution of the waspmote. << Look at Section 2.0 Hardware in the Waspmote Manual and you will get your answer.

Your Carbon Monoxide sensor needs to be calibrated before use. The graph provided on the datasheet is in log scale. Therefore you have to use the following equation: -

log(y) = mlog(x) + log(b) ; get the linear erlationship as I showed you above. and from Vout you can work out Rs. << Look at your Gas Board Manual.

Goodluck!!

libelium-dev · **Joined:** Mon Sep 28, 2009 1:06 pm **Posts:** 7487

Hi all,

Great post, AmroQuandour, thank you very much!

sayantanib, if you need further help, please let us know, but I think the post by AmroQuandour should be enough, it is very well explained.

Regards.

libelium-dev · **Joined:** Mon Sep 28, 2009 1:06 pm **Posts:** 7487

Hi sayantanib,

I'll try to clarify it a little bit:

The conversion you should use for humidity:

humidity = float(readValue) * 5000 / 1023;

humidity = (humidity - 800) / 31;

It has been extracted from the agriculture board API, you may take a look at the humidity_conversion function.

Regarding the gas sensor calibration, take a look at this posts, and let us know if you have any other question:

viewtopic.php?f=15&t=392

viewtopic.php?f=15&t=1159

viewtopic.php?f=15&t=1582

Regards.

Amin · **Joined:** Mon Jul 25, 2011 3:36 am **Posts:** 54

It is indeed a great post by AmroQuandour and can be useful for many viewers. However I noticed two small typos in Amro's post which is worth being addressed.

1 : In the post, it is mentioned that in order to calculate m:

the gradient m = ( Y2 - Y1 ) / ( X2 - X1) = ( 3 - 1.75 ) / (70 - 30) = 1.25 /50 = 0.025

which should be changed to :

the gradient m = ( Y2 - Y1 ) / ( X2 - X1) = ( 3 - 1.75 ) / (70 - 30) = 1.25 / 40 = 0.031

and hence :
b = y - mx = 3 - (0.031*70) = 0.81

2: I also think that in the Carbon Monoxide sensor data sheet, both the vertical and horizontal axis of a plot is scaled logarithmically, the plot is referred to as a log-log plot. The equation for a line on a log-log scale would be:

Log(y) = mlog(x) + b or
y = (x^m)(10^b)

instead of log(y) = mlog(x) + log(b)

Regards

libelium-dev · **Joined:** Mon Sep 28, 2009 1:06 pm **Posts:** 7487

Hi Amin,

thank you very much for the corrections, we hadn't noticed them when reviewing the post.

Best regards.

AmroQuandour · **Posted:** Thu Sep 01, 2011 3:28 pm

Dear Amin.

Thanks for the corrections.

You are right about the log equation: -

Log(y) = mlog(x) + b
y = (x^m)(10^b)
The proof is
--------------------------------------------

Log(y) - b = mlog(x)

=> 10^(log(y) - b) = 10^(mlog(x))

=> 10^(log(y)*10^(-b) = x^m

=> y * 10^-b = x^m

=> y / (10^b) = x^m

=> y = x^m * 10^b

---------------------------------------------

Amin · **Joined:** Mon Jul 25, 2011 3:36 am **Posts:** 54

Hi libelium-dev and AmroQuandour,

No worries, my pleasure :-)

Cheers

Cryptik · **Posted:** Thu Aug 02, 2012 5:07 am

So based on these equations, how do I convert a CO sensor value of 2.419355 to ppm? I have the CO sensor setup with a gain of 1 and a resistance value of 10 (10k Ohms).

The sensor is configured as follows:

Quote:

  // get the CO value
  g_gas_co = 0.0;
  USB.println(" - Configure CO sensor");
  SensorGas.configureSensor(SENS_SOCKET4B, DEF_GAIN_CO, DEF_RES_CO);
  USB.println(" - Waiting for CO sensor to stabilize");
  delay(1000);
  USB.print(" - Reading CO sensor ");
  g_gas_co = SensorGas.readValue(SENS_SOCKET4B);
  USB.println(g_gas_pres);
  delay(1000);

I am at a loss on how to convert this to a PPM value.

libelium-dev · **Joined:** Mon Sep 28, 2009 1:06 pm **Posts:** 7487

Hi Cryptik,

First of all, you have to calculate the sensor resistance, which you can get from the output voltage and the load resistance you loaded.

Then you have to intorduce this value in AmroQuandour's formula, or look at the graph. Take into account the you will need to calibrate the sensor to get an accurate measurement (this is, to get the m coefficeint used in the formula) because the ranges of the resistance in air and sensitivity are very wide.

Looking at the graph, and estimating a typical air resistance of 50kohms (be careful with this estimation, it is very rough) you would be having a concentration of around 150ppm, which is pretty high. Under which conditions are you performing your tests?

Regards.

Cryptik · **Posted:** Fri Aug 03, 2012 5:32 pm

So in this thread, three equations were offered to convert the humidity value from mV to %RH... but they all produce different values. For example the following equations are in this thread:

(1) Humidity = ((voltage_read - 1.25) / 0.025
(2) Humidity = ((voltage_read * 5000/3.3) - 800)/31
(3) Humidity = ((voltage_read * 5000/1023) - 800)/31

My humidity sensor returns a voltage of 1.319355 V while indoors in an air conditioned room (using the gas board). This voltage translates as follows using the equations above:

(1) = 2.7741
(2) = 38.6781
(3) = -25.5984

The only value that is close to correct is equation 2 (39% RH). Using a weather monitor, the monitor reports an indoor RH of 45%... so its more likely that equation 2 is the correct one.

My question is what am I missing with the other two equations?

Cheers,
Cryptik