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Carbon monoxide Detector Help

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pouchito

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Hi All,

I am designing a circuit that detects the CO in the air.
I bought the carbon monoxyde sensor called (MQ 307 hanwei manufacturer) but unfortunately i didn't find on the net any docs related to it.
i ve just found the datasheet which didn't help me too much analysing its operation sad.gif

All the docs i found are related to figaro sensor that have 4 pins while mine has 3 pins

Please can anyone help
Thanks in advance
 
Gather all the information for a project first, design it then order its parts.

Are you going to learn how to speak Chinese then go to the manufacturer hoping that somebody there knows anything about your sensor?
 
I assume this is the sensor you got.

Look here for the info on pins: **broken link removed**
And example circuits

Your device actually has 4 pins (kind of) but either pin A or pin B is connected to one end of the heater coil (pin 3)

See the attatched images

Hope this is of some help to you
 

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I ve checked the datasheet
but i m not able to fully understand its operation
can you help ??
what i understood is either the heater gonna work or the sensor, am i right ?
 
From what i can gather, you need to power the heater with 5V (either DC or AC) and the voltage between pins 2 and 3 will alter depending on the level of CO(carbon monoxide) in the air. Though I'm not 100% sure of this. The variable resistor it shows is used to adjust sensitivity.

The supply voltage needs to be accurate (to within 0.1V) I'd guess you'll be using DC power source because it's probably more convenient for this, and I'd recommend a voltage regulator (7805 or similar) in which case you need a supply voltage of at least 7V
 
what i understood is that the heater voltage must be equal to 0.9v
the power supply =5-6V
either the heater or the sensor gonna work, they will not work together

Am i right ????

what do u think the value of the variable resistor must be ???
 
If you post the datasheet here and we might be able to help you.
 
That link doesn't work for me.
 
Ok, things look different on the datasheet.

So you need 'a low temerature' to detect CO and a 'high temerature' to detect methane. I guess you need to power the heater with 0.2V for the low temp.

Looking at the test circuit, pin 1 is common ground for both the test and heater voltage, pin 2 is the sensor output, and pin3 is the +v for the heater.
Test voltage=5v

If you take a 5v supply, connect the negative to pin 1. connect +0.2v to pin 3, and connect load resistance between pin 2 and +5V, the voltage across the load resistance will alter depending on the level of CO.

If you did the same, but with a supply of +0.9V to pin 3, then the voltage across the load would alter depending on the level of Methane.

I "THINK" that's how it would work, though It will only let me view part of the data sheet.
 
Well I tested the MQ307 which requires a 3V instead for the 0.9V for the heater and i got these results :

in CO:
at t=0 and no heater voltage applied it was 1.74v

heater at 0.6v (for 120s) 1.84v ---- 1.82---- 1.81--- 1.8 ----- 1.73 ---- 1.57

heater at 3v (for 30 sec) 1.3v ---- 1.24 --- 1.22

again heater at 0.6 v 0.28 --- 0.24 ---0.22--- 0.16

again at 3v 1.18 --- 1.17 --- 1.2


at 0.6v 0.3 --- 0.1

I took 0.6V as the low voltage for the heater instead of 0.2V (MQ307A) since there is no datasheet for the MQ307 all we know it is the same exept for the heater voltage

Does the testings seems logical??
Do you think we can do the testings only once since we did not get the same results in the later case??


Thanks for helping Dudes :)
 
If you manage to get this working, how are you going to test it?

Do you have a bottle of CO handy or something?

Or you could burn some charcoal.

Anyway even though you know CO is highly toxic, I recommend you read the Wikipedia artical first before playing with the stuff. One breath of that stuff and you're history for sure!
https://en.wikipedia.org/wiki/Carbon_monoxide_poisoning
 
Hi Hero999,

Well I tested the CO and post the results, can u check them pls and tell me if it sounds logical ???

You r right the CO and his relatives (CO2....) are very toxic
I had an irresistable headache....vomitting....dizzyy....
Can you imagine that i m not able to analyze and interpret what i got from the testings ...still not able to fully concentrate :S
 
Hi

couldn't see page 2, it came up with a message about invalid page codes, then went blank.
Anyway, it seems logical that you'd get lower voltage readings at a higher heater voltage because at the higher heater voltage, the sensor detects methane rather that CO.

How have you tested it in CO? As hero999 suggested, you could burn some charcoal and expose the sensor to the fumes (though I recommend you do this in a well ventilated area)

Ideally you need to compare readings from the sensor being exposed to CO, and reading from the sensor being in 'clean' air (ie sensor outside) so that you can see the difference between the two. This will give you an idea of when CO is present, but because you don't know the exact quantity, you will not know whether it is a 'safe level'

Oh, don't expect to get exactly the same results every time, because you cannot accurately control the levels of CO in the air, and they may vary slightly.
 
I did my testing on a big motor
of course i didn't put directly the sensor on the motor exhaust in order not to damage it
i put it in a long tube then faced it to the CO
Well unfortunately i was poisoned ....cz i was explosed too much to the CO,Co2 ... without taking precautions
 
I tested the CO sensor in air before expolsing it to the CO too and i got normal results but i need ur comment about the results obtained while explosing the sensor to the CO
 
When you say you got normal results, what is normal? were the voltage readings higher? or lower?
There does seem to be quite a big difference between the two sets of readings with the heater voltage at 0.6V, though not much difference between the redings when heater is powered at 3V.

Maybe repeat the test (though in a well ventilated area!) and compare the results to the 2 sets you already have. . .
 
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