# MQ303A (alcohol sensor)

Hello

i bough a alcohol sensor MQ303A and just wanna double check so i don’t fry anything.

As far as i can see in the blurry schematic it wan’t 0,9 V between pin 1 and 3 and sensor out is pin 2.

In a 5 V environment i would need a 39 ohm resistor in series with the heater.

And some resistor to do a voltage divider i.e. 10k or so.

Have i forgotten anything?

So you had read the document carefully.I don’t think you forget something.

I tried it out now, and it seems to work just fine with a 32Ohm resistor in series with the 3:rd pin down to gnd

So at the moment the connection is as follows:

5 v (via 5v output on seeduino) to pin 1 (closest to the thing sticking out )
Seeduino AD pin 0 to Sensor pin 2 (middle) put a 10k to GND on this too.
GND to 32ohm to pin 3

Gives me values of 10-200 when using a rag of crappy whiskey (vat 69)

Hope this helps anyone that has bought one.

Yes, it gets hot (it’s a heater in it)
My voltage over pin 1 to pin 3: 0.88v (might differ on your since both resistors and sensor isn’t too precise made)

Improvements:
Change AREF to 0.88/0.90, better accuracy, though the sensor seem to be quite sensitive giving a lot of noise, probably fixed with a nozzle and some averaging.
Make a nozzle, since it will go haywire in a room with fumes.
Figure out how to take precise measurements, i’m guessing that blowing harder will give a different result.

Any questions?
real schematic will come upon request.

Sorry to drag up a dead post…

I’m not an EE and I’m having trouble figuring out the math on the resistor calculation. I’ve tried the circuit described in the post, and it works but my math doesn’t agree with the physical results. Please help me find the error in my logic.

I start with two resistors in series, the current limiting resistor and the heater and a 5 volt source. Applying ohms law, the equation is 5 = 0.12*(R + 4.5). Solve for R and you get approximately 37 ohms.

From here I want to find the voltage drop across the current limiting resistor so I apply ohms law again: V = 0.12*37 = 4.44 volts, which only leaves (5-4.44) or 0.56 volts for the heater.

I used a 33 ohm resistor and I measured the resistance on the heater to 4.3 ohms but with those resistors and a 5 volt source, I should see 130 mA but there’s only 120…

What gives?