I have an Arduino UNO. When I connect an MQ-9 sensor or an ESP8266MOD WiFi chip, the LCD backlight drops that it doesn't display anything, the red LEDs on the Arduino board dims, and the voltage regulator begins to get hot.

I am using a 12V -- 1.5A DC supply to power the Arduino as the USB port only works for uploading sketches, it doesn't even power the Arduino when nothing is connected. What could be wrong?

NOTE: When I unplug the gas sensor, the Arduino board LEDs get brighter and the LCD too, it works fine with some other sensors like an LM35 or a flame sensor.

3 Answers 3


You are overloading the Arduinos voltage regulator.

The Arduino uses a linear voltage regulator, which dissipates the excess voltages as heat.The higher the provided voltage, the lesser current you can draw from it (since it can only handle a specific power dissipation without a heatsink and power is voltage * current). If you have a genuine Arduino, the voltage regulator will go into thermal emergency shutdown, if you completely overload it. Most Arduino clones use a cheaper regulator, which will just burn out and possibly destroy the Arduino.

So if you require quite an amount of current (like the MQ-9 due to its heating element), you should buy an extra voltage regulator, in best case a switching "buck" regulator (since they are way more efficient than linear regulators). For example there are many rather cheap boards with a switching regulator, which take 12V and output a variable voltage controlled via a small potentiometer (which you can trim for 5V output).

the USB port only works for uploading sketches, it doesn't even power the arduino when nothing is connected.

Then most likely you have already overloaded the diode, that is in the USB power path (for protecting the PCs USB port and helping with supply voltage selection). If the diode were fried and failed open, then the connection from USB positive power to 5V rail on the Arduino is no longer there. A genuine Arduino Uno should also have a polyfuse in the USB line, which can save you from drawing too much current. It should connect again, after you removed the devices, that draw too much power.

  • Note that the Arduino Uno Rev 3 does not have a protection diode in the path of the USB, only a polyfuse and a mosfet controlled by an op-amp, which activates the mosfet if VIN is below 7V
    – Ferrybig
    Aug 12, 2021 at 17:54

The MQ-9 has a heating preparation cycle that draws more than a third of a watt which is more power than the Arduino can supply via a pin if I remember correctly. Have you taken that into allowance? You can design the heater power to be triggered by the Arduino without overloading the pin (e.g, using a transistor).

See https://electronics.stackexchange.com/questions/67092/how-much-current-can-i-draw-from-the-arduinos-pins

If your circuit already allows for the above it could also be a bad/shorted sensor? You could test it’s resistance with a multimeter and compare it to the data sheet.

  • 1
    I guess OP uses a sensor module which is powered from 5 V pin
    – Juraj
    Aug 12, 2021 at 8:14
  • 1
    I think your guess is right. Chrisl’s is a much better answer.
    – RowanP
    Aug 12, 2021 at 13:19

You are using a lot more power than an arduino pin can supply. The gas sensor takes a lot of power which the arduino's pin supply of 5v and 200mA MAX cannot tolerate. So I would suggest you power the gas sensor using ur 12v Powersupply by creating a simple potential divider, using a buck converter or simply one of those cheap 7805 regulators. For any other power related queries you can refer this excellent article by BILL from Dronebot Workshop. Good Luck!!

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.