4

I have this Arduino Pro Mini clone (ATmega328) 5v.

I gave it exactly 12 volts through the RAW power input, but it instantly got hot and started fuming. Fumes were coming out of the area close to the black S4 thingy. I don't know why, because I thought it can accept up to 12v through that input.

Now when I give it 5v only the middle led is lighting up and it's not connecting to the USB-UART interface anymore.

Is there any chance to save it?

arduino pro mini

3
  • By the description it sounds like you either, hooked the 12v in reverse (damaging the input diode [s5]) or accidentally burned out the power regulator. What happens when you put the 5v (regulated source) to the VCC & GND pins bypassing the regulator? Oct 3, 2020 at 17:55
  • @Tim_Stewart when I give it 5v through VCC input, the main red light is just lighting up. If I give it 5v through the RAW input, it also lights up, although it is dimmer. In neither case the built-in led (L) is lighting. And also it does nothing when I click on the red button.
    – Sergio
    Oct 3, 2020 at 18:22
  • @Sergio, what happens if you completely desolder the voltage regulator I circled in my answer I added? Can you then use the Arduino Pro Mini by powering it with a regulated 5V directly to Vcc? Oct 4, 2020 at 7:53

3 Answers 3

12

1. Your microcontroller is probably/maybe fine; your tiny little voltage regulator is dead; here's why:

Sergio, yes, you can probably/perhaps save it: your Arduino (the microcontroller--ie: processor at least) is likely just fine, assuming the linear regulator smoked and failed open, meaning it did NOT pass through the raw input voltage to your ATmega328 mcu. It's the linear voltage regulator that is fried! That's this circled component below:

enter image description here

So, desolder that part and throw it away and you can continue using the Arduino just fine. Just power it with a regulated 5V directly to VCC is all and you'll be fine. I've done exactly this on some parts I exploded or smoked the regulator on (see below), and they continued working just fine.

IMPORTANT: if this is a safety-critical part, or a product, throw away the whole Arduino just in case.

Details on the fixes:

  1. Desolder the voltage regular (circled in the image above, and shown desoldered on my Pro Mini in the photo below) and throw the desoldered voltage regulator away.

  2. Power the board now through Vcc with anything between 3.3V and 5V (recommended: a regulated 5V into Vcc, such as the 5V line coming straight out of your USB hub).

  3. Reflash the bootloader with a USBasp programmer (shown in 2nd of the two photos below).

  4. Upload a new blink sketch via the external USB serial UART to blink the built-in LED13. This works fine on my Pro Mini below shown with exploded and desoldered regulator:

    // the setup function runs once when you press reset or power the board
    void setup() {
      // initialize digital pin LED_BUILTIN as an output.
      pinMode(LED_BUILTIN, OUTPUT);
    }
    
    // the loop function runs over and over again forever
    void loop() {
      digitalWrite(LED_BUILTIN, HIGH);   // turn the LED on (HIGH is the voltage level)
      delay(500);                       // wait for a half-second
      digitalWrite(LED_BUILTIN, LOW);    // turn the LED off by making the voltage LOW
      delay(500);                       // wait for a half-second
    }
    
  5. Unplug the USBasp programmer when done, and continue development with just the external serial UART plugged in.

Here's some photos of my Arduino Pro Mini with exploded voltage regulator that I desoldered. I found it today. The Arduino still works fine! I desoldered the regulator around the year 2014 or so, but just re-flashed the bootloader, uploaded the code, tested it, and took these photos today:

Image 1: Arduino Pro Mini used as a data-logging altimeter in ~2014; photo from today. The voltage regulator exploded (read below), so I desoldered it. Both the power LED (top center of board), and LED13 (bottom-left of board) are ON. The power LED comes on automatically so long as you are powering the board through Vcc and GND from your USB port through your USB UART adapter, and the LED13 is blinking ON here due to the blink sketch above AND by powering the board with Vcc and GND from the USB serial UART to the Pro Mini. Note that the LED13 would NOT light up properly with just the USBasp plugged in, even when it was properly powered and programmed by the USBasp. I think it's because the USBasp was interfering with that pin. Plugging in the external serial UART to the Pro Mini's Vcc and GND fixed that, allowing LED13 to function properly as well while the Pro Mini ran the blink sketch above.

enter image description here

Image 2: External USB serial UART (in the USB hub top port), USBasp programmer wired to program the bootloader (in the USB hub under the USB UART), Pro Mini (bottom left) with power LED ON and LED13 (bottom-left of board) OFF.

enter image description here


On the surface, you didn't do anything wrong! You plugged 12V into the correct pin (RAW, which is the battery-in or voltage in line), and that is within the voltage limits of this part. BUT, what you neglected to consider is the power dissipation capability of this footprint or size of a part. The power dissipation capability is soooo low that realistically, 12V is too much, and will destroy the part.

6 years ago or so I made a couple little data-logging altimeters with the Arduino Pro Mini, connected 12.6V (a fully-charged 3S LiPo battery) to the RAW pin, and 15 seconds later: pop! They exploded! Literally, that little voltage regulator exploded into bits with 12.6V in. After that, I never put more than 8.4V in (a fully-charged 2S Lipo battery), and all was well. Since I was powering it from my main RC plane's 3S LiPo motor battery, this was accomplished by just tapping off of 2 of the 3 cells in the battery by plugging into the correct points on the battery's balance lead. So, I was able to use the same battery but just use 2 of its 3 cells instead of all 3 of them.

These regulators are just so tiny that they can't take much heat is all, and since they are linear regulators, all excess voltage is burned off as heat. Here's a couple example power dissipation calculations to make the point:

Imagine you are drawing 100mA total (0.1A) from this regulator. With 12V plugged in, you have to burn off 12V - 5V = 7V in that regulator chip.

Power_dissipated = Current*Voltage = I*V = 0.1A * 7V = 700mW

700mW is HUGE! Pop! Explosion! Dead voltage regulator chip.

But, if you feed it just 8V this same chip would have to burn off only 8V - 5V = 3V of excess voltage:

Power_dissipated = Current*Voltage = I*V = 0.1A * 3V = 300mW

That's 300mW/700mW = 43% as much power, or 57% less power. These numbers are just for example purposes, but you get the point.

Solution? 1) switch to Arduino Nanos, and/or 2) use a much lower input voltage (ex: 8~9V instead of 12V!). Nanos have MUCH bigger and beefier voltage regulators, and Nanos cost like $1 more when buying the Chinese clones.

2. How to calculate 1) how much power your regulator can dissipate without burning up, and 2) how much current you can therefore draw for different raw input voltages to the voltage regulator:

(Also: I wrote an article on my website covering more of these power, current, and voltage specs--see link here and at bottom.)

By the way, looking at the Pro Mini schematic here, I see that the original board used a MIC5205 linear regulator. The Chinese clone you have certainly uses a different regulator, but let's look up the MIC5205 datasheet for benchmarking purposes. A google search for "mic5205 datasheet" leads me here: https://ww1.microchip.com/downloads/en/DeviceDoc/20005785A.pdf. The device is rated up to 150mA current and up to +16V input voltage, but you must consider power!

The equation for that is hidden on pg 5 here:

enter image description here

Power_dissiplation_max =  P_D_max = (T_J_max – T_A)/theta_JA

theta_JA = thermal resistance from Junction to Ambient = 220 degC/W

T_A = ambient temperature = let's say, 40 deg C (104 deg F)

T_J_max = max junction (internal chip) temperature = +125 degC

So, we crunch the numbers:

P_D_max = (125C - 40C)/220 C/W = 85C / 220 C/W = 0.386W = 386mW max
power.

Final result: this tiny, linear voltage regulator, assuming absolute max junction temperature of +125 C (257 F) [HOT HOT HOT!--not good] and ambient air temperature of 40C (104 F) can dissipate 386mW max power before you risk making it go "pop"!

So, assuming you power it with 12V, how much current can you draw?

Voltage drop across the regulator = V_D = Vin - Vout = 12V - 5V = 7V

Power dissipation in the regulator = P_D = current * voltage drop = I*V_D

Solve for I:

I = P_D/V_D = 0.386W/7V = 0.055A = 55mA current! 

With 12Vin and 5Vout, any more than 55mA current draw goes "pop" or smokes the chip!

With 8Vin, however, you could draw:

V_D = 8V - 5V = 3V
I = P_D/V_D = 0.386W/3V = 0.129A = 129mA current

With 8Vin and 5Vout, any more than 129mA current draw goes "pop" or smokes the chip!

So, use a lower input voltage and you can draw much more current! How low can you go? Don't go lower than 5V + dropout voltage. The datasheet on p3 for this regulator shows a max dropout voltage of ~350mV, so stay > 5.35V input to get a regulated 5V output.

enter image description here

That's all for this MIC5205 chip, though. Note that other voltage regulators may have dropout voltages as high as 2~3V, so you'd have to have an input voltage of at least 7V or so to ensure a regulated 5V output. I can't see the numbers well enough on your chip to google for its actual Chinese or whatever datasheet to do the calculations for your chip, but you can use my example calculations above to redo these calculations for any datasheet or part.

Other reading:

  1. Google search for "sot 23-5 power dissipation"
    1. Microchip Application Note AN792: A Method to Determine How Much Power a SOT23 Can Dissipate in an Application
  2. I also wrote an article on my personal website on 6 Jan. 2014 called "Arduino Power, Current, and Voltage Limitations", here, for more info. on these limitations.
11
  • Please explain downvotes. This is one of my more well-thought-out answers, and one where I have real experience with this exact same part and exact same scenario at the exact same input voltage. Oct 4, 2020 at 5:30
  • Nevermind. I know who downvoted my answer. Enough said. It's not related to the answer quality, educational content, or correctness. Anyone other than that person, please consider the quality and correctness of this answer in making a judgement on whether or not to downvote it or upvote it. Oct 4, 2020 at 5:40
  • 2
    Do not worry about down-votes. The system lets people up or down vote answers as a way of eventually "grading" answers (or questions). A single down-vote means nothing.
    – Nick Gammon
    Oct 4, 2020 at 8:09
  • 1
    Thank you for such a thorough answer! I desoldered it completely as you suggested, it still lights up the middle light (but not the built-in one) and also it fails to connect to USB-UART adapter. Sad.
    – Sergio
    Oct 4, 2020 at 22:29
  • 1
    Downvote trolls here too huh? Is there a community on SE that doesn't have them? Oct 6, 2020 at 15:16
1

Unfortunately, it sounds like it is dead.

I did this early on when I started with arduinos. In my case I plugged in a "12v 1amp" wall adapter from a linksys router that I didn't check with a DMM to a uno-R3 barrel jack.

The adapter was putting out 22v @ god only knows how many amps. But it had the exact same symptoms I am sorry to say.

I wound up going with a cheap 5v/3.3v selectable breadboard power supply. You might want to look into those for future use, I got mine for less than $3 on eBay.

https://www.ebay.com/itm/1pcs-Breadboard-Power-Supply-Module-Shield-3-3V-5V-MB102-Solderless-Bread-Board/133519600754?hash=item1f16636c72:g:9NcAAOSwZzNfYHel

3
  • 1
    god only knows how many amps - adapters don't "put out amps". They have a capacity for current. That's like saying I dipped my cup into a pond "god knows how much water was in it". You get out the amount you require. The excess capacity is not a problem.
    – Nick Gammon
    Oct 6, 2020 at 7:21
  • @nick gammon, the point I was trying to make was the 12v adapter was damaged, I didn't check it. Once that part is blown, does it have the same resistance? (I.e volts / resistance = amps). Am I missing something? Oct 6, 2020 at 14:53
  • Unregulated adapters may well have a higher voltage under little load. They would be designed to output (in your case) around 12V under the load of the device it is designed for. Regulated adapters output the advertised voltage (or should do) regardless of the load.
    – Nick Gammon
    Oct 7, 2020 at 7:53
1

Wall adapters are not all the same. They can have AC and DC outputs. The DC outputs can be regulated or unregulated. Assuming 1 12V DC wall adapter unregulated you can get maybe 19Vdc from it depending on the design.Always check first, it is much more cost effective then going to the store to buy a new Arduino and what other devices.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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