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I am trying to power my Arduino Mini (which is connected to IR receiver and a relay board). The idea is to hide the entire setup behind the electrical board including the power supply. I have hacked more than a few power adapters (ranging from Mobile chargers 5-7V to 12 V adapters)

This is what I understand from what I have read so far about Arduino Mini power supply.

  1. I can power directly using Vcc pin if I supply 5V DC regulated. (not recommended)

  2. I can power Raw pin by using 5-12V DC and the onboard regulator will take care of converting it. 12V is the maximum, I should not use 12V it can fry the regular if used for a long period as the on-board regulator is a linear one, so a lot of heat.

This is what I understand from what I have tried so far about Arduino Mini power supply:

  1. Don't connect power supply EVER to Vcc, it will harm the board. ALWAYS use Raw.
  2. 12 V supply = puff. instantaneously. No more on-board regulator. Now that Arduino Mini only works if I connect it using Vcc. (the charger was rated 12V 500mA)
  3. Connecting 5V DC (300mA) is heating the regulator alot.

I have connected the Relay and IR receiver directly to the Arduino mini. Pin 10-14 are used for the relays Ins and PWD 3 for IR output.

The mini works with these 5V mobile chargers but the amount it heats up scares me. Any suggestions on what I should do?

Also how much current consumption would be needed. I have tried using a 5V 300mA charger and 5V 700mA charger. Both work almost the same. Why?

I am a computer engineer and very new to Arduino.

Linking a similar question: Most compact method of powering Arduino from wall socket

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  • What? Just take a USB cable apart and solder the positive lead to raw/vin/vcc. You'll get a lower onboard voltage (~4.5V) but it works fine for both relay and IR (rx and tx). You can probably go as low as 3.78V onboard before it gets too weird. Got such a device running here, it's working perfectly. If you've got a clean 5V (+/- 5%) you can solder the positive lead to 5V pin.
    – user400344
    Feb 15, 2017 at 4:41
  • Exactly what I have done as of now. The issue was the regular heating up. I have not measured the temperature, any good easy way of doing so?
    – karx
    Feb 15, 2017 at 6:38
  • I use an off-the-shelf laser IR thermometer with ~95% of temperature (so multiply by 1.05), it works well. The regulator heats up because you pull too much current, drive your loads with transistors. Only use 20mA per output pin, don't get greedy on 5V/3V3 pins, use 200mA, 100mA respectively (yes, they can handle more, I know)
    – user400344
    Feb 15, 2017 at 6:54

3 Answers 3

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I have connected the Relay and IR receiver directly to the Arduino mini. Pin 10-14 are used for the relays Ins and PWD 3 for IR output.

I hope you don't mean that you have connected the coils of the relays directly to the Arduino pins. If that's the case, the surprise won't be the regulator blowing up, but the MCU surviving. Please edit the post and add complete schematics and/or photos of your prototype.

The mini works with these 5V mobile chargers but the amount it heats up scares me. Any suggestions on what I should do.?

Use a small heatsink if you're concerned about the regulator heating up. But it really shouldn't be a problem unless you're driving too much current from it. Have you actually measured the temperature it reaches?

Also, don't feed the coil through the regulator, but directly from the 5V adapter (presuming your relays are 5V). This will save a lot of dissipation in the regulator. NOTE: don't do this if you use a higher (> 5) voltage adapter!!

Also how much current consumption would be needed. I have tried using a 5V 300mA charger and 5V 700mA charger. Both work almost the same. Why?

Those current ratings are maximum values, not constant values. So, if your circuit draws let's say 290 mA, both adapters will be able to do the job.

EDIT: additional information

To safely and properly use relays with arduino, you need some interfacing electronics like this:

MOSFET-controlled relay

Full article from the source, here. Flyback diode: 1N4148. You can use almost any "logic-level" N-channel MOSFET.

If you're using instead a relay module similar to the shown below, all the required interfacing electronics are already built into it and your digital pins won't suffer any overcurrent condition.

Commercial relay module

However, the regulator will (probably) still suffer from excessive current through it. In some of these relay modules you can optionally use a different power line (JD-VCC, see below) for the relay coil. If you own one of these, feed it directly from the 5V adapter, skipping the regulator.

Sample schematic for a real module channel

If the relay module doesn't have a separate power line, it may still work when connected to the 5V adapter if the dropout in the regulator is not too big (i.e., the "high" level from the digital pins is not too far away from the adapter voltage).

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  • Hi, thanks for answering one point after the other. When you say "connected the coils of the relay", you mean to say Vcc is connected from the boards 5V pin? Well the answer would be yes in any case because all 6 pins connected to relay are from the board itself and I will change it. Just want to understand what you mean by coil. Also as suggested I will measure the temperature and update you, as of now I don't have a good way measuring the temperature. I will see and get a laser temperature thermometer.
    – karx
    Feb 15, 2017 at 8:10
  • You're welcome. I'm concerned about how you may be wiring the relay, so I will update the post will some additional info about it.
    – Enric Blanco
    Feb 15, 2017 at 8:19
  • The coil is located between the input (control) terminals. The relay is activated by making a current flow through the coil, which is achieved by applying a voltage to it. Depending on the relay, this current can be as high as 100-120 mA per relay. The arduino pins can't neither source nor sink that much current. The arduino mini onboard regulator may also struggle if several relay are powered from it.
    – Enric Blanco
    Feb 15, 2017 at 9:09
  • If you have a multimeter, check whether it can do temperature readings. If it can, then you just need a temperature probe (they're not expensive) for it.
    – Enric Blanco
    Feb 15, 2017 at 9:12
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The Arduino website says you should not power your Arduino Mini with more than 9 volts.

Warning: Don't power the Arduino mini with more than 9 volts, or plug the power in backwards: you'll probably kill it.

https://www.arduino.cc/en/Main/arduinoBoardMini

The 5 to 12 Volts (not more than 12V for long time periods) refers to the Arduino UNO and probably others.

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  1. You certainly can connect a 5v or 3.3v to the Vcc pin for most boards. Most regulators can take a reverse voltage. To be sure, check the data sheet or place a reverse diode on the regulator.

  2. Again, depending on the particular regulator your board has. Most of them are good for 16 to 18v. But that can change drastly if more load current is involved.

  3. Well, that's your answer then. Put a reverse diode across the regulator, Between the Vcc and raw pins.

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