So I am powering my arduino with a 12V 5A adaptor. I was thinking what could happen if someone accidently connected a different specs adaptor (higher V higher A, Higher V Lower A, Lower V Higher A, Lower V Higher A). I was thinking of placing a fuse in line with the positive going to Vin to protect from high voltage. Which fuse should I search for? Would a glass type 12V 5A do (I can't find it somehow)?
Sorry, Even a "fast blow" fuse would not save your Arduino. Fuses are current operated, not voltage, and they have to heat up in order to melt the fuse material. Takes much too long to save solid state devices. Only thing that would work fast enough would be a circuit that senses the voltage and shorts the circuit and that will take out the fuse. Look for a "crowbar" circuit.
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Be careful when using the power plug or Vin. Powering through Vin or the power jack means that the Arduino and all peripherals that are on the 5V rail are powered by the onboard 5V regulator. The on board 5V regulator is not heat sinked so will supply limited current before it overheats and shuts down. The amount of current depends on the voltage input to Vin or the power jack. The higher the voltage the less current can by supplied. I would use a buck converter to drop the 12V to 5V and connect that to the 5V pin on the Arduino, bypassing the, weak, 5V regulator. Then the rated current of the DC DC converter is available on the 5V line.
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A fuse might not save the circuit but a blown fuse might signal that something with the power has gone wrong?
even a 5A adaptor wouldn't be enough? We an directly feed 5V into the 5V pin?
One more clarity please. I am powering a 5V LCD and a 5V relay from the arduino. When the relay is powered the LCD dims a little. Is it because of voltage drop or the current issue?
Yes, you can feed a well regulated 5V into the 5V pin.
It is not the current of the external supply that is the issue. The way that a linear voltage regulator (like the on board regulator on many Arduino boards) works is that the regulator drops the voltage at the input (12V) to 5V on the output. The power that must be dissipated (in Watts) is the voltage that is dropped times the current the circuit is drawing. References that I have seen put the maximum power that the onboard regulator can handle, before it gets too hot, is 1 Watt. With a 12V input the voltage that is dropped is 7V (12V - 5V = 7V). Thus the maximum current that can be drawn and stay under the 1W power dissipation is 1W / 7V = 143mA. An Uno or Nano will draw about 50mA by itself, leaving about 90mA (max) that can be drawn before the regulator shuts down.
The regulator struggles to supply the current required by the relay coil so the voltage drops.
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No, a blown fuse will indicate something on the board as died in a shorted condition. Extremely rare for any solid state device to die that way. They always die with an open circuit. A dead circuit board will tell you the power was wrong.
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Thanks a lot for that clarification...but two questions please...
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If I do not mind the voltage drop, is the system sustainable? Would there be any damages expected to the system in the long run?
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What qualifies as "Well Regulated"? Would something like this do?
No.
No, I would not trust that to supply the current you require.
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Thanks, I thought so too as the LCD screen I just learnt can itself draw as much as 200mA. If I've just got 90mA left by the calculation provided by @groundFungus then with the minimum 15mA required by relay its just too much.
But what's wrong with the module? I saw this too but it says its got 97.5% efficiancy...what does that mean? good or bad?
Buck converters are very hard to build and the layout is critical. That one you showed is a cheap nock off Chinese design. In my experience these can not always deliver the current you need and the current is claims is ambiguous.
If you must get a buck converter then get one from a quality supplier, but a normal regulated supply is all you need.
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But must I use a buck convertor? They come with a pot and I don't want to have to turn the pot and test the output voltage every time. I was trying to find a linear convertor stabilizer module that could step down 12v to 7v but couldn't find any. Can't I just use a resistance to loose 5V?
No. The voltage dropped by the resistor changes with the current through it. You really should learn about Ohm's Law.
Pololu is a reputable company that offers step-down voltage regulators. Each product has a page with specifications and instructions for use.
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Every time what? You set the pot once and then it is fixed.
No, you can simply use a regulated 5V supply. There are lots about for not very much $
This is a link to one. Yes it has a UK plug but then I am in the UK
5V power supply
Search for the same thing in the country you are in.
Note it has a barrel jack on it, don't plug this into the barrel socket of your Arduino, I normally use a barrel socket to wires connector to attach the output of the power supply to the 5V and ground pins.
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Is the system you are asking about:
12V 5A Supply --> Arduino Vraw, and Arduino Vin (aka 5v) ---> LCD and Relay.
So it depends on the relay. The board Vin is not meant to power much. The LCD is OK (if not back light) the relay is likely more that the on board regulator can handle.
What relay are you using?
You didn't say how you are controlling the relay? This can matter a lot.
If you touch the regulator on the board does it feel hot?
Regarding the " LM2596 DC-DC Buck Converter". It will work fine, however I personally try to stay away from such devices for projects I am expecting to work "forever".
I would need to better understand you control of the relay, however perhaps a 12V relay controlled with a MosFet might be a better option.
@JohnRob there is also a sensor involved being powered by Arduino, but I think you are right, my relay module might be the culprit, its this one I am using. The relay module is directly controlled and powered by my arduino.
@Grumpy_Mike I can't use a 5V supply as I am already using a 12V supply to power some hardware controlled through the relay switch (Arduino also powered by the same, but isolated with a diode and a capacitor). Don't want to use two adaptors. I think I'll look for ways to step down the voltage reliably to 5V and power the arduino directly through 5V point, or step down to 7V and power it through Vin.
I'll definitely be thinking of studying and using logic level MOSFETs more actively in my application. I had tried avoiding it with a relay module but I guess it is unavoidable now with such conditions. Using logic level MOSFETs scares me a little for I am not an electronics guy...can you share me a link to a sample application please, to get some ideas on how to use them to power high voltage and high power consuming devices without Arduino feeling anything?
Hey @JohnRob , so I tried the logic level MOSFET way. I used a IRFZ44N MOSFET by this config (I am using both pump and solenoid with the same wiring). I am still feeding Arduino direct 12V in Vin. It works fine, just 1 tiny concern. The moment the pump and solenoid activate, for a fraction of a second I can perhaps notice a voltage drop via dip in screen brightness. And when I power the system for the 1st time for a tiny moment the pump switches on. Why is it so?
The big problem here is that the IRFZ44N is simply NOT a logic level Mosfet.
Because the FET is not turning on fully so there is a limit to how much it can switch. This is caused by it being NOT a logic level Mosfet.
Please post images in the correct way. That link will not work for me.
You may want to read this before you proceed:-
how to get the best out of this forum
Got it, will post images next time. I thought I read somewhere it was logic level mosfet.