I have some of these flicker candles that use 3 AA batteries and I can use them but plugging them into an arduino, which happens to be there anyway, 5v and ground and they work fine. Now I have another that only takes 2 AA batteries and I am wondering if I can use some of these resistors that were in the newbie startup kit to reduce the voltage from 5v down to roughly 3v.
Is this possible?
Resistors can be used here only with a constant current. Try diodes instead, they drop a voltage (about 0.7V) widely independent of current flow.
No kidding, how would I do that. 5v --> positive part of diode--> positive part of load--> ground?
Use 3V3 output of Arduino.
Like you wanted to use a resistor. Use as many diodes as required.
In this case I could, but while I will always have the power supply anyway, I may decide to use the Arduino for something else. That seems to happen a lot. I get tired of one gadget I made, stop using it and then use the device for something else.
3 would do it I suppose.
I would also consider a resistor (1k or 10k) parallel to the LEDs to provide some load when the LEDs are off.
If you have a variable voltage source you can try measure the supply current while slowly increasing voltage. If the current does not increase it means a constant current source is driving the LED and it is probably safe to use the LED with 5 V. I would try it and take the risk but not everyone would do so because the LED may be damaged by this.
Not a good idea IMO, depending on the Arduino. Some Arduino boards (e.g. classic Nano) use the built-in 3V3 regulator of the CH340G chip. While this chip looks relatively big, in my experience it's quite flimsy electrically and overloading it can make it start corrupting data transfer.
It's a different story for boards that have an actual 3V3 regulator, especially if it's a somewhat more beefy AMS/LT1117 - and preferably not the once again quite flimsy 6202/6205.
great point. I should probably not be using the 5v arduino one either.
Anyone know if the arduino 5v is simply passed directly from the usb or does it go thru some other stuff?
Schematics of all Arduino's are published on their website.
Simple answer: when connected to USB, 5V passes through a Schottky diode to the rest of the board including the 5V pin(s). When connected to an external power source through the barrel jack, external power goes through an onboard linear regulator. Not on a Nano of course, as it doesn't have a barrel jack.
I have read about Schottky diodes superficially but what is the purpose of one here? Stop voltage from flowing if insufficient?
It's mostly intended for preventing damage to the USB port/computer in case a mild overvoltage condition occurs on the side of the Arduino. They call it the 'auto-selector' diode because it 'automatically' selects the power source the Arduino runs on - USB or 5V applied to the appropriate pin, without the external 5V source accidentally trying to power the USB port.
It's not a fool proof protection, but it's convenient as long as all voltages remain more or less within reasonable bounds.
I guess that's why you can have both plugged in and it still works!
Not all Arduinos have that 
Look at e.g. an Uno.
You're right of course, the Uno has a more fancy circuit for this. Although you could argue that the body diode of the mosfet functions similarly, but that would be stretching the argument
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