ofcourse i know how to use a transistor of mosfet (best option lower voltage drop) to be able to have more current from the pins of an arduino
but for speedy test project, i wonder if my next picture is a dumb solution
i want to double the current by using 2 pins!
will this help me????
I suggest using a diode on each output pin, to avoid feedback from one pin to the other.
Better would be a simple 2N2222 NPN transistor, with suitable base resistor. The net abounds with driver circuits with such transistors. Same number of resistors, one less output pin. Can also do this with any logic level mosfet, as you've inferred.
Yes, 'off the ark' some will say, but you wanted "simple test", not "modern best-in-class".
OR, drive the LED with two outputs, with a dedicated series resistance for each pin, not your sharing arrangement; that will work, but be sure to set the outputs high or low at the same time (look at port assignments, and port write options).
The 50K resistor is useless since there is a direct short across it.
D7 and D11 are on separate ports, making it impossible to switch the output levels simultaneously. This will lead to one being LOW while the other is HIGH, destroying the chip.
Even with pins on the same port, you would need to use direct port writes instead of digitalWrite().
You could put a diode in-line on each of the outputs, but with a UV led the voltage drop added to the drop from a diode is getting fairly close to Vcc.
This is why I'm recommending series resistances for EACH output. With the sum of those resistances exceeding 125 ohms there would be no scenario where there's an exceedance of 40 mA. If the OP calculates the resistances properly, there should be no risk.
For example, if each output were to source 15 mA(sum is 30 mA, the OP's intent) the resistances would be 120 ohms each ((5V-3.2V)/0.015); the outputs would therefore source/sink 20.8 mA, just at the limit of normal use expectations - but it only happens if the outputs are out of sync; good code practices should limit that to at most a few us.
i'm actually using a Seeeduino (but i don't want to cause that to make my question more confusing so i chose to use a scheme of the nano)
I was referring to the schematic in the initial post, where D7 and D11 are directly connected together.
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Technically, the 30mA for the UV led could be supplied by a single output and still be within the maximum specs.
If your Seeeduino use Atmega328 all post here are valid for it.
No, it would just make all responses irrelevant, if the processor is different. Unless you tell us the real deal(a link to the exact device being used), I'm out.
Yes it is.
To do it right and not destroy your Seeduino it only takes 1 transistor 1 resistor.
@stirf If you have an image to include, put it directly in your message. If that's a product link, I don't recognize the source, and I don't happily click on any link given.
my goal is to try/learn new stuff via arduino and repeating old and working solutions over and over again just isn't my intention, because it's not new, i don't learn anything i just repeat the old ways
cortex M0+ but to me it seems logic that whenever using 2 output pins to double the current works on system A will also work on system B
Start by reading the SAMD21 datasheet and see if what you want to do is even possible
atmega doesn't have a SLEEP modus, i wish to use this function so a regular arduino is not suitable, seeeduino is also programmable via the official arduino IDE, so quite the same
the powerled is already been removed, it's 0402 sized, very small!!
Datasheet and more.
right. As I said before, all answers based on '328s, for example, are meaningless.
go here:
SAMD21 Datasheet
And look for the IOH and IOL (current, output high and current, output low) specifications for your device. Much different, indeed.
Better work on that transistor/mosfet solution, I'm afraid.
