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Topic: Why arduino can fry? (Read 2 times) previous topic - next topic

AWOL

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Low (dc) resistance isn't the issue: you can actually short your avr's pins without damaging them

You've done this?
"Pete, it's a fool looks for logic in the chambers of the human heart." Ulysses Everett McGill.
Do not send technical questions via personal messaging - they will be ignored.

pedroply

Thanks for the reply's and the support.

you guy's answered my question very fast.

Thanks.

kg4wsv

Quote
Quote
Low (dc) resistance isn't the issue: you can actually short your avr's pins without damaging them.


You've done this?


I guess Atmel doesn't know what they're talking about when they give the current limits, both per-output and per-device, in the datasheet.


The simple fact is, you kill the arduino (actually the ATmega) by trying to draw more current than the chip is designed to handle.

-j

jrmcferren


Quote
Quote
Low (dc) resistance isn't the issue: you can actually short your avr's pins without damaging them.


You've done this?


I guess Atmel doesn't know what they're talking about when they give the current limits, both per-output and per-device, in the datasheet.


The simple fact is, you kill the arduino (actually the ATmega) by trying to draw more current than the chip is designed to handle.

-j



A short will either shut the regulator down for the DC input or open the PTC (fuse) for the USB. An overload could easily destroy the outputs. This is one of the reasons I cringe when I see people recommend that an LED being directly connected instead of through a resistor as an LED (or any other diode) is a short that just opens up below a certain voltage.

kg4wsv

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A short will either shut the regulator down for the DC input or open the PTC (fuse) for the USB.


Yes, but that happens well after you hit the per-pin limit of 20mA.

It protects the regulator, or the device on the other end of the USB cable, but it won't do all that much to protect an overloaded output pin.

-j

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