I'm a novice and I really don't understand max current draw when it comes to the 5V pins. I'm also unsure of the best place to post this as it's more of a general Arduino Mega question than it is a sensor question.
This link says 800mA is the max, or 200mA per pin.
This link says "At 5V, current = 5/33 = 0.151 A = 151 mA. The regulator on a typical Arduino will overheat if you try to draw this from the Vcc pin."
Granted, they don't mention the mega specifically, but I see the max for the Uno is 400mA. Do the regulators really overheat when drawing much less than their max?
Budvar10 seems to make a similar claim in this link. "The voltage about 12V and current above 200-300mA cause its heating."
I'm working with the various MQ sensors for fun, and I'm trying to figure out how many I can connect to one board. I'm not noticing any overheating issues with my MQ-3 contrary to the second link I cited.
And this may be a stupid question, but if the limit is 200mA per 5V pin, and I want to connect multiple sensors drawing around 150mA, am I able to connect the multiple 5V pins and grounds to the same circuit in a parallel fashion to reach the max?
None of your links work for me.
There are different ways to power the Mega and different ways to draw power from the Mega. If powered by USB, the internal 5V regulator is bypassed so you can draw about 500mA from the 5V pin. If powered through the Vin pin or power jack, the current is provided by the internal regulator and the current that you can draw from the 5V pin depends on the voltage supplied at the Vin pin or power jack. The higher the supply voltage the less current that the regulator will supply before overheating and (hopefully) shutdown. The recommend max power dissipation for the regulator is 1 Watt. So with 12V into the regulator the max current is about 140 mA (1W / (12V - 5V)). The Arduino uses around 50ma of that leaving less than 90mA (max) for everything else.
With 7.5V at Vin, 1W/ (7.5V - 5V) = 400mA.
Or you can use a 5V regulated supply (DC-DC convwerter) connected to the 5V pin. In that case the internal 5V regulator is bypassed and the current available through the 5V pin is the regulated supply current.
That is how much we can draw from the 5V pin. Also there are limits to what each chip pin can source or sink (recommended 20mA), and the chip can source or sink (200mA). There are other limits. Consult the data sheet.
Thanks! The links should be working now. I realized after I made my post that Budvar was commenting about the regulator overheating because of the voltage differential. The way you broke it down is extremely helpful, though.
Am I correct in understanding that I could use a 5V 2A DC power supply because it would bypass the regulator, but if I miswire something it would be most likely catastrophic?
Connect the 5V and ground from the supply to the 5V and ground of the Mega. Connect any peripherals that need 5V to the power supply 5V and ground, not the Mega 5V and ground. Do not run high currents through the Mega ground. That will provide protection to the Mega.
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The Mega chip s rated for 200mA on its VCC and Gnd pins. So the chip could in theory draw 800A to source as High signals, o or to sink as Low signals, across it's IO pins.
The 5V regulator is rated for 800 or 1000mA. It wont achieve that without overheating and shutting down, or failing.
For best results, feed no more than 7.5V into the barrel jack, and don't expect to get more than 300, 400mA before it overheats.
If you connect a 5V, 2A or more regulated to the 5V/Gnd on the power header, the board will be fine.
If you get it connected backwards, there is no protection. The reverse polarity protection diode is between the barrel jack and the 5V regulator, while the power header is After the regulator.