What is too big of current/voltage that will fry the voltage regulator on the arduino :~?
Over 1A.
The voltage will depend on the current you are drawing, see:-
http://www.thebox.myzen.co.uk/Tutorial/Power_Examples.html
I believe the proper answer is that if you attempt to draw too much current from the output on the on-board +5vdc voltage regulator, you won't 'fry' the regulator but rather just cause it go into shutdown protection mode as part of it's internal over current and over temperature internal protection circuity. However applying too much input voltage to the regulator is certainly a possible method to 'fry' it, check the datasheet of the specific regulator to find out that maximum input voltage rating.
Lefty
The answer will depend. High voltage (even with very small current) can kill it. Lots of current (even with very low voltage) can kill it. Middle current + middle voltage (lots of power dissipation) can kill it too.
dhenry:
The answer will depend. High voltage (even with very small current) can kill it. Lots of current (even with very low voltage) can kill it. Middle current + middle voltage (lots of power dissipation) can kill it too.
So the datasheet features of a typical (here a 7805)mean nothing and are simply lies?
The internal
current-limiting and thermal-shutdown features of these regulators essentially make them immune to overload.
Lefty
Depends on the definition of "fry".
- gets very hot.
- goes into thermal shutdown.
- gets so hot it's damaged.
In point of fact, even though the v.regs on Arduino boards tout 1 Amp, you usually
cannot draw anywheres near that amount of current without them getting
#1 = very hot, even if #2 = thermal shutdown doesn't occur, and #3 = not damaged.
IE, the tiny smt parts like the SOT-223 [the usual part found on an Arduino bd] are
rated at 15 degC/watt temperature rise, while the robust TO-220 parts are
3 degC/watt.
Even though they say max operating temp = 150 degC, you're not really gonna want to
let it run anywheres near so hot. 100 degC boiling water will scald your hand nicely,
after all.
If you limit to 80 degC [176 degF, which is still darn hot], ie 55 degC rise, this is
55/15 = 3.6 watts for SOT-223. With Vin = 9V, you get Pd = Vdrop * I, or
I = 3.6W/(9V-5V) = 0.9 Amp.
IOW, if you try to draw fully 1 Amp from the SOT-223, expect it to get very hot.
IE, the tiny smt parts like the SOT-223 [the usual part found on an Arduino bd] are
rated at 15 degC/watt temperature rise, while the robust TO-220 parts are
3 degC/watt.
Those figures assume infinitely large heatsinks.
Reality is quite different.
Those figures assume infinitely large heatsinks.
At this point, a lot of people will suggest a dc-dc converter or else say, "the pcb acts like a
heatsink", but of course it's not inifinite, and no one has probably tried to measure it's
effectiveness in any event.
Reality is quite different.
So, what is obviously clear is that the figures I gave are a 'best' case estimate, and in actual
practice, the v.reg will run even hotter yet. I personally don't like having my pcbs getting
so warm as 176 degF.
what is obviously clear is that the figures I gave are a 'best' case estimate, and in actual
practice, the v.reg will run even hotter yet.
What you have assumed is that the thermal resistance from the case to ambient is 0 (aka an infinitely large heatsink).
Most sot223 chips have no heatsink other than a puny copper trace on the board. You can check the datasheet to see the j-to-a thermal resistance of the package and the math is easy.
In general, it is good to assume 100c/w thermal resistance for those devices. That means in reality a 0.5w power dissipation is the most you should put on such a device.
Those 1amp ratings for such small packages are simply meaningless.
Helps to look in the right place, from LM1117 d/s,
Thermal Resistance
Junction-to-Case
3-Lead SOT-223 15.0 °C/W
3-Lead TO-220 3.0 °C/W
3-Lead TO-252 10 °C/WThermal Resistance
Junction-to-Ambient
(No air flow)
3-Lead SOT-223 (No heat sink) 136 °C/W
3-Lead TO-220 (No heat sink) 79 °C/W
3-Lead TO-252 (Note 9) (No heat sink) 92 °C/W
3-Lead TO-263 55 °C/W
8-Lead LLP(Note 10) 40 °C/W
From this, with poor heatsinking, the SOT-223 would experience a 68 degC rise at 0.5 Watt,
meaning gets rather hot with Iload = 0.5W/(9V-5V) = only 125 mA. Long way from 1 Amp.
Long way from 1 Amp.
That's the point. Unless you can put those things in liquid nitrogen, forget about going much above 0.5w.