40ma is absolute maximum, but 20-30 ma is a better maximum limit standard to stick to. Keep in mind due to internal resistance of the output pins the output voltage will sag as current draw from the output pin is increased. Just shows that there is a difference from fundamental principal and electrical reality of real world components. That is why the datasheet for devices are so important to study and understand a chip limits and capabilities.
Because there is no exact answer. If the board is being powered via USB then there is a 500ma thermofuse that sets a maximum limit for the board and anything you wire up to it. But if powered from an external DC voltage source then there might be somewhat more but it depends on the exact DC voltage input as there is a heat dissipation limit for the on-board 5vdc voltage regulator. So hard answers are desirable but often allusive in the real world.
No, totally different components, and a switch doesn't share all the characteristic of a switch and visa versa.
When performing Ohm's Law calculations, what specifies the values for V and I?V, to me, should be the supply voltage (3.3V or 5V would be what I'm using in this case).
QuoteWhen performing Ohm's Law calculations, what specifies the values for V and I?V, to me, should be the supply voltage (3.3V or 5V would be what I'm using in this case). No, V is actually the voltage "drop" across the resistor. Eg, if 1K is tied between 5V and 3.3V, the current I = Vdrop / R = (5V - 3.3V) / 1K = 1.7 mA, and not 5 mA or 3.3 mA.
32Mhz (forgive the capitalized M if it is incorrect please)
The algebraic sum of the products of the resistances of the conductors and the currents in them in a closed loop is equal to the total emf available in that loop.
No, V is actually the voltage "drop" across the resistor. Eg, if 1K is tied between 5V and 3.3V, the current I = Vdrop / R = (5V - 3.3V) / 1K = 1.7 mA, and not 5 mA or 3.3 mA.In your example, you mean a 1K(ohm) resistor, right? (just making sure--I'm really new, if you haven't figured it out yet. )So, in this LED (for example), it says "forward voltage drop" is 1.8V-2.2V (DC). Let's say I have a circuit that has 5V---LED---1K(ohm)---Gnd. I want to find the current. I = Vdrop / R = 2.2V / 1K = 2.2mA.Right? Or am I still confused... :/
or 1 / .2 = a 5 ? resistor?
The moment I hit post I knew I screwed up, but the dog asked to go out and you caught it before I could fix it .Oh, the old "the dog eat my correct posting" excuse, go to the corner. so it should be a 50? resistor... Do you know HOW I know I want to learn this? Because I keep on trying even though I have been wrong 100% of the time!
No with a forward voltage drop on the LED of 2.2V the drop across the resistor is 5 - 2.2 = 2.7 V
Thanks JimboZA,Although I got a great deal of information before your post, you gave me the piece of the puzzle I was lacking. I hadn't understood just WHERE the current I needed to power, in this case, an LED was coming from.
This is why controlling LEDs with resistors is a bad idea,
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