I read once that a 'load' gets only the current it needs from the battery / voltage supply
I read once that a 'load' gets only the current it needs from the battery / voltage supply.
If a led's spec says it draws 20mA @ 3.4V, then you can just as well apply Ohm's law:U=I*R, R = U/I, R = 3.4 / 0.02 = 3.4 * 50 = 170 Ohm
Note: Pulse width ≤0.1 msec, duty ≤1/10.
Note 1. Pulse Width = 100μs, 10% duty cycle (1/10th).
Now a led isn't a real resistor, @ 3.4V it acts as a resistor of 170 Ohm, i don't think the relation between voltage and current is as linear in a led as it's with a resistor.
i don't think the relation between voltage and current is as linear in a led as it's with a resistor.
I've seen many responses here, but half of them read as if people assume the resistance of a led is 0 Ohm?
@RACEMANIAC <lots of interesting stuff>
It's good you are trying to learn but I think you still don't see the magic of semiconductors.
Ok, I'll play your game.Take an odinary led and connect it with a 220 ohm resistor.Now connect your meter in series in current mode OR connect it across the resistor in VOLTAGE mode. Record the results. Now replace the led with a 170 ohm resistor and repeat the measurement.Now remove the 170 ohm resistor and replace the led.Verify that it still draws thd same current you measured before and ghen put a shorting jumper across the led, shorting the two led pins together and repeat the measurement.Which of the two substitutes draws an amount of current that closest approximates the current of led, the 170 ohm resistior you used in place of the led, or the shorting jumperyou put across the led ? How much of a difference in cureent did you see when you shorted the led ?
What are you using to type your messages that adds all those annoying new lines, making your text so narrow, it's really distracting.
Also, reading up on it a bit more, I'm starting to understand better why they keep using the voltage drop terminology . It does vary with current, but very little in diodes. So when you put too much voltage on a led, the current will indeed rise very quickly and basically short circuit .
I entirely agree. What is happening, is that a number of people have become so pi$$ed off with the unreliability of the forum software, that they are typing their responses in a separate editor and then using cut-and-past to insert them in the "reply" box, but in the process forgetting - or not comprehending - how to turn off the "line wrap" in that separate editor.
170 Ohm resistor + 220 Ohm: i would expect the same 20mA, and the same voltage drops in the circuit
here you short out the led, so only the 220 Ohm resistor is left, i'd expect the current to almost double as the 170 Ohm on the circuit disappears.
You need to reread the instructions
I reread them a few times before answering, but they weren't that clear ^^.
If when a voltage of 3.4V is on a led, it'll allow a current of 20mA trough it, it at that point is equivalent to a 170 Ohm resistor in the circuit, even if it's a magic semiconductor.
Atm my line of thought is if it sounds like a duck and quacks like one, it probably is one
What i'm currently thinking: If when a voltage of 3.4V is on a led, it'll allow a current of 20mA through it, it at that point is equivalent to a 170 Ohm resistor in the circuit, even if it's a magic semiconductor.
That is still expressed backwards.What you actually mean is: "If I am feeding a LED with 20 mA and at that current its voltage drop is 3.4V, it at that point is equivalent to a 170 Ohm resistor in the circuit".