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I disagree with this. But I'd suppose you'd have to specify what kinds of currents were talking about.
But consider diode dummy loads, for 405nm diodes at currents <500mA you'd use 6 rectifiers. Each has about .7v Vdrop, simulating a 4.2v Vdrop for the driver. But when the 445nm lasers first came out there was alot of confusion as to how many rectifiers to use. Dr. Lava said to use 4 even though 445nm diodes also run at ~4v. At high currents, >800mW, their Vdrop is more like 1v. So if you used 6, you'd really be stressing the driver to provide an amp of current at 6v! :yabbem:
Anyway, Mich what you gotta understand is how electricity works. Trying to apply these formulas without understanding what is actually going on is very hard, even though most of us find this stuff to be trivial.
What is electricity? It's the flow of electrons from atom to atom in a conditioning material, like copper aluminum silver.. ect.
Voltage is the force pushing the electrons along, or rather it's how much they are attracted to + end. They physically flow from negative to positive. It's counter intuitive, you can thank Benjamin Franklin for that. Anyways, think of voltage as water pressure. It's "how badly" the water/electrons want to flow. High voltage is harder to resist, low voltage is easier to resist.
Current is the amount of flow. Going back to the flowing water analogy higher current would be a wider pipe with more water passing by any one point at a given time. Current is measured in Amps. One amp is actually kind of alot of current for most circuits so the standard unit is the mA (milli amp, 1/1000th of an amp). 1000mA = 1A.
Resistance, measured in ohms, is the restive property of a material which only slightly conducts. Think of it as a kink in the hose. If you kink it, depending on the water pressure (voltage), you'll slow the current down considerably.
Ohms Law is the relationship between these.
1 ohm of resistance with 1v of voltage presented across it will cause 1 amp of current to flow.
Batteries have a fixed voltage (more or less), and the current that is drawn from them will depend on the resistance presented across them. They do not just output 30mA or whatever automatically.
Your LEDs have a voltage of 3v and a current of 20mA. That means that if you put 3v across the LED 20mA of current will flow. So, using Ohms Law, you can figure out that the LEDs essentially have a resistance of 150ohms each (3v / .02A = 150 ohms). At these low currents that resistance wont change much. So we can assume that won't change.
So as Toke pointed out, you usually don't want to wire LEDs in parallel (all + connected together, all - connected together) although I've never had a problem doing that in the my little projects over the years. Anyway, it's best to wire them in series (the + of one LED connects to the - of the other).
Now since each LED is 150ohm, two in series is 300ohms. Two kinks in the hose, double the resistance, so they're added together. Now your battery is 9v and you want 20mA for your LEDs. Even though you've got two in series you still only need 20mA. This is because they're connected together. When there is only one single loop for the electricity to flow through the current remain constant at all points. Remember that current is how much charge (electrons) are passing by any one point. Since there is only one path for them to take, the current must remain even.
So you want 20mA from your 9v, using ohms law... 9/.02 = 450 ohms. Your two LEDs already make up 300ohms so you only need a 150 ohm resistor in series with your LEDs.
battery+ ------- (150ohms) --- switch --- +LED1- ------ +LED2- ----- - battery
I hope this helps a bit. :beer:
-Tony
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