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Ohms law kinda fuzzy, asking for advice before frying LEDs

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Cheawick

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Hello folks, been years since I was serious about this subject so now cobwebs have got my brain a bit stuck.
Please consider the following:

Using 4 LEDs @2.1v, 20ma (each)
Using 2 D size batteries @ 1.2v, 4,500 mAh (each[not sure what mAh is])

Question:
What is mAh, what resistance is required to keep from frying the LEDs, and what formula calculations were used to come to this conclusion.

Basically I'm trying to make a string of LEDs like steady/blinking eyes for a halloween gag in the front yard. But i'd like to re-learn my old craft and build this project with as little help as possible. I promise to post my final design and even a few pics (if possible).
 
mAH is the amount of power that the battery holds at full charge ..
that is actually quite good..
essentially , one D battery could supply 4.5 A for an hour..or 2.25 A for two hours or 1.125 A for four hours..you get the idea..
ok 2.4V/.02A= 120 ohms V=IR or V/I=R .
 
check this topic out, it kinda pertains to what you want to do
 
I came up with a 15 ohm resistor. Assuming you want 2.1 volts to be across the LED.

2.4V - 2.1V = .3 or 300mV/20mA=15 Ohms

I would step it up to a 20 ohm or so for tolerance reasons.
 
Thanks ya'll. Another question if I understand ohms law correctly. In series circuit the voltage will always be the same no matter how many 2.1v LEDs I shove in there, yes? It is the ampere that is "changed" by adding the resistors, yes? But would the 20mA used by the LED act as an ammount of resistance that I need to account for?
The reason I ask this is that I may build a 200 ft string of the buggers, ofcourse after trying to figure out what resistance is built up from whatever guage wire I plan to use per foot.
:p Heck! I may even build a rectifier circuit just to plug the whole thing into an outlet!
 
the high current will smoke your LEDs if you put them all in series.

put them in parallel, with a resistor for each LED.

Or, put them in parallel, in groups, each parallel branch could have 1 resistor and 4 LEDs. This is a commonly used configuration among "do it yourselfers" that seems to work.
 
They would fry if the current is to high and well if the power supply can deliver such loads in series.

Considering the Ohmage load of the LED which is about 105 ohms at max rating if you series 4 LEDs of this type to a 2.4v source that roughly 420 Ohm load which turns a 5.7mA load through the series which would be very dim this would not fry them. If you put 1 in series with the source yes smokage because 2.4v would drop over the LED.

To get maximum brightness series a 15-20 ohm resistor with the LED then bank the Resistor/diode parallel to source.


Current will sooner or latter drop over a large resistive lead, for instance that is why you have to use larger gauge wire the farther you are from the source.

Parallel will keep the voltage more constant while increasing the load at the source.
 
Thanks again. So better to do a bit of series parallel action and watch my distance from the power source. Got it!
 
If your LEDs have a 2.1V drop (some of them might have) and your battery cells are exactly 1.2V (Ni-Cad or Ni-MH with a heavy load), then the current when using a 15 ohm current-limiting resistor will be 20mA. That was using typical values.

What if some of your LEDs have only a 1.7V drop that is still within their ratings? What if your battery cells are actually 1.3V each or more after just coming from the charger? Then the current when using a 15 ohm current-limiting resistor is 60mA! Fried LEDs. You don't have enough voltage difference between the LED voltage and the battery voltage to allow for variations in tolerance. Use 3 battery cell and re-calculate the resistor.

The transistors that you switch the LEDs with will also use-up some of that small voltage difference. What if your LEDs actually have a 2.4V drop which is still within their ratings? No light. Again, use 3 cells.

Some of the 2.0V-rated LEDs that I recently recieved actually measure only 1.75V at their rated current. My battery cells measure nearly 1.5V when they are removed from the charger. Since they are really bright when operating at only 5mA, I don't worry about frying them.
 
Hmm... More for thought... Actually, I was figuring on using two 555 timer IC's and a couple of capacitors so that I can control the entire blinking sequence. As for the additional power, you do have a very good point. I hadn't really thought that far along the line yet. Originally I was simply toying with the idea of using a separate battery pack of another 2 D cells and a reverse biased diode, to hold the battery pack at bay until the voltage on the otherside dropped lower than the diodes' rating. Another idea was to build a voltage sensor that would activate a latching relay to connect the other battery pack to the circuit.

As I said before, it has been years (actually about two decades :p ) since I was serious about this subject, so I damn near have to relearn everything.
 
Was just browsing some of the other LED discussions and think I figured the name of that diode: zener. :)
 
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