I thought the reason for not needing a resistor is due to having 3 in series, and therefore splitting the current into 1/3 for each LED.
EEP! The drawing you have shown us is 3 LEDs in 'parallel', not in series... You are right, in that the current will split 3 ways, but this is in no way a limiting factor... The only reason this design works is because the battery chosen is too weak due to it's high internal resistance to do any damage.
In a parallel circuit, each branch will pull as much current as it needs, and the entire circuit will be sum of these values If you use 4.5V worth of batteries, and each led drops 3.3V (As it appears to be a rough number you discovered during your test) then that leaves 0.8V unused so to speak...
Now the internal resistance of the batter can be imagined as a resistor in series with the battery. This extra 0.8V will drop across this internal resistance which will limit the current to something around what the LEDs operate at... but if this internal resistance is too small (usually from larger batteries, or simply adding more of those in parallel... even if their voltage is the same!) then more than the desired current will flow through the LEDs causing damage.
To resolve this problem, you use an external resistor in series with EACH LED, to ensure that too much current will not flow through when provided with an expected voltage.
There is a bit of a trick here... LEDs are not simply powered by voltage.... they are powered by current. This is usually a bit to understand, but it essentially boils down to the LED dropping a small range of voltage across itself, and the rest of the circuit compensating to keep the current in acceptable levels.
This resource may help you to better understand :
https://www.electronicspoint.com/resources/got-a-question-about-driving-leds.5/
I encourage you to give this a read and come on back
Please don't base your experiments on that dollar store keychain
Are you saying, I'd see a higher Volatge Drop across the LED with the switch Closed if there was a resistor in the circuit? (and where in the crcuit? in series with each LED?)
Because you have 3 LEDs in parallel, you would actually need to add 3 resistors; one in series with each LED.
In a circuit, there exists resistance. As current flows across this resistance a voltage difference will form. A voltage difference will also form across LEDs and other components even as the amount of current varies. What you are measuring is actually the voltage difference of the LED, and the voltage difference caused by the resistance of the wires (which is pretty much 0) .
The rest of the voltage is being dropped internally in the battery... by using a resistor in-line with the LED and using a larger battery, you'll find the voltage will be more stable.
So would you describe this is an efficient desing? given they use a battery which has a higher internal resistance (therefore limitting the current flow to the LED's slightly?) and they don't have to use a resistor?
For this particular design yes.
For any other design... no, never, not at all.
If you wanted longer battery life, you can put another battery in parallel which will keep the voltage the same... but the internal resistance of both batteries in parallel will be lower, which will result in more current through the LEDs. This design is only ever viable as a key-chain design. Scaling it up in ANY way will require one of two things:
- Resistor, for use in low, and medium power LED applications.
- Constant Current Switch mode power supply for use in high power LED applications.
Using resistors waste power due to heat, but sizing things properly can keep that to a minimum.
Using a switch mode supply is more effecient than the resistor approach, but is more complicated unless you buy pre-made modules.
I know I'm repeating everything but I just want to make sure I understand!
Cheers by the way
No worries. Read away and ask away.
Check the link I provided. Hopefully it will answer a few questions.
