Here... I will give you the chance to have the last word, all you have to do is explain exactly how the power gets from the supply to the load.

You make a good straight man.
That one's too easy.
Usually copper wires.

 

Here... I will give you the chance to have the last word, all you have to do is explain exactly how the power gets from the supply to the load.

wires - ?

 

wires - ?

wires is just the plural of that word, of course.

 

You picked one of the cases where it could matter. Flashlights don't usually take 1 AAA battery. They're more likely to use 2 or 3 because 1.5V isn't a high enough voltage for white LEDs.

In the case of cheap LED flashlights, it could matter because most don't follow conservative design practices and put LEDs in parallel without resistors for each parallel string. They don't care if LEDs burn out. They're hoping you'll buy another. I have half a dozen with more than half of the LEDs not working. Most of them were free and, while I would like them to work longer, they didn't cost me much. Some of the newer ones only have 1 LED. Haven't had any long enough to know how long they'll last.

Take the case of LEDs connected directly to CR2032 batteries. Those batteries can't burn out LEDs because they have so little capacity. If you took the same LEDs and connected them to a 3V source that could supply amps of current, the LEDs would likely burn out.

For your AAA replaced with D, the lower internal resistance of D could be a problem.

Thank you for the details! I used one AAA just for simplicity of my question but you also hit the nail on the head. The system I was going to hookup three D's to is a long string of LEDs, almost like those one may see on cars. It's roughly 20 feet long and has an LED every half foot and I just wanted to use D's to lessen the frequency of buying batteries lol.
If I could ask, how would having resisters amongst the string stop it from being overpowered?
I thank you for your answer, and everyone else's, I'm already enjoying this forum!

 

Thank you for the details! I used one AAA just for simplicity of my question but you also hit the nail on the head. The system I was going to hookup three D's to is a long string of LEDs, almost like those one may see on cars. It's roughly 20 feet long and has an LED every half foot and I just wanted to use D's to lessen the frequency of buying batteries lol.
If I could ask, how would having resisters amongst the string stop it from being overpowered?
I thank you for your answer, and everyone else's, I'm already enjoying this forum!

A diode is a highly nonlinear device, so a slight increase in applied voltage results in a large increase in current. LED's, due to the nature of their construction, aren't quite as nonlinear, but they have a significant offset voltage.

For instance, a red LED might conduct 10 mA with 1.8 V across it, but increasing this by just 200 mA mV (about a 10% increase in voltage) will likely result in 20 mA of current.

LED's also have quite a bit of variation from one LED to another, so 1.9 V across one LED might produce 5 mA through it while 1.9 mA through the one next to it, from the same package of LEDs, might produce 30 mA through it.

So it is common practice to use a resistor to set the current in an LED.

Let's see this with an example.

Imagine I have a 12 V supply and I want to put 20 mA through an LED that has a nominal forward voltage of 2.0 V. That means that I need a series resistor that drops 10 V when conducting 20 mA, which means a 500 Ω resistor.

Now imagine that I make a bunch of these and in one of them the LED actually only drops 1.6 V at 20 mA and in another it drops 2.4 V at that current (the are someone extreme variations, but outlandishly so). That means that in the first one I would have 10.4 V across my 500 Ω resistor, which would be a current of 20.8 mA (very close to the 20 mA target) and, in the other, a voltage of 9.6 V across it, resulting in 19.2 mA (again, pretty close to the 20 mA target). So my resistor is establishing close to the desired current despite major variations in the actual voltage drop across the LED.

EDIT: Fixed typo in voltage units.

 

A diode is a highly nonlinear device, so a slight increase in applied voltage results in a large increase in current. LED's, due to the nature of their construction, aren't quite as nonlinear, but they have a significant offset voltage.

For instance, a red LED might conduct 10 mA with 1.8 V across it, but increasing this by just 200 mA (about a 10% increase in voltage) will likely result in 20 mA of current.

LED's also have quite a bit of variation from one LED to another, so 1.9 V across one LED might produce 5 mA through it while 1.9 mA through the one next to it, from the same package of LEDs, might produce 30 mA through it.

So it is common practice to use a resistor to set the current in an LED.

Let's see this with an example.

Imagine I have a 12 V supply and I want to put 20 mA through an LED that has a nominal forward voltage of 2.0 V. That means that I need a series resistor that drops 10 V when conducting 20 mA, which means a 500 Ω resistor.

Now imagine that I make a bunch of these and in one of them the LED actually only drops 1.6 V at 20 mA and in another it drops 2.4 V at that current (the are someone extreme variations, but outlandishly so). That means that in the first one I would have 10.4 V across my 500 Ω resistor, which would be a current of 20.8 mA (very close to the 20 mA target) and, in the other, a voltage of 9.6 V across it, resulting in 19.2 mA (again, pretty close to the 20 mA target). So my resistor is establishing close to the desired current despite major variations in the actual voltage drop across the LED.

Hey, great example (through which I learn best) and thus the resister normalizes the voltage and secondarily the differential current amongst all the LED's.
Thank you greatly for the time spent in explaining all of this to me, it's become an asset of knowledge for me.
Cheers and have a great [evening | day]!

 

An ideal constant current source has infinite resistance.

Hence when driving an LED with a resistor in series, the larger you can make the resistance, the closer it comes to being an ideal constant current source.

 

The system I was going to hookup three D's to is a long string of LEDs, almost like those one may see on cars. It's roughly 20 feet long and has an LED every half foot and I just wanted to use D's to lessen the frequency of buying batteries lol.

Can you supply more info on the string of LEDs? What voltage was it designed for? What is the current draw? Do you have a datasheet?

 

The short answer to the question in post #1 is "YES".
A load connected to a power source of the correct voltage will only consume the power it needs, independent of the capability of the source. That is to say it will only draw the current that the load resistance will require.

 

If you are talking a flashlight that uses 1AAA battery, it is most likely using LED technology for illumination and not an incandescent bulb. There are many quality LED flashlights using 1 AAA. But white LEDs can't operate at 1.5/1.25V so there is a power supply circuit that takes the 1.5V and produces a higher voltage (step-up converter). You will probably have no problem connecting D batteries in parallel to power this type of flashlight and with 3-4 D's its runtime would be greater. An alkaline AAA provides about 1,200 mAh with a D alkaline at 10,000 mAh. Three D's in parallel give 30,000-60,000 mAh so your light would last 25 to 50 times longer. It however would be physically much larger.