Ah. It's starting to come together. That makes sense BackWoodsBrewer. I was actually pretty surprised that none of them did blow. On one of the smaller experiments the wire on the battery accidentally touched the lead of an led past the resistor and it was bright so that's what "sparked" the idea for the as many led's as I could fit. I have blown a led with this exact 9v just for kicks. I literally just tested the battery with the multimeter and its putting out 8.8V
bwoogie: I literally just tested the battery with the multimeter and its putting out 8.8V
As much as that? I'm surprised, I would have said more like 8V. The battery must be very new.
But again...try holding the battery in your hand for a minute while you admire all those LEDs.
Those 9V batteries are capable of putting out a couple of amps, but only for a few minutes before they’re completely drained.
The typical internal resistance of a PP3 is between 1? and 3? depending on manufacturer and chemistry. Let’s take 2? as a good mean value.
At 2? and a nominal 9V voltage, I=V/R, so I = 9 / 2 = 4.5A. That’s the short circuit current of a typical PP3.
Pass that through 117 LEDs in parallel (assuming the “perfect” LED - exactly the same voltage drop as all other LEDs and zero resistance) - 4.5/117 = 38mA. That’s normally above the IF of a typical LED, but won’t normally cause the LED to blow immediately.
Now, the world is not perfect. So what you have is equivalent to a perfect 9V power supply, with a 2? current limiting resistor, and a bunch of LEDs, each with its own VF. What that means in practice is that some LEDs will get a slightly higher current than others. You probably noticed that the LEDs may not have all been the same brightness.
The ones with the most current will blow first. That will mean there’s a greater proportion of the current available to the rest of the LEDs. The next ones to blow will do the same - increase the current available to the rest, and so on. A cascade effect will rapidly occur where just one LED blowing could cause the rest of the LEDs to blow.
Of course, all this relies on the battery holding out, which at 4.5A and the typical 565mAh of an alkeline PP3, would take (T=mAh/mA = 565/4500) = 0.12555…h, or about 7 and a half minutes.
majenko: Those 9V batteries are capable of putting out a couple of amps, but only for a few minutes before they're completely drained.
I accidentally shorted out a 3xAA battery pack yesterday and it melted the battery box.
Talking of that... Back to the PP3. 4.5A through 2? -
P=I²R = 4.5² * 2 = 20.25 * 2 = 40.5W
That's gonna get warm :)
bwoogie: So I learn best by trial and error. I don't know what I learned but I like it.
You learned that rather than researching and understanding what's going on, that you can simply wait for answers from others, nice!