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I have a pretty basic question. I put this little thing together but I don't understand why the POT is causing this result.

When powered, if the POT isn't fully open. LED 1 is bright and LED 2 is dim.

When POT is fully open LED 2 becomes VERY bright and LED 1 dims slightly.

Shouldn't they have equal brightness if the POT is fully open?

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My guess is that it's happening because that they are splitting the current.

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  • \$\begingroup\$ A pot connected as shown in your schematic is simply a resistor whose value can be easily changed - it doesn't do anything "magic". Consider the current flows in the circuit when the pot is set to zero, 1K, 5K... to see the effect that it has. \$\endgroup\$ – Peter Bennett May 17 '15 at 2:00
  • \$\begingroup\$ @PeterBennett thank you for your comment. I know potentiometers aren't "magic". I'm also aware they increase resistance. I am asking WHY LED 1 because dim when the POT is fully open. \$\endgroup\$ – Anthony Russell May 17 '15 at 2:04
  • \$\begingroup\$ Swap LED 1 and LED 2, what do you observe \$\endgroup\$ – geometrikal May 17 '15 at 2:21
  • \$\begingroup\$ @geometrikal the same. In fact I'm one step ahead. I removed all other components except for LED1 and LED2. When I manually touch LED2's contacts to LED1's contacts I get the same results. LED 1 dims and LED 2 is much brighter. \$\endgroup\$ – Anthony Russell May 17 '15 at 2:27
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Almost certainly, the two LEDs are not perfectly matched. For the same current (and brightness) level, LED1 has a greater voltage drop than LED2. Additionally, you are using confusing terms - what you call "open" on the pot is actually what everybody else calls "shorted", that is the resistance is almost zero. Just for illustration purposes, let's say that LED1 will operate well at 2 volts, while LED2 will shine at 1.5 volts.

So, at first most of the current is flowing through LED1, and it produces 2 volts. At the same time, LED 2 is being driven by LED1's 2 volts through the pot, to the nominal 1.5 volts. The pot is in the high-resistance position, so not a lot of current gets to LED 2. As a result, LED1 is bright and LED2 is dim.

Now turn the pot to what you call "open", which is the low-resistance setting. Now LED2 is hogging most of the current, since it operates at 1.5 volts, and this is not really enough to light up LED1. The result is that LED2 is bright and LED1 is dim.

If the two LEDs were identical, when the pot was set to "open", the two LEDs would be of equal brightness. But they're not.

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    \$\begingroup\$ You're entirely welcome. Also, if you change your pot to a 1k pot, rather than the 100k you're showing, you should be able to get a much more gradual transition of brightness. \$\endgroup\$ – WhatRoughBeast May 17 '15 at 3:16

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