# Challenge: can you light these LED's?

#### xox

Joined Sep 8, 2017
564
Here's the setup. You have an unlimited number of resistors, two LED's, a single push button switch, and plenty of wire. Your task is to have one LED light up when the switch is closed and the other to be lit when the switch is opened. In other words, at any given moment exactly one of them will be lit.

Can you solve the puzzle?

Joined Apr 16, 2011
532
Are we allowed a power source of some kind?

Or should we use the "plenty of wire" to harvest energy?

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#### xox

Joined Sep 8, 2017
564
Are we allowed a power source of some kind?

Or should we use the "plenty of wire" to harvest energy?
Well I'm guessing you could power yours with sheer sarcasm.

Yes, any DC power source is allowed.

#### Jony130

Joined Feb 17, 2009
5,230
It would be much easier if we could use a diode as well.

#### Beau Schwabe

Joined Nov 7, 2019
84
With a 5V DC supply and "typical" LED's, it can be done with 3 resistors.

How about adding to the challenge in that the LED's should have relatively the same amount of current to them when they are "ON"?

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Joined Apr 16, 2011
532
Well I'm guessing you could power yours with sheer sarcasm.
Apologies - you are right of course. I should have phrased it in a much more considerate way.

Sorry

#### peterdeco

Joined Oct 8, 2019
356
With a 5V DC supply and "typical" LED's, it can be done with 3 resistors.

How about adding to the challenge in that the LED's should have relatively the same amount of current to them when they are "ON"?

View attachment 224850View attachment 224851
I think with the switch open you have 2 LED's in series and will glow dimly.

#### Beau Schwabe

Joined Nov 7, 2019
84
peterdeco -
If the 100 Ohm resistor from the switch to the ground was not there then Yes, I would agree.

#### Tonyr1084

Joined Sep 24, 2015
5,966

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#### xox

Joined Sep 8, 2017
564
Apologies - you are right of course. I should have phrased it in a much more considerate way.

Sorry
No worries, no offense taken. I actually found it quite funny, honestly.

#### Tonyr1084

Joined Sep 24, 2015
5,966
@Beau Schwabe The two lower resistors form a parallel resistance along with the lower LED in series with the 200Ω. With the switch open there would be (and I know I'm going to screw this up) 66 1/3Ω, preventing the lower LED from lighting. Even then, the top LED would be at 2.5V across its junction.

However, with the switch closed (as shown) you have a voltage divider between the two lower resistors, putting the voltage across the junction of the lower LED at 1.66 volts.

#### xox

Joined Sep 8, 2017
564
I'm a little confused about your submission. SW1 always has to be closed for any power to flow through either LED. And SW2 simply turns the bottom LED on/off with no effect on the upper one, so...

#### Tonyr1084

Joined Sep 24, 2015
5,966
The addition of SW1 is so someone doesn't say LED1 would be always lit and drain the battery. As for SW2 adding in LED2, watch this video.
The key is the different forward voltages. As will be demonstrated in the video, the LED(s) with the lower forward voltage will draw all the current and prevent the LED(s) with the higher Vf from lighting.

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#### Tonyr1084

Joined Sep 24, 2015
5,966
In your initial supposition (post #1) you didn't specify whether both LED's had to be the same type or same Vf. If it is the case, where both LED's must be the same Vf then the addition of one more resistor on LED1 after the PBSW "Should" provide the same effect. However, the circuit I drew is one I've built and tested. The original intent of the video was to prove that you CAN use a single resistor on a set of parallel LED's as long as the Vf is very closely matched. However, if you mis-match LED's then the one with the lower Vf will draw all the current. IF your supply can deliver more than any single LED can handle then you run the risk of blowing out that first (lowest Vf) LED. The NEXT lowest Vf LED will then take over eating all the current till it dies. This leaves even more current available for the next victim. Like domino's they will all topple. But unlike domino's which fall at a specific rate, the failure mode of LED's will accelerate exponentially. In other words, when the first LED fails the rest will fall faster and faster.

xox

#### xox

Joined Sep 8, 2017
564
In your initial supposition (post #1) you didn't specify whether both LED's had to be the same type or same Vf. If it is the case, where both LED's must be the same Vf then the addition of one more resistor on LED1 after the PBSW "Should" provide the same effect. However, the circuit I drew is one I've built and tested. The original intent of the video was to prove that you CAN use a single resistor on a set of parallel LED's as long as the Vf is very closely matched. However, if you mis-match LED's then the one with the lower Vf will draw all the current. IF your supply can deliver more than any single LED can handle then you run the risk of blowing out that first (lowest Vf) LED. The NEXT lowest Vf LED will then take over eating all the current till it dies. This leaves even more current available for the next victim. Like domino's they will all topple. But unlike domino's which fall at a specific rate, the failure mode of LED's will accelerate exponentially. In other words, when the first LED fails the rest will fall faster and faster.
Okay, well that is an interesting approach. Of course, with any of these circuits, fairly specific values for the components would likely be a requirement. It's a balancing act!

#### ElectricSpidey

Joined Dec 2, 2017
1,653
Remove SW1 in post #9 and the circuit will work, because the color of the LEDs was not specified.

But I have a question...

Can you make an inductor with the wire?

#### Tonyr1084

Joined Sep 24, 2015
5,966
My guess would be that you can accomplish the same thing without any resistors IF you use a long enough length of wire to take advantage of the wire's linear resistance. OR maybe you could use a very very fine conductor to act as a higher resistance than a heavier gauge wire.

When we're not given restrictions we can come up with all kinds of things. Have you ever used an old carbon filter from a drinking system as a resistor? I have.

xox

#### xox

Joined Sep 8, 2017
564
My guess would be that you can accomplish the same thing without any resistors IF you use a long enough length of wire to take advantage of the wire's linear resistance. OR maybe you could use a very very fine conductor to act as a higher resistance than a heavier gauge wire.

When we're not given restrictions we can come up with all kinds of things. Have you ever used an old carbon filter from a drinking system as a resistor? I have.
Nice! Or like this guy, using water to replace both resistors and capacitors.

#### Tonyr1084

Joined Sep 24, 2015
5,966
For that fact, you could use a hotdog as a resistor. And when done you can eat your circuit.