Ok, forget about the hardware. Can we make this thing with nothing but hotdogs?OK, lets not take this thread off topic. The question is can you light two LEDS one at a time using a single push button, unlimited lengths of wire and unlimited numbers of resistors. To answer that question - as has been answered, "Yes".
I don't see how that could actually work. Closing the switch you'd get a short to ground, no?
Rule bender, eh? Submission rejected!LTSpice simulates. Closing the switch, LED1 lights and the battery for LED2 shorts out making LED2 off. Otherwise LED2 is on and LED1 is off.
Unfortunately, I am lousy at LTSpice and don't know how to simulate a push button switch. One could play with serial resistors to the batteries to prevent a dead short circuit. But as some might say, short circuiting a battery was not excluded in the problem definition.
What kind of push button switch is it? SPST? DPDT? Something else?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?
Or how about pickles or some other food that you can get to fluoresce?Ok, forget about the hardware. Can we make this thing with nothing but hotdogs?
Schematic?What kind of push button switch is it? SPST? DPDT? Something else?
Since there was no requirement that they be lit equally, there's a few simple ways to do it. One uses just two resistors in addition to the LEDs and a SPST switch. You could eliminate the resistors altogether if you are willing to use enough of the "plenty of wire".
I was kidding! Don't get me wrong though, I'd still love to see someone pull something like that off.Or how about pickles or some other food that you can get to fluoresce?
Didn't know if you wanted schematics or other outright spoilers.Schematic?
The bridge design is definitely much better than the one I came up with.Didn't know if you wanted schematics or other outright spoilers.
I don't have a schematic capture package installed on this machine, so Paint will have to do.
I'm assuming that the diodes are characterized as follows:
Vf = 2 V
If = 10 mA
Current changes by one decade every 120 mV.
Although I think that reducing the current in one diode by 90% would be good enough (especially when the other diode is on right next to it), I'll set the goal of reducing it by 99%.
View attachment 224919
With the switch open, we'll size the resistors to yield an I1 of 10 mA. When the switch is closed, the voltage across the other diode will be around 2 V and we want to drop about 240 mV across R1 at 0.01 mA. That means that R1 has to be 2.4 kΩ. When the switch is open, it needs 10 mA flowing in it so the voltage across it will be 24 V. Let's pick Vcc so that the voltage across R0 is 10x this, so it will be 24 kΩ and Vcc will be 266 V.
Since no constraint was placed on the supply voltage, I can pick whatever I want. I will need to dissipate nearly 3 W of power, but that's not outrageous in the pursuit of a brute force circuit like this.
When the switch is closed the nominal 2 V across the right LED would result in 11 mA flowing in it, which would increase the voltage across the diode by a pretty negligible 5 mV, making it about 2.05 mV.
If we want to call 10% of the current sufficient to be "off", then we can work with a much more reasonable supply voltage. Now we can use 120 Ω for R1, 1.2 kΩ for R0, and use a supply voltage of around 15 V.
Similarly, if we use diodes that are "on" with 1 mA of current we can scale things down (though not as much as you might think -- but we can get the power down to well under half a watt).
A much more elegant solution is a bridge arrangement, variations of which others have already posted.
View attachment 224922
Again assuming a desired diode current of 10 mA at a forward voltage drop of 2.0 V, I chose R1 = 100 Ω, R2 = 300 Ω, and R3 = 550 Ω.
Did you see post #9 ? Mine only has a switch-on-able power supply and a battery in the diagram. And a slightly different location for the resistor.I don't think you said the LED's had to be the same - so I chose the following:
Vf red = ~2V, Vf green ~3.2V
View attachment 225829
It's simply a modification to a circuit we tossed around her a day or so ago: Alternating green and red LEDs discrete circuit | Page 4 | All About Circuits
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