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Does the Periodic Table Of LED's schematic work?

Im not going to elaborate on my project, I dont want anyone to steal it. Basically, its a LED Array in the shape of the periodic table. The small black circles are 4 way connections of all wires, if they intersect with no circle, they do not connect.

Picture of Does the Periodic Table Of LED's schematic work?
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mathews7 years ago
No, as has been pointed out, you need to invert each LED, and give each row or column a protection resistor.

In my smaller diagram, I have attached the protection resistor to the rows. There are no values, but R=
(vcc-vled)
/imax

To have the LEDs off, all column inputs should be off, all row inputs should be on. To turn an LED on, only invert the corresponding column and row.
matrix.bmp
lemonie7 years ago
You'll need a different arrangement if you want one LED somewhere to light up. Some kind of microcontroller(s)?

L
PKTraceur (author) 7 years ago
When I turn two switches, I want one LED somewhere to light up. Or, have multiple ones on at the same time.
Well, as long as you reverse the orientation of your LEDs, (they are currently reverse biased, which simply won't work...diodes, (LED = Light Emitting Diode) can be thought of like a one-way valve. The schematic symbol points in the direction of current flow reverse the direction and it blocks flow) AND use a 5 or 12V LED, and corresponding voltage, some LEDs will definitely light when you flip a column and row switch... I'm just not sure if what lights up is what you actually intend.

Rgearding the LEDS and voltages, most LEDs operate with a Vf, or forward voltage drop of ~2V. So each would also require a limiting resistor in series to keep the LED from burning out with *most standard supply voltages. For instance, if the Supply voltage is 5V, and the diode current at Vf=2V is 10mA, then you'd need a (5-2)/0.01 = 300 (use a 330 standard) ohm limit resistor in series with the LED. That is, the resistor needs to dissipate the remaining voltage (3V) at a current of 10mA to prevent the LED from being damaged by the overvoltage condition.

The comments made about the lack of need for multiple supplies is also important. One supply should do the trick...you just attach the output of that supply to the various load points.

And, in general, the connections for a supply, resistor and led look like this (in crude form)
R D
(V+)----/\/\/\/\/-----|>-----(COM or V-)

It doesn't have to be, but the convention is to apply the LED so it's cathode is attached to COM, rather than attaching so it's anode is at V+, unless it is a special LED like the 5 and 12V LEDs I cited earlier...They are, of course, much more expensive than regular LEDs.

best of luck
kelseymh7 years ago
I must be feeling really stupid tonight....Aren't all the diodes backwards? I don't think you actually need separate batteries on each row. The row/column switches should be sufficent, with just a single power supply (e.g., at the upper left corner in the schematic). It looks to me like the whole column is going to light up at and above whichever row you switch on. To avoid that, I think you need the diodes to go in parallel up to their column-wise switch, rather than in series.
Shh...he's trying not to give the secret away. ;P yeah., they appear reversed to me as well. Of course, one has to also assume here that you're either omitting the series resistors or are using 5V or 12V LEDS and associated power. And no, as constructed, the circuit doesn't seem like it will work to isolate individual LEDs. Kelsey outlined the issue clearly
What you are trying to wire looks like an LED matrix. As KMH says, you should have the Leds strung from the columns to the rows, and you don't need the multiple batteries.
Wow that's one heck of a lot of LEDs. Are you trying to turn them on individually or in groups? It would appear that flipping any of the upper switches would result in turning many of the leds on based on current flow paths. It might be better to flip some of the leds around, so that they only turn on when current flows int the right direction. For the lower circuit, you would need to split up the two levels because closing one switch allows current to flow in both sets.