Using jumper cables to turn on LEDs

Question

Currently I have my LED's anode connected to the GPIO pin that controls it and the cathode is attached to a resistor and the other end of the resistor is on my breadboards - rail. How can I use jumper cables so I don't have to twist and ruin all my resistors? All of the tutorials I find online show how to do it without a ribbon cable, which is what I am using.

Current Wiring: https://i.sstatic.net/EUFN7.jpg

Also, a side question. One of the LEDs has 4 wires on it and it was not labelled what is what? How would I go about connecting that one up? I tried putting the longest of the 4 into the GPIO pin and then putting the smaller 3 connected to resistors onto the breadboards - side on the edge.

4 wire LED: https://i.sstatic.net/agxQk.jpg

EDIT: The white LED with 4 wires is this http://www.mpja.com/5mm-Red-Green-Blue-Full-Color-LED-4-Lead-Common-Anode/productinfo/17137%20OP/ and the longest wire is the anode and it now works fine. Now I need more than ever to figure out how to use the jumper cables so I have room to work with multiple GPIOs for this one light. Anyone know how to do that or could link to a guide?

Answer 1

Looking at your picture, why not make use of the extra 10 unused rows on your breadboard? I am going to assume that you've figured out that "row 1, holes a-e" are all interconnected, "row 1, holes f-j" are interconnected, "row 2, holes a-e" are interconnected, etc. Your unused rows 21-30 aren't used, so, for example, the anode of the left-hand blue LED is plugged into row 19, hole b. Plug the LED's cathode into row 21, hole D, plug one leg of the resistor into row 21, hole B, then plug the other resistor leg into your ground (-) strip.

For LEDs that are farther away from the GPIO pins, let's say the left-hand Red LED, if that was plugged into Row 3, hole D [[ I can't tell for sure in the picture ]] remove the LED from there, run a jumper wire from Row 3, hole D to Row 23, Hole D, put the anode of the Red LED In Row 23 Hole B, the cathode in Row 24, Hole D, one leg of the resistor in Row 24, Hole B and the other leg of the resistor into the ground strip.

If you don't have any decent jumper wire handy, see if you can find an old chunk of Cat5 solid ethernet cable. Strip the outer casing, separate the strand pairs into individual strands, smooth the "twisties" in the wire out (this can be rough on your fingertips if you're making a lot of jumper wires) and strip off each end of the wire. I've made lots of jumper wires from old Cat5 and even some old 25-pair phone wire (which has some really cool color combinations) and it works very well.

Be careful with Cat6 though - it's actually one gauge bigger and it might be a bit tougher to insert in your breadboard holes. If whatever cable you find, it there's a plastic '+' separating the 4 pairs of wires, it's probably Cat6 or higher.

Answer 2

Using a bread-board is usually a temporary means of allowing you to work out the circuit you want before transferring the design to Veroboard or just for something that you only need for a short while. For instance I recently needed to check that all the GPIO pins were working as outputs on a RPi that I had bought second-hand; I rigged up a circuit that enabled me to check that six pins at a time were operating and were not shorted together. I put something together with six instances of a NPN transistor, a series base resistor (to a RPi pin under test) and an LED and a series resistor in the collector:

Rather than soldering a resistor onto one lead of the LEDs I used an unused strip of five contacts for each "join" that the circuit needed - and where I needed to link a strip of such contacts (e.g. the emitters of each pair of transistors) a short length of insulated solid (single core) wire - actually from a length of household telephone cable that comes in four/six or eight core forms, provided that link. Sometimes the long leads can be too much, so a bit of trimming with side-cutters (or, at a push, the cutters in a pair of pliers) is justified - in this case I didn't need to because I was "recycling" resistors I had recovered from a scrap circuit board I had lying around...

With regard to the unknown pin-out on the RGB LED, I find that the "diode-check" on my DMM will measure the forward volt drop of a semiconductor junction up to around 3V but I can never remember which of the Red or Black leads indicates the Anode/Cathode terminals when I get a 0.7V reading on a forward biased normal silicon diode but the line or bar on a component lead is the CATHODE and an LED will glow faintly if connected in the same manner and thus then its CATHODE will be on the same lead. Not all DMM have a diode check but a resistance measurement setting will also produce a high-impedance (low-current) voltage across an LED in the same manner without causing damage to LEDs (which cannot tolerate more than a few volts the wrong way around).

Answer 3

The bottom one looks like an RGB LED, possibly common cathode. The cathode would be the longest lead. With a common cathode you connect the cathode to ground and connect each LED's anode through a resistor to the output pin.

EDIT: As indicated in the comments to the OP, it's a common anode RGB LED. In this case, the anode (longest leg) goes to V+, with the other three legs attached to the proper pins, again, through resistors.