How many 220 resistors do i need on a 16 LED 4x4 flat matrix?

Hi. I am ok with hardware, soldering, etc, but I want to understand something better.

I have a piece of cardboard and made a 4" x 4" grid and put an led into each one, with pins sticking out of the back of the cardboard. I then bent all the cathodes in each row together and soldered them in a line. 4 lines of cathodes with 4 pins sticking out of the end of each row. I want to control each led individually. So, can I solder those 4 cathode lines together into 1 then just plug it into the ground, or do I have to put 1 or more 220 resistors on each line or the one joined line? Thanks in advance for your time and replies.

Nickademus420: ….or do I have to put 1 or more 220uf resistors on each line or the one joined line?

Resistors are mesured in ohms.

Hey Nick,

I'm not expert, only been doing this for a short time myself, but I'll try my best to help.

While "Unsigned_Arduino" is correct in his statement, I'm guessing that did not help much.

To elaborate (just in case you did not know and that your statement wasn't a typo), Resistors are measured in Ohms. Capacitors are rated in farads. In your case, uf would be called a micro-farad.

Now, to answer your original question, Yes, you would want to put in line a 220 resistor with each remaining leg of your LED. Primarily because, an LED does not handle 5v very well over a period of time. most LEDs need between 2.1v and 3.3 volts. While there are exceptions, this is fairly normal.

By having the resistor in line with the remaining leg (called the Anode, BTW), the voltage is brought down by being shared across the resistor and LED.

Hope this helps.

Thanks for all the clarification. I updated my topic title. Lol. So I soldered all the short posts, anodes, into 4 rows, so at the end of each I soldered the resistor then wire all resistors together and plug that into the ground of the Arduino Uno? Then each cathode gets wired straight to the board in the posts I select. Sound correct?

So a further bit of clarification on your LED's.

The short leg would normally be the Cathode (and expected to be tied to ground) with a resistor. The Anode would be the long leg, and expecting positive voltage, and would go to the designated output pin.

So, reverse your statements below, and that would seem correct. Sometimes it helps if you can upload a schematic of your design.

Many folks here seem to use a program called "Fritzing" to do the schematic drawings. The link is embedded in the above word.

Personally, I am a Visio user, so I do not have any personal experience with Fritzing.

Many folks here seem to use a program called "Fritzing" to do the schematic drawings.

And they do it very badly or more often they produce a physical layout diagram and get castigated for it.

Just use a pen and paper to draw a schematic.

I updated my topic title.

You have to change the title of the first post so you can see the new title in the list of topics.

I am doing both actually, diagramming on paper and with Fritzing. I just didn't know if in Fritzing I could leave LED's off the breadboard and connect with wires, like I am doing in real life. I will try tonight. I will get it all straight here eventually. Thanks!

If you have a matrix, then you need one resistor and a MAX7219.

Easiest way is to get one of the crude matrix modules from eBay and if you want a different, "DIY" matrix, wire it to the module instead of the one supplied. |500x500 Be warned however, that if you at some stage wanted to stack those sort of small matrices, that one I linked is the wrong type. :astonished:

Hey James:

LEDs are rated by their current carrying capacity, typically measured in milliamperes, not by voltage. The voltage specification you have mentioned is not a rating, it is a characteristic.

If, by some means, you manage to get the correct current flowing through the LED then you can expect to measure the specified forward voltage.

Don

Hey floresta,

Obviously you are correct. I could have gone down the road of average of 20ma across an LED, talked up the Ohm's law, etc, but sort of just simpletoned it down for quick and dirty explanation.

Thanks

I got it to work. 1 resistor per string of led cathodes (-) then I wired each cathode together and soldered my ground wire to the end resistor. Thanks everyone!

James

I know you are trying to help, but you really aren't helping by simplifying things to the point where they become wrong.

In reply #9 you have added two more incorrect concepts.

(1) You do not measure current 'across' LEDs or any other device for that matter.

(2) Ohm's law does not apply to LEDs or to any other non-linear device.

Your request for a schematic was appropriate and Visio is also my choice for schematics but not very affordable for a lot of people.

Fritzing produces wiring diagrams which technically are quite different from schematics, but the price is right.

Don

Nick

One real advantage of using the MAX7219 approach is the fact that it provides constant current to each string (row or column - depending on the wiring) of LEDs no matter how many of the LEDs in that string are lit.

Your resistor approach will result in the intensity varying depending on the number of illuminated LEDs. This effect may or may not be annoying or even noticeable but it will be present.

As Paul mentioned, if you buy a module (either assembled or as a kit) make sure you get one with the input and output pins on the sides, not the top and bottom as shown in his picture. Basically, stay away from the modules where you can see the IC alongside the LED matrix. The desirable modules have the IC mounted out of sight, below the LED matrix.

Don

floresta: As Paul mentioned, if you buy a module (either assembled or as a kit) make sure you get one with the input and output pins on the sides, not the top and bottom as shown in his picture. Basically, stay away from the modules where you can see the IC alongside the LED matrix. The desirable modules have the IC mounted out of sight, below the LED matrix.

I explained it carefully and deliberately in post #7.

If you are making your own matrix, the one I cited is most appropriate. It comes as a kit with through-hole parts, so you assemble it but do not mount the socket headers for the display, or the display itself. You can then wire (solder) your display connections to the holes in the PCB. In the case of making your own, usually larger matrix, stackability is not a concern (you can stack those boards vertically or of course, turn them on their sides when so stacked).

There is the stackable module available as a kit also |500x500 but it uses surface mount components (and is a few cents cheaper), so is not so easy to work with.

The proper pre-assembled and stackable modules |500x500 have the matrices soldered in place so do not suit use with a custom assembled matrix.

For full "Times Square" display, there are gang modules. |500x500

James-WBF: Hey floresta,

Obviously you are correct. I could have gone down the road of average of 20ma across an LED

I am glad you didn’t, that average idea is one of the myths floating about the internet. It is totally wrong. The main idea is that if the LED is on for only half the time you can put twice the rated current through it. This is bollocks. Some LEDs have a higher pulse rating but if this is the case it is in the data sheet and only at the duty cycle specified in the data sheet. It normally only applies to IR LEDs where narrow pulse modulation is used in remote controls. Some mindless people have assumed this to mean an overall avrage, it does not.