Multiplexing LEDs

test153 · Joined: 09 Feb 2009 Posts: 28

How do I multiplex LEDs? I have been trying to find a solution on my own for quite some time, since I haven't come up with a good one i turn to you.

From what I understand one would have to use some kind of timer and update every LED 50 times a second to avoid flickering. Below is the circuit I'm using. The PIC I'm using is 12f629.

Sydney · Joined: 13 Feb 2009 Posts: 71

It looks like that would have to be done in 4 cycles, with 3 leds controllable in each cycle.

GP0 HIGH
GP1 FLOAT //D2 D6 D10 controllable

GP0 LOW
GP1 FLOAT //D1 D5 D9 controllable

GP0 FLOAT
GP1 HIGH //D4 D8 D12 controllable

GP0 FLOAT
GP1 LOW //D3 D7 D11 controllable

PCM programmer · Joined: 06 Sep 2003 Posts: 21708

Appnote on multiplexing LEDs:
http://ww1.microchip.com/downloads/en/AppNotes/00234a.pdf

Found with this search string at Google:

test153 · Joined: 09 Feb 2009 Posts: 28

Floating outputs is a nice approach I haven't tried, will indeed simplify the code. Thanks.

Sydney · Joined: 13 Feb 2009 Posts: 71

Start simple, just get it working with a delay_ms(5); between each cycle.

You could use an int16, with bits 0-11 representing your leds, then use bit_test() to see if the led should be on or off.

Here I have started you off:

PCM programmer · Joined: 06 Sep 2003 Posts: 21708

There are a lot of project examples to be found with Google
and they have source code:

Here's one:
http://www.micro-examples.com/public/microex-navig/doc/091-dymoclock
He has a function called dymoLight(), in which he writes to TRISB
and PortB to turn on the specified LED.

I don't know what the purpose of your project is. I don't know if this
example is completely applicable.

test153 · Joined: 09 Feb 2009 Posts: 28

Thanks for the code. I have one question. Wouldn't it be bad to return so much current through one pin, unless there is a special reason we use " output_float" instead of setting it as an input. Aslo if we drive three LEDs at once will they not light less bright?

My initial thought was to controll every LED one by one (lightining only on at a time), with approach is the best. As far as I know a single pin can only handle 25mA.

Sydney · Joined: 13 Feb 2009 Posts: 71

Sydney · Joined: 13 Feb 2009 Posts: 71

Here is a test program that should(might) work

test153 · Joined: 09 Feb 2009 Posts: 28

Thanks for the code. Will try it out when I get home.

The LEDs I'm using are blue ones with a VF of 3.8 under a 20mA load. Somehow I missread the datasheet and did indeed not a any resistors to GP0 and 1. The pins GP0 and 1 are the only ones I can add restistors to due to the PCB layout. I can't add a transitor either.

If I have three LEDs on I will fry the output... (20*3=60mA going back to GP0 or 1).

What is the difference between output_float() and input()?

Sydney · Joined: 13 Feb 2009 Posts: 71

output_float() makes the pin high impedance, by making it an input, all it does is set the corresponding tris bit high. input() reads the pin, and if you aren't using fast_io will also set the corresponding tris bit high.

Sydney · Joined: 13 Feb 2009 Posts: 71

If resistors aren't an option I would limit the duty cycle of the leds as I described, and you will prolly get away with it, either way the leds will appear as bright.

RLScott · Joined: 10 Jul 2007 Posts: 465

This is a very unusual circuit. Let's look at what it takes to turn on D1. If we understand that, we understand all the other LEDs because the diagram is symmetric.

To turn on D1, you must set GP5 HIGH and GP0 LOW. Just knowing those two facts has some important implications. Note that D1 is in parallel with the series combination of D3-D8-D5. Those three diodes point in the correct direction to conduct current whenever D1 is conducting current. The only thing that keeps D3-D8-D5 from lighting up is that it takes 3 forward diode drops of voltage to get current flowing through that series, and we only have one diode drop (the forward voltage of D1). But this does mean that the junctions of D3-D8 and D8-D5 must assume the voltages of .67Vcc and .33Vcc respectively. But that means GP1 must be .67Vcc and GP4 must be .33Vcc. In fact, by analyzing the series D3-D12-D9 in a similar way, you can see that GP2 must also be at .33Vcc.

So just starting from the minimal requirement to turn on D1, we find that all the other PIC pins (GP1, GP4, GP2) must be floating. If any of those three are forced either high or low, then some other diode besides D1 will be turned on. Therefore we can conclude that the only way to multiplex these 12 diodes in the circuit shown is to select only one row and one column (one high, one low) and make every other PIC pin floating.

As for adding resistors, yes, do so. Adding resistors in series with GP0 and GP1 is sufficient, as long as you don't try to turn on more than one LED at a time. Otherwise you have no real control over LED current, and are relying on the unspecified current-limiting properties of the PIC outputs.

Edit: I just realized that it is not strictly true that any non-participating output will turn on some other LED. In the example cited above for turning on D1, you could set GP1 high without turning on any other LEDs. That is because doing so would remove voltage from across D3 and divide the remaining voltage across the series of two diodes: D8-D5 and D12-D9. This would force GP2 and GP4 both to assume .50Vcc. So they definitely must be left floating. And the forward voltage drop of D1 is still not enough to light up a series of two diodes, so D8-D5 and D12-D9 will not light up, even though they have a forward voltage across them. The voltage is just not high enough.

But this does bring up another issue. Whenever you leave non-participating pins floating and they assume an intermediate voltage, you have to consider what the CMOS input structure does. For Schmitt trigger inputs, like GP2, it is no problem. They are made for tolerating intermediate voltages. But for all the TTL inputs (every other pin in the 12F629), putting an intermediate voltage on an input can cause the input stage to be biased into its linear region and cause excessive supply current to flow. It is not a good idea.
_________________
Robert Scott
Real-Time Specialties
Embedded Systems Consulting

Sydney · Joined: 13 Feb 2009 Posts: 71

test153 · Joined: 09 Feb 2009 Posts: 28

Would something like this work in practice? I have chosen to only light 1 LED at a time because of the risk of overloading an "output low" port. It works in the simulator and I think it will work on the hardware too. Unfortunately I can't test it right now and will have to wait until Monday.

It doesn't matter how many LEDs are lit it still takes 48ms (4*12) to cycle through all the LEDs - effectively keeping the brightness at a constant level. (at least that's how it's suppose to work).