Not a good battery design.

[William At MyBlueRoom]

I've seen this design before. How to light 2 LEDs on 1 I/O pin. It is an I/O efficent design but not a power efficent design. The load would be at least 100ohms (the rest of the schematic goes to a PIC10Fxsx) 4 I/O pin PIC.

https://www.electro-tech-online.com/custompdfs/2006/04/Amoeboid.pdf

IMO it would have been better to use a 12Fxxx PIC unless you were running it off the mains. *the Amoeboid is 9V battery powered.

I also like using a method that allows for LEDs = (pins * (pins-1))
eg: 3 pins can display 6 LEDs, 4 pins can display 12.

 

[eblc1388]

I also like using a method that allows for LEDs = (pins * (pins-1))
eg: 3 pins can display 6 LEDs, 4 pins can display 12.

It can be done without the pair of 100 ohm resistors between Vdd & Vss so no extra energy wastage either. The magic word is "Charlieplexing".

 

[DirtyLude]

It can be done without the pair of 100 ohm resistors between Vdd & Vss so no extra energy wastage either. The magic word is "Charlieplexing".

That's what he's talking about. There's two seperate concepts here. And the "method that allows for LEDs" he mentions at the end is not the same method shown in the original schematic posted. The 100ohm resistor method would only be useful for single pin 2 LEDs, though it would be a current hog.

Actually this goes with a question I have as well. Let me make up a scheme, it will be easier to explain, then.

 

[DirtyLude]

Would this be workable for 3.3v.

Assuming the forward voltage of both LED's together with resistors is greater than 3.3v, the LEDs will not turn on on their own. The microcontroller pin can be configured to either sink or source current, allowing one or the other to be turned on.

I know this works, I'm just wondering wether their will be current leakage through the LEDs.

Hmm.. I think the easiest way to answer this is to setup a test of the circuit myself, it's easy enough. Just thought I would throw it out there.

 

[eblc1388]

So what he wants is a method to drive 12 LEDs using 4 PIC pins and *not* using chalieplexing?

In your suggested circuit, I would replace the two 220R resistors with four 1N4148s so that none of the LED will light at 5V with the port pin configured as input. The only current limiting resistor will then be from the LED junction going to the port pin. A HIGH on the port pin will light the bottom LED and a LOW will light the top one.

 

[DirtyLude]

Actually I don't think he's asking any question. He's just showing a part of a schematic he's seen. I'm not certain the point of the post.

He says he also likes using the charlieplexing method, but you need 3 pins minimum for charlieplexing, unless you know another method.

In my own schematic, I have no requirement for 5v, so working at 5v is not needed.

 

[rockin_rick]

While I'm also not certain of a question here, there is a Microchip Technology "app" note regarding complementary LED drive for multiplexing LEDs on I/O pins. Search out TB029 (DS91029A)


Rick

 

[William At MyBlueRoom]

I guess what I was getting at was although the LEDs would work, the power consumption levels were wasteful. Using a 12Fxxx (6 I/O) instead of a 10Fxxx would have been a more sensible solution. Only pennies difference in price and the longer battery life would have made up for the difference.

I know driving lots of common anode or cathode LED displays is called charlieplexing, not sure if the name applies to driving many LEDs (like the tips & tricks app note on microchips site)

Here's a question:
What are the minimum number of digital I/O pins required for 2 LEDs and 2 switches?