Sorry Peter. But I'm sure you're going to enjoy a wonderful and exciting new adventure and experiences while you're learning.

Have fun. Best regards, Mike

 

OH!!

Ok mike I see now.. the 2XXX IC's are just an array of transistors!

I thought they were another type of logic chip!

I will try out your designs, also Nigel, I will try yours.

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Mike, your design works very well thank you!

Using an 8x8 matrix ... +v powered by the PICs B port and GNDs are switched on and off with a 4017 decade counter.

8 npn BJT are used to increase GND current!

All I need to do know is find the right UDN chip to replace the transistors which are switches on and off by the 4017

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PIC16F84a Game Module

 

Have you seen this? I modified it for a Parallax Basic Stamp 2 / 2px and a SX28 circuit and it works well.

http://drew.hickmans.net/movingsign.htm

 

You want a ULN2803 sinking driver. It'll accept the active high output from the 4017 and sink current to ground on your columns.

Have fun. Mike

 

On the cathodes I use ULN2803 chips (pos in to the ULN2803 provides neg out ot the LED cathodes).

I had been using PNP transistors / resistor circuits on my LED anodes (neg in to the transistor provides pos out into the LED anodes).

However, after reading this thread more yesterday, I see that a UDN2981 is a good substitute for the PNP transistor / resistor circuits. I ordered some samples from Allegro Microsystems Inc. yesterday.

 

Yes, the ULN2803 chips is the one I was going to get today.

Will the chip reduce the amount of resistors in my circuit?

Right now I have 16!

8 to bias the transistor from the 4017 outputs and 8 to limit current on the HIGH outputs of the PIC...

Does the ULN need resistors? I dont want to burn it out..

So that should take care of the first 8.. is there a way bring the other 8 down with out making the LEDS lose brightness.. well I know there will be some brightness lost when they are all on..

Thanks tdg,
Peter W

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PIC16F84a Game Module

 

The ULN2803 will allow more current to the LEDs that the PIC can not provide. You should have at least 500 ma to an 1 amp or more for your power supply. The resistors are current limiting resistors and they are only for the LEDs.

LED's will always need current limiting resistors added to them. The lower the value the brighter they will make the LED's appear. However, they also protect the LED's (make them last longer - no burn out) by using higher values like 220 or 470 ohms. I tend to make some sort of allowance and use 10 ohm (probably asking for trouble) to 100 ohm (little more safe) on every collumn or row.

I personnally believe that you only need to have a current limiting resistor for either every column or every row but you don't necessarily need to have them for both. So if you have 8 rows and 16 columns, I would put them on each of the 16 columns (to be safe) as there are more columns and more resistance to take in in excessive current / voltage. However, I have seen people use it both columns and rows - which you will probably see dimmer LEDs due also to duty cycle (refreshing rates for the LEDs). I think if you use LEDs with higher MCD ratings for brightness this will help too - but this would probably bring cost up too.

BTW: About LED's needing current limiting resistors, this is generally always the case EXCEPT for custom LED driver IC's which have them internally and use 1 external resistor for adjustment. An example of this would be the MAXIM 7219. The Maxim 7219 works with up to 64 LEDs of individual control and it works well. However, the only problem I have is figuring out how to cascade them in software (haven't quite mastered that yet).

Good luck.

 

Thanks and also...

Does the ULN need resistors? I dont want to burn it out..

Peter W

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9vDC Guitar Pedal PSU

PIC16F84a Game Module

 

Sorry - I was editing my post. See the first paragraph. Short answer no. Resistors are for the LEDs.

 

Careful Peter, you're getting some sloppy advice.

If you're scanning columns (only one column is on at any one time) then you'll use current limiting resistors on the row drivers. If you were to put current limiting resistors on the column lines when you're scanning columns you'd get a different voltage drop across each resistor depending on how many LEDs are lighted in that column. Your row drivers are only ever driving a single LED at a time as you scan columns.

 

Peter,

Mike is correct. I failed to mention more specifically it refers to what is not being scanned. In most cases the columns are scanned so the rows would get the current limiting resistors as Mike stated.

 

Just a thought: If you are scanning 8 rows and the LEDs are dim because of a 12.5% duty cycle, could you put a capacitor on each LED before the resistor to hold a charge and keep it lit a little longer?

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No, where's the current coming from to charge the capacitor? - if you had the excess current capability the LED would take it and be brighter.

If you need brighter, increase the current - adding drivers if required.

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And, you would need diodes to keep from back feeding the rows and columns.
Not practical.

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