Go Down

Topic: Two 4051's to drive 8x8 LED matrix - interference (Read 2 times) previous topic - next topic


I personally do not think that it would matter for my problem.

Yes but it would make other people understand what you are doing a lot better.

Sorry I don't know where you got that design from but it is very wrong.
First of all there is no current limiting components, a resistor is the simplest way of limiting the current.

Next you can only light up one LED at a time. This means that each LED is only on for 1/64th of the time. That means it will be very dim, remember you do fading with PWM by adjusting the on / off ratio.

I really do not get how leds that are in LOW@ +ve terminal and LOW@ -ve terminal can have any current passing through.

Yes you are right about this.


I was thinking that the pullup and pulldown resistors attached to pin 3 (common out/in) of mux/demux would limit current.
Leds are already dim if hooked up with those two resistors only.


Sorry but have you any idea how hard it is to see what a circuit is doing without a schematic.


Here is a new sketch. Hope it clarifies things.


Thanks now I can see exactly what is wrong.

1) You should not have any resistors in the +5V supply to the chip.
2) A series resistor of 2K2 in the common input to the selector is way too high. This needs to be in the order of 100R or so.

Having said that read my other concerns about this design, especial the bit about having a 1 to 64 on / off ratio of the LED the LED is always going to be dim.

Go Up