Re: Multiplexing LEDs - calculating resistor value


nospam wrote:
"roxlu" <diederickh@xxxxxxxxx> wrote:
My previous post about speeding up multiplexing resulted in a display
with
8x48 LEDs. As I suggested and read, the LEDs are becomming less bright
when more than 2 LEDs are on in a matrx of 8x48 >=.

Here is a small part of the circuit to show the current limiting
resistor of 100R.
http://imagebin.org/5461

It sounds like you are still driving the matrix wrong. With the circuit you
linked only one bit in the row shift register should be on so only one LED
connected to that 100R resistor will be on at a time and that LED gets the
whole 21 mA.

When you start driving the matrix the right way remember all 48 LEDs could
be on and the row drivers will have to source just over 1 Amp which the
transistors you chose may not be capable of.

--

Hi nospam and John

Thanx a lot for all your comments and suggestions.

It sounds like you are still driving the matrix wrong. With the circuit you
linked only one bit in the row shift register should be on so only one LED
connected to that 100R resistor will be on at a time and that LED gets the
whole 21 mA.

Just to make clear what you mean here...... I'm driving the matrix like
this: At start I make the first column (vertical) high. Than I set
which of the eight LEDs (the "row") must be turned on. This is how I'm
driving the matrix now.

Do you mean that I need to drive it like this:
Make the first column high, than lets say that LED 2 and 4 must be
turned on, I first shift this into the shift register: 0000 0010 (LED
2), than I put LED 2 off: 0000 0000 And than I put LED 4 on: 0000 1000.
Is this correct? If so, I think this will take a lot of processing
power from my uC.

Thanks,
Greetings

.

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