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LEDs light up when switching on MIDIbox


FantomXR
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Hey people,

everytime I switch on the MIDIbox, all LEDs will light up for a fraction of time. Is there any chance to avoid this behaviour? 

I think it has something to do with the initialization-process. I already tried to put a capacitor (100uF) between VDD and GND, but this didn't change anything...

Thanks,
Chris

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The state of outputs on 595 shift registers is undefined on power up. But it's possible to temporarily set the /OE pins (#13) high with an RC delay, meanwhile the Core will have sent a "clear all outputs" signal to the 595s. You could potentially do this with a free MCU pin too. Keep in mind that the /OE pin will be high impedance, which means any transistor bases used as current sinks will be floating = not a good idea. To avoid this, add a pull down resistor of 100k or so to the transistor base.

Edited by latigid on
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3 minutes ago, FantomXR said:

Hi!

Thanks for your reply!

I think using a RC delay sounds like the simplest solution. But why should I set this pin to high?

And do you have any idea what kind of values to choose?

thanks,

Chris

 

Hi,

/OE means output enable, active low (/ or a line on top of the pin name). When it's low, outputs are enabled. When it's high, outputs are tristated/high impedance/open circuit. Notice the DOUT pin 13 is normally connected to 0V/ground. For this to work, you have to charge a capacitor (e.g. 10uF) from +5V to /OE, then provide a resistance to 0V (10k, you choose the delay though). When the power is applied the /OE pin will "see" this as a high logic level, then fall to the low state once the capacitor has charged.

Finally, you add a diode to discharge quickly once the power turns off. Anode side to 0V, cathode side to /OE.

 

591a26bab74f8_RCdelay.thumb.PNG.307f661b

 

You can distribute this circuit to many /OE pins; I did the BLM 16x16 in groups of 4.

 

Best regards,
Andy

 

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  • 2 months later...

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