FantomXR Posted May 15, 2017 Report Share Posted May 15, 2017 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 Quote Link to comment Share on other sites More sharing options...
latigid on Posted May 15, 2017 Report Share Posted May 15, 2017 (edited) 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 May 15, 2017 by latigid on Quote Link to comment Share on other sites More sharing options...
FantomXR Posted May 15, 2017 Author Report Share Posted May 15, 2017 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 Quote Link to comment Share on other sites More sharing options...
latigid on Posted May 15, 2017 Report Share Posted May 15, 2017 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. You can distribute this circuit to many /OE pins; I did the BLM 16x16 in groups of 4. Best regards, Andy Quote Link to comment Share on other sites More sharing options...
FantomXR Posted May 16, 2017 Author Report Share Posted May 16, 2017 Great!! Thank you very much! Quote Link to comment Share on other sites More sharing options...
FantomXR Posted July 31, 2017 Author Report Share Posted July 31, 2017 I finally was able to test what you suggested! It works perfect! Thanks! Quote Link to comment Share on other sites More sharing options...
latigid on Posted July 31, 2017 Report Share Posted July 31, 2017 Quote Link to comment Share on other sites More sharing options...
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