Hi, I'm working with a 74AC74 D flip-flop. I wanted it to power up in a predictable state namely with the Q output low. The CLR pin is active low so at power up I have to keep the CLR pin low for a second, then set it high (5V) and keep it there.
Is there a simple way to do this with just a few discrete parts?
--zeb
...and giving yourself a real Username instead of putting your email address on that line would be wise.
Connect a capacitor (0.1 uF or so should do) between the CLR pin and ground, and a 10K resistor from that pin to +5V.
Incidently, all inputs to CMOS parts _must_ be connected to something. If an input is "not required" in your circuit, connect it to either ground or +5, whichever will allow the circuit to operate as you wish. With your 74AC74, the SET pins must be connected to +5, the D and clock may be connected to either ground of +5 if you don't use them.
The output pins may be left unconnected if not used, and must never be connected directly to ground or +5.
-- Peter Bennett, VE7CEI peterbb4 (at) interchange.ubc.ca GPS and NMEA info: http://vancouver-webpages.com/peter Vancouver Power Squadron: http://vancouver.powersquadron.ca
Thanks for the advice, but this prompts a question: I'm powering my circuit with a switching power supply. Do the outputs of switching power supplies typically ramp up or do the provide a nice square transition?
--zeb
I'd expect the supply voltage to rise fast enough for my suggestion to work. If not, use a higher value capacitor or resistor.
-- Peter Bennett, VE7CEI peterbb4 (at) interchange.ubc.ca GPS and NMEA info: http://vancouver-webpages.com/peter Vancouver Power Squadron: http://vancouver.powersquadron.ca
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