How do I initialize a D flip-flop?

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

 

[Peter Bennett]

On Thu, 15 Oct 2009 18:52:46 -0700 (PDT), zeb7k@hotmail.com wrote:

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

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

 

[JeffM]

ze...@hotmail.com wrote:

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

http://google.com/images?q=r+c+power-on-reset

....and giving yourself a real Username
instead of putting your email address on that line would be wise.
http://groups.google.com/groups/mysubs
From Google, that means signing up for a group before posting to it.

 

On Oct 15, 9:52 pm, ze...@hotmail.com wrote:

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

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

 

[Peter Bennett]

On Fri, 16 Oct 2009 05:46:37 -0700 (PDT), zeb7k@hotmail.com wrote:

On Oct 15, 9:52 pm, ze...@hotmail.com wrote:
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

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