I am trying to build a "simplified keyboard encoding circuit" found in my textbook, but I cannot understand one part. Namely, "there are two 74HC195 4-bit shift registers connected as an 8-bit ring counter with fixed bit pattern of seven 1s and one 0 preset into it when the powers turns on."

I know that the parallel-in/parallel-out register loads data when LOAD is HIGH, but how do I load the data only one time - when power is turned on? That is, when power is turned on, the counter loads data, but from that point on, it shift data indefinitely. I cannot think of any workable device that can be applied here.

Any ideas?
Thanks

 

While waiting for the experts, I've looked at the datasheet and it looks like you can load data on the parallel lines,
bring the shift/load line LOW (all of the parallel lines go to Q lines),
then change the shift/load line HIGH, and clock the serial data.

I think that's what you see in the function table, and in the Logic and Timing Diagrams.

So to load it one time, at startup, you just hold shift/load LOW for X microseconds, then keep it high until you turn it off.

Let's see what the experts say...

 

thanks for input

you just hold shift/load LOW for X microseconds

this is exactly where I am stuck. When I power up the circuit, I don't know how to invert the signal for just a short period of time. I could add a switch, but that'd be a wrong approach, because it should be automatic.
I feel like I am missing something very simple here... :|

 

Well, you could use a 555 in monostable mode.
http://en.wikipedia.org/wiki/555_timer_IC

In this one,
http://www.doctronics.co.uk/555.htm#monostable
you can change the amount of time that the signal stays low by changing R and C.

Again, the experts will give you better answers.
Maybe even a simple RC circuit...

 

Get a Power ON Reset chip.

 

A simple way is to connect a capacitor from the shift/load input to ground and a resistor from the shift/load to the V+ power. When power is applied the shift/load voltage will initially start at ground (logic low) due to the capacitor and charge to V+ (logic high) by the resistor with a time-constant of RC. You want the RC time-constant to be longer than the turn-on time of the power.

 

Thanks, guys!

A simple way is to connect a capacitor from the shift/load input to ground and a resistor from the shift/load to the V+ power. When power is applied the shift/load voltage will initially start at ground (logic low) due to the capacitor and charge to V+ (logic high) by the resistor with a time-constant of RC. You want the RC time-constant to be longer than the turn-on time of the power.

power-----resistor-----capacitor-----shift/load

Something like this? Sorry, I forgot how RC works..

 

Oh....very nice.
I've learned about the existence of a new chip! Power On Reset.
http://www.futurlec.com/Maxim/MAX699CPAa.shtml

It looks like for a change in power, you can have a fixed delay or a variable delay in "reset".

And yes! A simple RC circuit that you can vary the delay by varying R and C. And it is local to the chip, not affecting the other circuitry.

What a great forum!

 

So perhaps something like this?
This is similar to adjusting the frequency and duty cycle of the 555.
Thanks!

 

...................................

Sorry, I forgot how RC works..

Check the "Time-domain considerations" paragraph here.

 

So perhaps something like this?
...................

Yes, that's correct.

 

Thank you so much everyone!