Simplest shift register test circuit - clock

Question

I'm looking for the simplest way to test the operation of a 74HC595 shift register without complex circuitry.

Currently (on a breadboard) I have a 47K SIP resistor in a pull-up fashion connected to pins 9 through 14 to make all inputs a default high. Yes I'm aware pin 9 is meant as n output and I'd be wasting some current if output of that pin is high but that particular output is useless to me anyway.

I then select random Qx output pins to connect directly to LEDs through current limiting resistors.

So I begin my tests and determined that the data line (pin 14) works, and enable (pin 13) also works, and the store clock (pin 12) also works along with the reset (pin 10) works.

But the clock (pin 11) is hard to tell if it is working because I need to pulse it faster than the fastest typist which I can't do. Every time I pulse it (make clock line low then make clock line high about 1/2 second later) after inverting the input data line, it seems all data output changes to the new input value instead of just one data output line.

So then I added a 10uF capacitor between clock (pin 11) which is the capacitor across the "wire" button in an effort to eliminate "bouncing", but that method isn't working.

So is there a simple way to test this clock line so that the IC sees that only one clock pulse occurred rather than hundreds of pulses?

The only answer I can come up with so far is to make an oscillator and connect it's output to the clock but I want to avoid too many extra chips if possible. Heck, if I can rewire my capacitor or resistor or even change their values, then I'd rather do that just to make the clock respond correctly.

Any ideas?

Answer 1

the problem you have is probably contact bounce

here's a simple de-bounce circuit, it uses no semiconductors, but it reqires a SPDT switch,

^{simulate this circuit – Schematic created using CircuitLab}

Another possible problem is ground bounce, be sure to have a capacitor of at-least 100nF connected directly across the power pins of the chip, especially if you're using solderless breadboard.

Answer 2

**How hard is this? **

ground unused CMOS inputs... and add 0.1 to Vcc,Vee

Not exactly same chip

Clock uses a Schmitt Inverter. A switch can remove 100K from output and then jumper from 0 to 5V for debounced switch with suitable RC=T time of ~0.05s

Answer 3

^{simulate this circuit – Schematic created using CircuitLab}

To debounce a switch signal to the clock input, I use a quite large capacitor of 1uF, (between 0.1 and 10 uF depending how far you need to debounce) and two MOSFETS. The result is almost square wave. You can also replace the two MOSFETs by a Schmidtt Trigger (perfect square wave).

You can't place a capacitor of more than 0.01uF because the clock input needs a sharp rising edge. 0.01uF is not enough to debounce. Larger cap smooth the edge too much and it's not recognized as a clock pulse. The darlington circuit with the two MOSFETs or a Schmidtt Trigger corrects that.

You can also put a 0.1 or 0.01 uF cap (not in the schematic) in serie between the debouncer circuit and the input to create short pulses at button press. But not an obligation.

R3 and R4 avoid unwanted triggering from noise. Put them also with the Schmidtt Trigger.

The easiest experiment is to connect SHCP and STCP together and SD to Vcc.

You can also use only one MOSFET if you can cope with the resulting signal inversion, mentally. Using two MOSFETs gives a more intuitive ON/OFF perception.