0
\$\begingroup\$

Here is a simple circuit I found recommended on here for generating a very short pulse from longer pulses (in this example from a switch)

enter image description here

However it isn't behaving as expected when resetting a CD4017. The switch only successfully resets the CD4017 if there is long delay between presses, 4-5 seconds, pressing rapidly causes nothing to happen no matter how long I keep repeatedly pressing.

I tried adjusting value of resistor up to 200K to see if maybe the pulse was too short, this made no difference. I have no idea what's causing this strange behaviour.

\$\endgroup\$
6
  • \$\begingroup\$ Can you show a scope plot of pin 2 of the switch and the RST pin of the IC? Do they look right? What is the part number of that buffer? \$\endgroup\$ – Justin Sep 4 '20 at 17:41
  • 3
    \$\begingroup\$ There is no path for the capacitor to discharge when the button is not pressed. You have to wait till the capacitor discharges by its own leakage. May be add a pulldown resistor to the left plate of the capacitor just like the right plate? \$\endgroup\$ – AJN Sep 4 '20 at 17:43
  • 3
    \$\begingroup\$ Because you need to to discharge the cap before making the second contact. Put 200k across the cap. \$\endgroup\$ – Ale..chenski Sep 4 '20 at 17:45
  • \$\begingroup\$ Jay, I use a diode to discharge things and a slightly more complex circuit when the PB response needs to be solid and yet also responsive (from human vagaries of using the button and human perception of the responsiveness and accuracy in their opinions) and I don't want to use software on a MCU to get there. \$\endgroup\$ – jonk Sep 4 '20 at 18:25
  • 1
    \$\begingroup\$ I see it now, I originally thought the 10k resistor would discharge the cap but there's an open circuit with the switch open. I will try pull down resistor, thank you for the suggestions. \$\endgroup\$ – Jay Sep 4 '20 at 18:57
0
\$\begingroup\$

@Jay You forgot the resistor from VCC to your switch. Add 10k and everything works perfectly.enter image description here

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.