1
\$\begingroup\$

Schema of debouncing circuit

I just recently started dabbling with electronics, so probably I just made some beginner's mistake. Here's my problem: I want to debounce a switch in hardware. During my internet research, I found this circuit in several variations; but pretty much all of them have in common, that they make use of 2 resistors, a capacitor and an inverting Schmitt trigger. So I built one using 2 10k resistors, a 0.1 uF capacitor and a 74LS14N (see the attached schema). My problem is, that the input of the Schmitt trigger only drops to exactly 800mV when the button is pressed, and the inverter's output stays low.

I am desperate to know where those 800mV come from.

It is also possible that something went wrong during my wiring (I built this on a breadboard with jumper cables); but there's only so many ways you can mess it up, and I tried many times already, even on different boards, with different cables and different ICs - nothing helped.

\$\endgroup\$
0
2
\$\begingroup\$

TTL inputs source current when driven with a logic low. So as Solomon comments you must pull current from the input to ensure that the input goes low enough in voltage for it to be seen as a logic low. A resistor that is too high in value will not do this.

The 74LS14 typically requires a maximum of 0.4mA when held low, which would cause about 400mV across the 10kohm input resistor. Are you sure it is an LS device and not a standard 7414 that would require 0.8mA giving the 800mA that you see.

A CMOS version - 74HC14 would be a better choice of device. The input current is essentially zero for those.

As others comment a reduction in the resistor value would also sidestep the issue although the time constant would be reduced as well unless the capacitor is increased.

In general, it is not recommended to use pull-down resistors with TTL.

enter image description here 74LS14 Datasheet

\$\endgroup\$
3
  • \$\begingroup\$ The designation on the IC reads SN74LS14N; so if I am not mistaken, that is the TTL LS variant. Apart from that: redimensioning the RC circuit (1K resistors + 1uF capacitor) did the trick! I did some reading on the 74HC14 and it seems like using a CMOS output as input for TTL ICs does not seem to be a problem, so I'll probably switch to the CMOS variant. Thanks for the answer, now I just need to read up on why not to use pull-down resistors on TTL ICs. \$\endgroup\$ – pczora Sep 4 '20 at 17:31
  • \$\begingroup\$ Re, "TTL inputs source current when driven with a logic low." Yes, but for a different way of saying the same thing, which might have more impact on a newbie's ears: You must pull current from a TTL input in order to 'drive' it low. \$\endgroup\$ – Solomon Slow Sep 4 '20 at 21:53
  • \$\begingroup\$ I will try to keep that in mind :) \$\endgroup\$ – pczora Sep 4 '20 at 23:24
0
\$\begingroup\$

The 10k impedances are just too high for a LS TTL chip input, the chip input will source current out via the series 10k resistor when the pushbutton is pushed.

The values would be fine for a chip with CMOS input.

So the resitors should be in the 1k to 4k7 range. Note that if you change to smaller resistance by factor of 10x, you need to change to 10x larger capacitance to have same time constant.

\$\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.