Do u know?

Do you know how this works? This is a schmitt trigger and I am using it as one of my debouncer(used to filter signals) It doing good but I dont know how to solve the charging time of the capacitor. I am using a 1:mu: F capacitor, 10Volts and a 430hm: resistor,1/4 watts. Can you help me out with a formula or somewhat like an equation?tnx

Your circuit is a rather odd way to do bounce supression.

Here are 2 circuits that I use.

The time delay is approximately equal to the RC time constant.

It can't be predicted precisely because the CMOS Schmitt Trigger ICs have a fairly wide spread in their Vt+ and Vt- voltages.

The time constant with R = 1M and C = 100 nF is

1 * 0.1 = 0.1 sec. (if the resistance is in M and the capacitance is in uF then the result is seconds.

The IC can be a CMOS Hex Schmitt Trigger such as 74C14 or 40106, etc.
Or a 74HC version ie. 74HC14.

Don't they sell debouncing ICs?

Nope they don't. could this formula be ok? what do think? t = C * V / IR

I can swear that i have seen debouncin ICs...

If they don't, then what is this??

**broken link removed**

I don't know if its available here in our place

An IC to debounce a single switch?
The RC time constant is easy to calculate, it's R (in ohms) times C (in farads) and will give you a time in seconds required to charge the capacitor to 63.2% of the applied voltage.

Motorola/ON Semi make a debounce IC. Digikey has lots of surface-mount ones in stock. Newark stocks DIP but only non-lead ones. It is expensive for a Cmos IC.

The OP used a huge 1uF capacitor and a battery-killing resistor of only 430 ohms. A 0.001uF capacitor and a 470k resistor would do the same thing. But a 470k resistor doesn't provide much current for burning dirt from cheap switch contacts so a 47k resistor with a 0.01uF capacitor should be used.

You can use a 555 timer in it's monostable circuit configuration as a debouncer...all over the web on Google, and the math is basically just R*C.

IN your circuit, the charge time of the capacitor, it is negligible since there is no resistance in series with you cap.

But perhaps what you are really asking for is the discharge time? when the switch has been released, and the capacitor is fully charged (since the charging time was neglible we will assume this). The circuit is a C in series with an R essentially. Work out some math (or Google since this is a very basic capacitor demonstration circuit) and you get that R*C is the seconds that it will take for the capacitor to discharge to 1/3 it's initial level (and another R*C seconds it will be 1/3 of the aformentioned level, etc, etc).

paparts said:

Do you know how this works? This is a schmitt trigger and I am using it as one of my debouncer(used to filter signals) It doing good but I dont know how to solve the charging time of the capacitor. I am using a 1:mu: F capacitor, 10Volts and a 430hm: resistor,1/4 watts. Can you help me out with a formula or somewhat like an equation?tnx

Click to expand...

Change R1 to 47k and it will work. As it is, your time constant is much too short for the typical bounce characteristics of a switch.
However, it will be hard on your switch, because every time you close it, the 1uF cap dumps very high transient current through the switch. Ljcox's solutions are much better.