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Tactile Switch Trouble

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Meatmod

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I am trying to make a small LED turn on and off by pressing a very tiny Tactile Switch (Outer dimensions: 4.5 mm, Thickness: 0.55 mm)
My problem is that the Tactile switch I’m using is a Momentary SPST so the LED only lights when the button is held in.
I want it to work like a normal SPST (push to turn on, push again to turn off) but I have yet to find a Tact Switch that is non-momentary and that small!

I need a solution that keeps my project as small as possible. (Very tiny components, very compact space)
Any suggestions?
 
I have an Alternate Action circuit that is turned on then turns off then turns on again etc with a pushbutton. The CD4069 is 14 pins but makes 3 of these circuits. A surface-mount package is small. Each circuit needs 2 resistors and a transistor to drive an LED. The battery can be two 3V lithium cells to keep it small but an LED will quickly drain them.
 

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Thanks for the quick response!
That might do the trick, but does anyone manufacture this circuit or is it something I'll have to try to make on my own? (I'm a bit of a noob)

Any other ideas out there?
 
The tiny dual inverter chip will do the job but it is too small to solder to.
 
I'm going to give it a try!
Thanks for your help guys. I'm probably going to need more direction on how to wire everything together but I'll make a follow-up post when the time comes.
 
Hi, I am having the same problem as Meatmod and would like to try the circuit that audioguru provided. However, I quite like to understand what it is that I'm doing, and was wondering if anyone can help provide a quick and simple explanation?

Here's what I've got so far:

Switch off -> Switch on (momentary) -> Switch off
Input: low -> high -> low
Btwn inverters: high -> low -> high
Output: low -> high -> low << ???

The above suggests that after the first push of the button the output will still remain low, which shouldn't be the case. I think I'm missing out something with the feedback, but my circuit analysis is quite bad and I'm a bit confused.
 
Whatever state the circuit is in is maintained by the 4k7 resistor's feedback, that much you understood, Silverwolf.

1) With the switch open the only input to the first inverter is the 4k7 resistor, whose feedback maintains the current state of the flip flop.

2) While the switch is open, the 220k resistor charges the capacitor (slowly) to the level of the first inverter's output (the inverse of the current circuit output).

3) When the switch is closed, the capacitor voltage (which by now is the same as inverter 1's output) is applied to inverter 1's input, thus flipping the circuit state.

4) While the switch is closed, the 4k7 resistor, being small compared to the 220k, charges (quickly) the capacitor to the level of the second inverter's output, which makes the circuit stable again. Goto 1.

Any clearer now?
 
Thanks for your explanation Cabwood! I now understand how it works.

Check that the switch is not connected to ground. It should be connected to the capacitor instead.
I haven't built the circuit yet - I was just curious as to how it worked. Thanks for the tip though - I'll double check it when I come to building it.
 
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