Mark Ashley wrote:
Hi there
I am a novice at electronics so please bear with me!
I need to design a latching electronic switch with a momentary push switch
as a trigger. When the trigger is pushed, the switch will latch closed. W=
hen
the trigger is pushed again the switch will toggle back to the open
position. The switch needs to be able to break/supply power a 12v relay. =
If
the circuit loses and regains power the switch should default back to the
open position.
For reasons I won't go in to at this stage it is not possible just to use=
a
latching switch to supply the power to the relay.
Hopefully someone can help me if this design is possible!
Thanks
Mark
Hi, Mark. Novice and newbie questions always welcome here.
If you've got half of a CMOS inverter IC (4000-series would include
4049, 4069, 40106, 74C-series include 74C04, 74C14, &c) or three gates
you can set up as inverters (NAND, NOR, &c), this circuit might fill
the bill (view in fixed font or Notepad):
~ Logic Toggle Pushbutton With Power-On Reset
~ VCC VCC
~ + +
~ | |
~ 1N4002| |
~ VCC - C|
~ + ___ ^ C| RY1
~ | .----------|___|--------. | C|
~=2E33uF | | 22K | | |
~ --- | | '---o
~ --- 1N4002| | |
~ | |\ | |\ |\ | ___ |/
~ o--| O--|--o-----| O---o----| O--o--|___|-o-| Q1
~ | |/ | |/ | |/ 22K | |
~ .-. | .-. .-. |
~ 10K | | | 220K| | | | |
~ | | | | | 22K| | |
~ '-' | T '-' '-' |
~ | | --- | || | |
~ | '--o o-------o----||----. =3D=3D=3D =3D=3D=3D
~ =3D=3D=3D || | GND GND
~ GND SW1 .022uF |
~ |
|
=3D=3D=3D
GND
created by Andy=B4s ASCII-Circuit v1.24.140803 Beta
www.tech-chat.de
Make sure you use a darlington NPN transistor (like the TIP120) to
switch the relay. This keeps the load on the last inverter down to
around 1mA or so, which any CMOS inverter can handle at 12V. This
should be good for any relay coil that draws less than half an amp.
The first inverter with the diode is set up as MML (Mickey Mouse
Logic). It's only active at turn-on, and forces the input of the
second inverter to be low right after turn-on. That means the third
inverter will be low, and your transistor will be off. After a period
set by the R and C (something on the order of 3ms) the output of the
first inverter will be high, and the diode will effectively remove it
from the circuit.
This stuff was borrowed piecemeal from Don Lancaster's CMOS Cookbook,
which is a good intro to digital electronics for newbies. It's
available at Amazon, libraries, and Mr. Lancaster's website:
http://www.tinaja.com/
Good luck
Chris