Odd 'one-shot' requirement

[JMACgyver]

Howdy,

I want to thank everybody for the help I got with my first post a while ago, and hope that this request isn't too imposing.

I need to alter a momentarty pushbutton switch (SPST) circuit to work with a 'cherry' type snap-action lever switch, *however* the new switch will be in one of two states continuously at any given time.

I am needing a 'one-shot' type circuit that will output a single pulse (~500ms) only once when the lever switch is pressed, *and once again when it is released*. I can use the NO, C and NC contacts on the lever switch if it simplifies things. I need this extra circuit to draw it's power from the existing circuit, and use as little quiescent current draw as possible, as battery longevity is an issue (4.5v DC)

Just as an example application, lets say i want to indicate when a coffee mug is placed on or picked up from a coaster, but not while it's just sitting there, nor while it's not on there. Just a notification of the state change.

Thanks in advance!
--Electro--
aks David M.

PS: to clarify, the type of switch i HAVE to use will not be momentary, but that is the type of output I need to get from it.

[pebe]

Just so I fully understand the project.

Can you confirm you have a simple ON/OFF switch and you want it to generate a 500ms pulse whenever it changes state?
Is the pulse output level to be the same as the battery voltage?

[JMACgyver]

[quote="pebe"]Just so I fully understand the project.

Can you confirm you have a simple ON/OFF switch and you want it to generate a 500ms pulse whenever it changes state?
Is the pulse output level to be the same as the battery voltage?[/quote]

Umm, not necessarily a pulse. I mentioned the 500ms to factor in any built in debounce in the original circuit. I just need to make the on/off switch act like the original momentary switch. No voltage needed at the tie-in, only a brief shorting of the contacts.

Hope this helps,
--Electro--
aka David M.

[pebe]

This circuit should do what you want. Only one of the four XOR gates of the CD4070 is used. Its logic is that if the two inputs are both high or low, then the output is low. But if one of the inputs changes state then the output will go high.

With SW open, points A, B, and C will all be high and the output will be low. C1 will be charged via R1.

When SW closes, A goes low. B will go low but C will stay high for 1/2 sec until C1 discharges. During that time the output will go high. As soon as C1 has discharged so B and C are both low, the pulse will end.

The same thing will happen as the switch is opened again.

Tie the 6 unused inputs to either 0V or V+ as convenient. Leave the 3 unused outputs unconnected.

Check on the data sheet for pin connections and chack the gate will supply the current you need.

Best of luck.

[JMACgyver]

Pebe,

Thanks for the help. While I'm waiting on being able to pick up the parts needed, could you tell me about how much current it draws in its 'standby' or 'waiting-on-state-change' mode? (As I mentioned, I'll be using 4.5V)

thanks.
DYM

[pebe]

Like all the CMOS 4000 series gates, it takes peanuts of current. Less than .25microamp at 5V in fact. It is specified to operate down to 3V'

If you want it, the datasheet can be found here
http://www.datasheetcatalog.org/datashe ... 314_DS.pdf

[JMACgyver]

Sounds good, I look forward to trying it.

[bitsandpieces]

try using a cd4538 as this has both t- and t+ inputs