help on delay on power off

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bugspatrick

New Member

• Jan 20, 2005

• #1

Would somebody help me??

I need a simple circuit :
when power off, I want to give a "low" logic to a shutdown pin (low active), then 100-200ms later, turn off that IC.

Thank you very much for your help!

 

hotwaterwizard

Active Member

• Jan 20, 2005

• #2

How about this?

 

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• off_relay.gif

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B

bugspatrick

New Member

• Jan 20, 2005

• #3

hotwaterwizard said:

How about this?

Click to expand...

What about if only resistors and capacitors can be used for this purpose?

let me give more detail :
Input : 5V

in target IC: /shutdown pin, Vdd

the case is : if power disconnected accidentally, or manually power off, the sequence is : /shutdown pin will turn to LOW first, then Vdd of IC off.

Thank you.

 

TheOne

New Member

• Jan 20, 2005

• #4

You can do it with a diode and a capacitor, but then you will loose voltage over the diode. Using a schottky diode will reduce this drop to about 1/2 of a normal diode. If your application is sensitive to this drop, you can use a small transistor to do the switching. This will have an even lower drop.

 

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hotwaterwizard

Active Member

• Jan 20, 2005

• #5

Try this circuit with a much smaller capacitor.
It works from about 6v to 14v and may even work on 5vdc
Ommit the Relay and Protection Diode.

 

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hotwaterwizard

Active Member

• Jan 20, 2005

• #6

Here is another Circuit

 

Attachments

• delayedoff3v.gif

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