Time Delay Circuit

[AnthonyWow]

I have a computer system with a device on a 5V standby (two wires - ground and possitive). I need to delay the 5V standby power to this device when I power up the system. In effect, I want the system to have power for 2 to 3 second before the device receives the 5V standby power. Ideally, I would like to splice the delay ciruit in the positive wire. I believe I need and RC circuit, but I'm not sure. Can you recommend a circuit?

 

[Russlk]

How much current does your device draw? An RC delay would not be practical for more than a few microamps. A 555 timer could be configured to delay 2 seconds.

 

[AnthonyWow]

How much current does your device draw? An RC delay would not be practical for more than a few microamps. A 555 timer could be configured to delay 2 seconds.

I believe it is about 50-100mA. Not much.

 

[Roff]

This should work. If you want something simpler, but with looser tolerance, we could use a p-channel MOSFET, a resistor, and a capacitor (and a diode for fast recovery at turn-off).

 

[Optikon]

This should work. If you want something simpler, but with looser tolerance, we could use a p-channel MOSFET, a resistor, and a capacitor (and a diode for fast recovery at turn-off).

Don't you intend positive feedback for hysteresis? 1 Meg going to -ve.

 

[Roff]

This should work. If you want something simpler, but with looser tolerance, we could use a p-channel MOSFET, a resistor, and a capacitor (and a diode for fast recovery at turn-off).

Don't you intend positive feedback for hysteresis? 1 Meg going to -ve.

I don't understand your question. :?: :?:

 

[audioguru]

Its 1M resistor provides lots of positive feedback for hysteresis.

 

[Optikon]

This should work. If you want something simpler, but with looser tolerance, we could use a p-channel MOSFET, a resistor, and a capacitor (and a diode for fast recovery at turn-off).

Don't you intend positive feedback for hysteresis? 1 Meg going to -ve.

I don't understand your question. :?: :?:

whoops nevermind, I thought the 1 Meg was going to the inverting terminal.. upon closer inspection, I see that the it is going back to positive (for hysteresis). Very nice.