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Time Delay Circuit

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AnthonyWow

New Member
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?
 

AnthonyWow

New Member
Russlk said:
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

Well-Known Member
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).
 

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Optikon

New Member
Ron H said:
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

Well-Known Member
Optikon said:
Ron H said:
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

Well-Known Member
Most Helpful Member
Its 1M resistor provides lots of positive feedback for hysteresis.
 

Optikon

New Member
Ron H said:
Optikon said:
Ron H said:
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.
 
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