# Power cycling circuit

I am designing a circuit which may fail to start up. It has a power good output which is high when the supply is working. When the power good output goes low, the circuit has failed to start oscillating properly, and needs to be restarted. I thought about using a 555 timer but couldn't figure out how to make the output stay high when the input signal was low. Any ideas?

The circuit must do this because it fails to start under high current set points. It drives an LED. The idea is at higher currents the output folds back, but doesn't require an actual restart of the circuit to get it working again, and the fold-back mechanism makes the LED blink on and off.

Here's an image to show what I mean. Green shows the LED current. It has been set too high and the supply's overvoltage protection circuit kicks in to prevent more than 4.7V being applied across the LED (to prevent breakdown of the LED if in reverse.) The blue trace represents the power good output. During the oscillations the supply is outputting correctly, so it is high. It drops low after the oscillations stop. A short delay after it drops low, I need the supply to reset and try again - this would be accomplished by turning the supply controller IC on and off.

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You want a low-going "power good" to temporarily turn off the supply.

This circuit will do what you want. Choose R and C to get your desired delay.

Edit: there should probably be a resistor between the PNP base and NPN collector instead of a direct connection as shown.

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I love a man whom knows his analog design! –  Kortuk Nov 20 '10 at 1:36
I tried your circuit and couldn't get it working. The problem is when on, it only provides ~3V to the switcher IC, and it needs at least 6V to work properly. Any idea on why? –  Thomas O Nov 20 '10 at 10:24
What is Vcc? Which transistors did you use? What values of R and C? –  markrages Nov 20 '10 at 15:58
C = 1u, R = 1k, Vcc = ~6.5V. –  Thomas O Nov 21 '10 at 20:13
which transistors? –  markrages Nov 21 '10 at 20:50