Power-on reset - diode location?

[Michael Daly]

I see a lot of power-on reset circuits with the same general layout. A resistor connects Vcc to a capacitor which then connects to ground. The point where the resistor and capacitor are connected is tapped to a Schmitt trigger and that output is the reset signal.

In order to ensure the capacitor is discharged quickly when a power drop happens, a diode is placed in parallel with the resistor.

Why is the diode in parallel with the resistor? It seems to my simple view that if the diode is parallel with the capacitor, it will discharge the capacitor when power is removed. If it's parallel to the resistor, when the power drops, it allows a low-resistance bypass of the resistor, but I don't see how that discharges the capacitor.

I'm obviously not an electronics expert but I know just enough to fiddle with digital circuits in "Lego" mode.

 

[ronsimpson]

reset RCD

The diode needs to be across the resistor!
When power is first applied: VCC=5 volts and the reset node starts at 0 (reset) and slowly rises to 5 volts releasing reset.
When power is removes the capacitor needs to be discharged ASAP. VCC=0 and RESET=5 Power will flow from the capacitor to VCC through the diode much faster then it would if only the resistor was there.

 

[Michael Daly]

You know how you can puzzle over something for a while and the answer only comes when you ask a real person the question? I hit "post reply" and went for my lunchtime swim - no sooner did I get outside than the answer dawned on me.

I thought the diode across the capacitor would do it, since when power was on it acts like a non-conductor but when power is off, it would short the cap in the other direction. DUH! The cap will still have + on the Vcc side, so the diode will still be non-conductive - there is no "other direction".

Then I realized what you said. In a real world circuit, there'd be a sink for the current and the diode would flow in that direction, discharging the capacitor. I was confusing myself by thinking of the PoR circuit in isolation - if the power (and everything else) is removed instantly, the diode and resistor connect to nothing and there's no sink. That's not what happens 99.99999999% of the time.

Thanks for confirming!

I'll never get over how often asking a question can open one's mind up to a solution.

 

[Oznog]

I don't put a cap on the pin, just a 10k resistor to Vdd. Unless you have a really specific issue, this will do nicely.

BOR and WDT do a lot to protect against power problems as well as others.

 

[Hero999]

The diode across the capacitor won't help as (depending on which way you connect the diode) the capacitor will either never charge or the diode will never conduct.

It also depends on the delay you want and the other components in the circuit.

A 1nF capacitor and 10k resistor is a time constant of 100:mu:F in which case I wouldn't bother since it takes much longer than 10ms (100 time constants) to turn the power switch on and off so you can be reasonablly sure the capacitor will have discharged when you turn it back on again.

It also depends on what other capacitors are in the circuit, if you have a big fat 10,000:mu:F capacitor in the filter on the rectifier then adding a diode on the resistor in series with the 1nF capacitor won't make any difference. In this case I would switch the secondary side of the supply instead (or as well as) the primary.

One way would be to use a switch with additional contacts to short the capacitor when you turn it off or you could even use a p-channel or n-channel jugfet (depending on whether the capacitor is connected to +V or 0V) to short it when you turn the power off.