How to detect power cutout for PC?

Oh, MOVs are _so_ retro! (And they wear out.) Use a TransZorb. :-)

Cheers! Rich

In article , Mark Zenier wrote: [...]

The characters will be happening about 60 times per second. (600 Baud/10) The code could prescale the rate down so that the shared memory only gets updated about twice per second. This would save some CPU time. Also, I'd put some resistance between the transformer and the serial port and perhaps a MOV across it. If you unplug the transformer, the inductive kick could cause trouble.

Of course. John, you've got to read the thread. The cap discussed is a large electrolytic and the example given was 4,700 uf. The power supply specified was a 12 volt DC regulated wall wart.

Ed

And the current is?

Suppose it's 60 mA. That means the load resistance is 200 ohms. The time-constant with 4700 uF is 940 ms. After less than a second, the voltage is down to 0.37 x 12 = 4.4 V. After less than 3 s, it's 0.6 V.

No, fred bloggs is on to the answer. Open it, measure the voltage across it's internal battery, use that as a trigger. Compare it with a reference built out of a TL431 or something, use that to turn on an optoisolator, used to drive the input signal, perhaps a parallel port pin. This gives him a signal only when it is required.

The OP is in Turkey, so that would be 50 char/sec out of a possible 60 char/sec. On an American power circuit, you'd probably want to configure the the serial port to "7 bits no parity" so the UART has time to get reset after the stop bit, to avoid an occasional overrun error.

Oops, yea. Or maybe just put an incandescent lamp across the low voltage AC.

Mark Zenier snipped-for-privacy@eskimo.com Googleproofaddress(account:mzenier provider:eskimo domain:com)

you'll be wanting to open /dev/ttyS0 (or ttyS1 etc) and use ioctl() calls to get the status bits, or I think you can get the OS to send a SIGHUP signnal when the carrier detect line goes off if you first ctty() to the serial port.

Bye. Jasen

John, *READ* the thread!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! "If you use a relay with a high coil resistance, such as Radio Shack #275-248 and a 4,700 uF 16 volt (or higher) capacitor, you will get a few seconds delay. The higher the coil resistance and/or the higher the capacitance, the longer the delay. For example, and Omron G5V-1-2-DC12 relay has more than twice the coil resistance as the Radio Shack relay. If you used that relay, you would get a delay approaching 10 seconds."

The brief delay is all he needs. He can sense the relay with software once every desired_delay_time and won't need a routine that says if power bad, sense again in 4 seconds, if power still bad, signal.

However, since people seem to think the full delay needs to be in hardware, the cap amplifier is a good idea. I did that a long time ago, with bjt. Here are two circuits, aimed at the OP's self-described level of familiarity (and perhaps a bit beyond it). I think the first is the simplest of the two for the OP to get longer hardware delay - it yields about 1 1/2 minutes delay as drawn.

No it doesn't.

Funny that !

I've designed more power supplies than you could shake a stick at.

Indeed it won't.

It simply makes the reservoir cap bigger and reduces ripple.

To make a delay you need an *RC* network !

You don't seem equipped even to understand the very basics of electricity.

Graham

So ?

I can't find a note of the power rating but lets say it's 6W - so can supply 500mA.

dV/dt = I /C

And actually the current available to charge the C will be > the steady state figure making delta t shorter than above anyway.

Graham

You're not very good at this electricity stuff are you ?

Why not learn some basic science ?

Graham

Please *please* explain how the resistance of a relay coil across a 12V supply affects the charge time for a parallel cap !

Graham

Or use a PIC ! ;-)

Graham

Quite the bad mood ehh? A simple Radio Shack project would be like so- you don't like the direct coupling?- then use an optocoupler: View in a fixed-width font such as Courier.

. . 7805/TO-220 . ------- 1K . +----|IN OUT|---+----+--------+-----+ +-/\\/\\--CTS . | | |10V | | | | . | 0.1u | | | | | | 1N914 . +--||----+-------+ / | / +-|-+--/\\/\\--+---+---------------|- \\ | 10K |/ | . | | | | >---+-/\\/\\-|2N4401 | . Vbatt | | | +--|+ / |\\ | . ----- -> | | | | | / e | . 12 / +|2.2u / | | LM339 | | . 10K \\ === \\

It doesn't. It affects the _discharge_ time. :-)

Cheers! Rich

I guess you haven't read the whole thread either. The cap _charges_ right away, from the wall wart, and the relay pulls in, since its coil is in parallel with the cap.

When power is removed, the wall wart ceases to provide volts or current, so the capacitor proceeds to _discharge_ through the relay coil, holding it closed for some delay time.

Cheers! Rich

Speaking of not understanding the basics - suppose you put a BMF capacitor and a relay coil in parallel, across a 12V supply. Let the cap charge as long as it needs to, until you're at a steady state, with the relay pulled in.

Now, remove the 12V supply. What happens? When?

Thanks, Rich

Might as well help out that 7805 with input ripple rejection- this works for 12V or 24V: View in a fixed-width font such as Courier.

. . 7805/TO-220 . ------- 1K . +----|IN OUT|-+------+--------+-----+ +-/\\/\\--CTS . | 50V | |10V | | | | . | 0.1u | | | | | | 1N914 . +--||----+-----+ / | / +-|-+--/\\/\\--+---+---------------|- \\ | 10K |/ | . | | | | >---+-/\\/\\-|2N4401 | . Vbatt | | | +--|+ / |\\ | . ----- -> | | | | | / e | . 12 / +|2.2u / | | LM339 | | . 10K \\ === \\