How to detect power cutout for PC?

[Rich Grise]

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


113 milliseconds isn't much of a delay.

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

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

 

[Rich Grise]

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.

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

 

[Pooh Bear]

It doesn't. It affects the _discharge_ time.

Cheers!
Rich

I'd got confused over what kind of delay was wanted !

It still won't work for long - as you noted.

Graham

 

[Jasen Betts]

Now, if you really insist on letting a cap discharge, just amplify

it. Use a MOSFET, although you'd probably want to use another
transistor for a little positive feedback so you'd get a snap
action - I don't know if you'd want your MOSFET to be in the linear
region for 2-3 minutes.

yeah, but whay are you going to power the relay with, another plugpack
plugged into the UPS ?


back to the original design something can probably be done using an
optocoupler instead of a plugpack and relay, but a home made a cable
with a mains plug on one end and a serial plug on the other might get
some strange looks....


hmm, what's the best way to power the LED in an optocoupley from the mains,
a big honking resistor or something more subtle?

C1
|| ___
L--[680R]--||---~ /\ ~
|| / \
250v +--<->|->---+-----+
50Hz | \ / _L_ |
N-------------(--_ \/ ~T~ \ / LED
| ~~ ~ |10u ~T~ (in opttocoupler)
+-----------+-----+


now for C1

hmm 730V swing 100 times per second wanting to pass 10mA or slightly more

0.01/(730x100) = 0.139 uf rated for 250VAC

I can get 200nF seems to be a standard size

the 680 ohm resistor is there to stop the power-on surge (if ppower comes
on at one of the peaks of the AC cycle) from destroying the diodes in the
bridge... probably should be rated for 400V but only needs to be a 1/4
watt part...

the bridge needs only to be rated for 500ma and 5v1


maybe a fuse would be a good idea too, eg, 250AC 100ma

comments anyone?


Bye.
Jasen

 

[Ken Smith]

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

How about a Sidactor from Teccor. They are very spiffy spike catchers.

 

[Ken Smith]

The OP is in Turkey, so that would be 50 char/sec out of a possible 60
char/sec.
Yes

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.

If you are reading it at a hardware level it isn't much of a problem. At
a higher level, the OS may filter out overruns.