Re: Delay on a mains-powered relay?

ehsjr_at_bellatlantic.net
Date: 01/05/05 

Date: Wed, 05 Jan 2005 20:23:25 GMT

DaveC wrote:

>So the answer is... no.
>
>If power fails at the remote site, so does the control circuitry and its
>power. So a traditional delay relay won't work.
>
>The relay, powered by mains, simply opens the wireless sensor's sense circuit
>when power goes away. I just want it to do so after waiting a few minutes to
>see if power comes back on.
>
>The alarm isn't for security purposes, just for letting those responsible
>know that certain things have occurred, one of which is power fail. The
>sensor is a wireless device run on a lithium battery, and is a packaged deal
>so has no spare power available. The relay's not near the alarm CPU, so that
>power isn't available, either.
>
>Suggestions for a simple mains-powered circuit that will open a pair of
>contacts about 5 minutes after de-energizing? If it's to be battery-powered,
>it needs to be super-low drain. I rather keep battery replacements down to
>every 2 years, at most.
>
>An idea that just struck me: the presence of mains power could keep this
>circuit de-energized. When power fails, the battery is connected to the
>circuit which starts the countdown. When zero is reached (ie, 5 minutes have
>passed), a pair of contacts would open. And the energizing circuit
>(triggering these contacts) would only have to be a one-shot; once the alarm
>is triggered, the circuit could de-energize. That way the battery is utilized
>only briefly, during power-fail situations. How might I construct such a
>circuit?
>
>Thanks,
>
>
Refer to the post by Tony Williams. There might be a cheaper way.
As he mentioned, the capacitor is expensive - you'll pay over
$15.00 dollars for one. The alternative described below would cost
a lot less. The super caps would cost you $1.70 for 2. The opto costs
50 cents. The voltage regulator costs another 50 cents. You can
get 2 10 K pots for a dollar. Add another dollar for the 1000uf
cap and resistor. I assume same price for the transformer as in
the relay design. You'll save at least $10.00 per sensor, if
the idea works. Since you have a number of sensors, it seems worth at
least trying one as proposed below. I do not know if it is practical
for you, or if it will work, since I don't have specs on the
existing setup.

You *may* be able to use an optoisolator, a super cap, and a
potientiometer instead of the relay. The potientiometer would allow
you to adjust the delay time. You would need to use a lower voltage
transformer and a regulator IC. The cost would be a lot lower. The
key is the current sense loop - without specs, I don't know if an
optoisolator will work. Here's the details if you decide to try it.

Here's Tony's circuit, modified for an optoisolator:

   View in fixed width font.
               ____
 +----+ +---|AC +DC|--+--7809---+---+-------+
      | | | | | | +|C2 | __|__
      )||( | | +|C1 | === R1 | + |--+Sense
  120 )||(12 |Rect. | === | | | |Opto |
  Vac )||(Vac| | | | +|C3 P<-+ |__-__|--+Loop
      )||( | | | | === | | |
      | | | | | | | | | |
 +----+ +---|AC_-DC|--+---+-----+---+--+----+

C1 is a 1000 uf 25 volt cap. C2 and C3 are .1 farad 5 volt super caps.
(Allelectronics sells these for 85 cents each) R1 is a 100 ohm
1/4 watt resistor. P is a 10K potentiometer. The opto is a 4N35.
The 7809 is a 9 volt three terminal IC voltage regulator.
The sense loop is connected to the existing relay points at present.
I don't know which is + and which is minus, but polarity is important
when you connect it to the opto.

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