Delay on a mains-powered relay?

[Rob Gaddi]

Cheap microcontroller circuit, grounded to the low side of the wireless
sensor's sense contacts. Use the existing relay to signal the uC, and a
mosfet to open the connection. Depending on how much power the wireless
is providing to monitor the sense you may even be able to power just off
that, but if not, decently careful design on a ultra-low power
microcontroller should allow you to run for practically the shelf life
of the battery.

 

[Richard Henne]

If you're in the US, try http://www.Grainger.com and do a key word
search using 120V off delay relay. You should be able to find something
that would work.

 

[NSM]

| 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.

Look for a spring powered timer then. They use an AC clock motor to hold
them at, say, 5 minutes. When the power goes off the spring unwinds until
the clock goes to zero and the contacts close/open. When the power is
restored the motor rewinds the timer to 5 minutes. Looks a bit like a cheap
cooking timer on a frame.

http://www.thomasregisterdirectory.com/clocks/electric_timers_0014366_1.html

N

 

[NSM]

| It's an AC relay. Just how do you intend to attach that "big capacitor?"

Obviously use a light (0.5W) DC relay!

http://www.nteinc.com/relay_web/R16.html

N

 

[[email protected]]

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.

 

[legg]

http://www.Grainger.com

You're right. They appear to do it all for you for $85.00. Notice the
total power consumption is 180mW, so the 'relay' must be quite low
power to be included in the budget.

http://makeashorterlink.com/?M2F312B2A

RL

 

[[email protected]]

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,

Here's an even better mod to Tony's circuit. You need
4 parts - all available from Allelectronics:
5 volt regulated wall wart p/n DCTX-512 $3.50
Minature 4.5 volt relay p/n RLY-639 $0.75
1N4001 diode (1 needed) p/n 1N4001 15 for $1.00
1 farad 5.5 volt super cap p/n CBC-12 $3.50

D
Wall + ------|>---+-----+
Wart +|C |
=== Relay
| |
- -----------+-----+

I suspect the relay will stay energized longer
than 5 minutes. You can add a second diode to
reduce the time. You might need to add resistance
in parallel with the super cap to reduce it further.
This depends on the characteristics of the relay -
I don't have one to test.

Ed

 

[Bill Kaszeta / Photovoltaic Resources]

I've got several remote alarm sensors triggered by 120vac relays. When power
fails, the alarm calls the right folks.

Problem is that even the small power glitches get these guys out of bed. Not
a good thing.

How can I delay the de-energizing of these relays? 5 minutes without power is
OK. Will a series diode and shunt cap do the trick? If so, what size cap? If
not, what're my options?

The relay is similar to this one:

http://tinyurl.com/6pppw

Thanks,
--
Please, no "Go Google this" replies. I wouldn't
ask a question here if I hadn't done that already.

DaveC
[email protected]
This is an invalid return address
Please reply in the news group

It is likely that your wireless device has an input circuit with a dc voltage
across the inputs when the input is open (that is how the open contacts
are sensed). One side is likely common with one terminal of the
battery.

You should try putting a power FET device across the input (source
and drain) and using a capacitor from source to gate to hold the
FET on for a period after your relay opens. A resistor across the
capacitor sets the discharge period. I would start with 470 MF and
1meg resistor for 4-5 minutes delay (depends on gate voltage on
turn-off, device specific).

If the battery negative is common to one of the input terminals,
use a N-channel FET (I did something similar with an IRF520,
way over kill on current, but you are not likely to damage it).
If the wireless device voltage is only 3 volts, you will need a
FET designed for logic level gate voltages.

Connect as follows:
Source of FET to - input terminal
Drain of FET to the other input terminal
Capacitor negative (use 16V cap) to Source
Capacitor positive to Gate
1 meg resistor across capacitor
100K resistor to battery +, other side of 100K to one of your relay contacts
Other relay contact to Gate.

Use a grounded soldering iron as the FET gate can be
damaged by static electricity.

If the battery positive is common to one of the input terminals,
use a P-channel FET and reverse all polarities listed above.

When the relay contacts are closed, the 100K resistor uses the
battery to charge the capacitor. Voltage on the capacitor keeps
the FET ON.

When the relay contacts open (on power fail) the capacitor holds
a charge while the 1 meg resistor discharges the capacitor. At
some low voltage on the Gate, the wireless sensor will detect an
open circuit and send the alarm. Experiment with resistance and
capacitance to get an acceptable time delay.

If you shop the parts, likely cost per circuit is
USD $2.00-4 .00.
Bill Kaszeta
Photovoltaic Resources Int'l
Tempe Arizona USA
[email protected]

 

[CJT]

I've got several remote alarm sensors triggered by 120vac relays. When power
fails, the alarm calls the right folks.

Problem is that even the small power glitches get these guys out of bed. Not
a good thing.

How can I delay the de-energizing of these relays? 5 minutes without power is
OK. Will a series diode and shunt cap do the trick? If so, what size cap? If
not, what're my options?

The relay is similar to this one:

http://tinyurl.com/6pppw

Thanks,

Here's what you do. Put a cheap answering machine on the same power
circuit that you're trying to monitor, and connect it to the phone line
that the call gets made on.

Tell the guys that when they get a call (they'll still be awakened, so
perhaps this is not a perfect solution) to call back to that number.
If the answering machine answers, it has power, and hence so does the
equipment. If the answering machine fails to answer, get in the truck.

;-)

 

[[email protected]]

+----+ +---|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)

The 85 cent ones (CBC-125) are 1 F, 2.5 V. You'd need four in series to
handle 9 V. The manufacturer may recommend a resistor in parallel with
each cap, which I think makes a voltage divider to keep the voltage
about the same on each cap. Digi-Key has Cooper PowerStor capacitors
that come already packaged as two 2.5 V caps in series to get a 5 V
rating for $4.20 to $10 quantity 1.

Your second idea, using the CBC-11/CBC-12 1 F 5.5 V memory back-up
capacitors, may or may not work. Some of these capacitors are made for
very low current (microamps, CMOS memory) and not the tens to hundreds
of milliamps you'd need for a relay or the 10-20 milliamps you'd need
for an optoisolator or solid state relay.

Matt Roberds

 

[Paul Hovnanian P.E.]

I've got several remote alarm sensors triggered by 120vac relays. When power
fails, the alarm calls the right folks.

Problem is that even the small power glitches get these guys out of bed. Not
a good thing.

How can I delay the de-energizing of these relays? 5 minutes without power is
OK. Will a series diode and shunt cap do the trick? If so, what size cap? If
not, what're my options?

The relay is similar to this one:

http://tinyurl.com/6pppw

Thanks,

Look for a TDDO (Time Delay Drop Out) relay.

 

[James Sweet]

| I've got several remote alarm sensors triggered by 120vac relays. When
power
| fails, the alarm calls the right folks.
|
| Problem is that even the small power glitches get these guys out of bed.
Not
| a good thing.
|
| How can I delay the de-energizing of these relays? 5 minutes without power
is
| OK. Will a series diode and shunt cap do the trick? If so, what size cap?
If
| not, what're my options?

The normal method is a relay with a hydraulic delay. This means you don't
need a no fail power supply. However these days, if you have backup power,
there are solid state equivalents.

For your method you would need a large cap run off say, 12 VDC - maybe 1000
uF and up.

N

Just put a delay circuit in between the relay contacts and whatever
(assuming low voltage) signal the control to the alarm panel. A 555 IC
driving a small low voltage relay should do the trick.

 

[James Sweet]

| It's an AC relay. Just how do you intend to attach that "big capacitor?"

Obviously use a light (0.5W) DC relay!

http://www.nteinc.com/relay_web/R16.html

N

But it sounds like the 120v coil is used to detect mains failure, power goes
out, relay loses power, contacts open/close, triggering the alarm. The relay
can be left alone, you only have to mess with the low voltage signal it
switches.

 

[NSM]

| But it sounds like the 120v coil is used to detect mains failure, power
goes
| out, relay loses power, contacts open/close, triggering the alarm. The
relay
| can be left alone, you only have to mess with the low voltage signal it
| switches.

If he wants to use a cap to delay it the lower power the better. Even a reed
relay will work OK and keep the cap size down. That's the key - run from a
rectifier circuit.

N

 

[Ross Herbert]

| But it sounds like the 120v coil is used to detect mains failure, power
goes
| out, relay loses power, contacts open/close, triggering the alarm. The
relay
| can be left alone, you only have to mess with the low voltage signal it
| switches.

If he wants to use a cap to delay it the lower power the better. Even a reed
relay will work OK and keep the cap size down. That's the key - run from a
rectifier circuit.

N

It seems to me that the OP would prefer to use what hardware exists
now in order to minimise expense and the complexity involved in a
complete re-design of what he now has. The relay he is using now is a
3 pole changeover unit so depending upon what is currently connected
to the relay contacts will have some bearing on what he can do in the
easiest and most cost effective manner.

However, using a reed relay to detect the mains failure would involve
using an additional step-down transformer in order to supply the ac
sense input to the reed relay/rectifier/capacitor arrangement. While a
reasonable delay might be achieved it certainly would not be in the
order of more than a minute at most even if the reed relay coil was
10,000 ohms and a 10,000uF capacitor was used.

Assuming the OP did not want to alter the existing KUP relay or
wiring, the reed contact would be connected in series with the
existing 120Vac coil of the KUP relay so that the latter acted as a
slave off the reed relay. This would also mean that a separate
no-break 120Vac supply would have to be present to maintain the KUP
relay operated until after the reed relay contact had opened at the
end of the delay. Clearly, the need for a separate no-break 120Vac
supply was not on the OP's agenda, although a UPS could be used for
such a function.

Reed relay contacts are not intended to switch high voltages such as
120Vac and they don't like to switch ac current in inductive circuits.
The contacts quickly weld together or become pitted very quickly so a
reed relay contact would not be suitable to open the existing KUP coil
circuit. Even if the OP altered his existing KUP relay coil for a low
voltage dc unit capable of being switched by the reed contact, the dc
supply to hold the KUP relay operated AFTER the 120Vac mains had
failed would have to come from somewhere. Again, this would require a
separate no-break dc voltage source so this is even more complex than
using a UPS with the existing KUP relay. In this case a rechargeble
battery could be used but this would require that a battery charging
circuit would be required to maintain the battery in a fully charged
condition. All in all, it would be more expensive and there would be
no real advantage over sticking with the existing KUP relay and
providing a UPS.

If the OP really needs a delay exceeding 1 minute then the only way is
to incorporate a true delay-off relay as the ac sensing device and
slave the existing relay/s from this. He will still have to provide a
separate no-break 120Vac source (eg, UPS) to keep his existing KUP
relay operated though.

I would suggest the simplest method to achieve what the OP requires is
to first get a UPS of say 150 - 300VA rating and use this to supply
the existing KUP coils via the contact of a true delay-off relay such
as http://www.artisancontrols.com/pdf/4390.pdf A 4390A-E would give
him an adjustable 30 - 300 seconds delay and the allow its input to
sense the 120Vac mains supply. Several KUP relays (assuming more than
one) could be handled by the delay-off contact since it has an 8A
rating.

 

[NSM]

| It seems to me that the OP would prefer to use what hardware exists
| now in order to minimise expense and the complexity involved in a
| complete re-design of what he now has. The relay he is using now is a
| 3 pole changeover unit so depending upon what is currently connected
| to the relay contacts will have some bearing on what he can do in the
| easiest and most cost effective manner.

....

I remember small, inexpensive delays that used an AC clock motor to wind
them up. When the power went off they ran down and switched contacts. I
suggested that previously.

N

 

[[email protected]]

In alt.engineering.electrical [email protected] wrote:



The 85 cent ones (CBC-125) are 1 F, 2.5 V.

No, the CBC-125 is not the one I had in mind. I should have made it clearer.
It is CBC-131, which is 85 cents in qty < 10 and is .1 (point one) uf at
5 volts,
not 1 uf at 2.5 volts. The CBC-125 is $1.00 each in qty < 10

Some optos will work as low as 50 ua, from what I have read. The question
that is not answered is what his sense loop needs (how much current at
what sense voltage) and of course whether the opto will permit that current
over the full delay time.

Digi-Key has Cooper PowerStor capacitors
that come already packaged as two 2.5 V caps in series to get a 5 V
rating for $4.20 to $10 quantity 1.

That's good to know. I gave a quick try searching for super caps in
Mouser and came up empty. (Doesn't mean they don't have them -
I didn't give it much of an effort.)

Your second idea, using the CBC-11/CBC-12 1 F 5.5 V memory back-up
capacitors, may or may not work.

I agree. I don't know that it will work for sure, but I strongly suspect
it will. It is guaranteed to give *some* delay - the question is, how
much. At least it avoids the question of the sense loop voltage and
current requirements. (We know it has to be small from his brief
description, but not whether an opto's vce would prevent operation.)
What I didn't know is implied in your next sentence:

Some of these capacitors are made for
very low current (microamps, CMOS memory) and not the tens to hundreds
of milliamps you'd need for a relay or the 10-20 milliamps you'd need
for an optoisolator or solid state relay.

What happens if you place a 10 to 30 ma load on one of those caps?
The relay I specified will draw about 30 mA at ~ 4.4 volts and should
continue to stay energized down to 10 mA and possibly below that
as the voltage decays. The relay I *meant* to specify is RLY-635
(not 639). I had them both written down on my scratch sheet
when I looked up the parts. I think the 639 will work - but the
635 would give more delay. The 635 relay has a 500 ohm coil and
would draw 8.8 mA to start, and would probably still be energized
when the voltage dropped to 1.67, where it would draw about 3.3 mA.

Time will tell. As is typical with OP's we may never hear how he
makes out. But I already ordered all the parts I specified for the
second idea, and will test to see what kind of delay I can get. We
just have to wait a bit for the shipment to arrive.

 

[Ross Herbert]

| It seems to me that the OP would prefer to use what hardware exists
| now in order to minimise expense and the complexity involved in a
| complete re-design of what he now has. The relay he is using now is a
| 3 pole changeover unit so depending upon what is currently connected
| to the relay contacts will have some bearing on what he can do in the
| easiest and most cost effective manner.

...

I remember small, inexpensive delays that used an AC clock motor to wind
them up. When the power went off they ran down and switched contacts. I
suggested that previously.

N

Granted, such a device will work but it is hardly an elegant technical
solution. The whole idea of technology in industry is to make the
operation and maintenance of equipment reliable as possible without
having to worry about remembering to "wind up that timing relay".

 

[Ross Herbert]

Granted, such a device will work but it is hardly an elegant technical
solution. The whole idea of technology in industry is to make the
operation and maintenance of equipment reliable as possible without
having to worry about remembering to "wind up that timing relay".

PS. Not having seen one of these wind up relays I suppose it would be
possible for it to be automatically wound up again by a motor when
power was restored.

 

[No Spam]

PS. Not having seen one of these wind up relays I suppose it would be
possible for it to be automatically wound up again by a motor when
power was restored.

See http://www.automatictiming.com/pdf_atc/305E_data.pdf

No manual "wind up" required.