Power fail indicator circuit?

Two 120vac sump pumps do main and backup duty. Each is powered via dedicated circuit breaker from different phases of 3-phase 208 supply. If either
circuit fails the other pump will continue to function.
I need to flash an LED, or such, if one circuit fails, and to sound an audible alarm if both fail. A simple relay circuit, or such, will do. Simplicity is king.
Ideas?
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John E. wrote:

A neon across the breaker/fuse? There should be more than enough leakage through the motor to allow for a neon indicator.
If you need more than that, perhaps being a little more creative with an LDR or some such near the neon, giving isolation, and giving you all sorts of possibilities to alarms, indication, network, whatever.
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On Jul 16, 12:03 am, John Tserkezis

That will only tell you that the breaker is tripped not if the mains connection drops a phase.
A simple "normally closed" relay running on each pumps will do the trick. The horn and LED system can be low voltage and battery backed up to be sure it always works. ASCII art:
+9V------+--------------------------------- ! ! \ \ SW2 / O \ ! ! ! V LED1 ! --- ! ! ! +----------O HORN O---------------+ ! ! ! / ! \ ! / ! ! O V LED2 / SW1 --- ! ! GND GND
If either switch closes, one or the other LED lights. If both do, the horn blows. The relays would normally be energized by the mains holding both switches open.

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Looks very simple and reliable.
I'm having trouble locating small 120vac coil relays. Since all these 2 relays will do most of the time is remain open contact, and when closed, power an LED, I'd like them to be miniature. But all the 120vac relays are (relatively) huge.
Any suggestions for small line-voltage relays?
Or should I just half-wave rectify the AC and use a dropping resistor (and fuse!) with a 12vdc relay?
Thanks,
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Use a 12V wall wart and a 12VDC relay. Measure the voltage on the coil and if it's really high (more than 10-15%) add a series resistor.
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The circuit, as proposed by MoosFET uses 2 relays held open by 120vac.
How do you suggest doing this?
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John E. wrote:

What substitute a plug-in dc supply + dc relay for a supply-voltage ac relay? That's simple enough.
Personally, I'd go for a "working ok" lamp in preference to a "fault" lamp, almost every time. In which case a couple of mains small wattage lamps or neons would do fine.
Similarly, only differently, I'd want to sense the state of the sump, in preference to sensing whether the pumps have power - for operating the alarm. Pumps can still fail or block with all the power needed available.
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I see. Too many mains receptacles required for the limited space. I'm using a GFCI receptacle for each pump due to their -- by definition -- proximity to water. By taking a wire from the "load" side connector of the GFCI i can monitor if these trip.

... so as to reduce the possibility of a failed indicator not being noticed? I commend your approach. In my experience, a regularly lit indicator, if failed or turned off, draws less attention than does one that appears. Maybe add a "failed indicator" indicator? (c: But if that fails... Oh, the tragedy!

Good idea. I'll be adding a level sensor to the sump to trigger the alarm if the level goes above norms.
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John E. wrote:

I would agree, however:
One solution is to make the "ok" lamp indispensible. eg it provides illumination for something that has to be done on a regular basis. Like read a meter or provide (most of) the room light(s).
Another is to use the light to illuminate a semi-silvered mirror with a warning notice behind it. You won't see the warning sign whilst the light is on... it can be very effective..
"If you can read this notice, start praying"... ;)
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Ha!
I think I have the critical modes covered. The lights are only lit for the non-critical modes: if one pump's electrics fail, one or the other light goes on; If both fail, the piezo alarm sounds; if the sump level rises above normal max, another alarm screams bloody murder. Battery backed-up.
Equipment is to be used only during business hours (and sump is used only during these hours), so no "dial the manager in the middle of the night" mode is required.
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Substitute one wall wart + one low voltage DC relay for one 120VAC relay (plus R if required)
Multiply each of the above by the number of AC relays.
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Many DC relays will chatter when you try that.
D1 AC ---/\/\--->!-----+---- ! ) X ) ! ) ---------------------
X could be several uF of capacitor.
For some relays another diode will work for the X
Others have suggested the "wall wart + relay" option. You could also use the wall wart to charge a huge capacitor so that no batteries would be needed.
You could also use opto-isolators to keep some transistors turned off.
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If a circuilt fails you can't assume voltage is available to operate the neon lamp.
A simple relay that closes when power is lost on that circuit would do. A battery could operate the buzzer and LED. John
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If you go the route of the relay, (see my reply), you need two of them. I just though of perhaps a lower cost way.
Relays can be expensive compared to cheap "wall wart" power supplies. Assuming you have some of those consider this circuit:
1N400X DC IN ------+---->!-------+--------------- ! ! ! ! |/e --- +---/\/\----! PNP --- HUGE ! R2 !\ Q1 ! \ ----OUT ! / R1 ! \ ! ! ! ------+-----------------------------
Assuming the "wall wart" has some leakage, R1 may not be needed.
When the wpoer fails Q1 turns on. This could replace one relay contact in my suggested circuit. If you flip the 1N400X and use an NPN, you can replace the other switch.
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Two relays with two sets of normally closed contacts. one set of contacts of each relay supplies a lamp (independent power source for lamps-possibly battery). The other sets are in parallel supplying the hooter (also independently energised) so that when both are down you get an alarm. Now it may be cheaper to take this switch logic and do it electronically. The main thing is that the alarm and light sources be independent of the motor supply as there is no use shutting down the alarm system when you need it. The wall wart scheme suggested by MooseFET appears at first glance to be violating this necessity -one can't turn Q1 on or off if there is no DC supply from the wall wart and, if it is independent- then how does one sense the failure of a circuit? Possibly I am missing something?
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You only need one set of contacts per relay.
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