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safety if voltage drops

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ranariaz

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how can i make a safety ckt or any type of ckt that will give me an indication or contact when single phase AC voltage (220v) drops bellow 180 V. i want to make a tripping ckt that will cutt off the power and the dropped voltage will not effect electrical appliances
 
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Put a bridge on the mains and filter the output. Place a set of 1watt zeners on the output and take them to a 12v or 24v relay. When the voltage drops, the relay will drop out.
 
I think I'd like to see something like this:

**broken link removed**

The power-line voltage would be sensed by noninverting input to the comparator. A fixed voltage reference, independent of line voltage, would be applied to the inverting input. This voltage reference would be adjustable. When there was a negative imbalance (line voltage < reference voltage), the cutoff relay would be energized. (Probably the opposite would be better: the cutoff relay would be normally open and would be energized unless there was an imbalance, in which case it would open.)
 
You don't need anything complex. Just a bridge and set of zener diodes is all that is needed.

How would that work, exactly? How would you get a voltage reference if the whole thing is driven by the incoming power which could drop in voltage? Can you post the circuit?

And could you adjust your simple solution to a particular "setpoint' in voltage?
 
You can draw the circuit from my description above and I will correct it.

OK. Somewhat odd way to do things, but no problemo:

**broken link removed**

So I'm guessing the dropout voltage will be the zener voltage times the proportional reduction from line voltage to DC from the bridge and capacitor?
 
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This circuit will turn off a device if the main drops by a say 15v. The actual voltage is adjustable. The first thing to remember is this: The circuit detects the PEAK voltage and this is the voltage of the zener diodes.
For 240v mains, the peak is 338v.
For a voltage drop of about 12v(RMS), the zener diodes need to have a combined voltage of 320v (you will need 6 x 47v + 1 x 20v + 1 x 18v). The 10k resistor will have about 18v across it and the current will be nearly 2mA. The wattage will be 36mW.
For a voltage drop of about 27v(RMS), you will need zeners for a total of 300v by using 6 x 47v + 1 x 18v. The voltage across the 10k resistor will be 38v and the current will be nearly 4mA. The wattage dissipated by the 10k resistor will be 150mW.
The 10u prevents very sharp dips or drops from activating the circuit.
As the voltage drops, this drop in voltage will be passed directly to the top of the 10k resistor and as the voltage drops, the current into the base of the transistor will reduce. This current is amplified by the transistor and when it is not sufficient to keep the relay activated, it will drop-out.
 
I knew it was more complicated than "just a bridge and set of zener diodes" (your post #4 above).

For one thing, where does that nice 12 volts come from? Guess you need another power supply (or could you tap the zener string 12 volts or so from the bottom?).

These "oh-so-simple" circuits have a way of getting more and more complicated ...
 
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