simulate this circuit – Schematic created using CircuitLab

As you can see in the above circuit, I have converted AC to DC.

The transformer used in the above circuit is : If I apply 230V across the Primary, then I get 12V across the secondary coil of transformer.

Then I convert it to DC using 4 Diodes and a capacitor.

This DC voltage is applied to the coil of Relay.

This circuit works as expected.


When I decrease the voltage on Primary coil of transformer.

Suppose I apply 200V to the Primary coil of transformer then also the chances are that Relay coil gets energized.

What I want:

If there is 12V or higher applied at the Red box in the diagram, then only the coil of relay should get energized else I should get 0V across the two terminals of relay coil.

I asked one of my friend about this problem. He said I should use a Zener diode in this circuit to achieve what I want. But I don't know where do I place the Zener?

Can anybody give me a suggestion?

  • \$\begingroup\$ I you reduce the primary voltage and the rectified voltage on the secondary side drops below 12V (or whatever desired value), you would need a boost converter to get it higher. Note that your relay will probably still work fine down to maybe 8V. \$\endgroup\$
    – Rev
    Nov 9, 2016 at 7:17
  • \$\begingroup\$ @Rev1.0 I know that relay may work fine upto 8 volts or above. That is what I want to control. I will be applying variable AC voltage to the Primary Coil of Transformer between 180V and 240V AC. But the relay coil always gets energized. I don't want this behavior. I just want the relay coil to get energized if it gets 12V or higher. In short If there is less than 12V DC in the output of rectifier, then I want 0 Volts difference between 2 terminals of relay coil. But if Voltage is greater than or equal to 12V, then I would like to have that voltage difference between 2 terminals of relay coil. \$\endgroup\$
    – Vishal
    Nov 9, 2016 at 7:39
  • \$\begingroup\$ You can simplify your circuit by using the entire secondary of the transformer and eliminating two of the diodes, and you can do what you want by using a comparator with hysteresis, a reference, and a transistor to drive the relay. But... first things first. Please post the relay and transformer's data sheets or specifications. \$\endgroup\$
    – EM Fields
    Nov 9, 2016 at 8:14
  • \$\begingroup\$ @EMFields I have not yet purchased any components for this circuit. I will be using a 12-0-12 transformer @ 500mA as shown in this image: blogspot.tenettech.com/wp-content/uploads/2014/07/… and a standard 12V relay with output of 230V AC @ 20-30A \$\endgroup\$
    – Vishal
    Nov 9, 2016 at 8:24

3 Answers 3


In the schematic shown below, the entire secondary of the transformer has been used, two rectifiers have been eliminated, and C1 is used to smooth the full-wave rectified 16.5 volts.

U1 is a 12 volt linear regulator which drops the 16.5 VDC down to 12VDC for the relay and whatever else the 12 volts is being used for, R6 representing both the relay coil and the remaining parallel load, if there is one.

U2 is a voltage comparator with an internal 400 millivolt reference. The voltage divider R3 R4 is used to set the voltage at the non-inverting input of U2 to slightly over 400mV when the output of U1 is at 12 volts. That will force the output of U2 high, turning Q1 ON and energizing the relay coil.

If, for some reason, U1 OUT should fall below 12 volts, the voltage at U2+ will fall below 400mV, U2's ouput will go low, and the relay will become de-energized, opening its normally-open contacts.

U2 is supplied with about 6.5 millivolts of internal hysteresis and if more is needed it can be applied externally as shown on page 10 of the data sheet.

enter image description here

  • \$\begingroup\$ Would it be a useful idea to replace the 402 Ω resistor (R4, E96 series) with a 270 Ω resistor and a 1 kΩ pot in series for tweaking? \$\endgroup\$ Nov 9, 2016 at 20:02
  • \$\begingroup\$ @AndrewMorton -- if this is a one off you want to trim for accuracy, sure. For a repeatable design, you may not want a trim (it's more per-unit labor vs. having to source a more precise resistor). \$\endgroup\$ Nov 9, 2016 at 23:34
  • \$\begingroup\$ Thank you for the great answer. I have tried this circuit and it behaves as you mentioned. But this does not solve my problem. What I want is, If I supply 230V AC or very near to that to the Primary of the transformer, then only I want to energize the relay coil. Can you please update your answer accordingly? \$\endgroup\$
    – Vishal
    Nov 11, 2016 at 15:30
  • \$\begingroup\$ @Vishal: I can, but since I gave you exactly what you asked for and it turns out that you've wasted my time, I won't. If you want help with your new problem, I suggest you post a new question specifying what you really want, your country's power supply frequency and whether one end of your 230 volts is earthed, the required response time of the relay dropout circuitry, and the value of the mains dropout point and its tolerance. \$\endgroup\$
    – EM Fields
    Nov 11, 2016 at 17:53
  • \$\begingroup\$ Thanks for the answer. I will ask another question as per your suggestion. And sorry if I really wasted your time. \$\endgroup\$
    – Vishal
    Nov 11, 2016 at 18:53

A relay is an inappropriate component for a precision voltage detector.

The pull-in voltage depends on the spacing of the iron components in the magnetic circuit, that can take slightly different positions every time it operates. It also depends on the resistance of the copper coil, which if you've not met size of this variation before, may astonish you. Copper, and most other pure metals, have a large tempco of resistance, a 25C change in temperature results in a 10% change of resistance. As it's the coil current that pulls the relay in, the voltage sensitivity varies by 10% over 25C as well.

On drop-out, things get worse. The iron path in the relay has now closed, and much less current is needed to keep it closed. Possibly only a quarter. I'd expect it to stay closed down to 6v, it may stay closed right down to 3v.

You need an electronic solution. Depending how accurate and temperature insensitive it must be, you would use a voltage reference (a zener or something more stable) and a comparator, or maybe just the nominal 0.7v VBE of a transistor, it depends on your accuracy specifications.


Irrespective of your question, here's your basic problem...

The AC output is 12V and, after rectification and smoothing, the DC voltage will be about 16 V DC - this might be too much for your 12V relay coil.

OK that's the warning done with so you might want to put some series resistance in with your relay coil to ensure that it receives 12 volts instead of 16 volts.

Taking this one step further, you could put a little too much resistance in series so that the relay is only just operating at 230 V. Then when the voltage lowers a tad, the relay drops out and gives you the result you want.

If you want a more reliable solution you should use a 7805 voltage regulator powering a comparator - one input pin on the comparator can take (say) 2.5 volts by potential dividing the 5V and the other input pin can take a potential divided voltage from the 16 V dc output to produce 2.51 volts. The comparator output will switch when the 16 volts drops by maybe 100 mV to 200 mV. Use the comparator output to drive the relay.


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