Wednesday, July 2, 2014

Auto Power Off Circuit To Protect Home Appliance

Ramadan Mubarak to everyone. In this post I will be showing you guys a very simple circuit to protect your home appliance from sudden power failure for short time. What often times happen is when the power returns the voltage level can be a bit too high. So to protect the appliance we need to make a circuit that will cut off the conducting path to the appliance even after the return of AC power. You have to manually turn it on to restart the appliance. You can also use different modifications so that the circuit can be triggered at different AC main line voltage levels and essentially make a low and high voltage cut off circuit too.






What we will need?

1. Transformer to step down AC main line to something usable on relay like 15V.
2. Relay, should be chosen in a way that can handle the appliance.
3. Diodes for rectifying, protecting and Capacitors for smoothing and filtering.
4. Voltage Regulator to drive the relay without any unwanted noise or
5. Press button to activate the circuit.
6. Neon lamp and resistor as indicator.

Circuit Diagram


Auto Power Off Circuit

Component Description:
 
220V(110V) AC is provided via two wires, one of which is directly connected to the load(preferably the Neutral one) and the other(the Live one) is connected via a Relay and a press button.

Transformer that will drive the relay is also connected to the load in a parallel manner.

100Kohm resistor and the neon lamp will work as an indicator.

The 0.1uF capacitors are used for suppressing high frequency. The 470uF and 220uF capacitors are used as smoothing capacitor.

First four IN4007 diodes is connected as bridge rectifier, thus it will convert AC into DC. Another IN4007 diode will protect everything from the relay.

The 7812 is used for regulating purpose. Relays tend to generate noise or hum on unregulated power supply. Moreover it is better to feed a constant voltage to the relay to ensure it's proper working.


Mechanism:

Mechanism of this circuit is pretty simple and straightforward.

When the 220V AC IN is hooked up to power source, the load won't turn on because one of the wires is connected to the Normally open terminal of the relay. If we hit the "press button" once what it will do is provide power to the load and high side of the transformer momentarily. But in the mean time the relay will turn on as the power was available for the transformer. If the press button is released afterwards the load will continue to run.

But if the main 220V AC IN is lost the relay will automatically go to normally closed and thus if the power is back again the load won't run unless the press button is pressed again.

Warning: 

1. As this circuit has 220VAC connection with it, be sure to know what you are doing.
2. Before changing anything remove AC main power. 
3. Before connecting this circuit to AC main power check the circuit thoroughly for shorted connection or wrong connection.
4. Make sure to use components that will be enough for the load.
5. Double check relay's rating before connecting a load. Don't give higher than rated voltage on its electromagnet neither hook up more than rated load.
6. All capacitors should be rated higher than source voltage. 35V would be fine.
7. 7812 shouldn't be that hot. In case if its hot, attach a heat-sink to the regulator with some thermal paste.

That's all for now! Happy experimenting! Read my other entries here!

2 comments:

  1. what type of switch should i use at the relay side? is it the type used in automatic volume control? thanks

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    Replies
    1. You mean with the one in the right side of the relay? You can simply use a calling bell switch or some sort of press button that can withstand your load's current.

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