Schematic Volt

High Low voltage cut off with delay alarm

When the mains voltage is in the normal level ,the voltage at the negative terminal of zener diode D4 will be less than 5.6 Volts.At this condition transistor T1 will not conduct.The same time voltage at the negative terminal of zener diode D5 will be greater than 5.6 and so the transistor T2 will be conducting.The relay will be activated and the green LED wil be glowing.

When the mains voltage is higher than the set limit the transistor T1 becomes conducting since the voltage at the negative terminal of D4 is greater than 5.6 V.At the same time transistor T2 will be non conducting which results in the deactivation of relay to cut the mains supply from load.When the mains voltage is less than the set limit transistors T1 & T2 becomes non conducting making the relay to de- activate and cut the load from mains.Ω

The timer NE555 is wired as a monostable multivibrator with a pulse width of 10ms.When the power comes back after a cut off a negative voltage is obtained at the trigger pin which triggers the IC NE555.The transistor T3 gets forward biased and it drives the buzzer to produce a beep as an indication of power resumption.Also the transistor T1 is made on which in turn makes T2 off.As a result the relay will remain de- activate for 10ms and this provides the sufficient delay and the equipment is protected from surge voltages.




Notes.

* To calibrate the schema a autotransformer is needed.Connect the output of autotransformer to the transformer primary.
* Set the voltage to 260V and adjust VR1 to make the relay deactivated.
* Now set the autotransformer to 160V and adjust VR2 so that the relay is de-energized.
* VR3 can be used to vary the volume of buzzer.

Posted by Unknown at 3:45 AM

Email ThisBlogThis!Share to XShare to FacebookShare to Pinterest

Labels: alarm, cut, delay, high, low, off, voltage, with