Sunday, January 23, 2011

Overload protection circuit using 555 timer

-Vijay Kumar P.
Overload protection circuits are required in inverters and uninterrupted power supplies to make sure that the loads connected to them do not exceed the maximum power ratings. This project describes a simple overload protection circuit based on a 555 timer IC. The current drawn by a load is sensed through a 5 W 1 Ohm resistor. The voltage drop across this resistor is proportional to the current drawn and hence can be used to sense any short circuit of excess load current. An optocoupler is used to isolate the mains AC part from the rest of the circuit.



Overload current threshold can be varied through the potentiometer VR1. The mains AC supply to the load is connected through the N/C (normally closed) connection of relay RL1. When an overload is detected, the relay is energized to disconnect the load from the mains supply. The bypass capacitor C1 avoids any false trigger of the circuit because of fluctuations in the mains supply.

10 comments:

s. p. said...

How much is its max and min threshold of current on which it can operate.

Tittu Thomas said...
This comment has been removed by the author.
adelusi oluwatosin said...

sir thank very much. sir i build this circuit to handle a 3kva inverter, seem not working even to set the current threshold nothing was happened. please sir kindly help me if there is any adjustment on circuit. Thanks. sir reply to adelusioluwatosin01@gmail.com.

chintu raviteja said...

sir i built this circuit as a part of my mini project...i have connected to the resistive load and applied a 230v ac with help of auto-transformer but when i applied a load the 1ohm 5watt resister had burned out...am in very big confusion will u please help me

Shoaib Saleem said...

please design pcb of this circuit

V Designer said...

Several issues with this circuit:

This circuit relies on powering the relay to disconnect the load. This is a bad practice: if something happens to the 12V supply or anything goes wrong with the circuit, the load will not be disconnected on overload. A more robust design will disconnect the load if 12V disappears.

Disconnecting Neutral to the load while keeping the Hot connected is a bad practice. This is due to a possibility of the load developing a fault and drawing the current to the ground through the hot wire. In this case, the overload will be happening, but disconnecting Neutral will not do anything to prevent the potential meltdown.

The chosen 1 Ohm 5W resistor limits the usable overload to 400W. Controlling a larger load will cause the resistor to melt. Say you have a load that draws 7A at 230V, it will then be consuming 1610W. At 7A through the 1 Ohm resistor, the voltage drop will be 7V, dissipating 49W of power at the resistor. A 5W resistor will melt. A good overload protector should not be melting the circuit on overload.

This is not something that you’d want to build.

Hamed AL-hajri said...

am a student in the college and i tried to build this circuit and u succeed but as V Designer this is something cannot hold 3 min in case overloading , it will burn.another idea is to change the 1 ohm 5w resistor to a ceramic resistor not the small one . but over all it works but not with that efficiency .

Meguedmi Mourad said...
This comment has been removed by the author.
Meguedmi Mourad said...

I want to realize this circuit with triac BTA 41600 for protected in case of short circuit

sam sami said...

as V Designer said....
its not a reliable circuit.... don't pcb this circuit

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