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# Circuit Protector

## Description

Contributor:
D.Mohankumar

Prototype circuits always require a strict approach on power inputs. The main reason for this is the “raw” form of the circuit. This circuit gives an easy approach to protect the prototype against unwanted power inputs or those damaging power inputs. Restricted to quite a low power inputs, this circuit protector easily takes care of protecting the design.

This circuit can protect your breadboard or prototype circuit instantly when it detects a short circuit in the assembled circuit. This can save the valuable components from burning. It allows only maximum 150 mA current to pass through the circuit under test. When it senses more than 150 mA load current due to short circuit, it cut off the power supply to the load.

The circuit uses the following components:
1.      Relay Switch
2.      Diode IN4007
3.      Transistor BC547
4.      Push Switch
5.      Resistors, and

6.      9-12V power supply.
The circuit uses a 6 volt 100 Ohms relay and an NPN transistor BC 547 to control the power to the circuit under test. From the power supply, power to the relay and the circuit under test passes through the NO and Common contacts of the relay. So when S1 is pressed momentarily, relay activates and connects the NO and common contacts. So circuit under test gets power and the relay remains latched even if S1 is released.

R2 forms the current sensor. As long as a current of less than 150 mA passes through the circuit under test, T1 will be out of bias since its base voltage is less than 0.7 volts. When the current passing through R2 increases above 150 mA due to a short circuit in the circuit under test, T1 conducts. T1 then drains the relay current and the relay de-energize. Relay contacts open and breaks the power. Power supply can be established only after removing the short circuit and pressing S1 again.

Value of R2 is calculated as follows
Suppose the current through R2 is 120 mA or 0.12 A, then the voltage across it is
0.12 A x 4.7 R = 0.56 V
What happens, when this current increases to 150 mA?
0.15 A x 4.7 R= 0.71 V. This is enough to forward bias T1 to trip the relay.
By using the above formula, it is easy to select R2 for the required tripping current.

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