Power Supply With Relay Short Circuit Protection

In this instructable you will be making a power supply with short circuit protection that uses a relay.

In some electronic or electrical engineering courses they might tell you that mechanical components are not reliable and semiconductor switching should be used. However, not all engineers would agree with such opinion because semiconductor components might have to sustain high power, high current and high voltage for the duration switching periods. This might cause those components to burn and fail.

You will need:

- power source (you should use eight AA or eight AAA batteries only - do not use car battery unless you are an expert because the current is very high),

- two 4 AA/AAA battery holders or four 2 AA/AAA battery holders,

- soldering iron,

- solder,

- low power resistor (value will be explained in following paragraphs),

- high power resistor (value will be explained in following paragraphs),

- relay (value will be explained in following paragraphs),

- electrical tape,

- wires,

- SPST (Single Pole Single Throw) switch,

- load (DC motor, light bulb, buzzer, high power resistor),

- packaging material or box,

- wire stripper (optional),

- multi-meter (optional),

- matrix board (optional).

If you look at the circuit you will see that a short circuit across the output terminals will turn on the relay and thus disconnect the load from power supply. Thus if power supple is 12 V than the relay should turn on at least 12 V or above. If power supple if 5 V than you should use a 3 V relay instead because 3 V is less than 5 V. Keep in mind that all relays also have a maximum input voltage that should not be exceeded. If you look at the circuit you iwll see that the maximum relay input voltage should be above the value of power supply voltage to prevent failure of the relay.

The value of high power resistor is equal to: R1 = Power supply voltage/(Preferred maximum output current) - (Internal resistor of power source). The internal resistance is Rs. This value can be ignored if you do not know this value. With 12 V power supply the maximum current is 12 V / (Rs + R1) = 12 V / 20 ohms = 0.6 Amps.

In the circuit shown the LED is modelled with three general purpose diodes because the software used does not have an LED model. The old PSpice student edition software was used because it allows keyboard control and thus reduces schematic drawing time. New software simulation packages only allow computer mouse control thus increasing schematic drawing time. With the voltage across a typical LED equal to 2 V is approximately three times the voltage across a single diode, which is 0.7 V. The current across the LED should be 10 mA. Thus LED resistor is equal to: (12 V - 2 V) / 10 mA = 1 kohm resistance.

Also, not the general diode placed in parallel with the relay. This is very important to prevent high currents flowing out of the relay that might damage other circuit components.

In this diagram you see that I build a bird nest circuit because I could not be bothered drilling my matrix board for 14 relay terminals which would require 14 holes. Later you will see how I secured the components with electrical tape to prevent damage or short circuits due to mechanical movements.

You can see in the photo that I used low power resistors because I could not be bothered purchasing high power resistors. However, you must use high power resistors for this circuit.

Putting resistors in parallel will reduce power dissipation for each resistor. The value of total resistance of five resistors connected in parallel is equal to: Rp = 1/(1/R1+1/R2+1/R3+1/R4+1/R5) or for just two resistors: Rp = R1*R2/(R1+R2).

The maximum power across the resistor is equal to: P = V*I = 12 V*(12 V / 10 ohms) = 14 Watts. However, the relay turns on in less than a second after the output is shorted. Thus you might not need such a high power resistor. Maximum average power dissipation occurs when the load voltage is equal to the voltage across the R1 resistor if internal resistance of power source is zero. Thus if the voltage across R1 is 6 V then the average power will equal to: 6 V*(6 V / 10 ohms) = 3.6 Watts, which is a lot less than maximum power of 14 Watts.

You connect positive terminal to relay control input terminal and relay output terminal (solder those two terminals shown with two red circuits together). You need to make sure that the relay is conducting current between resistor and red wire when the relay is off. The other end of the red wire will be connected with SPST switch.

Next step is connecting relay output terminal, the green wire. Unlike the red wire the green wire is only connected to relay control input terminal and circuit output.

You will see in the next diagram that the other end of red wire is connected director power supply source (batteries).

You see that the LED resistor is attached to the relay with electrical tape.

I also soldered the general purpose diode across the relay input control terminals.

The switch is connected directly to power source as shown in the photo.

Electrical tape should only be used for temporary fixes and not long term fixes. However, this circuit is just proof of concept.

Next step is inserting the circuit into old packing materials. You can rip parts of the material with your hands or carefully and slowly cut with scissors to fit the circuit components in.

If the LED is turning off when the output terminals are shorted than the relay is not turning on and all power is drained from the source. Do not keep the terminals shorted for a long time if the LED is off or the batteries might explode.

You can see the circuit working in the video: