Categories: Power Supply

5V DC Regulated Power Supply with Short Circuit Protection

This is the circuit diagram of 5V DC regulated power supply which featured with short circuit protection system. There are 2 kind of output that are regulated 5V DC with short circuit protection and without circuit protection. The main circuit is protected from any damage due to short-circuit in the additional power supply circuit by cutting off the derived supply voltage. The derived supply voltage restores automatically when shorting is removed. An indicator LED is utilized to show whether short-circuit exists or not.

Power Supply Block:
This circuit works just like the ordinary DC power supply adapter, in the main power supply circuit, 230V AC from main home electric source is stepped down by transformer X1 (230V AC primary to 0-9V, 300mA secondary), then rectified by a fullwave rectifier comprising diodes D1 through D4 with bridge arrangement (you may use a single bridge diode), filtered by capacitor C1 and regulated by IC 7805 to give regulated 5V DC output (O/P1).

Short-Circuit Protection Block:
Transistors SK100 and BC547 are used to derive the secondary output of around 5V (O/P2) from the main 5V supply (O/P1).

The working of this circuit is quite simple. When the 5V DC output from regulator IC 7805 is available, transistor BC547 conducts through resistors R1 and R3 and LED1. As a result, transistor SK100 conducts and short-circuit protected 5V DC output appears across O/P2 terminals. The green LED (LED2)  lows to indicate the same, while the red LED (LED1) remains off due to the presence of the same voltage at both of its ends.

When O/P2 terminals short, BC547 cuts off due to grounding of its base. As a result, SK100 is also cut-off. Thus during short-circuit, the green LED (LED2) turns off and the red LED (LED1) glows. Capacitors C2 and C3 across the main 5V output (O/P1) absorb the voltage fluctuations occurring due to short-circuit in O/P2, ensuring disturbance-free O/P1. The design of the circuit is refer to the relationship given below:

RB = (HFE × VS) / (1.3 × IL)

where,

RB = Base resistances of transistors of SK100 and BC547
HFE = 200 for SK100 and 350 for BC547
Switching Voltage VS = 5V
1.3 = Safety factor
IL = Collector-emitter current of transistors

Build the circuit on a general purpose PCB and mount in a general circuit box. Connect O/P1 and O/P2 terminals on the front panel of the cabinet. Also connect the mains power cord to feed 230V AC to the transformer. Connect LED1 and LED2 for visual indication.

This circuit has been tested by the author, the circuit that already built is shown on the following image:

circuit diagram

View Comments

  • hay please if any 5-27V short detection components in fox or replace ( sorry I of the Indonesian state can not english )

  • Hi

    Great circuit diagram. I need to know the current output capabilities of the short circuit protected power supply.

    Will it be 500 mA because that's the max CE current of the SK100?
    And if it is, would I be able to replace the two transistors with ones that can withstand currents up to 1A?

    Regards
    Leighton

  • @Bill
    RB = Base resistances of transistors of SK100 and BC547
    So it will be R2 and R3

  • RB = (HFE × VS) / (1.3 × IL)
    What RB? There is no RB in the circuit.

    IL = Collector-emitter current of transistors
    And what is the UNIT???

    Also, what is the UNIT for RB??

    Assuming IL in Amps as it should be, the BC547 has CE current of 0.05 Amp for comfortable operation.
    Equation results is ~27 KOhm.
    Where is this resistor?

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