Flip Flop LED Flasher Using BC547 Transistors

This post presents circuit for blinking two LEDs alternatively using two BC547 transistors. This circuit is commonly called a flip flop LED flasher. Generally used in alarm systems to produce flashing lights effect.

Components

No.ComponentQuantity
1.Power supply (3V – 5V) 1
2.BC547 transistor2
3.330Ω resistor2
4.10kΩ resistor2
5.100μF electrolytic capacitor2
6.LED2

Circuit diagram

Flip Flop LED Flasher Using BC547 Transistors (Circuit Diagram)

The blinking rate of the LEDs can be increased or decreased by changing the values of R2, R3 resistors, and C1, C2 capacitors. The frequency of a flashing LEDs can be calculated by given formula:

Formula to calculate the frequency of a flashing LEDs

If R2, R3 resistors (R2=R3=R) and C1, C2 capacitors (C1=C2=C) have identical values then frequency can be calculated as follows:

Formula to calculate the frequency of a flashing LEDs when R2=R3 and C1=C2

The values of resistors are specified in ohms (Ω), values of capacitors are specified in farads (F).

By using values given in the circuit diagram (R2=R3=10kΩ and C1=C2=100μF), the frequency is about 0.72Hz and a time period is 1.386 seconds.

Example to calculate the frequency of a flashing LEDs

Designed circuit

Circuit is designed on a breadboard.

Flip Flop LED Flasher Using BC547 Transistors (Designed Circuit)

This article has 2 comments

  1. Tahir Bangash Reply

    halloo I Hope you are doing well I just wanted to conform that how to set the values of Resister R3 R2 or the capacitor c1 c2 to set the blinked frequency to 4 hz or set the time period to 0.25s

  2. lindevs Reply

    Hi, Tahir Bangash
    I updated the post description. I provided formulas and calculation example. You can choose 18kΩ R2, R3 resistors and 10μF C1, C2 capacitors. The frequency approximately will be 4Hz and a time period will be about 0.25 seconds.
    f = 1 / (1.386 * 18000 * 0.00001) ≈ 1 / 0.25 ≈ 4Hz

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