Fire Exit Sign using Shift Register

This circuit uses a 4-bit shift register and a d-type flip-flop to make a fire exit sign, which might be useful for a GCSE project. The shift register drives the LEDs to form a very eye-catching animation that shows the direction of the exit. In this circuit, a 555-timer circuit produces the clock input. The outputs from the shift register can drive LEDs directly, providing there are suitable series resistors to produce the voltage drop. Even if you do not make the circuit, it is important to understand how such circuits work.

We start the circuit by applying a momentary reset, which causes the output of the d-type flip-flop to reset so that Q=0 and /Q=1. As you can see the d-type flip-flop has two outputs where /Q is inverted. Therefore when Q=1, /Q=0, and when Q=0, then /Q=1.

The reset is also applied to the 4-bit shift register and by the same control line, and therefore its outputs will all be binary 0 as well.

When we feed the first clock pulse, at the rising edge, the value at the inverted output (/Q), which feeds the serial data input, appears at the output Q₀. Therefore, Q₀ has output of logic 1 and lights the first LED. Simultaneously, the input D₀ of the d-type flip-flop, which was logic 0, appears at its output and therefore Q=0, and /Q=1. Therefore, the serial input to the shift register will be 1 for each of the clock pulses, hence we will be feeding the register with binary 1 1 1 1. At each rising edge of the clock pulse, binary 1 appears at the output of successive stages, and therefore the arrows light up consecutively and remain lit

Once all the arrows are alight, the output at Q₃ becomes logic 1, which feeds the input D₀. On the rising edge of the next clock pulse, Q=1 and /Q=0. This time we will be feeding binary 0 0 0 0 into the shift register and therefore the LEDs will go OFF consecutively.

In the following pages of this multi-page article, you can see an animation of how that would appear.

Animations