# 3 simultaneous led chaser using only 555 and 4017

I've to design a circuit which consists of 30 LEDs in a row (RED, BLUE and GREEN) followed by a led matrix which spells out IN.

Conditions which I've to follow:

• When the circuit is powered, initially the 1st RED , BLUE and GREEN LEDs will be ON.

• 2nd RED LED should be ON when 1st RED LED goes OFF, 3rd RED LED should be ON when 2nd RED LED goes OFF, and so on. When 10th RED LED goes OFF, 2nd BLUE LED should be ON i.e for every 10 cycles of RED LEDs, a BLUE LED should be ON. This is analogous to a LED chasing circuit.

• Similarly for every 10 cycles of BLUE LEDs, a GREEN LED should be ON. Finally when 10th GREEN LED goes OFF, LEDs arranged in "IN" fashion should be ON.

Progress so far:

• I used 3 555 timers and to get 3 varied square pulses of 1kHz, 100Hz,10Hz for Red, blue and green respectively, and 4017 ic's to do the led chasing.
• I'm stuck at how do I initially light up all the first led's.
• This would be an easy one for a microcontroller, but since you're constraining yourself to the devices I used for similar projects as a child 30+ years ago, think about using a 555 as a monostable to create the initial conditions, and then to set off the chaser once it times out. Commented Feb 27, 2019 at 17:17
• Welcome to EE.SE. Draw a timing diagram instead of all that text. Commented Feb 27, 2019 at 17:22
• If you can't write a functional spec, you cannot optimize or realize the design. That includes a timing diagram. with inputs and output current, time and frequency, sequential triggers and etc Commented Feb 27, 2019 at 17:42
• Multiple independent 555's won't work. Two that generate pulses only when triggered by a third could barely work but is probably not the right solution. Likely you should make this all decoded from a counter running from a single clock; and yes, you don't have any of the edge conditions which would make software not be the best solution. This could also be entry into the concept of finite state machines - even in software, or especially if a hardware solution is being required for an academic class. Commented Feb 27, 2019 at 17:58
• the 1st of each RGB momentary at the same time? (OR logic) Is it unidirectional only , too bad Commented Feb 27, 2019 at 18:00

Rather than play 20 Questions, here is a solution based on my interpretation of what you posted.

The core of the circuit is one clock oscillator and three CD4017 Johnson Counters in series. The first counter is clocked by an external signal source, the second one is clocked by the Carry output of the first, and the third is clocked by the second. This gets you the 1001 step sequence described. The Q9 output of the third 4017 clocks a flipflop that drives the "IN" led array and inhibits further counting. An R-C power-on reset initializes the circuit, and a switch across the reset capacitor resets the circuit to start another sequence.

A 4017 output can source very little current, so you will need transistor arrays to drive the LEDs. You have 31 distinct outputs; a group of four ULN2803s has 32 outputs and can handle LED currents up to 300 mA reliably.

Update: Just caught this part: "When the circuit is powered, initially the 1st RED , BLUE and GREEN LEDs will be ON."

This is not as big a problem as it appears. When a 4017 is reset, the Q0 output goes high. This assures that the circuit always starts with the first red, green, and blue LEDs on. Either the Carry Out or the Q9 signal from the third 4017 can be used to clock the external flipflop that drives the IN sign.

The flipflop and the clock oscillator can be made with a single CD4093 quad Schmitt trigger NAND gate.

Having all colored LEDs off when IN is on will take a bit more, because there is no 4017 state where all 10 outputs are low. My first thought here is one PNP transistor, controlled by the IN flipflop, that sources power to the 30 LEDs. What is the power source for the circuit?

Here is a first pass solution. The clock is set to approx. 5 Hz so you can see the LEDs step along. For each of the three lines of LEDs, only one LED is on at a time. That is why there is only one current limiting resistor for 10 LEDs. The correct value for R8 depends on the current requirement of the "IN" array.