I working on a project to run some leds sequences.
I got a circuit (below) and want to expand it to have 3 off 4017 chip enableing to have 24 stage of counts
Making a 24 stage sequencer.
I wondering if you have any advice on this circuit for me.
Thankyou
Binh
18 Stage LED Sequencer
The question sometimes comes up of how to cascade 4017 decade counters for more than 10 sequencial
stages. The LED sequencer below shows a possible solution using a few extra parts.
When power is applied, the 15K resistor and 10uF cap at pin 15 will reset the counters to the zero count
where pin 3 is at +12 and all other outputs are at zero. The 2 diodes (1n914) and 15 resistor form a AND
gate so the clock pulse will be passed to the right side counter when the sequence starts. When the right
counter reaches the 10th count, pin 11 will move high enabling the AND gate on the right to pass the clock
pulse to the left side counter. As the left side counter advances, pin 3 will be low so that clock pulses
cannot advance the right counter. When the left counter turns over and pin 3 again moves high, the
sequence will repeat. Thus we get 18 total counts, 9 from the first counter, and 9 from the second.
Note that the 4017 counter will not deliver much current, and so the LED current is set to about 6mA using
a 1.5K resistor in series. For more current, you could use transistors on each output as shown in the
drawing above, (10 Channel LED Sequencer). But some of the newer bright LEDs are fairly bright at
6mA.
I got a circuit (below) and want to expand it to have 3 off 4017 chip enableing to have 24 stage of counts
Making a 24 stage sequencer.
I wondering if you have any advice on this circuit for me.
Thankyou
Binh
18 Stage LED Sequencer
The question sometimes comes up of how to cascade 4017 decade counters for more than 10 sequencial
stages. The LED sequencer below shows a possible solution using a few extra parts.
When power is applied, the 15K resistor and 10uF cap at pin 15 will reset the counters to the zero count
where pin 3 is at +12 and all other outputs are at zero. The 2 diodes (1n914) and 15 resistor form a AND
gate so the clock pulse will be passed to the right side counter when the sequence starts. When the right
counter reaches the 10th count, pin 11 will move high enabling the AND gate on the right to pass the clock
pulse to the left side counter. As the left side counter advances, pin 3 will be low so that clock pulses
cannot advance the right counter. When the left counter turns over and pin 3 again moves high, the
sequence will repeat. Thus we get 18 total counts, 9 from the first counter, and 9 from the second.
Note that the 4017 counter will not deliver much current, and so the LED current is set to about 6mA using
a 1.5K resistor in series. For more current, you could use transistors on each output as shown in the
drawing above, (10 Channel LED Sequencer). But some of the newer bright LEDs are fairly bright at
6mA.
