I want to build a strobe light sorta thing. I plan to have six lights each containing 20 or so LED's. I'll have 3 different colours.
But, I found a design in 'circuit simulator' that does what I want. Except you need to 'click' to make it change LED.
It has a decade counter and I was hoping there was some way to connect a 555 to make it auto clicking. Then adjust how fast the LED's change with a pot or something.
The animation is wrong. The diodes cause the LEDs to light back and forth, not just chase in one direction as shown.
The LEDs light 1,2,3,4,5,6,5,4,3,2,1,2,3 4... like the original Knight Rider car.
The CD4017 IC can drive only a few LEDs in series if its supply voltage is high enough. They won't be very bright. A darlington transistor can be added at each output to drive 20 LEDs.
It would be difficult to disconnect the LEDs that are driven and switch on a different colour LEDs. Just make a separate circuit for each colour and a switch powers which colour you want. Another CD4017 IC can drive darlington transistors in sequence and they power each colour circuit.
The 4017 will work from 3 to 15v and the 555 will work from about 4.5v up to 16v. So between 4.5 and 15 will keep both happy.
Audioguru is right about the need for some transistors if you want to light more then one small LED at a time. You mentioned 20 or so, in that case the 4017 won't be able to output enough current to light that many. But the output of the 4017 can be used to turn on a transistor, or darlington pair, and light the leds that way.
ok, but if the output of the 4017 is enough to power the LED with a forward voltage of about 2V and 20mA, why cant the same method of adding the correct resistors in parrallel work?
Texas Instruments datasheet for the CD4017 has detailed graphs of its minimum and typical output voltages and currents when its supply is 5V, 10V and 15V.
With a 5V supply its minimum and typical output current into a 2V LED is only 2mA.
With a 10V supply its minimum output current is 9mA and its typical current is 19mA.
With a 15V supply its minimum output current is 14mA and its typical current is 29mA.
A current-limiting resistor is needed when the power dissipation in an output transistor exceeds 100mW (a 9V supply with typical 15mA output current).
Which resistors do you want to connect in parallel?
well, i built a flasher on a breadboard, only problem is its kinda dim...
but now, how can i make a 16 led flasher. where alll leds go on and then off.
i used this circuit to make the flasher, can i modify it for 16 LED's and could someone please give me a schematic for this.
The LED has a fairly high current so it should be bright.
You have the 555 charging the timing capacitor for a long time when the LED is turned off, then you have the capacitor discharging for a short time when the LED is turned on.
Maybe the LED is not turned on long enough to appear bright.
That will work fine accept for LED brightness... I would recommend using a 74HC4017 since they are easier to get and more tolerant of abuse and a ULN2003 driver to up the current and voltage capacity.
Your 555 circuit driving a transistor has the LED turned on for 47ms which is long enough for your vision to see it at full brightness. If it is a 2V red LED then its current with a brand new 9V battery is about 26mA which would be bright, not dim.
Maybe you bought cheap dim LEDs.