4017

SgtWookie

Joined Jul 17, 2007
22,230
Bernard,
I see you are using 22k resistors on the ULN2804 inputs. Are you remembering that they have internal 10.5k resistors in series with the bases?
 

SgtWookie

Joined Jul 17, 2007
22,230
Thanks Bernard - well, 12v/10.5k ~= 1.14mA, so the internal base resistors are plenty large enough. With the extra 22k resistance, you have a total of 32.5k, and the base current is only ~ 369uA. While that should still allow the driver to work OK, they are 8 resistors that could be eliminated, saving on soldering and a bit on costs.
 

Wendy

Joined Mar 24, 2008
23,421
The 4071 gates you show are a good solution, thing about diode OR gates used here is the diode drop is cumulative, which makes them not usable.

I get the feeling there is another way though, maybe several, that are simpler overall. A 4017 isn't necessarily the right tool for this job.

Take the 40174 Hex D flip flop chip. The classic shift register made with D flip flops is practically made for this job.



You could do the same thing with the 4013, it would take more chips. I've never built this circuit, but it should work.

Next step is a driver for the 4017 that would do the same thing. It will be a little more complex, but would do the same job.
 

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Wendy

Joined Mar 24, 2008
23,421
OK, I've formalized Bernard's schematic, which means attention to detail.



Rather than adding a 9th CMOS gate I went with the diode version. The base design can live with one such gate, multiples would introduce major diode drop problems. Again, I've never built this, but am pretty sure it will work as is.

The scheme for driving the LED is explain in my LEDs, 555s, Flashers, and Light Chasers article, Chapter 10.


............................................Figure 10.3

The IRF510 is one of the poorer parts out there, but Radio Shack carries it. Most MOSFET in the TO220 case style have a conductance much less than 0.1Ω, which means they can carry significant currents without getting hot. The IRF540 is a good example of this.
 
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