Hi guys:
I'm working on an application where I'd like to have a two-bit counter that needs to continuously count 1, 2, 3, 1, 2, 3, 1, 2, 3... in a loop. I don't want zero. It is controlling three of the four channels of a mux and the zero channel is "off" (no signal), so I want to skip it.
Is there an easy way to do this, or do I need to build my own counter with logic?
Don
--- You could use an 8 bit asynchronously presettable counter, like a 4510 or a 4516, and have it load 0001 when it counts to 0100: (view in Courier)
+-------------+ | +---------+ | +-|LOAD | | |__ | | GND>-+-O|CE | | | | | | +--|MR | | | | | | +--|D4 Q4| | | | | | +--|D3 Q3|-+ | | | +--|D2 Q2|----->MSB OUT | | Vcc>-+--|D1 Q1|----->LSB OUT | | _ | +--|U/D | | | CLK>----|>CP | +---------+
If you need a 2-bit binary counter, you can do this with a pair of J-K flip flops and some wires. I don't know for sure, but I think you can get just about _any_ 2-bit sequence you want from a J-K flip-flop pair without additional logic.
(It would be interesting to know just how far this extends -- can you do this with any three-bit sequence and three J-Ks? Four? Five?)
-- Tim Wescott Control systems and communications consulting http://www.wescottdesign.com Need to learn how to apply control theory in your embedded system? "Applied Control Theory for Embedded Systems" by Tim Wescott Elsevier/Newnes, http://www.wescottdesign.com/actfes/actfes.html
Yes, and probably. Do you want a two bit binary representation of the count sequence (01, 10, 11, 01 etc.) or three individual bits that take turns turning on in sequence (001, 010, 100, 001, etc.)? Do you need a reset function to an initial count, or can the count start at any point in the sequence? It is the details that determine the simplest way to accomplish the counter.
-- Regards, John Popelish
/ / Oops, 4 bit----- JF
J-K flip-flops, what a blast from the past. Can't remember the last time I needed them. But you're right, that's a very clean solution, a single dual-FF package. I warmed up the long-forgotten neurons and came up with this solution
Ja =3D B, Ka =3D B Jb =3D 1, Kb =3D !A
I used to love solving those problems. :)
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