counter
Can I build a 3 bit modulo 7 counter using 2 flip flops ?, I'm thinking of using the clock as the third bit since it toggles from 1 to 0. Will that work ?
edit:
Found this circuit which uses two jk flip flops but counts down
Link, Circuit
Would it be possible to make it count up as well ?
You can create a 3-input, 3-output circuit that maps one state in the count sequence to the next. Then feed the 3 Q outputs of the FF's into this combinatorial circuit, and connect the output of the combinatorial circuit to the 3 D inputs of the FF's. Connect all 3 of the FF's clock lines together, and then whenever an active edge comes along the flip flops act like a 3 bit register, and just load up the next value.
A design simplification is to first design the up/down Johnson and ring counters independently. Then just place a mux that selects Q* of either one depending on y.
simulate this circuit – Schematic created using CircuitLab
Best Answer
Technically, that's a 2-bit counter which uses the clock state as a pseudo-bit. If the only problem you have with it is that it appears to count down and you want it to count up, you can simply change the LED connections, as so
simulate this circuit – Schematic created using CircuitLab
In the original, a high voltage produces a voltage across the LED/resistor and turns the LED on. In the new configuration a high produces no voltage and the LED is off, while a low voltage turns it on.
The requires that the logic you're using be able to sink currrent (suck it it) rather than source current (push it out), and you should be aware that if you're using the old 7400/74LS00 series ICs, they are not at all good about sourcing current, and the second approach is the preferred one if you want to get any brightness out of the LEDs.
With the LEDs changed you'll see that, in the original sequence, all 1's become 0's, and vice versa. Map out the new sequence and you'll see that it now counts up.