I found an interesting interview question for FPGA engineer online - Implement a counter with Mux, so I decided to try to do that. I tried to keep it simple and so implement a counter with two bits, so it can count only to 3 and then reset.

I came to a conclusion I need 3 multiplexers, two for each bit, and the other one to serve as a frequency divider for the clock (I'm sure there are other ideas, would love to hear them as well). Note that I wanted this frequency divider to work at the falling edge of the clock.

Basically this is what I tried to do:


simulate this circuit – Schematic created using CircuitLab

I used a clock of 1Hz just so it won't be too fast (though maybe it doesn't matter?), With mux1 being the frequency divider.

The idea here is that the counter is like this:

Q1 Q0
0  0
0  1
1  0
1  1

So, since Q0 changes with every clock edge, I connected the clock to mux3. Since Q1 toggles every two clocks, I tried to design a frequency divider which will make sure that Q1 will toggle every two clock (clk1) falling edges.

I thought it would work well, but in a simulation (used Multisim for that, because I don't know how to simulate it work circuitlab, would love if you could show me how I can do that), I got a very strange result - the simulation just stopped. Here is how it looked (the output of mux1):

Output of mux1

When I zoomed in at the falling edge (at 500ms) it looked like this:

output of mux1 zoomed in

So my questions are:

  1. Why doesn't my frequency divider work?
  2. Is this implementation of a counter even good, if that frequency divider worked? Is this even possible?

1 Answer 1


Think about what each part of your circuit does.


All this is doing is saying if CLK1 is high, then output 1. If CLK1 is low, output 0. What is the purpose of this mux? Simply MUX3 = CLK1


Again, if MUX1 is high, output 0, if MUX1 is low, output 1. So basically MUX2 = !MUX1


What for example happens when CLK1 is low? It selects In0 of MUX1. In0 is simply the inverse of the output of MUX1.

So in other words MUX1 = !MUX1 - you've made an oscillator that will toggle at whatever the propagation delay of the not gate is. Uncontrolled inverter loops like this are almost never a good idea.

What happens if CLK1 is high? Well, you get MUX1 = MUX1, so basically the output will hold whatever state it had previously.

I think the theory of the interview question lies in the fact that a 2-input MUX is functionally complete - you can use multiple muxes to make any logic gate.

One of those is that a mux can be configured as an AND gate with one input inverted by simply tying the IN1 pin to GND. You can then simply use two of these to create an S-R latch.

Once you have an S-R latch it is very easy to build registers, and then counters.

  • \$\begingroup\$ About the last line - I expected that when clk1 goes low, the output of mux1 will be the inverse of itself, since it is now In0. If the output was '1', the input In0 would be '0', right? and once clk1 goes low, the output should go to '0'... \$\endgroup\$
    – nettek
    Commented Oct 30, 2018 at 8:34
  • \$\begingroup\$ What I also don't understand, is why the simulation just stopped (more precisely, the scope just stopped showing the graph). \$\endgroup\$
    – nettek
    Commented Oct 30, 2018 at 8:36
  • 3
    \$\begingroup\$ @Eran the output will indeed be the inverse of itself. However as soon as the output changes, the in0 will change, which will change the output again. Essentially you have a not gate with input and output connected. The simulation probably crashed as it tries to oscillate. \$\endgroup\$ Commented Oct 30, 2018 at 8:38
  • \$\begingroup\$ Thanks for your answer! I now understand the problem with what I did. \$\endgroup\$
    – nettek
    Commented Oct 30, 2018 at 9:01

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