Build a JK Flip Flop Using multiplexers

Discussion in 'Homework Help' started by agentofdarkness, Nov 6, 2007.

  1. agentofdarkness

    Thread Starter Active Member

    Oct 9, 2007
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    0
    I need to build an electronic light dimmer for my Logic Design class. We are given a state table and told to build the circuit using JK Flip Flops. We are suppose to build the JK Flip Flop using 2 2 to 1 multiplexers. So far I've only be able to implement a JK FF using 3 2 to 1 MUXs. Is it possible to implement this using 2 2 to 1 MUXs?
     
  2. sax1johno

    Member

    Oct 20, 2007
    17
    0
    Are there any other requirements to your problem? Is it that you can only use 2 2x1 multiplexers, or can you use basic logic gates as well?

    I will try to work it out when I get a chance later today.
     
  3. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
    you can use the enable input of the multiplexer as an input.(if u are having a problem with the i/p terminals limitation)
     
  4. agentofdarkness

    Thread Starter Active Member

    Oct 9, 2007
    42
    0
    We can use anything in our lab kit to build it. We have quad input AND,OR,NAND,NOR,XOR gates as well as other ICs such as counters. This lab forces to use a multiplexer because it would be far too complicated to build a JK FF with gates (and I don't have enough gates). I'm not sure who I can implement the JK using the enable as an input. I have worked out how I can build the circuit with a 4 to 1 MUX (except the clock input). After that, I realized that I only had 2 to 1 MUX to use. I figured out how to do it using a 3 2 to 1 MUX. (Table is hard to see, the | seperate the columns)
    |MUX1: | MUX2: | MUX3
    0 |Q | 0 | Output of MUX1
    1 |1 | Q' | Output of MUX2
    Enable |J | J | K

    If I had a 4 to 1 MUX, I could do it like this

    JK Q+
    00 Q
    01 0
    10 1
    11 Q'

    Where Q is the present state and Q+ is the next state. Is there a simple way using logic gates to do this without building another MUX from gates? Thanks for the help. Also, If recca2 could write up how I should wire up the 2 MUXs to implement the JK FF in some sort of table that would really help.
     
  5. agentofdarkness

    Thread Starter Active Member

    Oct 9, 2007
    42
    0
    I checked the lab manual and the experiment worksheet we are given and we are suppose to use 4 4 to 1 MUXs. My question is, why does it take 2 4 to 1 MUX to implement a JK flip flop? Each MUX handles a different variable. Like J1 would have a MUX, K1 has a MUX, J2 has a MUX, and K2 has a MUX. The only thing I can think of is that one input is for the variable (J or K) and the other input is for the clock. Is this correct? In this case, how would I have to wire the MUXs together to implement a JK flip flop?
     
  6. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
    perhaps a schematic (rough one will do) might be able to help.
    i m not sure we can build a clocked FF using only two 2:1 mux.
    hopefully someone with more experience here might be able to help you further.
     
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