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I'm trying to implement a shift register and therefore need values to be stored on the downwards edge of the clock signal (otherwise the whole register just sets to the input), so I am using a master-slave D-type flip-flop to store each bit. The design also requires a control line that resets the value stored in each flip-flop to 0 (low voltage) regardless of the clock value. How would I implement this by editing the below circuit?

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Any reason why you haven't used a DFF logic IC but have made it out of NAND gates instead? It would be straightforward to implement reset otherwise and you'd be making a simpler circuit. \$\endgroup\$
    – TonyM
    Commented Nov 21, 2016 at 12:46
  • \$\begingroup\$ @TonyM I'm not an electrical engineer, I'm a computer scientist! I need to use just logic gates for my circuit. \$\endgroup\$
    – Rob Murray
    Commented Nov 21, 2016 at 15:12
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    \$\begingroup\$ There's no difference between using a logic IC or using individual gates to get from point A to point B except that if you use a chip you'll have a lot less wiring - and the joys that go with it - to contend with. :) \$\endgroup\$
    – EM Fields
    Commented Nov 21, 2016 at 15:19
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    \$\begingroup\$ But you will have to make your circuit out of many more logic ICs than if you used a D-type flip-flop logic IC. Sorry, it seemed a fair question. \$\endgroup\$
    – TonyM
    Commented Nov 21, 2016 at 15:19
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    \$\begingroup\$ Good for you, to be encouraged :-) I did electronics at college and computer science at uni, found that they complemented each other well. Electronics design became the forte. Anyway, let me think about this question. \$\endgroup\$
    – TonyM
    Commented Nov 21, 2016 at 21:05

2 Answers 2

Reset to default
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An asynchronous reset can be implemented by adding a third input to the lower NAND gate in each of the cross-coupled pairs in your diagram. Connect them together and drive this input low to reset the output; otherwise, drive it high.

BTW, it would have been easier to talk about the individual components of your diagram if you had left the reference designators on them.

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Paddding

Here's how TI does it:

enter image description here

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  • \$\begingroup\$ That's how TI did it in a design entirely different than a conventional master-slave flip-flop. The SN7474 uses three S-R latches and is not a master-slave design. If one is into such things, it is quite interesting to analyze the three S-R design and see how it works; it's much less obvious than a conventional master-slave FF. \$\endgroup\$
    – Eric Smith
    Commented Jul 21, 2017 at 6:35

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