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I'm trying to design an electronically toggled switch(one single switch turning an LED on and off) using a T-Flip Flop(a 74HC73, to be precise). I believe I've connected all the inputs and outputs properly, yet, when I'm sending a clock pulse, the output is not getting toggled.

Please take a look at the circuit and try to figure out the problem:

wrong circuit

Here's what the circuit is like on paper:

schematic

simulate this circuit – Schematic created using CircuitLab

I think it is worthy to mention here that I haven't tried the circuit in reality yet, but, the Autodesk Simulator(which is pretty accurate) is not giving appropriate results. So, I believe the problem is with the circuit only.

Thank you for your time!

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  • \$\begingroup\$ The top circuit has 2 buttons? while the bottom one only has 1? \$\endgroup\$
    – Ferrybig
    Apr 16, 2016 at 12:18
  • \$\begingroup\$ No pullup resistor... add one (100K or so) from SW1 to 5V. (You will also have to debounce the switch, in reality) \$\endgroup\$
    – user16324
    Apr 16, 2016 at 12:23
  • \$\begingroup\$ @Ferrybig The other(the grey one) button is for Reset, there's no Reset pin in the schematic. \$\endgroup\$
    – Swap
    Apr 16, 2016 at 12:31
  • \$\begingroup\$ @BrianDrummond Didn't help. \$\endgroup\$
    – Swap
    Apr 16, 2016 at 12:36
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    \$\begingroup\$ your 0V line is not connected \$\endgroup\$ Apr 16, 2016 at 12:41

2 Answers 2

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Just to correct your faults

(1) You need to add a wire from the bottom 0V rail to the top 0V rail otherwise you are not powering the chip.

(2) Your pull up resistors need to be on the high side of the switches with the other end of the switch to 0V.

(3) Your circuit diagram should reflect the actual circuit you are building.

enter image description here

Also, as st2000 points out, you'll find that push switches are very 'bouncy' and for a single push you get several pulses. This is not a problem in simulation because they are treated as perfect components. In real life you should add a de-bounce circuit to the clock input. The simplest way is to add a 10uF capacitor across the switch (SW1).

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  • \$\begingroup\$ It works perfectly, thanks for putting in all the effort! :-) \$\endgroup\$
    – Swap
    Apr 16, 2016 at 17:15
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After addressing the errors, you need to understand how to debounce a mechanical switch. You can read about debouncing switches here and here. Basically you can use logic:

enter image description here

Or an RC timed circuit cleaned up with a Schmidt trigger:

enter image description here

Or (not as clean) just the RC timed circuit:

enter image description here

In all cases note how the switch is connected to one side of the power rail and a resistor to the other. When the switch is engaged the switch over rides the resistor pulling the voltage to the opposite power rail.

With few exception, the input to logic gates should not be ambiguous. That is, they should be clearly at or near one or the other power rail.

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