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Found this circuit for a 555 timer. I can't understand the use of the d-type flip-flop. Please explain.enter image description here. And is the variable resistor used to adjust the frequency?

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The D-types just divides the 555 output by two generating a square wave output from both.

The issue with that though is there is nothing in the circuit to ensure the two D-Types do not start out 180 degrees out of phase.

As you suspect the pot adjusts the frequency.

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  • \$\begingroup\$ still unclear to me. Isn't the output of the 555 timer already a square wave. \$\endgroup\$ – JOHNDOE Mar 3 '18 at 18:03
  • \$\begingroup\$ Not when you want to have a variable frequency. \$\endgroup\$ – Oldfart Mar 3 '18 at 18:05
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    \$\begingroup\$ @JOHNDOE getting an exact square wave out of a 555 is actually a bit of an art especially with variable frequency. \$\endgroup\$ – Trevor_G Mar 3 '18 at 18:16
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    \$\begingroup\$ @Trevor_G. Not to be picky, but I would point out that a flip-flop can ONLY put out a perfect 50% duty cycle, as long as the frequency is not changing. \$\endgroup\$ – Sparky256 Mar 3 '18 at 20:46
  • \$\begingroup\$ @Sparky256 true that :) \$\endgroup\$ – Trevor_G Mar 4 '18 at 13:55
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The D-flipflops are connected to provide a divide-by 2. I assume the main reason for those is so that the output duty cycle remains 50/50 even when the frequency varies (Which is not what comes out of the 555.)

The FET gives and inverted output but they could have gotten that by using the Q-bar output so maybe they needed a specific drive characteristic.

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  • \$\begingroup\$ so when you vary the frequency the duty cycle of the 555 does not remain constant \$\endgroup\$ – JOHNDOE Mar 3 '18 at 18:07
  • \$\begingroup\$ one more question I want the output square pulse to have a pnp output and a npn output how can I modify the circuit to achieve this? \$\endgroup\$ – JOHNDOE Mar 3 '18 at 18:12

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