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This question already has an answer here:

How can i get 50% duty cycles?enter image description here

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marked as duplicate by dim, laptop2d, Daniel Grillo, JRE, brhans Oct 26 '16 at 21:02

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    \$\begingroup\$ Funnily enough the linked "duplicate" doesn't have the classic answer, which is Andy's second circuit, which works very well if the output isn't heavily loaded. \$\endgroup\$ – Brian Drummond Oct 24 '16 at 9:13
  • \$\begingroup\$ Is there other methods to do it? can i change the resistor value or capacitor? \$\endgroup\$ – Vincci Oct 24 '16 at 9:20
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Try this: -

enter image description here

R1 dictates charge time of C1 and R2 dictates the discharge time.

Or this: -

enter image description here

Taken from here

Alternatively put a d type flip-flop on the output to divide the frequency by two and deliver a very good 50% duty cycle.

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  • \$\begingroup\$ Is there other methods to do it? For example by changing the resistor value or capacitor value? \$\endgroup\$ – Vincci Oct 24 '16 at 9:21
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    \$\begingroup\$ You cannot ever achieve a perfect 50% duty cycle with the standard circuit because charging the capacitor is done via two resistors whilst discharge is done via one resistor so, duty can be nearly 50% to nearly 100% with the standard circuit no matter how hard you try to change any value. \$\endgroup\$ – Andy aka Oct 24 '16 at 9:26

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