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How do I design a circuit to convert a 100 mV and 50 Hz sine wave to a square wave? [duplicate]



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Triangular waveform to square waveform circuitSquare wave / Sine wave is more audibleSine wave to square wave - Schmitt triggerWhat is the best way to get a sine wave from a square wave?How to build a circuit that generates a sine wave?High-current capable (±250A) AC power-supply, sine wave/square wave (±20V)Why sine wave not square wave?Sine to Square only with a MCUHow to check (with DIY methods) if an Inverter returns a Square or a Sine Wave?Need a circuit to convert 230V sine wave into 5V square waveSine to square wave converter



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7












$begingroup$



This question already has an answer here:



  • Triangular waveform to square waveform circuit

    2 answers



I have a sine wave of 100 mV and 50 Hz. I want to design a circuit that converts this sine wave into a square wave as shown in a figure.



Enter image description here










share|improve this question









New contributor




Umangcern is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$



marked as duplicate by Eugene Sh., Sunnyskyguy EE75, Finbarr, RoyC, Bimpelrekkie yesterday


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.

















  • $begingroup$
    You would almost certainly want to add hysteresis to your solution for a low level low frequency application
    $endgroup$
    – sstobbe
    Apr 12 at 15:04










  • $begingroup$
    Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
    $endgroup$
    – analogsystemsrf
    Apr 12 at 21:23

















7












$begingroup$



This question already has an answer here:



  • Triangular waveform to square waveform circuit

    2 answers



I have a sine wave of 100 mV and 50 Hz. I want to design a circuit that converts this sine wave into a square wave as shown in a figure.



Enter image description here










share|improve this question









New contributor




Umangcern is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$



marked as duplicate by Eugene Sh., Sunnyskyguy EE75, Finbarr, RoyC, Bimpelrekkie yesterday


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.

















  • $begingroup$
    You would almost certainly want to add hysteresis to your solution for a low level low frequency application
    $endgroup$
    – sstobbe
    Apr 12 at 15:04










  • $begingroup$
    Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
    $endgroup$
    – analogsystemsrf
    Apr 12 at 21:23













7












7








7


1



$begingroup$



This question already has an answer here:



  • Triangular waveform to square waveform circuit

    2 answers



I have a sine wave of 100 mV and 50 Hz. I want to design a circuit that converts this sine wave into a square wave as shown in a figure.



Enter image description here










share|improve this question









New contributor




Umangcern is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$





This question already has an answer here:



  • Triangular waveform to square waveform circuit

    2 answers



I have a sine wave of 100 mV and 50 Hz. I want to design a circuit that converts this sine wave into a square wave as shown in a figure.



Enter image description here





This question already has an answer here:



  • Triangular waveform to square waveform circuit

    2 answers







circuit-design sine square






share|improve this question









New contributor




Umangcern is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




Umangcern is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited Apr 12 at 20:43









Peter Mortensen

1,60031422




1,60031422






New contributor




Umangcern is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked Apr 12 at 14:47









UmangcernUmangcern

344




344




New contributor




Umangcern is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Umangcern is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Umangcern is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




marked as duplicate by Eugene Sh., Sunnyskyguy EE75, Finbarr, RoyC, Bimpelrekkie yesterday


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.









marked as duplicate by Eugene Sh., Sunnyskyguy EE75, Finbarr, RoyC, Bimpelrekkie yesterday


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.













  • $begingroup$
    You would almost certainly want to add hysteresis to your solution for a low level low frequency application
    $endgroup$
    – sstobbe
    Apr 12 at 15:04










  • $begingroup$
    Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
    $endgroup$
    – analogsystemsrf
    Apr 12 at 21:23
















  • $begingroup$
    You would almost certainly want to add hysteresis to your solution for a low level low frequency application
    $endgroup$
    – sstobbe
    Apr 12 at 15:04










  • $begingroup$
    Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
    $endgroup$
    – analogsystemsrf
    Apr 12 at 21:23















$begingroup$
You would almost certainly want to add hysteresis to your solution for a low level low frequency application
$endgroup$
– sstobbe
Apr 12 at 15:04




$begingroup$
You would almost certainly want to add hysteresis to your solution for a low level low frequency application
$endgroup$
– sstobbe
Apr 12 at 15:04












$begingroup$
Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
$endgroup$
– analogsystemsrf
Apr 12 at 21:23




$begingroup$
Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
$endgroup$
– analogsystemsrf
Apr 12 at 21:23










1 Answer
1






active

oldest

votes


















17












$begingroup$

The easiest way to do this would be to use a comparator.



enter image description here



Picture taken from linked site



All you have to do is set your Vref level to where you want your square wave to trigger. When the sine wave crosses the Vref level, the comparator output will go high. As it approaches it again and goes below the Vref level, the comparator output goes low.



You will then get yourself a square wave.



Be aware the example shown in the graphic is of a non-inverting comparator. An inverting comparator works with the same principle, but the output is inverted






share|improve this answer









$endgroup$












  • $begingroup$
    The OP seems to want to cut off the negative half of the sine wave, so a diode may be needed at the input.
    $endgroup$
    – JimmyB
    Apr 12 at 15:01






  • 3




    $begingroup$
    @JimmyB Not if you pick a comparator that can handle a negative input voltage.
    $endgroup$
    – Hearth
    Apr 12 at 15:03










  • $begingroup$
    A diode for 100mV will be hard to find...
    $endgroup$
    – Eugene Sh.
    Apr 12 at 15:08

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









17












$begingroup$

The easiest way to do this would be to use a comparator.



enter image description here



Picture taken from linked site



All you have to do is set your Vref level to where you want your square wave to trigger. When the sine wave crosses the Vref level, the comparator output will go high. As it approaches it again and goes below the Vref level, the comparator output goes low.



You will then get yourself a square wave.



Be aware the example shown in the graphic is of a non-inverting comparator. An inverting comparator works with the same principle, but the output is inverted






share|improve this answer









$endgroup$












  • $begingroup$
    The OP seems to want to cut off the negative half of the sine wave, so a diode may be needed at the input.
    $endgroup$
    – JimmyB
    Apr 12 at 15:01






  • 3




    $begingroup$
    @JimmyB Not if you pick a comparator that can handle a negative input voltage.
    $endgroup$
    – Hearth
    Apr 12 at 15:03










  • $begingroup$
    A diode for 100mV will be hard to find...
    $endgroup$
    – Eugene Sh.
    Apr 12 at 15:08















17












$begingroup$

The easiest way to do this would be to use a comparator.



enter image description here



Picture taken from linked site



All you have to do is set your Vref level to where you want your square wave to trigger. When the sine wave crosses the Vref level, the comparator output will go high. As it approaches it again and goes below the Vref level, the comparator output goes low.



You will then get yourself a square wave.



Be aware the example shown in the graphic is of a non-inverting comparator. An inverting comparator works with the same principle, but the output is inverted






share|improve this answer









$endgroup$












  • $begingroup$
    The OP seems to want to cut off the negative half of the sine wave, so a diode may be needed at the input.
    $endgroup$
    – JimmyB
    Apr 12 at 15:01






  • 3




    $begingroup$
    @JimmyB Not if you pick a comparator that can handle a negative input voltage.
    $endgroup$
    – Hearth
    Apr 12 at 15:03










  • $begingroup$
    A diode for 100mV will be hard to find...
    $endgroup$
    – Eugene Sh.
    Apr 12 at 15:08













17












17








17





$begingroup$

The easiest way to do this would be to use a comparator.



enter image description here



Picture taken from linked site



All you have to do is set your Vref level to where you want your square wave to trigger. When the sine wave crosses the Vref level, the comparator output will go high. As it approaches it again and goes below the Vref level, the comparator output goes low.



You will then get yourself a square wave.



Be aware the example shown in the graphic is of a non-inverting comparator. An inverting comparator works with the same principle, but the output is inverted






share|improve this answer









$endgroup$



The easiest way to do this would be to use a comparator.



enter image description here



Picture taken from linked site



All you have to do is set your Vref level to where you want your square wave to trigger. When the sine wave crosses the Vref level, the comparator output will go high. As it approaches it again and goes below the Vref level, the comparator output goes low.



You will then get yourself a square wave.



Be aware the example shown in the graphic is of a non-inverting comparator. An inverting comparator works with the same principle, but the output is inverted







share|improve this answer












share|improve this answer



share|improve this answer










answered Apr 12 at 14:54









MCGMCG

6,87331851




6,87331851











  • $begingroup$
    The OP seems to want to cut off the negative half of the sine wave, so a diode may be needed at the input.
    $endgroup$
    – JimmyB
    Apr 12 at 15:01






  • 3




    $begingroup$
    @JimmyB Not if you pick a comparator that can handle a negative input voltage.
    $endgroup$
    – Hearth
    Apr 12 at 15:03










  • $begingroup$
    A diode for 100mV will be hard to find...
    $endgroup$
    – Eugene Sh.
    Apr 12 at 15:08
















  • $begingroup$
    The OP seems to want to cut off the negative half of the sine wave, so a diode may be needed at the input.
    $endgroup$
    – JimmyB
    Apr 12 at 15:01






  • 3




    $begingroup$
    @JimmyB Not if you pick a comparator that can handle a negative input voltage.
    $endgroup$
    – Hearth
    Apr 12 at 15:03










  • $begingroup$
    A diode for 100mV will be hard to find...
    $endgroup$
    – Eugene Sh.
    Apr 12 at 15:08















$begingroup$
The OP seems to want to cut off the negative half of the sine wave, so a diode may be needed at the input.
$endgroup$
– JimmyB
Apr 12 at 15:01




$begingroup$
The OP seems to want to cut off the negative half of the sine wave, so a diode may be needed at the input.
$endgroup$
– JimmyB
Apr 12 at 15:01




3




3




$begingroup$
@JimmyB Not if you pick a comparator that can handle a negative input voltage.
$endgroup$
– Hearth
Apr 12 at 15:03




$begingroup$
@JimmyB Not if you pick a comparator that can handle a negative input voltage.
$endgroup$
– Hearth
Apr 12 at 15:03












$begingroup$
A diode for 100mV will be hard to find...
$endgroup$
– Eugene Sh.
Apr 12 at 15:08




$begingroup$
A diode for 100mV will be hard to find...
$endgroup$
– Eugene Sh.
Apr 12 at 15:08



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