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





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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






circuit-design sine square







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  • $begingroup$
    Possible duplicate of Triangular waveform to square waveform circuit
    $endgroup$
    – Eugene Sh.
    7 hours ago










  • $begingroup$
    You would almost certainly want to add hysteresis to your solution for a low level low frequency application
    $endgroup$
    – sstobbe
    7 hours ago










  • $begingroup$
    Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
    $endgroup$
    – analogsystemsrf
    47 mins ago


















4












$begingroup$


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






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.







$endgroup$












  • $begingroup$
    Possible duplicate of Triangular waveform to square waveform circuit
    $endgroup$
    – Eugene Sh.
    7 hours ago










  • $begingroup$
    You would almost certainly want to add hysteresis to your solution for a low level low frequency application
    $endgroup$
    – sstobbe
    7 hours ago










  • $begingroup$
    Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
    $endgroup$
    – analogsystemsrf
    47 mins ago














4












4








4





$begingroup$


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






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.







$endgroup$




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






circuit-design sine square




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 1 hour ago









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 7 hours ago









UmangcernUmangcern

213




213




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.












  • $begingroup$
    Possible duplicate of Triangular waveform to square waveform circuit
    $endgroup$
    – Eugene Sh.
    7 hours ago










  • $begingroup$
    You would almost certainly want to add hysteresis to your solution for a low level low frequency application
    $endgroup$
    – sstobbe
    7 hours ago










  • $begingroup$
    Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
    $endgroup$
    – analogsystemsrf
    47 mins ago


















  • $begingroup$
    Possible duplicate of Triangular waveform to square waveform circuit
    $endgroup$
    – Eugene Sh.
    7 hours ago










  • $begingroup$
    You would almost certainly want to add hysteresis to your solution for a low level low frequency application
    $endgroup$
    – sstobbe
    7 hours ago










  • $begingroup$
    Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
    $endgroup$
    – analogsystemsrf
    47 mins ago
















$begingroup$
Possible duplicate of Triangular waveform to square waveform circuit
$endgroup$
– Eugene Sh.
7 hours ago




$begingroup$
Possible duplicate of Triangular waveform to square waveform circuit
$endgroup$
– Eugene Sh.
7 hours ago












$begingroup$
You would almost certainly want to add hysteresis to your solution for a low level low frequency application
$endgroup$
– sstobbe
7 hours ago




$begingroup$
You would almost certainly want to add hysteresis to your solution for a low level low frequency application
$endgroup$
– sstobbe
7 hours ago












$begingroup$
Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
$endgroup$
– analogsystemsrf
47 mins ago




$begingroup$
Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
$endgroup$
– analogsystemsrf
47 mins ago










1 Answer
1






active

oldest

votes


















10












$begingroup$

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
    7 hours ago






  • 2




    $begingroup$
    @JimmyB Not if you pick a comparator that can handle a negative input voltage.
    $endgroup$
    – Hearth
    7 hours ago










  • $begingroup$
    A diode for 100mV will be hard to find...
    $endgroup$
    – Eugene Sh.
    7 hours ago










  • $begingroup$
    There are a dozen better ways to produce an accurate narrow pulse centred around a sine peak which may have noise or change in amplitude
    $endgroup$
    – Sunnyskyguy EE75
    6 hours ago






  • 15




    $begingroup$
    @SunnyskyguyEE75, yes there are. Hence why my opening line was the easiest way. Not the best way. A beginner question deserves a beginners answer. Once the OP has built up knowledge and experience, then better, but more involved ways can be looked at. There is no need to try and belittle my answer by saying there are a dozen better ways just because of what I said in the comments to the question. By all means, write your own overcomplicated answer which I am sure will be useful to more experienced engineers but useless to a beginner
    $endgroup$
    – MCG
    6 hours ago












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









10












$begingroup$

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
    7 hours ago






  • 2




    $begingroup$
    @JimmyB Not if you pick a comparator that can handle a negative input voltage.
    $endgroup$
    – Hearth
    7 hours ago










  • $begingroup$
    A diode for 100mV will be hard to find...
    $endgroup$
    – Eugene Sh.
    7 hours ago










  • $begingroup$
    There are a dozen better ways to produce an accurate narrow pulse centred around a sine peak which may have noise or change in amplitude
    $endgroup$
    – Sunnyskyguy EE75
    6 hours ago






  • 15




    $begingroup$
    @SunnyskyguyEE75, yes there are. Hence why my opening line was the easiest way. Not the best way. A beginner question deserves a beginners answer. Once the OP has built up knowledge and experience, then better, but more involved ways can be looked at. There is no need to try and belittle my answer by saying there are a dozen better ways just because of what I said in the comments to the question. By all means, write your own overcomplicated answer which I am sure will be useful to more experienced engineers but useless to a beginner
    $endgroup$
    – MCG
    6 hours ago
















10












$begingroup$

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
    7 hours ago






  • 2




    $begingroup$
    @JimmyB Not if you pick a comparator that can handle a negative input voltage.
    $endgroup$
    – Hearth
    7 hours ago










  • $begingroup$
    A diode for 100mV will be hard to find...
    $endgroup$
    – Eugene Sh.
    7 hours ago










  • $begingroup$
    There are a dozen better ways to produce an accurate narrow pulse centred around a sine peak which may have noise or change in amplitude
    $endgroup$
    – Sunnyskyguy EE75
    6 hours ago






  • 15




    $begingroup$
    @SunnyskyguyEE75, yes there are. Hence why my opening line was the easiest way. Not the best way. A beginner question deserves a beginners answer. Once the OP has built up knowledge and experience, then better, but more involved ways can be looked at. There is no need to try and belittle my answer by saying there are a dozen better ways just because of what I said in the comments to the question. By all means, write your own overcomplicated answer which I am sure will be useful to more experienced engineers but useless to a beginner
    $endgroup$
    – MCG
    6 hours ago














10












10








10





$begingroup$

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 7 hours ago









MCGMCG

6,77331851




6,77331851












  • $begingroup$
    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
    7 hours ago






  • 2




    $begingroup$
    @JimmyB Not if you pick a comparator that can handle a negative input voltage.
    $endgroup$
    – Hearth
    7 hours ago










  • $begingroup$
    A diode for 100mV will be hard to find...
    $endgroup$
    – Eugene Sh.
    7 hours ago










  • $begingroup$
    There are a dozen better ways to produce an accurate narrow pulse centred around a sine peak which may have noise or change in amplitude
    $endgroup$
    – Sunnyskyguy EE75
    6 hours ago






  • 15




    $begingroup$
    @SunnyskyguyEE75, yes there are. Hence why my opening line was the easiest way. Not the best way. A beginner question deserves a beginners answer. Once the OP has built up knowledge and experience, then better, but more involved ways can be looked at. There is no need to try and belittle my answer by saying there are a dozen better ways just because of what I said in the comments to the question. By all means, write your own overcomplicated answer which I am sure will be useful to more experienced engineers but useless to a beginner
    $endgroup$
    – MCG
    6 hours ago


















  • $begingroup$
    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
    7 hours ago






  • 2




    $begingroup$
    @JimmyB Not if you pick a comparator that can handle a negative input voltage.
    $endgroup$
    – Hearth
    7 hours ago










  • $begingroup$
    A diode for 100mV will be hard to find...
    $endgroup$
    – Eugene Sh.
    7 hours ago










  • $begingroup$
    There are a dozen better ways to produce an accurate narrow pulse centred around a sine peak which may have noise or change in amplitude
    $endgroup$
    – Sunnyskyguy EE75
    6 hours ago






  • 15




    $begingroup$
    @SunnyskyguyEE75, yes there are. Hence why my opening line was the easiest way. Not the best way. A beginner question deserves a beginners answer. Once the OP has built up knowledge and experience, then better, but more involved ways can be looked at. There is no need to try and belittle my answer by saying there are a dozen better ways just because of what I said in the comments to the question. By all means, write your own overcomplicated answer which I am sure will be useful to more experienced engineers but useless to a beginner
    $endgroup$
    – MCG
    6 hours ago
















$begingroup$
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
7 hours ago




$begingroup$
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
7 hours ago




2




2




$begingroup$
@JimmyB Not if you pick a comparator that can handle a negative input voltage.
$endgroup$
– Hearth
7 hours ago




$begingroup$
@JimmyB Not if you pick a comparator that can handle a negative input voltage.
$endgroup$
– Hearth
7 hours ago












$begingroup$
A diode for 100mV will be hard to find...
$endgroup$
– Eugene Sh.
7 hours ago




$begingroup$
A diode for 100mV will be hard to find...
$endgroup$
– Eugene Sh.
7 hours ago












$begingroup$
There are a dozen better ways to produce an accurate narrow pulse centred around a sine peak which may have noise or change in amplitude
$endgroup$
– Sunnyskyguy EE75
6 hours ago




$begingroup$
There are a dozen better ways to produce an accurate narrow pulse centred around a sine peak which may have noise or change in amplitude
$endgroup$
– Sunnyskyguy EE75
6 hours ago




15




15




$begingroup$
@SunnyskyguyEE75, yes there are. Hence why my opening line was the easiest way. Not the best way. A beginner question deserves a beginners answer. Once the OP has built up knowledge and experience, then better, but more involved ways can be looked at. There is no need to try and belittle my answer by saying there are a dozen better ways just because of what I said in the comments to the question. By all means, write your own overcomplicated answer which I am sure will be useful to more experienced engineers but useless to a beginner
$endgroup$
– MCG
6 hours ago




$begingroup$
@SunnyskyguyEE75, yes there are. Hence why my opening line was the easiest way. Not the best way. A beginner question deserves a beginners answer. Once the OP has built up knowledge and experience, then better, but more involved ways can be looked at. There is no need to try and belittle my answer by saying there are a dozen better ways just because of what I said in the comments to the question. By all means, write your own overcomplicated answer which I am sure will be useful to more experienced engineers but useless to a beginner
$endgroup$
– MCG
6 hours ago










Umangcern is a new contributor. Be nice, and check out our Code of Conduct.










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