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.
circuit-design sine square
edited 8 hours ago
Peter Mortensen
1,60031422
asked 13 hours ago
UmangcernUmangcern
323
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Umangcern is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
7
$begingroup$
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.
circuit-design sine square
edited 8 hours ago
Peter Mortensen
1,60031422
asked 13 hours ago
UmangcernUmangcern
323
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.
13 hours ago
$begingroup$
You would almost certainly want to add hysteresis to your solution for a low level low frequency application
$endgroup$
– sstobbe
13 hours ago
$begingroup$
Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
$endgroup$
– analogsystemsrf
7 hours ago
add a comment |
7
7
7
1
$begingroup$
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.
circuit-design sine square
edited 8 hours ago
Peter Mortensen
1,60031422
asked 13 hours ago
UmangcernUmangcern
323
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.
circuit-design sine square
circuit-design sine square
edited 8 hours ago
Peter Mortensen
1,60031422
asked 13 hours ago
UmangcernUmangcern
323
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.
edited 8 hours ago
Peter Mortensen
1,60031422
asked 13 hours ago
UmangcernUmangcern
323
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.
edited 8 hours ago
Peter Mortensen
1,60031422
edited 8 hours ago
Peter Mortensen
1,60031422
edited 8 hours ago
Peter Mortensen
1,60031422
1,60031422
asked 13 hours ago
UmangcernUmangcern
323
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 13 hours ago
UmangcernUmangcern
323
asked 13 hours ago
UmangcernUmangcern
323
323
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.
13 hours ago
$begingroup$
You would almost certainly want to add hysteresis to your solution for a low level low frequency application
$endgroup$
– sstobbe
13 hours ago
$begingroup$
Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
$endgroup$
– analogsystemsrf
7 hours ago
add a comment |
$begingroup$
Possible duplicate of Triangular waveform to square waveform circuit
$endgroup$
– Eugene Sh.
13 hours ago
$begingroup$
You would almost certainly want to add hysteresis to your solution for a low level low frequency application
$endgroup$
– sstobbe
13 hours ago
$begingroup$
Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
$endgroup$
– analogsystemsrf
7 hours ago
$begingroup$
Possible duplicate of Triangular waveform to square waveform circuit
$endgroup$
– Eugene Sh.
13 hours ago
$begingroup$
Possible duplicate of Triangular waveform to square waveform circuit
$endgroup$
– Eugene Sh.
13 hours ago
$begingroup$
You would almost certainly want to add hysteresis to your solution for a low level low frequency application
$endgroup$
– sstobbe
13 hours ago
$begingroup$
You would almost certainly want to add hysteresis to your solution for a low level low frequency application
$endgroup$
– sstobbe
13 hours ago
$begingroup$
Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
$endgroup$
– analogsystemsrf
7 hours ago
$begingroup$
Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
$endgroup$
– analogsystemsrf
7 hours ago
add a comment |
1 Answer
1
active
oldest
votes
15
$begingroup$
The easiest way to do this would be to use a comparator.
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
answered 13 hours ago
MCGMCG
6,82431851
$endgroup$
$begingroup$
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
13 hours ago
2
$begingroup$
@JimmyB Not if you pick a comparator that can handle a negative input voltage.
$endgroup$
– Hearth
13 hours ago
$begingroup$
A diode for 100mV will be hard to find...
$endgroup$
– Eugene Sh.
13 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
13 hours ago
21
$begingroup$
@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
13 hours ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
15
$begingroup$
The easiest way to do this would be to use a comparator.
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
answered 13 hours ago
MCGMCG
6,82431851
$endgroup$
$begingroup$
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
13 hours ago
2
$begingroup$
@JimmyB Not if you pick a comparator that can handle a negative input voltage.
$endgroup$
– Hearth
13 hours ago
$begingroup$
A diode for 100mV will be hard to find...
$endgroup$
– Eugene Sh.
13 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
13 hours ago
21
$begingroup$
@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
13 hours ago
add a comment |
15
$begingroup$
The easiest way to do this would be to use a comparator.
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
answered 13 hours ago
MCGMCG
6,82431851
$endgroup$
$begingroup$
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
13 hours ago
2
$begingroup$
@JimmyB Not if you pick a comparator that can handle a negative input voltage.
$endgroup$
– Hearth
13 hours ago
$begingroup$
A diode for 100mV will be hard to find...
$endgroup$
– Eugene Sh.
13 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
13 hours ago
21
$begingroup$
@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
13 hours ago
add a comment |
15
15
15
$begingroup$
The easiest way to do this would be to use a comparator.
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
answered 13 hours ago
MCGMCG
6,82431851
$endgroup$
The easiest way to do this would be to use a comparator.
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
answered 13 hours ago
MCGMCG
6,82431851
answered 13 hours ago
MCGMCG
6,82431851
answered 13 hours ago
MCGMCG
6,82431851
answered 13 hours ago
MCGMCG
6,82431851
6,82431851
$begingroup$
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
13 hours ago
2
$begingroup$
@JimmyB Not if you pick a comparator that can handle a negative input voltage.
$endgroup$
– Hearth
13 hours ago
$begingroup$
A diode for 100mV will be hard to find...
$endgroup$
– Eugene Sh.
13 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
13 hours ago
21
$begingroup$
@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
13 hours ago
add a comment |
$begingroup$
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
13 hours ago
2
$begingroup$
@JimmyB Not if you pick a comparator that can handle a negative input voltage.
$endgroup$
– Hearth
13 hours ago
$begingroup$
A diode for 100mV will be hard to find...
$endgroup$
– Eugene Sh.
13 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
13 hours ago
21
$begingroup$
@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
13 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
13 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
13 hours ago
2
2
$begingroup$
@JimmyB Not if you pick a comparator that can handle a negative input voltage.
$endgroup$
– Hearth
13 hours ago
$begingroup$
@JimmyB Not if you pick a comparator that can handle a negative input voltage.
$endgroup$
– Hearth
13 hours ago
$begingroup$
A diode for 100mV will be hard to find...
$endgroup$
– Eugene Sh.
13 hours ago
$begingroup$
A diode for 100mV will be hard to find...
$endgroup$
– Eugene Sh.
13 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
13 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
13 hours ago
21
21
$begingroup$
@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
13 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
13 hours ago
add a comment |
Umangcern is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Possible duplicate of Triangular waveform to square waveform circuit
$endgroup$
– Eugene Sh.
13 hours ago
$begingroup$
You would almost certainly want to add hysteresis to your solution for a low level low frequency application
$endgroup$
– sstobbe
13 hours ago
$begingroup$
Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
$endgroup$
– analogsystemsrf
7 hours ago