How do I design a circuit to convert a 100 mV and 50 Hz sine wave to a square wave? The 2019 Stack Overflow Developer Survey Results Are In 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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How do I design a circuit to convert a 100 mV and 50 Hz sine wave to a square wave?
The 2019 Stack Overflow Developer Survey Results Are In
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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$begingroup$
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 1 hour ago
Peter Mortensen
1,60031422
asked 7 hours ago
UmangcernUmangcern
113
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.
$endgroup$
add a comment |
$begingroup$
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 1 hour ago
Peter Mortensen
1,60031422
asked 7 hours ago
UmangcernUmangcern
113
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
6 hours ago
$begingroup$
Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
$endgroup$
– analogsystemsrf
28 mins ago
add a comment |
$begingroup$
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 1 hour ago
Peter Mortensen
1,60031422
asked 7 hours ago
UmangcernUmangcern
113
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 1 hour ago
Peter Mortensen
1,60031422
asked 7 hours ago
UmangcernUmangcern
113
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 1 hour ago
Peter Mortensen
1,60031422
asked 7 hours ago
UmangcernUmangcern
113
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 1 hour ago
Peter Mortensen
1,60031422
edited 1 hour ago
Peter Mortensen
1,60031422
edited 1 hour ago
Peter Mortensen
1,60031422
1,60031422
asked 7 hours ago
UmangcernUmangcern
113
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
113
asked 7 hours ago
UmangcernUmangcern
113
113
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
6 hours ago
$begingroup$
Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
$endgroup$
– analogsystemsrf
28 mins ago
add a comment |
$begingroup$
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
6 hours ago
$begingroup$
Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
$endgroup$
– analogsystemsrf
28 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
6 hours ago
$begingroup$
You would almost certainly want to add hysteresis to your solution for a low level low frequency application
$endgroup$
– sstobbe
6 hours ago
$begingroup$
Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
$endgroup$
– analogsystemsrf
28 mins ago
$begingroup$
Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
$endgroup$
– analogsystemsrf
28 mins ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 6 hours ago
MCGMCG
6,75431850
$endgroup$
$begingroup$
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
6 hours ago
1
$begingroup$
@JimmyB Not if you pick a comparator that can handle a negative input voltage.
$endgroup$
– Hearth
6 hours ago
$begingroup$
A diode for 100mV will be hard to find...
$endgroup$
– Eugene Sh.
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
12
$begingroup$
@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
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 6 hours ago
MCGMCG
6,75431850
$endgroup$
$begingroup$
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
6 hours ago
1
$begingroup$
@JimmyB Not if you pick a comparator that can handle a negative input voltage.
$endgroup$
– Hearth
6 hours ago
$begingroup$
A diode for 100mV will be hard to find...
$endgroup$
– Eugene Sh.
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
12
$begingroup$
@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
add a comment |
$begingroup$
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 6 hours ago
MCGMCG
6,75431850
$endgroup$
$begingroup$
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
6 hours ago
1
$begingroup$
@JimmyB Not if you pick a comparator that can handle a negative input voltage.
$endgroup$
– Hearth
6 hours ago
$begingroup$
A diode for 100mV will be hard to find...
$endgroup$
– Eugene Sh.
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
12
$begingroup$
@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
add a comment |
$begingroup$
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 6 hours ago
MCGMCG
6,75431850
$endgroup$
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 6 hours ago
MCGMCG
6,75431850
answered 6 hours ago
MCGMCG
6,75431850
answered 6 hours ago
MCGMCG
6,75431850
answered 6 hours ago
MCGMCG
6,75431850
6,75431850
$begingroup$
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
6 hours ago
1
$begingroup$
@JimmyB Not if you pick a comparator that can handle a negative input voltage.
$endgroup$
– Hearth
6 hours ago
$begingroup$
A diode for 100mV will be hard to find...
$endgroup$
– Eugene Sh.
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
12
$begingroup$
@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
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
6 hours ago
1
$begingroup$
@JimmyB Not if you pick a comparator that can handle a negative input voltage.
$endgroup$
– Hearth
6 hours ago
$begingroup$
A diode for 100mV will be hard to find...
$endgroup$
– Eugene Sh.
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
12
$begingroup$
@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
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
6 hours ago
1
1
$begingroup$
@JimmyB Not if you pick a comparator that can handle a negative input voltage.
$endgroup$
– Hearth
6 hours ago
$begingroup$
@JimmyB Not if you pick a comparator that can handle a negative input voltage.
$endgroup$
– Hearth
6 hours ago
$begingroup$
A diode for 100mV will be hard to find...
$endgroup$
– Eugene Sh.
6 hours ago
$begingroup$
A diode for 100mV will be hard to find...
$endgroup$
– Eugene Sh.
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
$begingroup$
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
12
12
$begingroup$
@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
add a comment |
Umangcern is a new contributor. Be nice, and check out our Code of Conduct.
Umangcern is a new contributor. Be nice, and check out our Code of Conduct.
Umangcern is a new contributor. Be nice, and check out our Code of Conduct.
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.
7 hours ago
$begingroup$
You would almost certainly want to add hysteresis to your solution for a low level low frequency application
$endgroup$
– sstobbe
6 hours ago
$begingroup$
Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
$endgroup$
– analogsystemsrf
28 mins ago