How do I design a circuit to convert a 100 mV and 50 Hz sine wave to a square wave? The 2019...
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How do I design a circuit to convert a 100 mV and 50 Hz sine wave to a square wave?
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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 28 mins ago
Peter Mortensen
1,60031422
asked 6 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.
$endgroup$
|
show 8 more comments
$begingroup$
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 28 mins ago
Peter Mortensen
1,60031422
asked 6 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.
6 hours ago
$begingroup$
Define peak sine Vp and f tolerances of these variables and tolerance of PW in % of period T=1/f. Can you do that?
$endgroup$
– Sunnyskyguy EE75
6 hours ago
4
$begingroup$
@SunnyskyguyEE75 someone asking a beginner question such as this would clearly not be able to provide the information you asked for. 1. That information is not critical to being able to answer and 2. it looks to me like a deliberate attempt to either confuse the OP or to pretty much be a show off. So really, those are pointless requests
$endgroup$
– MCG
6 hours ago
2
$begingroup$
Overloading someone who is clearly a beginner with too much information with overly technical language than is required is either just to show off your knowledge or intending to confuse. OP asked how to get a square wave from a sine wave. Simple as that. That is the question that needs to be addressed.
$endgroup$
– MCG
5 hours ago
1
$begingroup$
@SunnyskyguyEE75 If someone asks how to calculate a voltage drop over a resistor, you don't then ask for resistor tolerance, ambient room temperature, input voltage ripple and all sorts of other stuff. You simply say "Ohm's Law". It provides an answer to a simple question without over complicating it.
$endgroup$
– MCG
5 hours ago
|
show 8 more comments
$begingroup$
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 28 mins ago
Peter Mortensen
1,60031422
asked 6 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 28 mins ago
Peter Mortensen
1,60031422
asked 6 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 28 mins ago
Peter Mortensen
1,60031422
asked 6 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 28 mins ago
Peter Mortensen
1,60031422
edited 28 mins ago
Peter Mortensen
1,60031422
edited 28 mins ago
Peter Mortensen
1,60031422
1,60031422
asked 6 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 6 hours ago
UmangcernUmangcern
113
asked 6 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.
6 hours ago
$begingroup$
Define peak sine Vp and f tolerances of these variables and tolerance of PW in % of period T=1/f. Can you do that?
$endgroup$
– Sunnyskyguy EE75
6 hours ago
4
$begingroup$
@SunnyskyguyEE75 someone asking a beginner question such as this would clearly not be able to provide the information you asked for. 1. That information is not critical to being able to answer and 2. it looks to me like a deliberate attempt to either confuse the OP or to pretty much be a show off. So really, those are pointless requests
$endgroup$
– MCG
6 hours ago
2
$begingroup$
Overloading someone who is clearly a beginner with too much information with overly technical language than is required is either just to show off your knowledge or intending to confuse. OP asked how to get a square wave from a sine wave. Simple as that. That is the question that needs to be addressed.
$endgroup$
– MCG
5 hours ago
1
$begingroup$
@SunnyskyguyEE75 If someone asks how to calculate a voltage drop over a resistor, you don't then ask for resistor tolerance, ambient room temperature, input voltage ripple and all sorts of other stuff. You simply say "Ohm's Law". It provides an answer to a simple question without over complicating it.
$endgroup$
– MCG
5 hours ago
|
show 8 more comments
$begingroup$
Possible duplicate of Triangular waveform to square waveform circuit
$endgroup$
– Eugene Sh.
6 hours ago
$begingroup$
Define peak sine Vp and f tolerances of these variables and tolerance of PW in % of period T=1/f. Can you do that?
$endgroup$
– Sunnyskyguy EE75
6 hours ago
4
$begingroup$
@SunnyskyguyEE75 someone asking a beginner question such as this would clearly not be able to provide the information you asked for. 1. That information is not critical to being able to answer and 2. it looks to me like a deliberate attempt to either confuse the OP or to pretty much be a show off. So really, those are pointless requests
$endgroup$
– MCG
6 hours ago
2
$begingroup$
Overloading someone who is clearly a beginner with too much information with overly technical language than is required is either just to show off your knowledge or intending to confuse. OP asked how to get a square wave from a sine wave. Simple as that. That is the question that needs to be addressed.
$endgroup$
– MCG
5 hours ago
1
$begingroup$
@SunnyskyguyEE75 If someone asks how to calculate a voltage drop over a resistor, you don't then ask for resistor tolerance, ambient room temperature, input voltage ripple and all sorts of other stuff. You simply say "Ohm's Law". It provides an answer to a simple question without over complicating it.
$endgroup$
– MCG
5 hours ago
$begingroup$
Possible duplicate of Triangular waveform to square waveform circuit
$endgroup$
– Eugene Sh.
6 hours ago
$begingroup$
Possible duplicate of Triangular waveform to square waveform circuit
$endgroup$
– Eugene Sh.
6 hours ago
$begingroup$
Define peak sine Vp and f tolerances of these variables and tolerance of PW in % of period T=1/f. Can you do that?
$endgroup$
– Sunnyskyguy EE75
6 hours ago
$begingroup$
Define peak sine Vp and f tolerances of these variables and tolerance of PW in % of period T=1/f. Can you do that?
$endgroup$
– Sunnyskyguy EE75
6 hours ago
4
4
$begingroup$
@SunnyskyguyEE75 someone asking a beginner question such as this would clearly not be able to provide the information you asked for. 1. That information is not critical to being able to answer and 2. it looks to me like a deliberate attempt to either confuse the OP or to pretty much be a show off. So really, those are pointless requests
$endgroup$
– MCG
6 hours ago
$begingroup$
@SunnyskyguyEE75 someone asking a beginner question such as this would clearly not be able to provide the information you asked for. 1. That information is not critical to being able to answer and 2. it looks to me like a deliberate attempt to either confuse the OP or to pretty much be a show off. So really, those are pointless requests
$endgroup$
– MCG
6 hours ago
2
2
$begingroup$
Overloading someone who is clearly a beginner with too much information with overly technical language than is required is either just to show off your knowledge or intending to confuse. OP asked how to get a square wave from a sine wave. Simple as that. That is the question that needs to be addressed.
$endgroup$
– MCG
5 hours ago
$begingroup$
Overloading someone who is clearly a beginner with too much information with overly technical language than is required is either just to show off your knowledge or intending to confuse. OP asked how to get a square wave from a sine wave. Simple as that. That is the question that needs to be addressed.
$endgroup$
– MCG
5 hours ago
1
1
$begingroup$
@SunnyskyguyEE75 If someone asks how to calculate a voltage drop over a resistor, you don't then ask for resistor tolerance, ambient room temperature, input voltage ripple and all sorts of other stuff. You simply say "Ohm's Law". It provides an answer to a simple question without over complicating it.
$endgroup$
– MCG
5 hours ago
$begingroup$
@SunnyskyguyEE75 If someone asks how to calculate a voltage drop over a resistor, you don't then ask for resistor tolerance, ambient room temperature, input voltage ripple and all sorts of other stuff. You simply say "Ohm's Law". It provides an answer to a simple question without over complicating it.
$endgroup$
– MCG
5 hours ago
|
show 8 more comments
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
6,73431850
$endgroup$
$begingroup$
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
5 hours ago
9
$begingroup$
@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
5 hours ago
add a comment |
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$begingroup$
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
6,73431850
$endgroup$
$begingroup$
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
5 hours ago
9
$begingroup$
@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
5 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
6,73431850
$endgroup$
$begingroup$
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
5 hours ago
9
$begingroup$
@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
5 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
6,73431850
$endgroup$
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
6,73431850
answered 6 hours ago
MCGMCG
6,73431850
answered 6 hours ago
MCGMCG
6,73431850
answered 6 hours ago
MCGMCG
6,73431850
6,73431850
$begingroup$
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
5 hours ago
9
$begingroup$
@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
5 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
5 hours ago
9
$begingroup$
@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
5 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
5 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
5 hours ago
9
9
$begingroup$
@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
5 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
5 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.
6 hours ago
$begingroup$
Define peak sine Vp and f tolerances of these variables and tolerance of PW in % of period T=1/f. Can you do that?
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– Sunnyskyguy EE75
6 hours ago
4
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@SunnyskyguyEE75 someone asking a beginner question such as this would clearly not be able to provide the information you asked for. 1. That information is not critical to being able to answer and 2. it looks to me like a deliberate attempt to either confuse the OP or to pretty much be a show off. So really, those are pointless requests
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– MCG
6 hours ago
2
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Overloading someone who is clearly a beginner with too much information with overly technical language than is required is either just to show off your knowledge or intending to confuse. OP asked how to get a square wave from a sine wave. Simple as that. That is the question that needs to be addressed.
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– MCG
5 hours ago
1
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@SunnyskyguyEE75 If someone asks how to calculate a voltage drop over a resistor, you don't then ask for resistor tolerance, ambient room temperature, input voltage ripple and all sorts of other stuff. You simply say "Ohm's Law". It provides an answer to a simple question without over complicating it.
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– MCG
5 hours ago