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
New contributor
$endgroup$
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
New contributor
$endgroup$
1
$begingroup$
Possible duplicate of Triangular waveform to square waveform circuit
$endgroup$
– Eugene Sh.
yesterday
$begingroup$
You would almost certainly want to add hysteresis to your solution for a low level low frequency application
$endgroup$
– sstobbe
yesterday
$begingroup$
Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
$endgroup$
– analogsystemsrf
yesterday
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
New contributor
$endgroup$
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
New contributor
New contributor
edited yesterday
Peter Mortensen
1,60031422
1,60031422
New contributor
asked yesterday
UmangcernUmangcern
344
344
New contributor
New contributor
1
$begingroup$
Possible duplicate of Triangular waveform to square waveform circuit
$endgroup$
– Eugene Sh.
yesterday
$begingroup$
You would almost certainly want to add hysteresis to your solution for a low level low frequency application
$endgroup$
– sstobbe
yesterday
$begingroup$
Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
$endgroup$
– analogsystemsrf
yesterday
add a comment |
1
$begingroup$
Possible duplicate of Triangular waveform to square waveform circuit
$endgroup$
– Eugene Sh.
yesterday
$begingroup$
You would almost certainly want to add hysteresis to your solution for a low level low frequency application
$endgroup$
– sstobbe
yesterday
$begingroup$
Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
$endgroup$
– analogsystemsrf
yesterday
1
1
$begingroup$
Possible duplicate of Triangular waveform to square waveform circuit
$endgroup$
– Eugene Sh.
yesterday
$begingroup$
Possible duplicate of Triangular waveform to square waveform circuit
$endgroup$
– Eugene Sh.
yesterday
$begingroup$
You would almost certainly want to add hysteresis to your solution for a low level low frequency application
$endgroup$
– sstobbe
yesterday
$begingroup$
You would almost certainly want to add hysteresis to your solution for a low level low frequency application
$endgroup$
– sstobbe
yesterday
$begingroup$
Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
$endgroup$
– analogsystemsrf
yesterday
$begingroup$
Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
$endgroup$
– analogsystemsrf
yesterday
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
$endgroup$
$begingroup$
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
yesterday
3
$begingroup$
@JimmyB Not if you pick a comparator that can handle a negative input voltage.
$endgroup$
– Hearth
yesterday
$begingroup$
A diode for 100mV will be hard to find...
$endgroup$
– Eugene Sh.
yesterday
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
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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
$endgroup$
$begingroup$
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
yesterday
3
$begingroup$
@JimmyB Not if you pick a comparator that can handle a negative input voltage.
$endgroup$
– Hearth
yesterday
$begingroup$
A diode for 100mV will be hard to find...
$endgroup$
– Eugene Sh.
yesterday
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
$endgroup$
$begingroup$
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
yesterday
3
$begingroup$
@JimmyB Not if you pick a comparator that can handle a negative input voltage.
$endgroup$
– Hearth
yesterday
$begingroup$
A diode for 100mV will be hard to find...
$endgroup$
– Eugene Sh.
yesterday
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
$endgroup$
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 yesterday
MCGMCG
6,84431851
6,84431851
$begingroup$
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
yesterday
3
$begingroup$
@JimmyB Not if you pick a comparator that can handle a negative input voltage.
$endgroup$
– Hearth
yesterday
$begingroup$
A diode for 100mV will be hard to find...
$endgroup$
– Eugene Sh.
yesterday
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
yesterday
3
$begingroup$
@JimmyB Not if you pick a comparator that can handle a negative input voltage.
$endgroup$
– Hearth
yesterday
$begingroup$
A diode for 100mV will be hard to find...
$endgroup$
– Eugene Sh.
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
3
3
$begingroup$
@JimmyB Not if you pick a comparator that can handle a negative input voltage.
$endgroup$
– Hearth
yesterday
$begingroup$
@JimmyB Not if you pick a comparator that can handle a negative input voltage.
$endgroup$
– Hearth
yesterday
$begingroup$
A diode for 100mV will be hard to find...
$endgroup$
– Eugene Sh.
yesterday
$begingroup$
A diode for 100mV will be hard to find...
$endgroup$
– Eugene Sh.
yesterday
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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-circuit-design, sine, square
1
$begingroup$
Possible duplicate of Triangular waveform to square waveform circuit
$endgroup$
– Eugene Sh.
yesterday
$begingroup$
You would almost certainly want to add hysteresis to your solution for a low level low frequency application
$endgroup$
– sstobbe
yesterday
$begingroup$
Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
$endgroup$
– analogsystemsrf
yesterday