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

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7

1

$begingroup$

This question already has an answer here:

• 
  Triangular waveform to square waveform circuit
  
  2 answers
  
  

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

share|improve this question

edited Apr 12 at 20:43

Peter Mortensen

1,60031422

asked Apr 12 at 14:47

UmangcernUmangcern

344

$endgroup$

marked as duplicate by Eugene Sh., Sunnyskyguy EE75, Finbarr, RoyC, Bimpelrekkie Apr 15 at 11:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You would almost certainly want to add hysteresis to your solution for a low level low frequency application
  $endgroup$
  – sstobbe
  Apr 12 at 15:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
  $endgroup$
  – analogsystemsrf
  Apr 12 at 21:23
  

  
  
  
  

add a comment | 

7

1

$begingroup$

This question already has an answer here:

• 
  Triangular waveform to square waveform circuit
  
  2 answers
  
  

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

share|improve this question

edited Apr 12 at 20:43

Peter Mortensen

1,60031422

asked Apr 12 at 14:47

UmangcernUmangcern

344

$endgroup$

marked as duplicate by Eugene Sh., Sunnyskyguy EE75, Finbarr, RoyC, Bimpelrekkie Apr 15 at 11:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You would almost certainly want to add hysteresis to your solution for a low level low frequency application
  $endgroup$
  – sstobbe
  Apr 12 at 15:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does the OP want a 50% duty cycle? in which case, some zero-crossing is needed.
  $endgroup$
  – analogsystemsrf
  Apr 12 at 21:23
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  Triangular waveform to square waveform circuit
  
  2 answers
  
  

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

share|improve this question

edited Apr 12 at 20:43

Peter Mortensen

1,60031422

asked Apr 12 at 14:47

UmangcernUmangcern

344

$endgroup$

share|improve this question

edited Apr 12 at 20:43

Peter Mortensen

1,60031422

asked Apr 12 at 14:47

UmangcernUmangcern

344

share|improve this question

edited Apr 12 at 20:43

Peter Mortensen

1,60031422

asked Apr 12 at 14:47

UmangcernUmangcern

344

edited Apr 12 at 20:43

Peter Mortensen

1,60031422

edited Apr 12 at 20:43

Peter Mortensen

1,60031422

asked Apr 12 at 14:47

UmangcernUmangcern

344

asked Apr 12 at 14:47

UmangcernUmangcern

344

1 Answer
1

active

oldest

votes

17

$begingroup$

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

share|improve this answer

answered Apr 12 at 14:54

MCGMCG

6,87131851

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Apr 12 at 15:01
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  @JimmyB Not if you pick a comparator that can handle a negative input voltage.
  $endgroup$
  – Hearth
  Apr 12 at 15:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A diode for 100mV will be hard to find...
  $endgroup$
  – Eugene Sh.
  Apr 12 at 15:08
  

  
  
  
  

add a comment | 

17

$begingroup$

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

share|improve this answer

answered Apr 12 at 14:54

MCGMCG

6,87131851

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Apr 12 at 15:01
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  @JimmyB Not if you pick a comparator that can handle a negative input voltage.
  $endgroup$
  – Hearth
  Apr 12 at 15:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A diode for 100mV will be hard to find...
  $endgroup$
  – Eugene Sh.
  Apr 12 at 15:08
  

  
  
  
  

add a comment | 

17

$begingroup$

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

share|improve this answer

answered Apr 12 at 14:54

MCGMCG

6,87131851

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Apr 12 at 15:01
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  @JimmyB Not if you pick a comparator that can handle a negative input voltage.
  $endgroup$
  – Hearth
  Apr 12 at 15:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A diode for 100mV will be hard to find...
  $endgroup$
  – Eugene Sh.
  Apr 12 at 15:08
  

  
  
  
  

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

share|improve this answer

answered Apr 12 at 14:54

MCGMCG

6,87131851

$endgroup$

share|improve this answer

answered Apr 12 at 14:54

MCGMCG

6,87131851

answered Apr 12 at 14:54

MCGMCG

6,87131851

answered Apr 12 at 14:54

MCGMCG

6,87131851