Yhyjyk

Didn't get enough time to take a Coding Test - what to do now?

Accepted by European university, rejected by all American ones I applied to? Possible reasons?

Button changing its text & action. Good or terrible?

What can I do if neighbor is blocking my solar panels intentionally?

For what reasons would an animal species NOT cross a *horizontal* land bridge?

Did the new image of black hole confirm the general theory of relativity?

What is the padding with red substance inside of steak packaging?

Do I have Disadvantage attacking with an off-hand weapon?

Why can I use a list index as an indexing variable in a for loop?

"... to apply for a visa" or "... and applied for a visa"?

Do working physicists consider Newtonian mechanics to be "falsified"?

Simulating Exploding Dice

Solving overdetermined system by QR decomposition

Match Roman Numerals

Can each chord in a progression create its own key?

Variable with quotation marks "$()"

How to handle characters who are more educated than the author?

How do you keep chess fun when your opponent constantly beats you?

Why don't hard Brexiteers insist on a hard border to prevent illegal immigration after Brexit?

Am I ethically obligated to go into work on an off day if the reason is sudden?

Can withdrawing asylum be illegal?

Could an empire control the whole planet with today's comunication methods?

Why are PDP-7-style microprogrammed instructions out of vogue?

Deal with toxic manager when you can't quit

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

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

2

$begingroup$

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

share|improve this question

edited 30 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

2

$begingroup$

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

share|improve this question

edited 30 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

share|improve this question

edited 30 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.

$endgroup$

share|improve this question

edited 30 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.

share|improve this question

edited 30 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 30 mins ago

Peter Mortensen

1,60031422

edited 30 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.

asked 6 hours ago

UmangcernUmangcern

113

1 Answer
1

active

oldest

votes

6

$begingroup$

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 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 | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("schematics", function ()
StackExchange.schematics.init();
);
, "cicuitlab");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "135"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

Umangcern is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f432206%2fhow-do-i-design-a-circuit-to-convert-a-100-mv-and-50-hz-sine-wave-to-a-square-wa%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

6

$begingroup$

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 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 | 

6

$begingroup$

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 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

share|improve this answer

answered 6 hours ago

MCGMCG

6,73431850

$endgroup$

share|improve this answer

answered 6 hours ago

MCGMCG

6,73431850

answered 6 hours ago

MCGMCG

6,73431850

answered 6 hours ago

MCGMCG

6,73431850