Mdthbs

How can I deal with a player trying to insert real-world mythology into my homebrew setting?

Is Arc Length always irrational between two rational points?

Shortest distance around a pyramid?

Can I call 112 to check a police officer's identity in the Czech Republic?

Single word for "refusing to move to next activity unless present one is completed."

Cops: The Hidden OEIS Substring

Rhombuses, kites etc

Why does Hellboy file down his horns?

Why was hardware diversification an asset for the IBM PC ecosystem?

How to hide what's behind an object in a non destructive way / give it an "invisibility cloak"

Robbers: The Hidden OEIS Substring

Get ids only where one id is null and other isn't

Turning arguments into exponents

Was I subtly told to resign?

Comparing two limsup's

Simple LED driver, transistor and GPIO

What's the minimum number of sensors for a hobby GPS waypoint-following UAV?

How to know whether a Tamron lens is compatible with Canon EOS 60D?

Managing and organizing the massively increased number of classes after switching to SOLID?

I wrote two alternate fugue expositions for one subject do both follow good harmonic conventions?

For a hashing function like MD5, how similar can two plaintext strings be and still generate the same hash?

Is there any word for "disobedience to God"?

Setting MAC field to all-zero to indicate unencrypted data

What are some examples of special things about Russian?

Does a potentiometer need a resistor in series between power and ground?

How to buffer an ATMEGA for millisecond-level power interruption without strange behavior?Can you damage an iPhone by charging it using a (homemade) solar panel?Detect Arduino power failure and save dataPull up does not workControlling 90 LEDs using one WS2811Basic N-MOSFET Circuit Simulation ResultsHelp with basic transimpedance amplifier circuitSD card initialisation problem with Atmega32u4How to find current through series of potentiometers, and if it is safe for potentiometers? (Arduino)LDR breaks over time what to do?

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

1

$begingroup$

I have a potentiometer I use with a circuit to get data readings. This circuit is connected to an arduino mega 2560. I was wondering if I need a resistor in series with the potentiometer? Attached are two picture of what I mean, sorry if the circuits are a bit rough first time I have used an online schematic maker. I would also like to add that the potentiometer values are likely wrong, it was a generic one a friend gave me. All I know is that it goes from 5 ohm to 10k ohm.

arduino voltage

share|improve this question

edited 8 hours ago

Huisman

3,92722 gold badges44 silver badges2727 bronze badges

asked 9 hours ago

dragonstorm24dragonstorm24

822 bronze badges

New contributor

dragonstorm24 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 | 

1

$begingroup$

I have a potentiometer I use with a circuit to get data readings. This circuit is connected to an arduino mega 2560. I was wondering if I need a resistor in series with the potentiometer? Attached are two picture of what I mean, sorry if the circuits are a bit rough first time I have used an online schematic maker. I would also like to add that the potentiometer values are likely wrong, it was a generic one a friend gave me. All I know is that it goes from 5 ohm to 10k ohm.

arduino voltage

share|improve this question

edited 8 hours ago

Huisman

3,92722 gold badges44 silver badges2727 bronze badges

asked 9 hours ago

dragonstorm24dragonstorm24

822 bronze badges

New contributor

dragonstorm24 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 potentiometer I use with a circuit to get data readings. This circuit is connected to an arduino mega 2560. I was wondering if I need a resistor in series with the potentiometer? Attached are two picture of what I mean, sorry if the circuits are a bit rough first time I have used an online schematic maker. I would also like to add that the potentiometer values are likely wrong, it was a generic one a friend gave me. All I know is that it goes from 5 ohm to 10k ohm.

arduino voltage

share|improve this question

edited 8 hours ago

Huisman

3,92722 gold badges44 silver badges2727 bronze badges

asked 9 hours ago

dragonstorm24dragonstorm24

822 bronze badges

New contributor

dragonstorm24 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 8 hours ago

Huisman

3,92722 gold badges44 silver badges2727 bronze badges

asked 9 hours ago

dragonstorm24dragonstorm24

822 bronze badges

New contributor

dragonstorm24 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 8 hours ago

Huisman

3,92722 gold badges44 silver badges2727 bronze badges

asked 9 hours ago

dragonstorm24dragonstorm24

822 bronze badges

New contributor

dragonstorm24 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 8 hours ago

Huisman

3,92722 gold badges44 silver badges2727 bronze badges

edited 8 hours ago

Huisman

3,92722 gold badges44 silver badges2727 bronze badges

asked 9 hours ago

dragonstorm24dragonstorm24

822 bronze badges

New contributor

dragonstorm24 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 9 hours ago

dragonstorm24dragonstorm24

822 bronze badges

3 Answers
3

active

oldest

votes

4

$begingroup$

First, a couple of CircuitLab tips. Double-click a component to edit its properties. 'R' = rotate, 'H' = horizontal flip. 'V' = vertical flip. Note that when you use the CircuitLab button on the editor toolbar an editable schematic is saved in your post. That makes it easy for us to copy and edit in our answers. You don't need a CircuitLab account, no screengrabs, no image uploads, no background grid.

The important thing is that your potentiometer circuit shares the same ground as your microcontroller. If you leave out the ground connection then you have an open circuit and no current can flow from the potentiometer to the micro.

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. Both circuits share a common ground with the microcontroller.

There is a difference between the two circuits.

• The wiper on Figure 1a can go from 0 V at the bottom to 5 V at the top.


• The wiper on Figure 1b can go from 0 V at the bottom to 5/4 V (1.25 V) at the top because R1 and R3 form a potential divider with the maximum voltage give by $ frac {R3}{R1 + R3} V_2 $.

If your micro's analog input is 0 to 5 V then (a) uses the full scale (typically 1024 counts on a 10-bit ADC). (b) would give a maximum of 1024/4 = 256 counts for 1.25 V.

share|improve this answer

answered 8 hours ago

TransistorTransistor

95k88 gold badges9595 silver badges207207 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your help, I do have a shared ground just didn't know how to draw it online. I'll keep your advice in mind for any future questions.
  $endgroup$
  – dragonstorm24
  8 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

No, the wiper pin is the part of the pot which "sees" a different resistance based on its position. There will always be the max resistance between power and ground pins on the potentiometer.

share|improve this answer

answered 8 hours ago

Tanner OleksiukTanner Oleksiuk

1855 bronze badges

$endgroup$

add a comment | 

0

$begingroup$

No, you don't need a resistor between the positive side of the potentiometer and the power, assuming that it's the same voltage level as the arduino runs on.

But, taking your circuits litterally, note that you do need a ground connection between the battery/potentiometer and the arduino.

share|improve this answer

answered 8 hours ago

Wouter van OoijenWouter van Ooijen

45.1k11 gold badge5252 silver badges121121 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The arduino provides both the power and the ground I just did a bad job of drawing out the circuit.
  $endgroup$
  – dragonstorm24
  8 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
});


}
});

dragonstorm24 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%2f447924%2fdoes-a-potentiometer-need-a-resistor-in-series-between-power-and-ground%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

4

$begingroup$

First, a couple of CircuitLab tips. Double-click a component to edit its properties. 'R' = rotate, 'H' = horizontal flip. 'V' = vertical flip. Note that when you use the CircuitLab button on the editor toolbar an editable schematic is saved in your post. That makes it easy for us to copy and edit in our answers. You don't need a CircuitLab account, no screengrabs, no image uploads, no background grid.

The important thing is that your potentiometer circuit shares the same ground as your microcontroller. If you leave out the ground connection then you have an open circuit and no current can flow from the potentiometer to the micro.

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. Both circuits share a common ground with the microcontroller.

There is a difference between the two circuits.

• The wiper on Figure 1a can go from 0 V at the bottom to 5 V at the top.


• The wiper on Figure 1b can go from 0 V at the bottom to 5/4 V (1.25 V) at the top because R1 and R3 form a potential divider with the maximum voltage give by $ frac {R3}{R1 + R3} V_2 $.

If your micro's analog input is 0 to 5 V then (a) uses the full scale (typically 1024 counts on a 10-bit ADC). (b) would give a maximum of 1024/4 = 256 counts for 1.25 V.

share|improve this answer

answered 8 hours ago

TransistorTransistor

95k88 gold badges9595 silver badges207207 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your help, I do have a shared ground just didn't know how to draw it online. I'll keep your advice in mind for any future questions.
  $endgroup$
  – dragonstorm24
  8 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

First, a couple of CircuitLab tips. Double-click a component to edit its properties. 'R' = rotate, 'H' = horizontal flip. 'V' = vertical flip. Note that when you use the CircuitLab button on the editor toolbar an editable schematic is saved in your post. That makes it easy for us to copy and edit in our answers. You don't need a CircuitLab account, no screengrabs, no image uploads, no background grid.

The important thing is that your potentiometer circuit shares the same ground as your microcontroller. If you leave out the ground connection then you have an open circuit and no current can flow from the potentiometer to the micro.

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. Both circuits share a common ground with the microcontroller.

There is a difference between the two circuits.

• The wiper on Figure 1a can go from 0 V at the bottom to 5 V at the top.


• The wiper on Figure 1b can go from 0 V at the bottom to 5/4 V (1.25 V) at the top because R1 and R3 form a potential divider with the maximum voltage give by $ frac {R3}{R1 + R3} V_2 $.

If your micro's analog input is 0 to 5 V then (a) uses the full scale (typically 1024 counts on a 10-bit ADC). (b) would give a maximum of 1024/4 = 256 counts for 1.25 V.

share|improve this answer

answered 8 hours ago

TransistorTransistor

95k88 gold badges9595 silver badges207207 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your help, I do have a shared ground just didn't know how to draw it online. I'll keep your advice in mind for any future questions.
  $endgroup$
  – dragonstorm24
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

First, a couple of CircuitLab tips. Double-click a component to edit its properties. 'R' = rotate, 'H' = horizontal flip. 'V' = vertical flip. Note that when you use the CircuitLab button on the editor toolbar an editable schematic is saved in your post. That makes it easy for us to copy and edit in our answers. You don't need a CircuitLab account, no screengrabs, no image uploads, no background grid.

The important thing is that your potentiometer circuit shares the same ground as your microcontroller. If you leave out the ground connection then you have an open circuit and no current can flow from the potentiometer to the micro.

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. Both circuits share a common ground with the microcontroller.

There is a difference between the two circuits.

• The wiper on Figure 1a can go from 0 V at the bottom to 5 V at the top.


• The wiper on Figure 1b can go from 0 V at the bottom to 5/4 V (1.25 V) at the top because R1 and R3 form a potential divider with the maximum voltage give by $ frac {R3}{R1 + R3} V_2 $.

If your micro's analog input is 0 to 5 V then (a) uses the full scale (typically 1024 counts on a 10-bit ADC). (b) would give a maximum of 1024/4 = 256 counts for 1.25 V.

share|improve this answer

answered 8 hours ago

TransistorTransistor

95k88 gold badges9595 silver badges207207 bronze badges

$endgroup$

share|improve this answer

answered 8 hours ago

TransistorTransistor

95k88 gold badges9595 silver badges207207 bronze badges

answered 8 hours ago

TransistorTransistor

95k88 gold badges9595 silver badges207207 bronze badges

answered 8 hours ago

TransistorTransistor

95k88 gold badges9595 silver badges207207 bronze badges

1

$begingroup$

No, the wiper pin is the part of the pot which "sees" a different resistance based on its position. There will always be the max resistance between power and ground pins on the potentiometer.

share|improve this answer

answered 8 hours ago

Tanner OleksiukTanner Oleksiuk

1855 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

No, the wiper pin is the part of the pot which "sees" a different resistance based on its position. There will always be the max resistance between power and ground pins on the potentiometer.

share|improve this answer

answered 8 hours ago

Tanner OleksiukTanner Oleksiuk

1855 bronze badges

$endgroup$

add a comment | 

$begingroup$

No, the wiper pin is the part of the pot which "sees" a different resistance based on its position. There will always be the max resistance between power and ground pins on the potentiometer.

share|improve this answer

answered 8 hours ago

Tanner OleksiukTanner Oleksiuk

1855 bronze badges

$endgroup$

share|improve this answer

answered 8 hours ago

Tanner OleksiukTanner Oleksiuk

1855 bronze badges

answered 8 hours ago

Tanner OleksiukTanner Oleksiuk

1855 bronze badges

answered 8 hours ago

Tanner OleksiukTanner Oleksiuk

1855 bronze badges

0

$begingroup$

No, you don't need a resistor between the positive side of the potentiometer and the power, assuming that it's the same voltage level as the arduino runs on.

But, taking your circuits litterally, note that you do need a ground connection between the battery/potentiometer and the arduino.

share|improve this answer

answered 8 hours ago

Wouter van OoijenWouter van Ooijen

45.1k11 gold badge5252 silver badges121121 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The arduino provides both the power and the ground I just did a bad job of drawing out the circuit.
  $endgroup$
  – dragonstorm24
  8 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

No, you don't need a resistor between the positive side of the potentiometer and the power, assuming that it's the same voltage level as the arduino runs on.

But, taking your circuits litterally, note that you do need a ground connection between the battery/potentiometer and the arduino.

share|improve this answer

answered 8 hours ago

Wouter van OoijenWouter van Ooijen

45.1k11 gold badge5252 silver badges121121 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The arduino provides both the power and the ground I just did a bad job of drawing out the circuit.
  $endgroup$
  – dragonstorm24
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

No, you don't need a resistor between the positive side of the potentiometer and the power, assuming that it's the same voltage level as the arduino runs on.

But, taking your circuits litterally, note that you do need a ground connection between the battery/potentiometer and the arduino.

share|improve this answer

answered 8 hours ago

Wouter van OoijenWouter van Ooijen

45.1k11 gold badge5252 silver badges121121 bronze badges

$endgroup$

share|improve this answer

answered 8 hours ago

Wouter van OoijenWouter van Ooijen

45.1k11 gold badge5252 silver badges121121 bronze badges

answered 8 hours ago

Wouter van OoijenWouter van Ooijen

45.1k11 gold badge5252 silver badges121121 bronze badges

answered 8 hours ago

Wouter van OoijenWouter van Ooijen

45.1k11 gold badge5252 silver badges121121 bronze badges