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Trying to use an LCD 1602 to display, but I don't have a 10 KOhm potentiometer

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Trying to use an LCD 1602 to display, but I don't have a 10 KOhm potentiometer

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1

I'm following this tutorial page and I don't have a 10k ohm potentiometer on hand. Could I replace it with resistors to simulate a manual setting and replace the resistors to adjust to a new settings.

From what I was reading this pin is for contrast and I'm sure I could manually set the contrast once and be happy with it for quick prototyping. I'll be ordering a 10k ohm potentiometer on Monday.

potentiometer

share|improve this question

edited 1 hour ago

Michel Keijzers

6,69441938

asked 2 hours ago

TolureTolure

1083

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add a comment | 

1

I'm following this tutorial page and I don't have a 10k ohm potentiometer on hand. Could I replace it with resistors to simulate a manual setting and replace the resistors to adjust to a new settings.

From what I was reading this pin is for contrast and I'm sure I could manually set the contrast once and be happy with it for quick prototyping. I'll be ordering a 10k ohm potentiometer on Monday.

potentiometer

share|improve this question

edited 1 hour ago

Michel Keijzers

6,69441938

asked 2 hours ago

TolureTolure

1083

New contributor

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

add a comment | 

I'm following this tutorial page and I don't have a 10k ohm potentiometer on hand. Could I replace it with resistors to simulate a manual setting and replace the resistors to adjust to a new settings.

From what I was reading this pin is for contrast and I'm sure I could manually set the contrast once and be happy with it for quick prototyping. I'll be ordering a 10k ohm potentiometer on Monday.

potentiometer

share|improve this question

edited 1 hour ago

Michel Keijzers

6,69441938

asked 2 hours ago

TolureTolure

1083

New contributor

Tolure 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 1 hour ago

Michel Keijzers

6,69441938

asked 2 hours ago

TolureTolure

1083

New contributor

Tolure 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 1 hour ago

Michel Keijzers

6,69441938

asked 2 hours ago

TolureTolure

1083

New contributor

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

edited 1 hour ago

Michel Keijzers

6,69441938

edited 1 hour ago

Michel Keijzers

6,69441938

asked 2 hours ago

TolureTolure

1083

New contributor

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

asked 2 hours ago

TolureTolure

1083

3 Answers
3

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2

Yes. All a potentiometer is, is two resistors end to end. The "wiper" is the join between the two resistors.

It doesn't matter too much what value potentiometer (and hence what value resistors) you use, as long as it's "around" the 10kΩ mark (100kΩ should be fine, 50kΩ, 1kΩ etc - anything below 1kΩ may not work) - what matters is the ratio of the values of the resistor.

share|improve this answer

answered 1 hour ago

Majenko♦Majenko

68.1k43277

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Would this work + ---[10K]--[connection to wiper]--[100K]--- -
  
  – Tolure
  1 hour ago
  

  
  
  
  

add a comment | 

1

Yes, you always can do that, depending on the resistors you have at hand, you can combine them to make e.g. 1 KOhm, 2 KOhm etc resistors.

Probably you need a linear potentiometer (which means halfway is 5 KOhm).

In a linear potentiometer, you can use the following resistance (made up by a single or a combination of resistors in series or parallel):

share|improve this answer

answered 1 hour ago

Michel KeijzersMichel Keijzers

6,69441938

add a comment | 

-1

Google "resistor voltage divider" to understand how this potentiometer works. It would be a shame to waste the cost of buying a 10K potentiometer if you don't need it. As other have responded, a pair of fixed resistors would work fine.

share|improve this answer

answered 1 hour ago

John P.John P.

121

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John P. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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2

Yes. All a potentiometer is, is two resistors end to end. The "wiper" is the join between the two resistors.

It doesn't matter too much what value potentiometer (and hence what value resistors) you use, as long as it's "around" the 10kΩ mark (100kΩ should be fine, 50kΩ, 1kΩ etc - anything below 1kΩ may not work) - what matters is the ratio of the values of the resistor.

share|improve this answer

answered 1 hour ago

Majenko♦Majenko

68.1k43277

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Would this work + ---[10K]--[connection to wiper]--[100K]--- -
  
  – Tolure
  1 hour ago
  

  
  
  
  

add a comment | 

2

Yes. All a potentiometer is, is two resistors end to end. The "wiper" is the join between the two resistors.

It doesn't matter too much what value potentiometer (and hence what value resistors) you use, as long as it's "around" the 10kΩ mark (100kΩ should be fine, 50kΩ, 1kΩ etc - anything below 1kΩ may not work) - what matters is the ratio of the values of the resistor.

share|improve this answer

answered 1 hour ago

Majenko♦Majenko

68.1k43277

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Would this work + ---[10K]--[connection to wiper]--[100K]--- -
  
  – Tolure
  1 hour ago
  

  
  
  
  

add a comment | 

Yes. All a potentiometer is, is two resistors end to end. The "wiper" is the join between the two resistors.

It doesn't matter too much what value potentiometer (and hence what value resistors) you use, as long as it's "around" the 10kΩ mark (100kΩ should be fine, 50kΩ, 1kΩ etc - anything below 1kΩ may not work) - what matters is the ratio of the values of the resistor.

share|improve this answer

answered 1 hour ago

Majenko♦Majenko

68.1k43277

share|improve this answer

answered 1 hour ago

Majenko♦Majenko

68.1k43277

answered 1 hour ago

Majenko♦Majenko

68.1k43277

answered 1 hour ago

Majenko♦Majenko

68.1k43277

1

Yes, you always can do that, depending on the resistors you have at hand, you can combine them to make e.g. 1 KOhm, 2 KOhm etc resistors.

Probably you need a linear potentiometer (which means halfway is 5 KOhm).

In a linear potentiometer, you can use the following resistance (made up by a single or a combination of resistors in series or parallel):

share|improve this answer

answered 1 hour ago

Michel KeijzersMichel Keijzers

6,69441938

add a comment | 

1

Yes, you always can do that, depending on the resistors you have at hand, you can combine them to make e.g. 1 KOhm, 2 KOhm etc resistors.

Probably you need a linear potentiometer (which means halfway is 5 KOhm).

In a linear potentiometer, you can use the following resistance (made up by a single or a combination of resistors in series or parallel):

share|improve this answer

answered 1 hour ago

Michel KeijzersMichel Keijzers

6,69441938

add a comment | 

Yes, you always can do that, depending on the resistors you have at hand, you can combine them to make e.g. 1 KOhm, 2 KOhm etc resistors.

Probably you need a linear potentiometer (which means halfway is 5 KOhm).

In a linear potentiometer, you can use the following resistance (made up by a single or a combination of resistors in series or parallel):

share|improve this answer

answered 1 hour ago

Michel KeijzersMichel Keijzers

6,69441938

share|improve this answer

answered 1 hour ago

Michel KeijzersMichel Keijzers

6,69441938

answered 1 hour ago

Michel KeijzersMichel Keijzers

6,69441938

answered 1 hour ago

Michel KeijzersMichel Keijzers

6,69441938

-1

Google "resistor voltage divider" to understand how this potentiometer works. It would be a shame to waste the cost of buying a 10K potentiometer if you don't need it. As other have responded, a pair of fixed resistors would work fine.

share|improve this answer

answered 1 hour ago

John P.John P.

121

New contributor

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

add a comment | 

-1

Google "resistor voltage divider" to understand how this potentiometer works. It would be a shame to waste the cost of buying a 10K potentiometer if you don't need it. As other have responded, a pair of fixed resistors would work fine.

share|improve this answer

answered 1 hour ago

John P.John P.

121

New contributor

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

add a comment | 

Google "resistor voltage divider" to understand how this potentiometer works. It would be a shame to waste the cost of buying a 10K potentiometer if you don't need it. As other have responded, a pair of fixed resistors would work fine.

share|improve this answer

answered 1 hour ago

John P.John P.

121

New contributor

John P. 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 answer

answered 1 hour ago

John P.John P.

121

New contributor

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

answered 1 hour ago

John P.John P.

121

New contributor

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

answered 1 hour ago

John P.John P.

121