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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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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
edited 3 hours ago
Michel Keijzers
6,69441938
asked 3 hours ago
TolureTolure
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Tolure is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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
edited 3 hours ago
Michel Keijzers
6,69441938
asked 3 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
edited 3 hours ago
Michel Keijzers
6,69441938
asked 3 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.
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
potentiometer
edited 3 hours ago
Michel Keijzers
6,69441938
asked 3 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 3 hours ago
Michel Keijzers
6,69441938
asked 3 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 3 hours ago
Michel Keijzers
6,69441938
edited 3 hours ago
Michel Keijzers
6,69441938
edited 3 hours ago
Michel Keijzers
6,69441938
6,69441938
asked 3 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 3 hours ago
TolureTolure
1083
asked 3 hours ago
TolureTolure
1083
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.
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.
Tolure is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
add a comment |
3 Answers
3
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oldest
votes
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.
answered 3 hours ago
Majenko♦Majenko
68.1k43277
Would this work + ---[10K]--[connection to wiper]--[100K]--- -
– Tolure
3 hours ago
It may. It depends if that is the resistor ratio that you need. You will have to try it and see. If it's not right then change the resistors. But that arrangement of resistors is the right idea, yes.
– Majenko♦
1 hour ago
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):
answered 3 hours ago
Michel KeijzersMichel Keijzers
6,69441938
add a comment |
Disregard. My reply didn't add anything useful. I'm glad to see this forum is so watchful regarding the quality of answers and help.
answered 2 hours ago
John P.John P.
81
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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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3 Answers
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active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
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active
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votes
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.
answered 3 hours ago
Majenko♦Majenko
68.1k43277
Would this work + ---[10K]--[connection to wiper]--[100K]--- -
– Tolure
3 hours ago
It may. It depends if that is the resistor ratio that you need. You will have to try it and see. If it's not right then change the resistors. But that arrangement of resistors is the right idea, yes.
– Majenko♦
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.
answered 3 hours ago
Majenko♦Majenko
68.1k43277
Would this work + ---[10K]--[connection to wiper]--[100K]--- -
– Tolure
3 hours ago
It may. It depends if that is the resistor ratio that you need. You will have to try it and see. If it's not right then change the resistors. But that arrangement of resistors is the right idea, yes.
– Majenko♦
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.
answered 3 hours ago
Majenko♦Majenko
68.1k43277
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.
answered 3 hours ago
Majenko♦Majenko
68.1k43277
answered 3 hours ago
Majenko♦Majenko
68.1k43277
answered 3 hours ago
Majenko♦Majenko
68.1k43277
answered 3 hours ago
Majenko♦Majenko
68.1k43277
68.1k43277
Would this work + ---[10K]--[connection to wiper]--[100K]--- -
– Tolure
3 hours ago
It may. It depends if that is the resistor ratio that you need. You will have to try it and see. If it's not right then change the resistors. But that arrangement of resistors is the right idea, yes.
– Majenko♦
1 hour ago
add a comment |
Would this work + ---[10K]--[connection to wiper]--[100K]--- -
– Tolure
3 hours ago
It may. It depends if that is the resistor ratio that you need. You will have to try it and see. If it's not right then change the resistors. But that arrangement of resistors is the right idea, yes.
– Majenko♦
1 hour ago
Would this work + ---[10K]--[connection to wiper]--[100K]--- -
– Tolure
3 hours ago
Would this work + ---[10K]--[connection to wiper]--[100K]--- -
– Tolure
3 hours ago
It may. It depends if that is the resistor ratio that you need. You will have to try it and see. If it's not right then change the resistors. But that arrangement of resistors is the right idea, yes.
– Majenko♦
1 hour ago
It may. It depends if that is the resistor ratio that you need. You will have to try it and see. If it's not right then change the resistors. But that arrangement of resistors is the right idea, yes.
– Majenko♦
1 hour ago
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):
answered 3 hours 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):
answered 3 hours 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):
answered 3 hours ago
Michel KeijzersMichel Keijzers
6,69441938
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):
answered 3 hours ago
Michel KeijzersMichel Keijzers
6,69441938
answered 3 hours ago
Michel KeijzersMichel Keijzers
6,69441938
answered 3 hours ago
Michel KeijzersMichel Keijzers
6,69441938
answered 3 hours ago
Michel KeijzersMichel Keijzers
6,69441938
6,69441938
add a comment |
add a comment |
Disregard. My reply didn't add anything useful. I'm glad to see this forum is so watchful regarding the quality of answers and help.
answered 2 hours ago
John P.John P.
81
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John P. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Disregard. My reply didn't add anything useful. I'm glad to see this forum is so watchful regarding the quality of answers and help.
answered 2 hours ago
John P.John P.
81
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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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Disregard. My reply didn't add anything useful. I'm glad to see this forum is so watchful regarding the quality of answers and help.
answered 2 hours ago
John P.John P.
81
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.
Disregard. My reply didn't add anything useful. I'm glad to see this forum is so watchful regarding the quality of answers and help.
answered 2 hours ago
John P.John P.
81
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John P. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 min ago
answered 2 hours ago
John P.John P.
81
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answered 2 hours ago
John P.John P.
81
answered 2 hours ago
John P.John P.
81
81
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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.
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 |
add a comment |
Tolure is a new contributor. Be nice, and check out our Code of Conduct.
Tolure is a new contributor. Be nice, and check out our Code of Conduct.
Tolure is a new contributor. Be nice, and check out our Code of Conduct.
Tolure is a new contributor. Be nice, and check out our Code of Conduct.
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