How can I change step-down my variable input voltage?Lowering voltage linearRGB LED Strip - Variable Voltage Vs. PWMHow do I measure ~16V battery voltage in an ultra low power system?How do I amplify an input signal to match the peak of another signalDual power supply for Op AmpAdvice and critical remarks regarding the manufacturing of a DC-DC step-down converterBetter way to divide voltage for input to an ADC DAQ?Precision voltage dividerHow to calibrate Unipolar ADC in softwarePulling voltage on a “pull-up” input down to match voltage of voltage output from another device
How do I hide Chekhov's Gun?
Min function accepting varying number of arguments in C++17
Instead of Universal Basic Income, why not Universal Basic NEEDS?
How to explain that I do not want to visit a country due to personal safety concern?
Fantasy series where a Vietnam vet is transported to a fantasy land
How do anti-virus programs start at Windows boot?
What's causing this power spike in STM32 low power mode
What approach do we need to follow for projects without a test environment?
How to make healing in an exploration game interesting
What do Xenomorphs eat in the Alien series?
Site Collection Administrator has left the building
It's a yearly task, alright
Could the Saturn V actually have launched astronauts around Venus?
How to deal with taxi scam when on vacation?
At what level can a dragon innately cast its spells?
If I can solve Sudoku can I solve TSP? If yes, how?
A sequence that has integer values for prime indexes only:
Know when to turn notes upside-down(eighth notes, sixteen notes, etc.)
Gravity magic - How does it work?
Recruiter wants very extensive technical details about all of my previous work
Professor being mistaken for a grad student
Why you should not leave finger prints on bulbs and plugs?
Define, (actually define) the "stability" and "energy" of a compound
Can anyone tell me why this program fails?
How can I change step-down my variable input voltage?
Lowering voltage linearRGB LED Strip - Variable Voltage Vs. PWMHow do I measure ~16V battery voltage in an ultra low power system?How do I amplify an input signal to match the peak of another signalDual power supply for Op AmpAdvice and critical remarks regarding the manufacturing of a DC-DC step-down converterBetter way to divide voltage for input to an ADC DAQ?Precision voltage dividerHow to calibrate Unipolar ADC in softwarePulling voltage on a “pull-up” input down to match voltage of voltage output from another device
$begingroup$
Ok so here's the deal:
I have a variable DC Voltage source from 0-10V.
I need to step that down to a variable source of 0-3V.
This 0-3V DC will be fed to an Analog to Digital converter in a microcontroller.
I know I can potentially use a Voltage Divider (using resistors) but apparently, that's not a good solution.
Op-amps don't provide a gain < 1.
So I'm just struggling as to how I can accomplish this.
The microcontroller: https://www.microchip.com/wwwproducts/en/PIC18F47K42
microcontroller voltage power dc variable
edited 3 hours ago
pipe
10.2k42657
asked 18 hours ago
Alee321Alee321
112
New contributor
Alee321 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 |
$begingroup$
Ok so here's the deal:
I have a variable DC Voltage source from 0-10V.
I need to step that down to a variable source of 0-3V.
This 0-3V DC will be fed to an Analog to Digital converter in a microcontroller.
I know I can potentially use a Voltage Divider (using resistors) but apparently, that's not a good solution.
Op-amps don't provide a gain < 1.
So I'm just struggling as to how I can accomplish this.
The microcontroller: https://www.microchip.com/wwwproducts/en/PIC18F47K42
microcontroller voltage power dc variable
edited 3 hours ago
pipe
10.2k42657
asked 18 hours ago
Alee321Alee321
112
New contributor
Alee321 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
How can you use a voltage of 0 to 3V as the supply voltage for a microcontroller? Your description of this doesn't make any sense. You should draw a schematic.
$endgroup$
– Elliot Alderson
18 hours ago
$begingroup$
@ElliotAlderson you're right I'm sorry. It's not to power the microcontroller. The application is to simulate dimming, based on the 0-10V from the DC Source the voltage must be stepped down from 0-3V. This is the voltage range readable by the uC which will determine the duty cycle of a PWM used to dim an LED.
$endgroup$
– Alee321
18 hours ago
2
$begingroup$
A resistive divider is fine then. You might want to put an op-amp voltage follower in between the divider and ADC input to buffer things but it's not always necessary. Search this site. There are many existing answers addressing your question.
$endgroup$
– Toor
18 hours ago
3
$begingroup$
A voltage divider may be a perfectly good solution provided (a) the source impedance isn't too high and (b) you can tolerate the small energy consumption. Your question doesn't include details of either.
$endgroup$
– Transistor
18 hours ago
add a comment |
$begingroup$
Ok so here's the deal:
I have a variable DC Voltage source from 0-10V.
I need to step that down to a variable source of 0-3V.
This 0-3V DC will be fed to an Analog to Digital converter in a microcontroller.
I know I can potentially use a Voltage Divider (using resistors) but apparently, that's not a good solution.
Op-amps don't provide a gain < 1.
So I'm just struggling as to how I can accomplish this.
The microcontroller: https://www.microchip.com/wwwproducts/en/PIC18F47K42
microcontroller voltage power dc variable
edited 3 hours ago
pipe
10.2k42657
asked 18 hours ago
Alee321Alee321
112
New contributor
Alee321 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Ok so here's the deal:
I have a variable DC Voltage source from 0-10V.
I need to step that down to a variable source of 0-3V.
This 0-3V DC will be fed to an Analog to Digital converter in a microcontroller.
I know I can potentially use a Voltage Divider (using resistors) but apparently, that's not a good solution.
Op-amps don't provide a gain < 1.
So I'm just struggling as to how I can accomplish this.
The microcontroller: https://www.microchip.com/wwwproducts/en/PIC18F47K42
microcontroller voltage power dc variable
microcontroller voltage power dc variable
edited 3 hours ago
pipe
10.2k42657
asked 18 hours ago
Alee321Alee321
112
New contributor
Alee321 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
pipe
10.2k42657
asked 18 hours ago
Alee321Alee321
112
New contributor
Alee321 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
pipe
10.2k42657
edited 3 hours ago
pipe
10.2k42657
edited 3 hours ago
pipe
10.2k42657
10.2k42657
asked 18 hours ago
Alee321Alee321
112
New contributor
Alee321 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 18 hours ago
Alee321Alee321
112
asked 18 hours ago
Alee321Alee321
112
112
New contributor
Alee321 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Alee321 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Alee321 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
How can you use a voltage of 0 to 3V as the supply voltage for a microcontroller? Your description of this doesn't make any sense. You should draw a schematic.
$endgroup$
– Elliot Alderson
18 hours ago
$begingroup$
@ElliotAlderson you're right I'm sorry. It's not to power the microcontroller. The application is to simulate dimming, based on the 0-10V from the DC Source the voltage must be stepped down from 0-3V. This is the voltage range readable by the uC which will determine the duty cycle of a PWM used to dim an LED.
$endgroup$
– Alee321
18 hours ago
2
$begingroup$
A resistive divider is fine then. You might want to put an op-amp voltage follower in between the divider and ADC input to buffer things but it's not always necessary. Search this site. There are many existing answers addressing your question.
$endgroup$
– Toor
18 hours ago
3
$begingroup$
A voltage divider may be a perfectly good solution provided (a) the source impedance isn't too high and (b) you can tolerate the small energy consumption. Your question doesn't include details of either.
$endgroup$
– Transistor
18 hours ago
add a comment |
$begingroup$
How can you use a voltage of 0 to 3V as the supply voltage for a microcontroller? Your description of this doesn't make any sense. You should draw a schematic.
$endgroup$
– Elliot Alderson
18 hours ago
$begingroup$
@ElliotAlderson you're right I'm sorry. It's not to power the microcontroller. The application is to simulate dimming, based on the 0-10V from the DC Source the voltage must be stepped down from 0-3V. This is the voltage range readable by the uC which will determine the duty cycle of a PWM used to dim an LED.
$endgroup$
– Alee321
18 hours ago
2
$begingroup$
A resistive divider is fine then. You might want to put an op-amp voltage follower in between the divider and ADC input to buffer things but it's not always necessary. Search this site. There are many existing answers addressing your question.
$endgroup$
– Toor
18 hours ago
3
$begingroup$
A voltage divider may be a perfectly good solution provided (a) the source impedance isn't too high and (b) you can tolerate the small energy consumption. Your question doesn't include details of either.
$endgroup$
– Transistor
18 hours ago
$begingroup$
How can you use a voltage of 0 to 3V as the supply voltage for a microcontroller? Your description of this doesn't make any sense. You should draw a schematic.
$endgroup$
– Elliot Alderson
18 hours ago
$begingroup$
How can you use a voltage of 0 to 3V as the supply voltage for a microcontroller? Your description of this doesn't make any sense. You should draw a schematic.
$endgroup$
– Elliot Alderson
18 hours ago
$begingroup$
@ElliotAlderson you're right I'm sorry. It's not to power the microcontroller. The application is to simulate dimming, based on the 0-10V from the DC Source the voltage must be stepped down from 0-3V. This is the voltage range readable by the uC which will determine the duty cycle of a PWM used to dim an LED.
$endgroup$
– Alee321
18 hours ago
$begingroup$
@ElliotAlderson you're right I'm sorry. It's not to power the microcontroller. The application is to simulate dimming, based on the 0-10V from the DC Source the voltage must be stepped down from 0-3V. This is the voltage range readable by the uC which will determine the duty cycle of a PWM used to dim an LED.
$endgroup$
– Alee321
18 hours ago
2
2
$begingroup$
A resistive divider is fine then. You might want to put an op-amp voltage follower in between the divider and ADC input to buffer things but it's not always necessary. Search this site. There are many existing answers addressing your question.
$endgroup$
– Toor
18 hours ago
$begingroup$
A resistive divider is fine then. You might want to put an op-amp voltage follower in between the divider and ADC input to buffer things but it's not always necessary. Search this site. There are many existing answers addressing your question.
$endgroup$
– Toor
18 hours ago
3
3
$begingroup$
A voltage divider may be a perfectly good solution provided (a) the source impedance isn't too high and (b) you can tolerate the small energy consumption. Your question doesn't include details of either.
$endgroup$
– Transistor
18 hours ago
$begingroup$
A voltage divider may be a perfectly good solution provided (a) the source impedance isn't too high and (b) you can tolerate the small energy consumption. Your question doesn't include details of either.
$endgroup$
– Transistor
18 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
R divider works fine as long as R is not too high.
If you know the conversion rate and Hold Cap acquisition time
EQUATION 36-1: ACQUISITION TIME EXAMPLE in the datasheet provides the formula and example of choosing R values < 10k.
Rs is the source equivalent // resistance of the R divider, R1//R2.
answered 17 hours ago
Sunnyskyguy EE75Sunnyskyguy EE75
68.8k22598
$endgroup$
add a comment |
$begingroup$
A resistor divider works fine as long as the source impedance is low and the ADC impedance is high (compared to the resistors used for the divider).
If your source impedance is high, use an opamp in voltage follower mode (gain = 1) before the divider. If your ADC impedance is low (unlikely) use a voltage follower after the divider.
answered 17 hours ago
evildemonicevildemonic
2,378721
$endgroup$
$begingroup$
ADCs can have low input impedances, because those sample-and-hold circuits have a capacitor that needs to be charged! A tiny one, but it's still there. Too high a resistance can mean you need to sample for longer, slowing your reads, which may or may not be acceptable.
$endgroup$
– Hearth
16 hours ago
add a comment |
$begingroup$
In addition to @Toor suggestion (voltage follower and voltage divider) and in response to OP statement
Op-amps don't provide a gain < 1.
follows my alternative configuration using a single supply difference amplifier. The output is reversed - if you do not mind correct it in software applying $V_o=3 - 0.3V_i$.
answered 16 hours ago
Dirceu Rodrigues JrDirceu Rodrigues Jr
1,838612
$endgroup$
1
$begingroup$
As opposed to my voltage follower suggestion, this one has better common-mode rejection at the expense of inverting the signal.
$endgroup$
– Toor
16 hours ago
$begingroup$
Are those resistor values correct? I think it's supposed to be more along the lines of 6.66K/3.33K, or 10K/5K or something like that.
$endgroup$
– Toor
16 hours ago
$begingroup$
I think that it's correct. Try to apply superposition to inputs 10 V and Vi.
$endgroup$
– Dirceu Rodrigues Jr
15 hours ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
);
);
, "mathjax-editing");
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
);
);
Alee321 is a new contributor. Be nice, and check out our Code of Conduct.
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
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f427305%2fhow-can-i-change-step-down-my-variable-input-voltage%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
R divider works fine as long as R is not too high.
If you know the conversion rate and Hold Cap acquisition time
EQUATION 36-1: ACQUISITION TIME EXAMPLE in the datasheet provides the formula and example of choosing R values < 10k.
Rs is the source equivalent // resistance of the R divider, R1//R2.
answered 17 hours ago
Sunnyskyguy EE75Sunnyskyguy EE75
68.8k22598
$endgroup$
add a comment |
$begingroup$
R divider works fine as long as R is not too high.
If you know the conversion rate and Hold Cap acquisition time
EQUATION 36-1: ACQUISITION TIME EXAMPLE in the datasheet provides the formula and example of choosing R values < 10k.
Rs is the source equivalent // resistance of the R divider, R1//R2.
answered 17 hours ago
Sunnyskyguy EE75Sunnyskyguy EE75
68.8k22598
$endgroup$
add a comment |
$begingroup$
R divider works fine as long as R is not too high.
If you know the conversion rate and Hold Cap acquisition time
EQUATION 36-1: ACQUISITION TIME EXAMPLE in the datasheet provides the formula and example of choosing R values < 10k.
Rs is the source equivalent // resistance of the R divider, R1//R2.
answered 17 hours ago
Sunnyskyguy EE75Sunnyskyguy EE75
68.8k22598
$endgroup$
R divider works fine as long as R is not too high.
If you know the conversion rate and Hold Cap acquisition time
EQUATION 36-1: ACQUISITION TIME EXAMPLE in the datasheet provides the formula and example of choosing R values < 10k.
Rs is the source equivalent // resistance of the R divider, R1//R2.
answered 17 hours ago
Sunnyskyguy EE75Sunnyskyguy EE75
68.8k22598
answered 17 hours ago
Sunnyskyguy EE75Sunnyskyguy EE75
68.8k22598
answered 17 hours ago
Sunnyskyguy EE75Sunnyskyguy EE75
68.8k22598
answered 17 hours ago
Sunnyskyguy EE75Sunnyskyguy EE75
68.8k22598
68.8k22598
add a comment |
add a comment |
$begingroup$
A resistor divider works fine as long as the source impedance is low and the ADC impedance is high (compared to the resistors used for the divider).
If your source impedance is high, use an opamp in voltage follower mode (gain = 1) before the divider. If your ADC impedance is low (unlikely) use a voltage follower after the divider.
answered 17 hours ago
evildemonicevildemonic
2,378721
$endgroup$
$begingroup$
ADCs can have low input impedances, because those sample-and-hold circuits have a capacitor that needs to be charged! A tiny one, but it's still there. Too high a resistance can mean you need to sample for longer, slowing your reads, which may or may not be acceptable.
$endgroup$
– Hearth
16 hours ago
add a comment |
$begingroup$
A resistor divider works fine as long as the source impedance is low and the ADC impedance is high (compared to the resistors used for the divider).
If your source impedance is high, use an opamp in voltage follower mode (gain = 1) before the divider. If your ADC impedance is low (unlikely) use a voltage follower after the divider.
answered 17 hours ago
evildemonicevildemonic
2,378721
$endgroup$
$begingroup$
ADCs can have low input impedances, because those sample-and-hold circuits have a capacitor that needs to be charged! A tiny one, but it's still there. Too high a resistance can mean you need to sample for longer, slowing your reads, which may or may not be acceptable.
$endgroup$
– Hearth
16 hours ago
add a comment |
$begingroup$
A resistor divider works fine as long as the source impedance is low and the ADC impedance is high (compared to the resistors used for the divider).
If your source impedance is high, use an opamp in voltage follower mode (gain = 1) before the divider. If your ADC impedance is low (unlikely) use a voltage follower after the divider.
answered 17 hours ago
evildemonicevildemonic
2,378721
$endgroup$
A resistor divider works fine as long as the source impedance is low and the ADC impedance is high (compared to the resistors used for the divider).
If your source impedance is high, use an opamp in voltage follower mode (gain = 1) before the divider. If your ADC impedance is low (unlikely) use a voltage follower after the divider.
answered 17 hours ago
evildemonicevildemonic
2,378721
answered 17 hours ago
evildemonicevildemonic
2,378721
answered 17 hours ago
evildemonicevildemonic
2,378721
answered 17 hours ago
evildemonicevildemonic
2,378721
2,378721
$begingroup$
ADCs can have low input impedances, because those sample-and-hold circuits have a capacitor that needs to be charged! A tiny one, but it's still there. Too high a resistance can mean you need to sample for longer, slowing your reads, which may or may not be acceptable.
$endgroup$
– Hearth
16 hours ago
add a comment |
$begingroup$
ADCs can have low input impedances, because those sample-and-hold circuits have a capacitor that needs to be charged! A tiny one, but it's still there. Too high a resistance can mean you need to sample for longer, slowing your reads, which may or may not be acceptable.
$endgroup$
– Hearth
16 hours ago
$begingroup$
ADCs can have low input impedances, because those sample-and-hold circuits have a capacitor that needs to be charged! A tiny one, but it's still there. Too high a resistance can mean you need to sample for longer, slowing your reads, which may or may not be acceptable.
$endgroup$
– Hearth
16 hours ago
$begingroup$
ADCs can have low input impedances, because those sample-and-hold circuits have a capacitor that needs to be charged! A tiny one, but it's still there. Too high a resistance can mean you need to sample for longer, slowing your reads, which may or may not be acceptable.
$endgroup$
– Hearth
16 hours ago
add a comment |
$begingroup$
In addition to @Toor suggestion (voltage follower and voltage divider) and in response to OP statement
Op-amps don't provide a gain < 1.
follows my alternative configuration using a single supply difference amplifier. The output is reversed - if you do not mind correct it in software applying $V_o=3 - 0.3V_i$.
answered 16 hours ago
Dirceu Rodrigues JrDirceu Rodrigues Jr
1,838612
$endgroup$
1
$begingroup$
As opposed to my voltage follower suggestion, this one has better common-mode rejection at the expense of inverting the signal.
$endgroup$
– Toor
16 hours ago
$begingroup$
Are those resistor values correct? I think it's supposed to be more along the lines of 6.66K/3.33K, or 10K/5K or something like that.
$endgroup$
– Toor
16 hours ago
$begingroup$
I think that it's correct. Try to apply superposition to inputs 10 V and Vi.
$endgroup$
– Dirceu Rodrigues Jr
15 hours ago
add a comment |
$begingroup$
In addition to @Toor suggestion (voltage follower and voltage divider) and in response to OP statement
Op-amps don't provide a gain < 1.
follows my alternative configuration using a single supply difference amplifier. The output is reversed - if you do not mind correct it in software applying $V_o=3 - 0.3V_i$.
answered 16 hours ago
Dirceu Rodrigues JrDirceu Rodrigues Jr
1,838612
$endgroup$
1
$begingroup$
As opposed to my voltage follower suggestion, this one has better common-mode rejection at the expense of inverting the signal.
$endgroup$
– Toor
16 hours ago
$begingroup$
Are those resistor values correct? I think it's supposed to be more along the lines of 6.66K/3.33K, or 10K/5K or something like that.
$endgroup$
– Toor
16 hours ago
$begingroup$
I think that it's correct. Try to apply superposition to inputs 10 V and Vi.
$endgroup$
– Dirceu Rodrigues Jr
15 hours ago
add a comment |
$begingroup$
In addition to @Toor suggestion (voltage follower and voltage divider) and in response to OP statement
Op-amps don't provide a gain < 1.
follows my alternative configuration using a single supply difference amplifier. The output is reversed - if you do not mind correct it in software applying $V_o=3 - 0.3V_i$.
answered 16 hours ago
Dirceu Rodrigues JrDirceu Rodrigues Jr
1,838612
$endgroup$
In addition to @Toor suggestion (voltage follower and voltage divider) and in response to OP statement
Op-amps don't provide a gain < 1.
follows my alternative configuration using a single supply difference amplifier. The output is reversed - if you do not mind correct it in software applying $V_o=3 - 0.3V_i$.
answered 16 hours ago
Dirceu Rodrigues JrDirceu Rodrigues Jr
1,838612
answered 16 hours ago
Dirceu Rodrigues JrDirceu Rodrigues Jr
1,838612
answered 16 hours ago
Dirceu Rodrigues JrDirceu Rodrigues Jr
1,838612
answered 16 hours ago
Dirceu Rodrigues JrDirceu Rodrigues Jr
1,838612
1,838612
1
$begingroup$
As opposed to my voltage follower suggestion, this one has better common-mode rejection at the expense of inverting the signal.
$endgroup$
– Toor
16 hours ago
$begingroup$
Are those resistor values correct? I think it's supposed to be more along the lines of 6.66K/3.33K, or 10K/5K or something like that.
$endgroup$
– Toor
16 hours ago
$begingroup$
I think that it's correct. Try to apply superposition to inputs 10 V and Vi.
$endgroup$
– Dirceu Rodrigues Jr
15 hours ago
add a comment |
1
$begingroup$
As opposed to my voltage follower suggestion, this one has better common-mode rejection at the expense of inverting the signal.
$endgroup$
– Toor
16 hours ago
$begingroup$
Are those resistor values correct? I think it's supposed to be more along the lines of 6.66K/3.33K, or 10K/5K or something like that.
$endgroup$
– Toor
16 hours ago
$begingroup$
I think that it's correct. Try to apply superposition to inputs 10 V and Vi.
$endgroup$
– Dirceu Rodrigues Jr
15 hours ago
1
1
$begingroup$
As opposed to my voltage follower suggestion, this one has better common-mode rejection at the expense of inverting the signal.
$endgroup$
– Toor
16 hours ago
$begingroup$
As opposed to my voltage follower suggestion, this one has better common-mode rejection at the expense of inverting the signal.
$endgroup$
– Toor
16 hours ago
$begingroup$
Are those resistor values correct? I think it's supposed to be more along the lines of 6.66K/3.33K, or 10K/5K or something like that.
$endgroup$
– Toor
16 hours ago
$begingroup$
Are those resistor values correct? I think it's supposed to be more along the lines of 6.66K/3.33K, or 10K/5K or something like that.
$endgroup$
– Toor
16 hours ago
$begingroup$
I think that it's correct. Try to apply superposition to inputs 10 V and Vi.
$endgroup$
– Dirceu Rodrigues Jr
15 hours ago
$begingroup$
I think that it's correct. Try to apply superposition to inputs 10 V and Vi.
$endgroup$
– Dirceu Rodrigues Jr
15 hours ago
add a comment |
Alee321 is a new contributor. Be nice, and check out our Code of Conduct.
Alee321 is a new contributor. Be nice, and check out our Code of Conduct.
Alee321 is a new contributor. Be nice, and check out our Code of Conduct.
Alee321 is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Electrical Engineering Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f427305%2fhow-can-i-change-step-down-my-variable-input-voltage%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
How can you use a voltage of 0 to 3V as the supply voltage for a microcontroller? Your description of this doesn't make any sense. You should draw a schematic.
$endgroup$
– Elliot Alderson
18 hours ago
$begingroup$
@ElliotAlderson you're right I'm sorry. It's not to power the microcontroller. The application is to simulate dimming, based on the 0-10V from the DC Source the voltage must be stepped down from 0-3V. This is the voltage range readable by the uC which will determine the duty cycle of a PWM used to dim an LED.
$endgroup$
– Alee321
18 hours ago
2
$begingroup$
A resistive divider is fine then. You might want to put an op-amp voltage follower in between the divider and ADC input to buffer things but it's not always necessary. Search this site. There are many existing answers addressing your question.
$endgroup$
– Toor
18 hours ago
3
$begingroup$
A voltage divider may be a perfectly good solution provided (a) the source impedance isn't too high and (b) you can tolerate the small energy consumption. Your question doesn't include details of either.
$endgroup$
– Transistor
18 hours ago