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How can I change step-down my variable input voltage? [Microcontroller]

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How can I change step-down my variable input voltage? [Microcontroller]

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1

$begingroup$

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

Thanks!

microcontroller voltage power dc variable

share|improve this question

edited 5 hours ago

Alee321

asked 5 hours ago

Alee321Alee321

62

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
  5 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
  5 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
  5 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  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
  5 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

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

Thanks!

microcontroller voltage power dc variable

share|improve this question

edited 5 hours ago

Alee321

asked 5 hours ago

Alee321Alee321

62

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
  5 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
  5 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
  5 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  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
  5 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

Thanks!

microcontroller voltage power dc variable

share|improve this question

edited 5 hours ago

Alee321

asked 5 hours ago

Alee321Alee321

62

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.

$endgroup$

share|improve this question

edited 5 hours ago

Alee321

asked 5 hours ago

Alee321Alee321

62

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.

share|improve this question

edited 5 hours ago

Alee321

asked 5 hours ago

Alee321Alee321

62

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 5 hours ago

Alee321Alee321

62

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 5 hours ago

Alee321Alee321

62

3 Answers
3

active

oldest

votes

2

$begingroup$

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.

share|improve this answer

answered 4 hours ago

Sunnyskyguy EE75Sunnyskyguy EE75

68.7k22598

$endgroup$

add a comment | 

1

$begingroup$

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.

share|improve this answer

answered 4 hours ago

evildemonicevildemonic

2,368721

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  3 hours ago
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

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 $V_o=3 - 0.3V_i$.

share|improve this answer

answered 3 hours ago

Dirceu Rodrigues JrDirceu Rodrigues Jr

1,828612

$endgroup$

• 
  
  
  

  
  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
  3 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
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think that it's correct. Try to apply superposition to inputs 10 V and Vi.
  $endgroup$
  – Dirceu Rodrigues Jr
  2 hours ago
  

  
  
  
  

add a comment | 

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2

$begingroup$

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.

share|improve this answer

answered 4 hours ago

Sunnyskyguy EE75Sunnyskyguy EE75

68.7k22598

$endgroup$

add a comment | 

2

$begingroup$

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.

share|improve this answer

answered 4 hours ago

Sunnyskyguy EE75Sunnyskyguy EE75

68.7k22598

$endgroup$

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.

share|improve this answer

answered 4 hours ago

Sunnyskyguy EE75Sunnyskyguy EE75

68.7k22598

$endgroup$

share|improve this answer

answered 4 hours ago

Sunnyskyguy EE75Sunnyskyguy EE75

68.7k22598

answered 4 hours ago

Sunnyskyguy EE75Sunnyskyguy EE75

68.7k22598

answered 4 hours ago

Sunnyskyguy EE75Sunnyskyguy EE75

68.7k22598

1

$begingroup$

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.

share|improve this answer

answered 4 hours ago

evildemonicevildemonic

2,368721

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  3 hours ago
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

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.

share|improve this answer

answered 4 hours ago

evildemonicevildemonic

2,368721

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  3 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.

share|improve this answer

answered 4 hours ago

evildemonicevildemonic

2,368721

$endgroup$

share|improve this answer

answered 4 hours ago

evildemonicevildemonic

2,368721

answered 4 hours ago

evildemonicevildemonic

2,368721

answered 4 hours ago

evildemonicevildemonic

2,368721

0

$begingroup$

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 $V_o=3 - 0.3V_i$.

share|improve this answer

answered 3 hours ago

Dirceu Rodrigues JrDirceu Rodrigues Jr

1,828612

$endgroup$

• 
  
  
  

  
  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
  3 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
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think that it's correct. Try to apply superposition to inputs 10 V and Vi.
  $endgroup$
  – Dirceu Rodrigues Jr
  2 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

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 $V_o=3 - 0.3V_i$.

share|improve this answer

answered 3 hours ago

Dirceu Rodrigues JrDirceu Rodrigues Jr

1,828612

$endgroup$

• 
  
  
  

  
  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
  3 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
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think that it's correct. Try to apply superposition to inputs 10 V and Vi.
  $endgroup$
  – Dirceu Rodrigues Jr
  2 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 $V_o=3 - 0.3V_i$.

share|improve this answer

answered 3 hours ago

Dirceu Rodrigues JrDirceu Rodrigues Jr

1,828612

$endgroup$

share|improve this answer

answered 3 hours ago

Dirceu Rodrigues JrDirceu Rodrigues Jr

1,828612

answered 3 hours ago

Dirceu Rodrigues JrDirceu Rodrigues Jr

1,828612

answered 3 hours ago

Dirceu Rodrigues JrDirceu Rodrigues Jr

1,828612