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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

2

$begingroup$

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

share|improve this question

edited Mar 15 at 12:45

pipe

10.2k42658

asked Mar 14 at 22:00

Alee321Alee321

112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 14 at 22:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  Mar 14 at 22:10
  
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 14 at 22:12
  
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 14 at 22:12
  

  
  
  
  

add a comment | 

2

$begingroup$

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

share|improve this question

edited Mar 15 at 12:45

pipe

10.2k42658

asked Mar 14 at 22:00

Alee321Alee321

112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 14 at 22:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  Mar 14 at 22:10
  
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 14 at 22:12
  
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 14 at 22:12
  

  
  
  
  

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

share|improve this question

edited Mar 15 at 12:45

pipe

10.2k42658

asked Mar 14 at 22:00

Alee321Alee321

112

$endgroup$

share|improve this question

edited Mar 15 at 12:45

pipe

10.2k42658

asked Mar 14 at 22:00

Alee321Alee321

112

share|improve this question

edited Mar 15 at 12:45

pipe

10.2k42658

asked Mar 14 at 22:00

Alee321Alee321

112

edited Mar 15 at 12:45

pipe

10.2k42658

edited Mar 15 at 12:45

pipe

10.2k42658

asked Mar 14 at 22:00

Alee321Alee321

112

asked Mar 14 at 22:00

Alee321Alee321

112

3 Answers
3

active

oldest

votes

3

$begingroup$

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 Mar 14 at 22:55

Sunnyskyguy EE75Sunnyskyguy EE75

69.2k22598

$endgroup$

add a comment | 

2

$begingroup$

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 Mar 14 at 22:26

evildemonicevildemonic

2,478822

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 14 at 23:33
  
  

  
  
  
  

add a comment | 

1

$begingroup$

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 Mar 14 at 23:50

Dirceu Rodrigues JrDirceu Rodrigues Jr

1,868612

$endgroup$

• 
  
  
  

  
  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
  Mar 14 at 23:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 14 at 23:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think that it's correct. Try to apply superposition to inputs 10 V and Vi.
  $endgroup$
  – Dirceu Rodrigues Jr
  Mar 15 at 0:18
  

  
  
  
  

add a comment | 

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3

$begingroup$

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 Mar 14 at 22:55

Sunnyskyguy EE75Sunnyskyguy EE75

69.2k22598

$endgroup$

add a comment | 

3

$begingroup$

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 Mar 14 at 22:55

Sunnyskyguy EE75Sunnyskyguy EE75

69.2k22598

$endgroup$

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 Mar 14 at 22:55

Sunnyskyguy EE75Sunnyskyguy EE75

69.2k22598

$endgroup$

share|improve this answer

answered Mar 14 at 22:55

Sunnyskyguy EE75Sunnyskyguy EE75

69.2k22598

answered Mar 14 at 22:55

Sunnyskyguy EE75Sunnyskyguy EE75

69.2k22598

answered Mar 14 at 22:55

Sunnyskyguy EE75Sunnyskyguy EE75

69.2k22598

2

$begingroup$

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 Mar 14 at 22:26

evildemonicevildemonic

2,478822

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 14 at 23:33
  
  

  
  
  
  

add a comment | 

2

$begingroup$

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 Mar 14 at 22:26

evildemonicevildemonic

2,478822

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 14 at 23:33
  
  

  
  
  
  

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 Mar 14 at 22:26

evildemonicevildemonic

2,478822

$endgroup$

share|improve this answer

answered Mar 14 at 22:26

evildemonicevildemonic

2,478822

answered Mar 14 at 22:26

evildemonicevildemonic

2,478822

answered Mar 14 at 22:26

evildemonicevildemonic

2,478822

1

$begingroup$

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 Mar 14 at 23:50

Dirceu Rodrigues JrDirceu Rodrigues Jr

1,868612

$endgroup$

• 
  
  
  

  
  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
  Mar 14 at 23:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 14 at 23:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think that it's correct. Try to apply superposition to inputs 10 V and Vi.
  $endgroup$
  – Dirceu Rodrigues Jr
  Mar 15 at 0:18
  

  
  
  
  

add a comment | 

1

$begingroup$

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 Mar 14 at 23:50

Dirceu Rodrigues JrDirceu Rodrigues Jr

1,868612

$endgroup$

• 
  
  
  

  
  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
  Mar 14 at 23:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 14 at 23:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think that it's correct. Try to apply superposition to inputs 10 V and Vi.
  $endgroup$
  – Dirceu Rodrigues Jr
  Mar 15 at 0:18
  

  
  
  
  

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 Mar 14 at 23:50

Dirceu Rodrigues JrDirceu Rodrigues Jr

1,868612

$endgroup$

share|improve this answer

answered Mar 14 at 23:50

Dirceu Rodrigues JrDirceu Rodrigues Jr

1,868612

answered Mar 14 at 23:50

Dirceu Rodrigues JrDirceu Rodrigues Jr

1,868612

answered Mar 14 at 23:50

Dirceu Rodrigues JrDirceu Rodrigues Jr

1,868612