Same Value of DC Voltage Drop Achieved for Each Turn of the Rotary Switch

Thread Starter

howie4024795

Joined Jan 30, 2011
10
Hi, I was hoping someone would suggest how to approach this voltage division inquiry with a series circuit of resistors. Each time I turn the rotary switch (12 positions) the output would be 50mV less than the position before.
I was thinking this might be a stepped attenuator.
Or maybe a window comparator as far as the math is concerned?
I see example circuits that monitor the voltage of your car battery and display it on a LM3914 with LED's as indicators. The many trigger voltages are set by the designer with a string of resistors and then sets up a comparator to pit against the trigger points.

For example: I start with 2.0v DC that would be at the top of the series string of resistors.
1st rotary position = 2.000v then next would be 1.950v then 1.900v etc.
In the car battery examples, the current value is not mentioned that I can decipher.
I am hoping the current would not matter with my inquiry because I would like to use this divider on any value of voltage as long as I can drop 50mV. I would calculate resistor wattages when appropriate.
So I would be looking for the last (12th) position on the rotary to output 0v.
Thank you for any suggestions and help you may offer.
 

Papabravo

Joined Feb 24, 2006
13,926
I believe you have too many conditions to satisfy them all simultaneously.
With the 12 position switch if you need one position for 2.0 V and one position for 0 V, then you have ten positions to cover the interval. At 50 mV per interval you don't have enough positions.
 

MrChips

Joined Oct 2, 2009
21,329
A simple solution is a chain of 12 resistors in series, 11 identical resistors and a 12th resistor for the first step,

Select how many amps you want to flow along the resistor chain.
For example, select 1mA.
Hence 11 resistors each have to drop 50mV. R = V / I = 50mV / 1mA = 50Ω

The first resistor has to drop 1.45V. R = V / I = 1.45V / 1mA = 1450Ω

Edit: Just read pb's reply. If you want the last position to be 0V then you need 10 resistors at 50Ω and the first resistor is 1500Ω.
 

Thread Starter

howie4024795

Joined Jan 30, 2011
10
A simple solution is a chain of 12 resistors in series, 11 identical resistors and a 12th resistor for the first step,

Select how many amps you want to flow along the resistor chain.
For example, select 1mA.
Hence 11 resistors each have to drop 50mV. R = V / I = 50mV / 1mA = 50Ω

The first resistor has to drop 1.45V. R = V / I = 1.45V / 1mA = 1450Ω
I will try that now.
If the current changes in the next application I use it on, I assume the 50mV may not be achieved?
 

Thread Starter

howie4024795

Joined Jan 30, 2011
10
I believe you have too many conditions to satisfy them all simultaneously.
With the 12 position switch if you need one position for 2.0 V and one position for 0 V, then you have ten positions to cover the interval. At 50 mV per interval you don't have enough positions.
Thank you. I did not catch that.
if there were enough positions would that change the value of the first resistor?
 

crutschow

Joined Mar 14, 2008
25,119
The step change must be the voltage range divided by the number of steps.
So, for example, with a 0 to 2V range and 12 steps then the voltage step per change would be 2V / 12 steps = 166.7mV per step.
 
Last edited:

MrChips

Joined Oct 2, 2009
21,329
Just remember that 0V counts as one position.
Hence 0-1V in 0.1V steps actually requires 11 positions not 10.
With 12 positions you can have 0-1.1V in 0.1V steps.
 

MrChips

Joined Oct 2, 2009
21,329
One other thing to note.
The input and output impedance will change with each position.
If you want fixed input and output impedance such as in an audio pad attenuator then you need to go with a T-pad or π-pad attenuator.
 

Thread Starter

howie4024795

Joined Jan 30, 2011
10
One other thing to note.
The input and output impedance will change with each position.
If you want fixed input and output impedance such as in an audio pad attenuator then you need to go with a T-pad or π-pad attenuator.
Fixed input impedance would be preferred. Would a buffered op amp take care of the input?
 

AnalogKid

Joined Aug 1, 2013
8,487
What is the minimum voltage you want out of the stepped attenuator?

How many steps total, including the top and bottom positions?

Yes, you will need something to prevent the downstream circuit from loading the attenuator. A classic approach is an opamp configured as a voltage follower. Note that low cost opamps can have 10 mV or more of input offset voltage error, and you want 50 mV per step. That is a possible error of 10% - 20%. Do you have access to (or a budget for) higher grade opamps?

ak
 

MrChips

Joined Oct 2, 2009
21,329
Voltage will be fed to control VCO.
That is not the big picture. What are you building? Is it a music synthesizer, for example?
Why do you need switched positions?
Will a variable pot work for your application?
There is such a thing as a digital potentiometer or you can use a DAC (digital to analog converter).
PWM (pulse-width modulation) can give stepped analog voltages.
You can control the frequency of an oscillator in other ways beside VCO.
PLL (phase-locked loop) is one solution.
MCU (micro-controller unit) is another solution.

I have given you about 7 possible options. We can help you if we only knew what you are trying to build.
 

dendad

Joined Feb 20, 2016
3,564
You may find it better to use an R2R network.
https://en.wikipedia.org/wiki/Resistor_ladder
If it were me, I'd use a rotary encoder, processor and ADC with a display to give much finer steps.
But really, you do need to give us more info. Why is it so hard to get this often?
For your VCO, what frequency range and step size do you need?
It may be a lot better to use a digital generator of some sort, like a DDS or processor driven DAC.
More info PLEASE!!!!
 
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