Multiple voltages question

Thread Starter

Archaeus

Joined Sep 10, 2016
34
Good day,

I was wondering how what the input on an lm339n comparator would read voltage wise if two independent potentiometer wiper voltages were applied. For instance, in the schematic if both are set to 2.5v would the input see 2.5 or 5v? If one was 3v and the other 2v, would it be combined, an average or the largest voltage? Any info is appreciated.

P.S. In actuality, the two potentiometers are a part of an arduino joystick I'd like to configure using a comparator. They have maximum wipe of 5%-95%.

20170428_130529_HDR-1.jpg
 

dl324

Joined Mar 30, 2015
9,626
I was wondering how what the input on an lm339n comparator would read voltage wise if two independent potentiometer wiper voltages were applied.
You need to label the inputs for us to be able to tell you what the output voltage will be. We'll also need the resistances of the two pots; that's an unusual configuration...

The comparator output is open collector, so you need a pull-up resistor to get a HIGH output voltage.
 

Thread Starter

Archaeus

Joined Sep 10, 2016
34
I understand that, but thank you for pointing it out. The output stage is not what I'm wondering about. The Input voltage after combing the two potentiometers with various voltages is what I'm unsure of. What would the input 'see'?
 

dl324

Joined Mar 30, 2015
9,626
The Input voltage after combing the two potentiometers with various voltages is what I'm unsure of. What would the input 'see'?
To determine that, we need to know the resistances and pot wiper positions.

It's a bad idea to do this. Consider the case when the left pot is at the top (5V) and the right is at the bottom (0V); you have a short across the power supply. At least briefly...
 

dendad

Joined Feb 20, 2016
3,261
The input voltage will be dependent on the series parallel combination of all the resistances.You will not be able to separate the two inputs but they will interact to produce a single voltage.
If you have them both at 2.5V (or both any voltage the same), that is what you will get, but it will be more complicated when they are different.
An extreme example is if one is set to 5V and the other is set to 0V, you will get smoke ;)
A series resistor from the wipers will help prevent that.
Why do you want both pots to one output anyway?
 

crutschow

Joined Mar 14, 2008
24,125
The output of the two wipers in parallel is a non-linear function of the relative rotations.
Below is the LTspice simulation for pot U2 at the 10% point (green) and 50% point (yellow) with pot U1 going from the 5% to 95% point (your stated limits).

Note that the pot current can get high if one is at the top extreme and the other is at the bottom extreme, depending upon the pot resistance, so you may want to add some resistance is series with one of the wiper outputs.

upload_2017-4-28_16-8-22.png
 

Thread Starter

Archaeus

Joined Sep 10, 2016
34
They are 10k pots and range from 5% either side. There are no shorts or total resistance. Could you give an example of say 4v and 2v?
 

Thread Starter

Archaeus

Joined Sep 10, 2016
34
I see that. Thanks for the explanation crutschow. I will keep that in mind or reconfigure better to isolate components.
 

Thread Starter

Archaeus

Joined Sep 10, 2016
34
Thank you for the response. The other posts had some more information on the wiper ability. Apparently it does not go from from one extreme to the other.
 

Thread Starter

Archaeus

Joined Sep 10, 2016
34
Forgive me if you thought I was being argumentative. I was not. Simply stating that apparently the wipers are designed to never produce zero resistance. Looks like it is best to redesign regardless. I am still new and was just wondering how that combination would work.
 

crutschow

Joined Mar 14, 2008
24,125
If you truly want the linear sum of the two voltages (or a voltage proportional to that sum) then you could use a summing op amp circuit.
That will prevent any significant interaction between the two outputs.

For example, here's a circuit using one quad RR op amp package to sum the two pot voltages and output 1/2 the sum of their values.
Simulation is for 5% through 95% position of pot U1 for 5%, 50%, and 95% positions of pot U2.

upload_2017-4-29_10-47-42.png
 
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