# Voltage control 0-10V

Discussion in 'General Electronics Chat' started by johnnyhandsome, Jan 19, 2016.

1. ### johnnyhandsome Thread Starter New Member

Jan 19, 2016
8
1
Hey all!

I'm trying to create a circuit with witch i can control the voltage and change it from 0-10v linearley!

first idea was a simple voltage divider but that will give me a curved output voltage.
the whole idea is to pull a wire that rotates a variable resistor to increase the voltage (and opposite)

Any ideas?

// Johnny

Jul 18, 2013
10,827
2,495
Sounds like you have the answer, rotate a linear pot?
Or do you want to move the wire in a linear fashion?
Max.

3. ### johnnyhandsome Thread Starter New Member

Jan 19, 2016
8
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I'd like the voltage output to be linear. Just using voltage divider vill give this equation: Vout=(Vin*R1)/(R1+R2) And lets say that either R1 or R2 is variable the equation is rational and will give me a plotted curve instead of a line

Last edited: Jan 19, 2016
4. ### ScottWang Moderator

Aug 23, 2012
4,924
777
What's the purpose for that 0~10V power supply?
Why you need it from 0V~...?
If you just using voltage divider, does the current enough?
Or maybe you can using LM317 and in series with 3 rectifier diodes on the output pin.

5. ### johnnyhandsome Thread Starter New Member

Jan 19, 2016
8
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Voltage divider*****

6. ### johnnyhandsome Thread Starter New Member

Jan 19, 2016
8
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Purpose is to send the voltage signal into compact-rio convert to digital signal and run an Engine. The given task was to make the voltage signal linear. Current are not gonna be much of a problem i was told.
0v is to go from no throttle to 10v(full throttle)

Jul 18, 2013
10,827
2,495
I get (I think) you have a 0-10v analogue input to the RIO but what do you want to control WITH?
Manual? a Mechanism? a MCU?
Max.

8. ### johnnyhandsome Thread Starter New Member

Jan 19, 2016
8
1
I Have a 12V Battery. I want to be able to send a voltage signal varying linearly between 0-10v by turning potentiometer. 50%turn on the pot gives me 5v. 75%turn on the pot gives me 7,5V. Thats about it

Actually it could vary between 0-12v as well. the issue is to make linear.

Jul 18, 2013
10,827
2,495
So you need a stepper motor on the Pot shaft this will give you 1.8° resolution gearing will make it finer.
A PMW open loop DC motor may not be precise enough.
Max.

10. ### johnnyhandsome Thread Starter New Member

Jan 19, 2016
8
1
My plan was to adjust the pot by hand. Don't think you really get me. the potentiometer is linear so it doesn't matter how fine or fast i turn it. The problem is to get an output voltage looking like y=kx+0-400x355.png not image001.gif

Should i use a LT3080 with a pot on the adjust leg? will it be linear? maybe some condensers to make it more smooth?

11. ### ScottWang Moderator

Aug 23, 2012
4,924
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If you want to use the voltage as A to D and current is not important then using a rail to rail op amp as a voltage follower, and using a 2K resistor in series with a 10K pot connecting to +12V and gnd , and pin 2 of 10K pot connecting to the input of voltage follower.

(+12V) → 2K → 10K pot → gnd
Pin 2 of pot → input of voltage follower.

12. ### GopherT AAC Fanatic!

Nov 23, 2012
6,300
4,020
I think you are confusing a potentiometer and a potentiometer wired as a rheostat (variable resistor).

A true potentiometer is a fixed (lets say, 10K pot connecting 0 to 12V. Then you have a wiper that slides from the zero point to the 12v point. Any place the wiper is located, you get the output as a/10k. The total sum of resistance above plus below the wiper is always a constant so no curve.

A variable resistor (rheostat) can be made by connecting the wiper of a potentiometer to one leg and essentially by-passing part of the resistance of the potentiometer. Coupling this with a fixed resistor makes a potentiometer of sorts - not a good one.

So, wire a potentiometer across your voltage supply and the wiper will give you the linear voltage reference you are looking for. This solution will be particularly precise if there is no load (current flow) on the wiper. Sending it to the input of an ADC for an arduino or other high input impedance chip will do the trick. What are you connecting to? The higher the current flow out of the wiper, the less accuracy you will see (and less linear your voltage vs. wiper position).

Last edited: Jan 19, 2016
13. ### crutschow Expert

Mar 14, 2008
13,472
3,361
Gopher T is correct.
You need to connect the pot in a potentiometer configuration, not a rheostat one.
That will give you the linear voltage with rotation that you want.

johnnyhandsome likes this.
14. ### WBahn Moderator

Mar 31, 2012
18,079
4,917
Using a potentiometer that is turn some fraction of its travel, say x (which varies from 0.0 to 1.0) your two resistors in your voltage divider are Rpot·x and Rpot·(1-x).

Vout = Vcc · (Rpot·x) / [Rpot(x) + Rpot(1-x)] = x·Vcc

Seems pretty linear with respect to the pot position.

#12 likes this.
15. ### johnnyhandsome Thread Starter New Member

Jan 19, 2016
8
1
Thats way to easy you are right and i made it more difficult than it should be. thank you!
I think that we are connecting the signal to a NI 9401 module but I'm not totally sure

GopherT likes this.
16. ### Darrell Teegarden New Member

Sep 8, 2015
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mcgyvr and #12 like this.
17. ### hp1729 Well-Known Member

Nov 23, 2015
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Not just a pot? Pot with op amp buffer on the output?

18. ### ScottWang Moderator

Aug 23, 2012
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777
He seems didn't want that buffer, did you see my posted on #11.
I even thought about to protects the ADC when its input voltage exceed 10V could be damaged itself and I used 2K as the voltage divider to make sure the input keep less than 10V.

19. ### hp1729 Well-Known Member

Nov 23, 2015
2,083
232
I don't remember the LT3080 going up to 10 Volts. Check out the data sheet carefully. The idea is good. An op amp with a buffer transistor on the output configured as a buffer. How much current do you need?

20. ### johnnyhandsome Thread Starter New Member

Jan 19, 2016
8
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The currents I need should not be to large so it would damage the compact rio, and not to small so it will be disturbed by the engine and other things I would say max is about 200mA. No need to burn to much power.

Simplest idea is now to have 12v battery connected to a resistor(voltage drop 2v) resistor connected with a pot parallel with a resistor. And my signal is from the tap on the pot. With this I can regulate the voltage from 0-10v and not risking a to high current. I'll post the circuit later. But all ideas are welcome!