Designing an adjustable voltage source using only resistors and a pot

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  • Thread starter garthenar
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Homework Statement:

In this project, you will design the simplest form of voltage source with specifications as follow:

a. It should be possible to adjust the voltage between -2V and +2V using one potentiometer. It should not be possible to accidentally obtain a voltage outside this range.

b. The circuit should use as little power as possible, should not exceed 50mA.

c. The default resistance load (RL) is 10MΩ.

Available components:
• Potentiometers: 5KΩ, 10KΩ, 50KΩ
• Assorted resistors with values from 1.1Ω to 10MΩ
• Analog Discovery Power supply: +5V and -5V

Relevant Equations:

Voltage division, KCL, KVL, Node, and Mesh analysis. Those are the tools I have in my toolbox. But right now if feels like I'm trying to hammer in a nail with a screwdriver.
1576162328493.png


I need to create a powersupply that takes in a +5 and -5v and has an adjustable output that varies between -2 and +2 v using one potentiometer.
I already came up with a design myself but it did not work. I need some help figuring this out.

I think my idea for how to get specific positive or negative voltages I want are decent but something is going wrong in the middle. I think I have current going where I dont want it to go. I'd love to use a zenner diode but that is not allowed.

I have attatched my original circuit design.
1576162888563.png

It's my first time using a potentiometer but I was able to adjust the voltage, just not to what I need too.

And once again, Thank you for your time.
 

Answers and Replies

  • #2
BvU
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Hi,

There is a symmetry in the exercise text that I miss in your solution. Can you post your considerations ?

From my end: suppose I want to start with limiting the ##\pm## 5 V to ##\pm## 2 V, what values (value ratios) for the resistors in a simple series circuit are needed ? Combine this with the 50 mA constraint and replace the middle resistor with a potentiometer !

1576164734041.png
 
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  • #3
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The problem is that I need a single output that will range from -2 to +2 v. Unfortunately I don't know how to accomplish that with a single potentiometer. That's why I had that strange setup before. How do I combine the separate voltages into a single output?
 
  • #4
berkeman
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How do I combine the separate voltages into a single output?
Hint1: What if the middle resistance in the diagram by @BvU is a potentiometer? (EDIT -- I see BvU already mentioned that...)

Hint2: Be sure to include the load resistor attached to the wiper of the potentiometer when calculating the values of the pullup and pulldown resistances...
 
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  • #5
BvU
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They are not separate: you can assume both are wrt ground
 
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  • #6
gneill
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Following @BvU 's suggestion (which I heartily support!) Suppose we envision the power supply as follows:

1576205527693.png

When the potentiometer wiper is at the very top (as shown above), we expect +2 V to be applied to the load resistance.

By symmetry, if the wiper were at the bottom of the potentiometer the load resistance would see -2 V, right?

Now, consider that your question requires the total current drawn be less than 50 mA (the question statement mentioned minimal power, yet stated a current value. I'm not sure that I'm happy with that way of presenting such a requirement in a homework question). I propose that you could determine the minimum power required between the choices of potentiometer values. This might require ##I^2 R## calculations for all of the resistances involved. :frown:

From a practical engineering standpoint the 200 nA drawn by the ##10 M\Omega## load resistor at 2 V is essentially insignificant compared to the allowed 50 mA total current supplied by the voltage sources. But if you're required to minimize the power drawn, you may have to consider situations where it's not insignificant if you can make the total current drawn much less than 50 mA. You'll have to see what can be achieved with each choice of potentiometer value.

You have been given several values of potentiometer resistances to select from. This seems to be your determining point. For each of these values can you find values of R that will provide the required load voltage for those pot values? How does the power consumption fair with each pot and R value? Methinks an application of nodal analysis might prove fruitful.
 
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  • #7
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They are not separate: you can assume both are wrt ground
Thank you. It turns out I had the potentiometer wired wrong. I figured out how to use it right after I checked the resistance between each leg. My circuit is working properly now and I'm having a great time with it.
 
  • #8
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Hint1: What if the middle resistance in the diagram by @BvU is a potentiometer? (EDIT -- I see BvU already mentioned that...)

Hint2: Be sure to include the load resistor attached to the wiper of the potentiometer when calculating the values of the pullup and pulldown resistances...
Thank you. It turns out I had the potentiometer wired wrong. It was my first time using one and I didn't realize that the resistance between the two inputs was the total value of the internal resistor. After some sleep I got my circuit working and I'm having a lot of fun with it.
 
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  • #9
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Following @BvU 's suggestion (which I heartily support!) Suppose we envision the power supply as follows:

View attachment 254062
When the potentiometer wiper is at the very top (as shown above), we expect +2 V to be applied to the load resistance.

By symmetry, if the wiper were at the bottom of the potentiometer the load resistance would see -2 V, right?

Now, consider that your question requires the total current drawn be less than 50 mA (the question statement mentioned minimal power, yet stated a current value. I'm not sure that I'm happy with that way of presenting such a requirement in a homework question). I propose that you could determine the minimum power required between the choices of potentiometer values. This might require ##I^2 R## calculations for all of the resistances involved. :frown:

From a practical engineering standpoint the 200 nA drawn by the ##10 M\Omega## load resistor at 2 V is essentially insignificant compared to the allowed 50 mA total current supplied by the voltage sources. But if you're required to minimize the power drawn, you may have to consider situations where it's not insignificant if you can make the total current drawn much less than 50 mA. You'll have to see what can be achieved with each choice of potentiometer value.

You have been given several values of potentiometer resistances to select from. This seems to be your determining point. For each of these values can you find values of R that will provide the required load voltage for those pot values? How does the power consumption fair with each pot and R value? Methinks an application of nodal analysis might prove fruitful.
I solved the problem after I got a little bit of sleep. I didn't fully understand how to wire the potentiometer but I got it figured out. Thank you for your help.
 
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