Designing an adjustable voltage source using only resistors and a pot

In summary, the conversation discusses the need for a power supply that takes in +5 and -5v and has an adjustable output ranging from -2 to +2 v using a single potentiometer. The individual seeking help has designed a circuit but it did not work and is now looking for assistance in figuring out the problem. Suggestions are made to use a potentiometer as the middle resistance and to include the load resistor when calculating values for the pullup and pulldown resistances. After some trial and error, the individual realizes they had wired the potentiometer incorrectly and their circuit is now functioning properly.
  • #1
garthenar
35
8
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 don't 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.
 
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  • #2
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
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
garthenar said:
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
They are not separate: you can assume both are wrt ground
 
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  • #6
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
BvU said:
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
berkeman said:
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
gneill said:
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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  • #10
I need help in this
garthenar said:
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.

View attachment 254034

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 don't want it to go. I'd love to use a zenner diode but that is not allowed.

I have attatched my original circuit design.
View attachment 254035
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.
Can you please help me in this i am unable to understand as to how you solved it so?
 
  • #11
TforTortilla said:
I need help in this

Can you please help me in this i am unable to understand as to how you solved it so?
Yeah, gladly.

My big misconception was on how the potentiometer worked. It can be treated as a wire with two resistors. The middle leg is connected between the resistors. Changing the position of the potentiometer will increase resistance on one side and decrease it on the other. ( this video helped me a lot)

1576205527693-png.png

the wire with the arrow is connected to the middle leg of the potentiometer. That'll get your started if you have any more questions I can give you more but I want to keep in the rules of the forum.

How long until your project/lab is due?
 
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FAQ: Designing an adjustable voltage source using only resistors and a pot

1. How can I design an adjustable voltage source using only resistors and a pot?

To design an adjustable voltage source using only resistors and a pot, you will need to create a voltage divider circuit. This circuit will consist of two resistors in series, with the potentiometer (pot) connected in between them. By adjusting the resistance of the pot, you can vary the output voltage of the circuit.

2. What is the purpose of using only resistors and a pot to create an adjustable voltage source?

The purpose of using only resistors and a pot to create an adjustable voltage source is to have a simple and cost-effective solution for varying the output voltage. By using a pot, you can easily adjust the voltage without the need for additional components or complex circuitry.

3. How do I choose the appropriate resistors and pot for my adjustable voltage source?

The values of the resistors and pot you choose will depend on the desired output voltage range and the maximum current that will be drawn from the circuit. You will need to calculate the resistance values using Ohm's law and choose resistors and a pot with appropriate wattage ratings to handle the current.

4. Can I use this adjustable voltage source for high current applications?

Yes, you can use this adjustable voltage source for high current applications, but you will need to ensure that the resistors and pot you choose can handle the current. You may also need to add additional components, such as transistors, to handle the high current.

5. Are there any limitations to using only resistors and a pot to create an adjustable voltage source?

One limitation of using only resistors and a pot is that the output voltage may not be as precise as using other components, such as voltage regulators. The output voltage may also vary with changes in temperature or load. Additionally, the maximum output voltage may be limited by the maximum resistance value of the pot.

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