Using a potentiometer to constrain voltage between -9 and 9V

[dadadeee23]

I am stumped on a problem I have been working on. I am told to build a circuit that will have an output voltage between -9 and 9V.

We are given two 12 V batteries as power sources, and a choice between 3 potentiometers with values of 10k ohm, 20k ohm, and 50k ohm. We are allowed to use any resistor within the range of 10 ohms to 100Mohms.

How the heck to I use the potentiometer to get a negative voltage? I used voltage division to get a resistor value for a positive 9 volts. I'm stumped

 

[davenn]

hi there
welcome to PF

if you connect the 2 12V batteries in series, then you can produce a -12V, 0V and +12V rails

now figure out how you will drop it to + and - 9V :)

Dave

 

[dadadeee23]

so I need to add my potentiometer and resistor. If I put them in series and leave R as a variable setting it equal to 9V, I can solve for the resistance? I chose the 10k pot first, and I calculated using voltage division, the resistance to be 3333.33 ohms. Does this sound right?

 

[vk6kro]

No, you would need two resistors with a pot between them. You are almost right though.

This is to get the positive and negative 9 volts relative to the centre point of the two batteries.

 

[dadadeee23]

okay, so I think I see what your saying. I would use the same valued resistor on the opposite side (relative to the center pt of 0V) which would be my voltage divider for the -12V. I am a little confused on how that battery is producing a negative voltage though. Should I switch the polarity of the bottom one? I guess from KVL that would give me a negative voltage.

 

[vk6kro]

If you measure relative to the junction of the two batteries, the other terminal of the lower battery is negative 12 volts.

That is all they mean.