"see-saw" pot output (beginner's question)

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Marcaroni
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"see-saw" pot output (beginner's question)

Post by Marcaroni » Thu Sep 22, 2011 8:09 am

I haven't actually attempted this yet. I'm not exactly sure where to start.
I'd like to have a circuit with a single potentiometer that drives 2 outputs operating like a "see-saw".
so as i adjust the pot "up," one output increases in voltage as the other decreases and vice versa.
so whenever the output is max voltage on one, the other is zero volts.
preferably with a 5k or 10k pot.
too much for a beginner? i don't really want a solution, just a push in the right direction. unless it's trivial.

thanks,
marcaroni

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VernGraner
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Re: "see-saw" pot output (beginner's question)

Post by VernGraner » Thu Sep 22, 2011 8:51 am

Marcaroni wrote:I haven't actually attempted this yet. I'm not exactly sure where to start.
I'd like to have a circuit with a single potentiometer that drives 2 outputs operating like a "see-saw".
so as i adjust the pot "up," one output increases in voltage as the other decreases and vice versa.
so whenever the output is max voltage on one, the other is zero volts.
preferably with a 5k or 10k pot.
This would be simplest to answer with a schematic, but i dont have the drawing tools on this machine so let me see if i can describe what you would build. To accomplish what you want to do, it seems to me you would connect the "wiper" of the potentiometer to a V+ source, then connect each of the other two connections to your loads and from the load to ground. as you rotate the potentiometer from one end to the other the voltage will rise on one side and drop on the other - essentially a "balance" control. Is that enough of a hint? :smile:

Also- a word of advice - Most potentiometers cannot handle much amperage so if you are planning to do this to cause one motor to speed up while another slows down, or cause one light to brighten as another one dims, you would have to build a slightly more complex circuit where the potentiometer would be driving some sort of buffer that shields it from excessive current draw that could destroy it.

Hope this helps :)

Vern
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Vern Graner

Marcaroni
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Re: "see-saw" pot output (beginner's question)

Post by Marcaroni » Thu Sep 22, 2011 8:59 am

VernGraner wrote:
Marcaroni wrote:I haven't actually attempted this yet. I'm not exactly sure where to start.
I'd like to have a circuit with a single potentiometer that drives 2 outputs operating like a "see-saw".
so as i adjust the pot "up," one output increases in voltage as the other decreases and vice versa.
so whenever the output is max voltage on one, the other is zero volts.
preferably with a 5k or 10k pot.
This would be simplest to answer with a schematic, but i dont have the drawing tools on this machine so let me see if i can describe what you would build. To accomplish what you want to do, it seems to me you would connect the "wiper" of the potentiometer to a V+ source, then connect each of the other two connections to your loads and from the load to ground. as you rotate the potentiometer from one end to the other the voltage will rise on one side and drop on the other - essentially a "balance" control. Is that enough of a hint? :smile:

Also- a word of advice - Most potentiometers cannot handle much amperage so if you are planning to do this to cause one motor to speed up while another slows down, or cause one light to brighten as another one dims, you would have to build a slightly more complex circuit where the potentiometer would be driving some sort of buffer that shields it from excessive current draw that could destroy it.

Hope this helps :)

Vern


that sounds simple enough. even for me. no schematic necessary. thanks, Vern.
-marcaroni

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Janitor Tzap
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Re: "see-saw" pot output (beginner's question)

Post by Janitor Tzap » Thu Sep 22, 2011 11:50 am

Sorry,
if I'm a bit late answering this.

But what a bout using something like an L-Pad instead of a standard carbon potentiometer?
I've used L-Pads in 150 watt RMS speakers, and each L-Pad was rated at 200 watts.
Here is how they are used with speakers.
http://www.colomar.com/Shavano/lpad.html

But like Vern said:
VernGraner wrote:Also- a word of advice - Most potentiometers cannot handle much amperage so if you are planning to do this to cause one motor to speed up while another slows down, or cause one light to brighten as another one dims, you would have to build a slightly more complex circuit where the potentiometer would be driving some sort of buffer that shields it from excessive current draw that could destroy it.

Hope this helps :)

Vern
Thus, you need to know what the maximum load in the circuit will be.
Before you can just put in any old L-Pad, and expect it to hold up.
Also, L-Pads aren't cheap.
They can range from $10 to $30 each depending upon the type you need. :sad:


Signed: Janitor Tzap

Marcaroni
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Re: "see-saw" pot output (beginner's question)

Post by Marcaroni » Thu Sep 22, 2011 12:25 pm

Janitor Tzap wrote:Sorry,
if I'm a bit late answering this.

But what a bout using something like an L-Pad instead of a standard carbon potentiometer?
I've used L-Pads in 150 watt RMS speakers, and each L-Pad was rated at 200 watts.
Here is how they are used with speakers.
http://www.colomar.com/Shavano/lpad.html

But like Vern said:
VernGraner wrote:Also- a word of advice - Most potentiometers cannot handle much amperage so if you are planning to do this to cause one motor to speed up while another slows down, or cause one light to brighten as another one dims, you would have to build a slightly more complex circuit where the potentiometer would be driving some sort of buffer that shields it from excessive current draw that could destroy it.

Hope this helps :)

Vern
Thus, you need to know what the maximum load in the circuit will be.
Before you can just put in any old L-Pad, and expect it to hold up.
Also, L-Pads aren't cheap.
They can range from $10 to $30 each depending upon the type you need. :sad:


Signed: Janitor Tzap

thanks for the tip, J.T.
I'm planning on using slide pots though.
but this early in the process it's all still up in the air....i'll keep it in mind.

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haklesup
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Re: "see-saw" pot output (beginner's question)

Post by haklesup » Thu Sep 22, 2011 6:31 pm

you'll want to use a stereo pots. with the ends wired opposite. The two wipers will have complementary voltages.

Take VernGraner warning seriously, Pots are designed to deliver a reference voltage to a circuit. If you try to draw much current out of the wiper, when you dial down to one end or the other, you have a very high risk of burning out the resistance element near to the end as its segment of R gets small and I goes way up. A higher wattage pot is not the solution.

For motor speed control, there are many IC based solutions depending on the type of motor. Most need a totem pole transistor arrangement on the output to deliver the power to the load.

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Bob Scott
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Re: "see-saw" pot output (beginner's question)

Post by Bob Scott » Fri Sep 23, 2011 9:34 am

My idea uses only one potentiometer. It works like a "balance" control in a stereo.
Both signals should first be fed through a 10K resistor, so you need a couple of resistors. Both resistors are wired to the end terminals of a linear 100K pot. The wiper of the pot is grounded. The outputs are taken from the end terminals of the pot.
When the pot is turned completely one way, one signal is grounded out. Turned he other way, the other signal is grounded and has no output.
When the pot is in the middle, both signals are equal but attenuated about ~20% from the original.

If you need lower impedance outputs to drive any significant current, add amplifiers to the outputs.
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