# Scaling A Pots Voltage

Discussion in 'Electronic Basics' started by Michael, Oct 13, 2007.

1. ### MichaelGuest

Hi,

I have a motor controller board that uses a 5K pot for speed
control/setting.

The problem I have is that my joystick only moves the pot 20% either side of
center so I only have 40% of the total range.

Is there anyway to scale this voltage up to the full range? I was thinking
about using an amplifier but don't know of one that will increase a voltage
above a certain point and decrease it below another.

Can anyone suggest something suitable?

Michael

2. ### NinjaGuest

If the amplifier approach appeals to you, it's simply gain and offset that
you need to do the job. See http://focus.ti.com/lit/an/sloa097/sloa097.pdf
for an application note on how to design the amplifier.

3. ### Stephen J. RushGuest

An op-amp circuit would be easier to adjust, but there is an old trick
used by early R/C model builders back when they had to build their own
joysticks: get a wirewound pot and bypass part of the element with silver
paint. The paint was a General Cement product intended for repairing
printed circuit boards. I suspect that getting it right would take as
long as building an op-amp scaler with trimpots for gain and offset.

4. ### John FieldsGuest

---
The brute-force way would be to use two supplies: (View in Courier)

.. E1
.. |
.. [R1]
.. |
.. E2--+
.. |
.. [R2]<--- EOUT
.. |
.. E3--+
.. |
.. [R3]
.. |
.. E4

E1 and E4 are the supply voltages, EOUT is the wiper voltage (which
can vary from E2 to E3), R1 and R3 are the inaccessible regions of
the pot, and R2 is the 40% of the total resistance of the pot you
can access.

Since you have a 5k pot and you can only use 40% of it, that
resistance is:

R2 = 0.4 Rt = 0.4 * 5000R = 2000 ohms.

That makes:

Rt - 2000R 3000R
R1 = R3 = ------------ = ------- = 1500 ohms
2 2

Let's say that your motor control voltage needs to be 0 to 10V.

Then that means your circuit now looks like this:

.. E1
.. |
.. [1500R] R1
.. |
.. E2--+---10V
.. |
.. R2 [2000R]<--- EOUT
.. |
.. E3--+---0V
.. |
.. [1500R] R3]
.. |
.. E4

Since you want 10V across 5k, that means (with no load on the wiper)
that the current through the pot will be:

E2 - E3 10V
I = --------- = ------- = 0.005A = 5 milliamperes.
R2 2000R

Now, since the current in a series circuit is everywhere the same,
R1 and R3 will also have 5mA through them and the voltages they will
each drop will be:

E = IR = 0.005A * 1500R = 7.5V

So, E1 needs to be 7.5V more positive than E2 and E4 needs to be
7.5V more negative than E3, reducing the circuit to:

.. 17.5V
.. |
.. [R1]<--- EOUT
.. |
.. -7.5V

Finally, since there's 25V across the pot and 5mA through it, it'll
be dissipating:

P = IE 0.005A * 25V = 0.125 watt, which is very little.