# OpAmp circuit question.

Discussion in 'Electronic Design' started by [email protected], Nov 7, 2005.

1. ### Guest

Hi All,

My problem could maybe trivial to a good analog engineer, but i'm just
unable to come up with a easy to build solution:

- The input of my circuit should be a potentiometer.
- The output of my circuit should be Vcc/2 while the input
potentiometer is at its center, moving the potmeter towards its
minimum, the output should drive towards vcc, moving to its minimum
value it should drive the output to GND.

I've tryed to coming up with some circuits, using a Vcc/2 reference and
adding/substracting it with an opAmp. But i imagine that it could be
done in an more easy way.

Thanks for getting me on the right track.

2. ### TomGuest

- The input of my circuit should be a potentiometer.
Are you serious?

pot low = VCC
pot high = GND
pot wiper = output (buffer as needed)

greetings,
Tom

3. ### John WoodgateGuest

I read in sci.electronics.design that wrote
The potentiometer itself will do that for you. Just connect Vcc to the
'minimum' end of the track and GND to the 'maximum' end. If the pot
resistance gives you too much voltage drop at the centre position,
connect the slider to a simple op-amp buffer and feed its output to the

Of course, you my not have meant quite what you wrote, in which case
disregard the above.

4. ### Guest

Oeps,

I incorrectly described my problem.

Background:

I have a joystick that outputs at rest a 2.5V center voltage. Moving
the joystick up moves the voltage to 5V, moving it down to 0V.

I need to replace this joystick by a 'Potentiometer + Switch'

-The switch in position 'A' should add the potentiometer's output to
the 2.5V
-The switch in positon 'B' should substract the potentiometer's output
from the 2.5V

Quit different from what i first described.

Sorry for the confusion and thanks for any hints.

5. ### Rich GriseGuest

So, you just need a reversing switch. An ordinary DPDT switch, with
the two wipers connected to the two ends of the pot, and +5 and Gnd
supplied to opposite poles for the two positions.

Cheers!
Rich

6. ### John FieldsGuest

---
There seems to be something wrong with your requirement since on the
one hand you say that moving the pot toward its minimum should drive
the circuit's output toward Vcc, but then when it actually _gets_
to its minimum the output should suddenly flip to GND? That can be
done, but is it really what you want?

Sounds to me like a homework assignment where the instructor gave out what
sounds like a pretty contrived excercise to invert the slope of an input
signal.

About the only real application I can think of is some sort of audio
application where you want a audio (log) taper and you're trying to avoid
having to come up with a reverse taper pot (pretty hard to find these days)...
although that wouldn't jive with Vcc/2 being at the mid-point of the pot...

8. ### Guest

In fact, i'm retrofitting an old control with something new. One
constraint is the use of a momentary action SPDT switch and the use of
a single tap potentiometer.

The summery:

The switch at rest : 2.5V output.
The switch pushed in position A: the potentiometers position should be
added to the 2.5V (for example , if the potmeter is at 50% of its value
the output should be 3.75V)
The switch pushed in position B: the potentiometers position should be
substracted from the 2.5V (for example , if the potmeter is at 50% of
its value ther output should be 1.25V)

Thx

9. ### TomGuest

In fact, i'm retrofitting an old control with something new. One
Momentary action SPDT switch? So normally the switch is in position A,
and you need to activate it to put it in position B? Or is there a
third "OFF" position?
(view in fixed width font such as courier)

5V ---o
\ H POT L
+----------/\/\/\/\----- 2.5V
^
0V ---o SPDT | W
|
+--------- OUT

greetings,
Tom

10. ### Guest

5V ---o
Tom,

Thx, amazing how sometimes sucs a simple circuit just won't come up in
my brain...

But this is what i wanted to do.

Thanks a lot for helping my out of my to-far-away thoughts.