Hi,
I have a electronics problem, ( well its a problem for a mechanical engineer anyway). I have the need to convert a 12vdc analogue signal to a 5vdc analogue signal so an amplification of approx 0.4. I've listed all the details below.

I'd appriciate an explanation of the cct but to be honest just a circuit would do.

Input 0vdc to 12vdc. This is produced from a 0 to 6.5 K ohms pot, which feeds another control system, clearly this means that I cannot affect this voltage by the cct that I attach.

Output This need to be 0 to 5vdc which is feed into a micro, which has a pull up resistor at 100K ohms fitted ( & I cannot remove this resitor. )

Many Thanks in advance

Arron

 

Sounds like you just need a resistive divider, with a ratio of 1.4:1

As we know, the voltage at the midpoint of the divider is Vin(R2/R1+R2).

Because of your pull-up resistor, we cannot use very high values of R1 & R2. I would guess that 1.4k and 1k would be ok, and would not affect the accuracy too much. I assume your 12V source can supply 5mA.

 

Thanks for your reply, I have a few questions.

If my Pot is 1/2 way ( 3k either side of wiper) then voltage at wiper = 6 volts.
If the potential diver has a combined resistance of 1.4K and shares a common ground with the pot then this will make the lower side of the pot 3K in parallel with 1.4K giving a combined resistance of approx 1.3K pulling the wiper voltage down to approx 2.5 volts. This would cause a large error on the control system that it feeds. I have not accounted for the 100k resistor that comes off the potential divider that is connected to 5 volts as I couldent get my head around the problem.

I'm sorry that I may not have described the problem adequatly. The pot is connected to a shaft and the pot acts as a sensor for shaft position, and is continually moving.

 

The ratio of the values quoted was correct, but the values themselves were too low (actually 2.4K in total, not 1.4K as you suggested).

You need to use higher values to prevent it pulling the voltage down.

Unfortunately the 100K you can't remove would tend to upset things, an easy solution would be an opamp buffer feeding from the output of the potential divider (which could then use high values) to the A2D.

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Perhaps I misinterpreted your question. I was under the impression that the control system you mentioned was to supply the 0-5V signal, instead of being taken directly from the pot.

In that case, Nigel is on the right track, using an opamp to buffer the voltage at the pot wiper, and then using a resistive divider (one resistor 14k, the other 10k. Still, the 100k pullup will affect the accuracy a little, so if you want to improve that, buffer the output from the divider again before taking it to the uP.

Attached Images

12V-5V.gif (3.9 KB, 487 views)

 

Hi dansoarr,

It is not a good idea to have the wiper
continually moving.

The track will soon wear away.

Maybe a light sensor could be found that covers
this sort of range of resistance,
possibly a lamp could be obscured by something on
the shaft.

There are other non-contact possibilities,
but obscuring a light dependant resistor
is probably the most straight forward.

Best of luck, John

 

That's a great help, & I'm sure that I can get it running now. I assume that a 741 would be ok as the opamp ?

 

A 741 would be fine, you actually only need the second opamp, the value of the potential divider at 140K/100K is more than high enough to connect directly to the slider of the pot. Although if you did use both opamps you could reduce their values if you wanted to.

__________________
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http://www.winpicprog.co.uk

 

Quote:
The pot is connected to a shaft and the pot acts as
a sensor for shaft position, and is continually moving.


Well, we shall see how long it runs for.
My guess is about two hours for a wire-wound pot,
and about half an hour for a thin film track.
One of those solid carbon 'Morganite' types (very rare
these days) would probably last four or five hours
before it became unusable.

Personally i think this method of determining the shaft
position is not practical, unless its purely for
demonstration and not expected to run for more than a
few minutes at a time.

Best of luck with it, John

 

First off thanks for the help, the pot in question is used on an output shaft that has a 120° movement and is in constantly moving for 6 hours a day and has been in service for 5 years. I belive that the pot orgionally came from a kawasaki motorcycle.

 

Life span of pot.

Hi dansoarr,
thank you for your reply.

I am clearly quite wrong about that pot,
however i am surprised that a potentiometer
can last that long.

It is a surprise to me that a pot can run
continuously for years (@ six hrs a day)

Is there something special about this type ?

John

 

It has been suggested that it may well be a resistive polymer rather than a wire wound device. The unit would live in a fairly harsh environment in its natural habitat.

Arron

 

I am intrigued by this component.
previously my attempts at mechanical feedback have failed
at the point of interface.
I have had to resort to non-contact feedback on various
items, such as lights and sensors.

I haven't heard of 'resistive polymers' used for pots before.

Any idea which Kawasaki motorbike it may have come from ?
i will try to look it up ...

Regards, John