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[Reply #15 on:]

This is the data sheet

http://media.digikey.com/pdf/Data%20Sheets/Spectra%20Symbol/TSP%20Series%20ThinPot.pdf

I bought the circular sensor

-Resistance - Standard: 10k Ohms

[Reply #16 on:]

Hi Richard and thank for the reply

like you said "Connecting 5V to the top and bottom of a 10K pot should never burn it out"

This is why I just cant understand why this happens - If you look at the last page of the datasheet you see

Electrical Schematic

PIN 3 (GND)  (LEFT BUS BAR)
PIN 2            (COLLECTOR)  or wiper
PIN 1 (V+)    (RIGHT BUS BAR)


This is exactly how I plug it ( I have also tried plugging it with the V+ and GND reversed (but that gives the same result - only in reverse; that is, only 1/4 of the sensors gives out any reading if I put 5v trough a 20-50 ohm resistors and If I dont have the resistor it overheats and eventually burns.)

P.s  when the sensor starts burning; small burning holes start burning up and getting bigger at the end of the circle on each side (that is on the touch surface; right before electricity comes from or goes to the pins)

[Reply #17 on:]

No. I have never plugged V to the midle (collector)

I might have touched it but I cant remember the pot burning up straight after I did so - It burns up after having been connected for about a minute and half  if plugged pin 1 straigt to 5v and pin 3 to ground

what I meant is that I have tried routing V through low ohm resistors 20 to 50 ohm, and that keeps the pot from burning or overheating, but I get a wrong reading from the sensor

When I  connect the positive and negative leads of an ohmmeter to pin 1 and middle pin of the SoftPot  I get an accurate reading from the softpot  (that is 0 to 10 k if I remember correctly)

[Reply #18 on:]

When I  connect the positive and negative leads of an ohmmeter to pin 1 and middle pin of the SoftPot  I get an accurate reading from the softpot  (that is 0 to 10 k if I remember correctly)

No you should get 10K if you connect the two ends. If you connect pin 1 and the middle then you will get a varying resistance depending on whee you touch the pot.

There is no way you can get what you are saying if the data sheet is correct. So you have either misinterpreted what you are doing, the device is faulty or the data sheet that is posted doesn't match your device.

[Reply #19 on:]

I noticed that the datasheet specifies recommended operating power as 0.5W (under electrical specifications). For a 10k "pot" you would then need to supply about 7mA at 70V. For 5V this would be 100mA which may indeed burn the sensor if it expects 7mA only.

All in all I think the datasheet is at best unclear (why label plus and ground on a pot - they should be interchangeable for pure resistive loads). Perhaps you need to give the supplier a call to find out what the real interface requirements are.

[Reply #20 on:]

For 5V this would be 100mA which may indeed burn the sensor if it expects 7mA only.

No power doesn't work like that, it is the product of voltage and current.
Power = Voltage X Current

so at the current at 5V would need to be:-
Current = Power / Voltage = 0.5 / 5 = 100mA so that means 100mA is OK
BUT if you connect 5V across 10K the current flowing is 0.5mA which is a long way from 100mA

Anyway the data sheet says 1W not 0.5W so it is even less of an issue.

[Reply #21 on:]

Maybe I was unclear Mike, but my point is that "current" is killing the sensor - not power. Given 0.5W then at 70V - current would be as low as 7mA - at 5V it would be 100mA. The product (I*V) and so the power (0.5W) is the same.

0.5W is not listed as a maximum, but a recomended level - why would they do that? With a supply voltage of 5V, power would be as low as 2.5mW for a 10k resistor. Something is "fishy" and I'm suggesting it's not a true "pot" (resistance is not 10k unless it sees 0.5W) and perhaps requirement calls for a much higher voltage.

[Reply #22 on:]

I don't think there's anything wrong with my math Richard.

You wrote:

Assuming that the softpot is 10K as advertised, putting 5V across it will result in 0.0005 amps (0.5 mA) which equals 0.0025 watts (2.5 mW)

I wrote:

With a supply voltage of 5V, power would be as low as 2.5mW for a 10k resistor

If you want to expose a 10k resistor to 0.5W you will need to feed it with just above 70V. Current is then about 7mA (70/10k) and power (70V*7mA) is 0.49W.

I'm merely suggesting that the power requirement stays fixed and that the sensor requires "heating" (0.5W) before it will reach its advertized characteristic. When "heating" the sensor however there is a big difference between 7mA and 100mA.

[Reply #23 on:]

@BenF
I think you reasoning is that there might be a hidden maximum current that is independent of power ratings that is doing the damage.

If this were a semiconductor then this could be true, but this is a passive linear component and will operate at 0.5W with a variate of currents and voltages. This device is overheating therefore it must be experiencing a power dissipation over it's rating. With a 5V supply the only way to do this is to connect between the wiper and one end. I blew up two pots like this when I was 12, I have never forgotten that lesson.

I think this is a problem of either misidentifying the wiper or having some mechanical fault with the soft pot.

[Reply #24 on:]

.. but this is a passive linear component and will operate at 0.5W with a variate of currents and voltages ..

That's the question I would say - Is it a passive linear component?

If it's a semiconductor made from some substrate that requires 0.5W or thereabouts  to operate within specifications, interface requirements would be different than for a linear pot.

I do agree however that a more likely explanation is incorrect wiring (connecting the wiper and a bus bar to 5V/GND respectively). If the sensor reads low ohms when not touched - that would also explain why it burns out so fast.

Measuring the resistance between bus bars (should be 10k) and also looking at the actual current through the bus bars (should be no more than 0.5mA at 5V) in a live circuit ought to give him useful answers.

[Reply #25 on:]

I havent had  a chance to measure the sofpot but I think I have an idea of what is happening

I plug the softpot to arduino mega  

(http://arduino.cc/en/Main/ArduinoBoardMega)    

  and the analog input (that I plug the wiper collector to can be programmed as an ouput also and therfore give out voltage - it must be that the programming I am using for the arduino board (Standard Firmata for Max Msp)  has the analog input ouputs programmed so that they give out voltage  and therfore burn the wiper

[Reply #26 on:]

Thanks

Ill try that tonight

 

[Reply #27 on:]

Looks like someone also had the problem of the softpots burning up for no reason (my pinouts are correct: +5V on the triangle, GND on Pin 3 and voltage out from the center into the arduno. I have read through all of the replies in this thread and looks like the consensus is to put either a 100 or 1000 ohm resistor between the softpot output pin and the arudino analog-in pin to make sure that high-current does not go through the softpots.

Unfortunately, these problems (and this discussion) were not in sparkfun's item comment before I bought these things.

[Reply #28 on:]

Hy evtsai.km2

I have sort of got these sensors working.  

If I plug 5v into either right or left pin and then middle pin to analog in on the arduinio  (without connecting anything to ground) then the softpot doesnt burn up, but only about half of the sensor works (gives out a value)

then I plugged a 9v battery to the sensor with the same wireing  and then the whole touch sensor works - but when you touch (for a short while) the sensor at the end where 9v go in, it heats up (just at that position).

So the sensor sort of works like that.

[Reply #29 on:]

can one use these 'SoftPots' to obtain accurate distance measurements? Like around 0.01mm?