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Topic: Receive data from linear potentiometer? (Read 6 times) previous topic - next topic

Grumpy_Mike

Quote
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.

BenF

#21
Feb 11, 2010, 11:31 pm Last Edit: Feb 11, 2010, 11:32 pm by borref Reason: 1
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.

BenF

#22
Feb 12, 2010, 06:29 am Last Edit: Feb 12, 2010, 07:04 am by borref Reason: 1
I don't think there's anything wrong with my math Richard.

You wrote:
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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:
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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.

Grumpy_Mike

@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.

BenF

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.. 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.

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