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I've been testing some capacitors off a treadmills PCB using a 9 volt battery to charge them up, then taking voltage readings with a multimeter and watching as they discharge.

Most seemed to release the 9v very slowly but one 25v 470uF capacitor leaked it away considerably faster than the rest. Is this a sign of a bad capacitor that needs replacing?

Is it worth buying those capacitor testers for £10-12 off eBay for further tests?
 
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PEG

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AAAAAArrrggghhhh!!!!!!!


........right....NOW,it's discharged........;) (circa 0.5 secs)
 
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AAAAAArrrggghhhh!!!!!!!


........right....NOW,it's discharged........;) (circa 0.5 secs)
Well, hopefully the 9v didn't hurt you much!:p

I've been discharging mine with a screwdriver which makes a small spark. No idea if that's best practice or not, just trying to figure out if the quick one is knackered or not, it loses about 1v per second while the others take about 3-4 seconds to lose a volt those times being guestimates rather than stopwatch figures.
 

telectrix

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try the screwdriver on a 300uF charged to 350V. i still have my 10" blade driver from the 70's with the scars.
 

marconi

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A capacitor C charged to a voltage Vc and then discharged through a resistor R reduces in voltage according to an exponential law:

V = Vc e exp (-t/RC) where e is approximately 2.718.....

What this says is if t = CR ie the value of the capacitor times the value of the resistance a quantity called the time constant, the voltage will have fallen to Vc e exp(-1) which is about 0.37 of the initial Vc.

If you know the resistance of your multimeter,you can use this value for R. A moving coil multimeter with a needle and scale has a sensitivity value - so many Ohms per Volt - which depends on the voltage range scale selected. If it was 10000 Ohms/Volt and you had the 10V scale selected the multimeter 'looks like' a resistor of 100000 Ohm.

Thus knowing the marked up value of C, the R of the m'meter, Vc of the battery and what happens to V after a period equal to one time constant (CR seconds) you can do some experiments which I leave you to think about and do to determine the true state of the capacitor.
 

PEG

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try the screwdriver on a 300uF charged to 350V. i still have my 10" blade driver from the 70's with the scars.
Is that a TV doctor's war story? ;)
 
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try the screwdriver on a 300uF charged to 350V. i still have my 10" blade driver from the 70's with the scars.
Try holding the plastic handle next time, innit! I could always charge the 450v 470uF capacitor via the mains and see how that goes, maybe discharge it on an aluminium ladder that somebody is up?
 
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A capacitor C charged to a voltage Vc and then discharged through a resistor R reduces in voltage according to an exponential law:

V = Vc e exp (-t/RC) where e is approximately 2.718.....

What this says is if t = CR ie the value of the capacitor times the value of the resistance a quantity called the time constant, the voltage will have fallen to Vc e exp(-1) which is about 0.37 of the initial Vc.

If you know the resistance of your multimeter,you can use this value for R. A moving coil multimeter with a needle and scale has a sensitivity value - so many Ohms per Volt - which depends on the voltage range scale selected. If it was 10000 Ohms/Volt and you had the 10V scale selected the multimeter 'looks like' a resistor of 100000 Ohm.

Thus knowing the marked up value of C, the R of the m'meter, Vc of the battery and what happens to V after a period equal to one time constant (CR seconds) you can do some experiments which I leave you to think about and do to determine the true state of the capacitor.
I'll have to try and decipher this when I'm more awake, reminds me of the science modules that were not my forte in college. I figured it out back then though...
 

marconi

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If the capacitors are electrolytic types it's important to connect the leads the right way around. They are marked with a + or - or both.
 
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If the capacitors are electrolytic types it's important to connect the leads the right way around. They are marked with a + or - or both.
Yes the capacitors I have, have a silver stripe down the side to denote negative and the PCB has a shaded side denoting the side the negative pin pokes through. Shouldn't be any issues there but I'll suck it and see...

I'll report back if anything catches fire. :coldsweat:
 

DPG

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If you do happen to try connecting one up with reverse polarity, then please do not have your eyes anywhere near it!

The smaller value electrolytics can be quite fun, as they don't have the 'weak-point' at the top of the can which ruptures under pressure. The small ones quite often shoot several feet up in the air.

Obviously I don't recommend doing anything like this and would never do it.

I have also never connected up tantalum capacitors the wrong way to see the purple smoke.
 
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Back to the main issue as it were, I've reinstalled my treadmills circuit board after installing a few new capacitors and it came on but give an error code E02. After a bit of Googling I found 2 answers to fixing this error code, one was to swap the circuit board for a new one, about 125 dollars on alibaba or was it AliExpress? The other, check the lead from the circuit board to the motor, which I did as part of the reinstallation process.

Anyone worked on treadmills or in electrical repair?
 
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Oh, I also found out what the small black push button on the circuit board does, it seems to test the incline motor, it takes it up 1 level of incline and back down again. All the time I was working on the circuit board I wondered what that was for...

The E02 error comes after the count down to start the treadmill motor by the way.
 
I used to test using an ohmmeter to charge, the voltage charged the capacitor - so 0 ohms becomes infinity as a short circuit becomes an open circuit - the capacitor is charged. Then discharge using voltmeter, voltage decreases to zero. Very basic but worked in the day (moving coil meters).
 
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