Hi there,
Somehow I feel kind of lost at the moment. I read through my old physics notes from highschool, but couldn't quite get it all back to mee.
Assume I have a capacitor, C=10.000 uF and want to discharge it (e.g I'm testing a new design on a powersupply, but don't have it connected to anything and wants to discharge my cap). If I use a 6k ohm resistor, how long time will it take to discharge it completely?
Have a good day,
//magnus
Somehow I feel kind of lost at the moment. I read through my old physics notes from highschool, but couldn't quite get it all back to mee.
Assume I have a capacitor, C=10.000 uF and want to discharge it (e.g I'm testing a new design on a powersupply, but don't have it connected to anything and wants to discharge my cap). If I use a 6k ohm resistor, how long time will it take to discharge it completely?
Have a good day,
//magnus
Magnus,
The technically correct answer (the kind that would make the Number Are Everything crowd happy) is: Forever.
It's an asymptotic process in the sense that the voltage gets lower and lower, approaching zero, but never actually gets there. However, numbers aside, in the real world the cap will get to some low-enough voltage (say, 1V, which is low enough that it won't hurt you or the circuit) in a rational period of time.
More particularly:
T=RC
where,
T= time in seconds (technically, T should be the Greek letter tau)
R= resistance in ohms
C= capacitance in Farads
So, in your case, we plug in the numbers to yield:
T= 6k*.01F
which leads us to T= 60 seconds
Okay, so what happens in 60 seconds? Is the capacitor discharged completely?
Er, well, no. The voltage will have dropped to 63% of its original value. So if you started at, say, 50V then you'd be reading about 31.5V after 60 seconds. So then what? We're still at too high a voltage to start soldering anything. Well, you plug in the numbers and crank the handle again. Starting with 31.5V, after another 60 seconds the voltage will have dropped another 63% to about 19.8V, and so on. So every time 60 seconds pass, you'll drop another 63% of the voltage that you had at the beginning of that time period. If you plot all this out on a graph, you'll see a curving line that drops towards 0V, but it is also starting to flatten out as it gets closer.
As a practical matter, I keep a couple of wire wound 10W resistors nearby (the one nearest me looks like it's 500 ohms) that I use to discharge caps. Just grab one and stick it across the terminals. If it gets too hot to hold, don't be a hero, put the silly thing down before you burn yourself, or use a higher value next time. I'm content to get the voltage down to, say, 2-3V before I start working on the circuit. Yes, it will make a tiny spark. Don't worry about it.
Note that this is not the same thing as a permanently installed bleeder resistor. But that doesn't sound like what you're after.
Grey
The technically correct answer (the kind that would make the Number Are Everything crowd happy) is: Forever.
It's an asymptotic process in the sense that the voltage gets lower and lower, approaching zero, but never actually gets there. However, numbers aside, in the real world the cap will get to some low-enough voltage (say, 1V, which is low enough that it won't hurt you or the circuit) in a rational period of time.
More particularly:
T=RC
where,
T= time in seconds (technically, T should be the Greek letter tau)
R= resistance in ohms
C= capacitance in Farads
So, in your case, we plug in the numbers to yield:
T= 6k*.01F
which leads us to T= 60 seconds
Okay, so what happens in 60 seconds? Is the capacitor discharged completely?
Er, well, no. The voltage will have dropped to 63% of its original value. So if you started at, say, 50V then you'd be reading about 31.5V after 60 seconds. So then what? We're still at too high a voltage to start soldering anything. Well, you plug in the numbers and crank the handle again. Starting with 31.5V, after another 60 seconds the voltage will have dropped another 63% to about 19.8V, and so on. So every time 60 seconds pass, you'll drop another 63% of the voltage that you had at the beginning of that time period. If you plot all this out on a graph, you'll see a curving line that drops towards 0V, but it is also starting to flatten out as it gets closer.
As a practical matter, I keep a couple of wire wound 10W resistors nearby (the one nearest me looks like it's 500 ohms) that I use to discharge caps. Just grab one and stick it across the terminals. If it gets too hot to hold, don't be a hero, put the silly thing down before you burn yourself, or use a higher value next time. I'm content to get the voltage down to, say, 2-3V before I start working on the circuit. Yes, it will make a tiny spark. Don't worry about it.
Note that this is not the same thing as a permanently installed bleeder resistor. But that doesn't sound like what you're after.
Grey
Thanks alot!
I usually build my stuff at work, where people with varying background likes to look at and touch my different projects. Actually... why do I bother in this case... ;=) This could actually teach them to keep their hands off... ;=) Though it's really hard to find good salespeople nowdays.
Thanks again four your explanaition,
//magnus
I usually build my stuff at work, where people with varying background likes to look at and touch my different projects. Actually... why do I bother in this case... ;=) This could actually teach them to keep their hands off... ;=) Though it's really hard to find good salespeople nowdays.
Thanks again four your explanaition,
//magnus
Pass DIY Addict
Joined 2000
Paid Member
Eric,
I'd choose a time constant (the formula above) of several minutes. Too low a value will give you an unnecessarily high current drain to ground, and most folks want all that juice going to the circuit, not wasted as heat.
As a practical matter, I'd use something like 47-100k (wattage according to your chosen rail voltage). Others will differ, I'm sure, as to what constitutes a reasonable value.
The only real reason for a bleeder resistor is as a safety issue (okay, okay, some of the old-timers used them as a sort of 'ballast' on the power supply to even things out, but I haven't seen that in quite some time--natch, that means it's retro & cool again). I don't use them, myself. I figure that if I shock myself (and I do every so often, t'ain't no big thing, I just cuss and go on with my life), that it's my fault. None of the commercial equipment I'm familiar with uses them, either.
Grey
I'd choose a time constant (the formula above) of several minutes. Too low a value will give you an unnecessarily high current drain to ground, and most folks want all that juice going to the circuit, not wasted as heat.
As a practical matter, I'd use something like 47-100k (wattage according to your chosen rail voltage). Others will differ, I'm sure, as to what constitutes a reasonable value.
The only real reason for a bleeder resistor is as a safety issue (okay, okay, some of the old-timers used them as a sort of 'ballast' on the power supply to even things out, but I haven't seen that in quite some time--natch, that means it's retro & cool again). I don't use them, myself. I figure that if I shock myself (and I do every so often, t'ain't no big thing, I just cuss and go on with my life), that it's my fault. None of the commercial equipment I'm familiar with uses them, either.
Grey
Pass DIY Addict
Joined 2000
Paid Member
Since the bleeder thread is closed
A capacitor will discharge 90% of its value in 2.5 RC time constants and 99% of its value in 5 RC time constants. The tradeoff is between the energy you want to burn in the bleeder and the time that you want to wait to get into your gear.
Here's a simple example of the discharge:
A capacitor will discharge 90% of its value in 2.5 RC time constants and 99% of its value in 5 RC time constants. The tradeoff is between the energy you want to burn in the bleeder and the time that you want to wait to get into your gear.
Here's a simple example of the discharge:
Attachments
Without a bleeder, DA can also zap you unexpectedly from a supposedly discharged cap. Throw away the two watts or so it'll cost you to stay safe- use a bleeder with a reasonably quick time constant- say, half the time it takes you to unscrew the access plate (or less).
Death is very permanent. And it doesn't always come directly from the shock- you're around a lot of dangerous and fragile objects for a guy who's jerking around in a large spasm.
Death is very permanent. And it doesn't always come directly from the shock- you're around a lot of dangerous and fragile objects for a guy who's jerking around in a large spasm.- Status
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