Discharging Capacitors

[AnnabellaRenee87]

What would be the proper process to discharge a capacitor on a PCB?

The PCB is a power supply in a Samsung TV and I want to go ahead and replace the overly cheep capacitors in it but the large 400 volt 2000 uf one isn't popped, yet.

I was just gonna try an desolder it but was afraid something would short out as I am gonna use solder wick. I have a desoldering Iron with a air sucker too as well as a solder sucker (the think you push a plunger down and hit the button to suck away the solder real quick. Come to think of it, it may be better to use the solder sucker as its plastic but IDK.

Also, Been a fan of Dave's videos on YouTube for a while and decided to make an account today to!

[Kryoclasm]

Well, this may not be the proper way but I just use an old screwdriver and short the pins on the bottom of the PCB. 


ZAAP!! , discharged. 

*not a screwdriver, but cool image.

[walshms]

How about a slightly safer way that does less harm to the capacitor itself?

A resistor across the terminals (to limit the current) will do nicely.

[AnnabellaRenee87]

Well, this may not be the proper way but I just use an old screwdriver and short the pins on the bottom of the PCB. 


ZAAP!! , discharged. 

At first I thought about doing that but I'm worried about damaging the board, would it?

[c4757p]

We've discussed this many, many times. Answers range from "short that motherf-r with a screwdriver" to "discharge gently with a resistor". I'm usually in the former group...

1) 400V 2000uF is a whale of a lot of energy (160J). If it's fully charged, you probably should not just short it out.
2) If it hasn't been plugged in for a while, it's not going to be fully charged. These caps have internal leakage, and besides, the power supply would keep running until it has drained the capacitor significantly.

So bearing in mind #2, I'd just short it out. Take a leak first, though If you think it might be fully charged, pick a resistor: the capacitor will discharge in approximately 5*R*C seconds, and the resistor will dissipate V^2/R peak. You have V (400V) and C (2000uF), so pick a resistor that can drain it in a reasonable amount of time without exceeding its power rating.*

*A massive wirewound power resistor will not mind if you exceed its rating briefly.

Also keep in mind dielectric absorption - if you discharge it quickly, it can regain charge when you remove the load. You should keep a load/short attached for a minute or so to drain it off completely.

Oops, I typed too slowly, looks like you two beat me to it! Oh well, can't let these words go to waste...

How about a slightly safer way that does less harm to the capacitor itself?

I was under the impression that he was going to replace the capacitor.

[Alex]

...or using an e.g. 10k power resistor with insulated terminals connected to the cap. Like this:



Monitor final voltage before touching. Voltage will rise a bit after you disconnect the resistor.

[madires]

You would use a resistor to discharge the cap. Use a power resistor in the 33-47k range with 5W at least (P = V^2 / R).

[flolic]

230V 40W light bulb. My standard piece of equipment, and best way to discharge large capacitors from switching power supplies and camera flashes.

[AnnabellaRenee87]

Thank you all for all the wonderful answers, I so don't regret signing up now.

I love your all's sense of humor. LOL

^I should clarify, I never regretted signing up

[Dave]

I don't think the capacitor will discharge itself if you play it a Rolling Stones song.

Although, this song might just scare it into submission.

[AG6QR]

I like the lightbulb idea.  The especially nice thing is that, because of the temperature coefficient, it lowers its resistance as the voltage drops off, thus speeding up the job compared to a fixed resistance.

Though usually, I just pick a resistor such that V^2/R is less than or equal to the resistor's power rating.  If I've got time, a simple 1/4 watt resistor will do the trick, though a resistor with a higher power rating and lower resistance can do the job faster.

The formula for the resistance is R=V^2/P . For example, assuming your capacitor had 400V (I bet its actual charge is much less than this), you could use a 1/4W resistor of about 640K or higher.  R=400*400 / (0.25).  Draining a big cap at 1/4 watt won't be fast; watch the voltage with a meter before touching the capacitor terminals.

[Gall]

My contraption: an old 10W wire-wound resistor connected to a set of cheap salvaged multimeter probes via a piece of audio cable. (I replaced probes of one of my multimeters with alligator clips and thus got a set of extra probes without wires).

[Gall]

BTW, such a resistor may withstand about 100W (10 times its rating) for a short time. Wire-wound resostors of any kind are generally hard to burn.

[AmmoJammo]

Just measure the voltage across it! it may have little to no charge anyway :p

You sure its 2000uf, 400v? cos that's a big capacitor!

[rolycat]

I like the lightbulb idea.  The especially nice thing is that, because of the temperature coefficient, it lowers its resistance as the voltage drops off, thus speeding up the job compared to a fixed resistance.

You've got that backwards. The resistance of a filament bulb increases as it heats up, so it will actually be marginally slower than a fixed resistor with the same cold resistance.

The advantages of using a light bulb are that they are cheap for a given power capability and give a visible indication if there's a lot of energy in the capacitor.

[AG6QR]

I like the lightbulb idea.  The especially nice thing is that, because of the temperature coefficient, it lowers its resistance as the voltage drops off, thus speeding up the job compared to a fixed resistance.

You've got that backwards. The resistance of a filament bulb increases as it heats up, so it will actually be marginally slower than a fixed resistor with the same cold resistance.

The advantages of using a light bulb are that they are cheap for a given power capability and give a visible indication if there's a lot of energy in the capacitor.

No, I didn't have it backwards.  The cold resistance of the light bulb is so low that it would dissipate too much power at the high voltage.  So you couldn't use that resistance at the high voltage.

If you're going to use a fixed resistor, you'd have to use one that worked OK at the beginning of the discharge cycle.  For a fixed "hot" resistance, that is, a fixed maximum power dissipation, the light bulb is better than a fixed resistor, because the light bulb's resistance becomes lower as the charge drains.

[AG6QR]

I thought  light bulb filaments had a higher resistance when running hot which is why they blow when switched on from cold. The cold filament has a lower resistance and hence a higher current surge at switch on blows them. This is why slow start circuits can prolong the life of a bulb.

Yes, I believe you have that exactly right.  If you operate a bulb at a constant voltage, there's a brief initial surge, followed by a heating which increases resistance.  The hot filament both limits the current and emits light.

As the charge drains the filament will be heated at a lower rate but is it low enough to cool down in the time it takes for the capacitor to discharge and offset the initial heating?

The discharge time depends on the capacitance and voltage you're talking about, but usually I'd say yes.  In any case, you can always look at the bulb.  If the bulb dims, it's doing so because it's getting cooler, and it's decreasing its resistance compared to the resistance when it was bright.

[calexanian]

I like the lightbulb idea.  The especially nice thing is that, because of the temperature coefficient, it lowers its resistance as the voltage drops off, thus speeding up the job compared to a fixed resistance.

You've got that backwards. The resistance of a filament bulb increases as it heats up, so it will actually be marginally slower than a fixed resistor with the same cold resistance.

The advantages of using a light bulb are that they are cheap for a given power capability and give a visible indication if there's a lot of energy in the capacitor.

True. A light bulb filament in vacuum will increase in impedance at temperature.

[rolycat]

I like the lightbulb idea.  The especially nice thing is that, because of the temperature coefficient, it lowers its resistance as the voltage drops off, thus speeding up the job compared to a fixed resistance.

You've got that backwards. The resistance of a filament bulb increases as it heats up, so it will actually be marginally slower than a fixed resistor with the same cold resistance.

The advantages of using a light bulb are that they are cheap for a given power capability and give a visible indication if there's a lot of energy in the capacitor.

No, I didn't have it backwards.  The cold resistance of the light bulb is so low that it would dissipate too much power at the high voltage.  So you couldn't use that resistance at the high voltage.

If you're going to use a fixed resistor, you'd have to use one that worked OK at the beginning of the discharge cycle.  For a fixed "hot" resistance, that is, a fixed maximum power dissipation, the light bulb is better than a fixed resistor, because the light bulb's resistance becomes lower as the charge drains.

You are still confused - or at least your explanation is. The maximum power dissipation occurs when the light bulb is cold, and a power resistor with a similar cold resistance is perfectly usable with the energy levels encountered in switching power supplies.

In practice a light bulb such as the one flolic uses discharges significantly more slowly than a fixed resistor of a similar value, and this is due to the increase in resistance when it is hot:

A 470uF 400V capacitor was charged to 300V. A 60W 240V light bulb and a scope were connected and the voltage discharge curve was measured:


The light bulb had a cold resistance of around 75 ohms, so a 100 ohm 50W power resistor was substituted and the measurement was repeated:


Despite having a higher initial resistance the fixed resistor discharged the capacitor about four times faster than the lamp.

[vk6zgo]

We've discussed this many, many times. Answers range from "short that motherf-r with a screwdriver" to "discharge gently with a resistor". I'm usually in the former group...

1) 400V 2000uF is a whale of a lot of energy (160J). If it's fully charged, you probably should not just short it out.
2) If it hasn't been plugged in for a while, it's not going to be fully charged. These caps have internal leakage, and besides, the power supply would keep running until it has drained the capacitor significantly.

So bearing in mind #2, I'd just short it out. Take a leak first, though If you think it might be fully charged, pick a resistor: the capacitor will discharge in approximately 5*R*C seconds, and the resistor will dissipate V^2/R peak. You have V (400V) and C (2000uF), so pick a resistor that can drain it in a reasonable amount of time without exceeding its power rating.*

*A massive wirewound power resistor will not mind if you exceed its rating briefly.

Also keep in mind dielectric absorption - if you discharge it quickly, it can regain charge when you remove the load. You should keep a load/short attached for a minute or so to drain it off completely.

Oops, I typed too slowly, looks like you two beat me to it! Oh well, can't let these words go to waste...

How about a slightly safer way that does less harm to the capacitor itself?

I was under the impression that he was going to replace the capacitor.

I'm also of the former school,for the reasons you give,& also because the alternatives lend themselves to less than ideal implementation.
I'd rather short the thing with a well-insulated screwdriver than try to hold a resistor across it without "zapping" myself.
I'm not really sure if it does damage the capacitor,but if you're going to replace it,who cares?