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Discharge caps????? How to....

badbadbad

badbadbad

Well-Known Member
Got a amp, it whines when plugged in, so after about five minutes I unplugged it and it has sat for 3-4 hours.

Do I need to discharge these caps or is it safe to get in and tear it down?

If so, whats the best way? SHort the cap to ground?

Thanks
 
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Put a multimeter across the terminals and measure there DC voltage.Safest way to discharge is with a 5 watt resistor clipped across the terminals 150 ohm or so.
 
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so, if across the cap terminals shows no dc, its safe?

or the speaker terminals?

Thanks for the input
 
badbadbad said:
so, if across the cap terminals shows no dc, its safe?

or the speaker terminals?

Thanks for the input
Click to expand...

If there's no voltage across the caps, it's pretty much safe. Keep in mind that you are at ground potential. A cap could be discharged, but common to another reference voltage, so measure across the cap and from one terminal of the cap to ground. If all is zero or low, you're safe. You won't see this a lot with audio circuits, but it's a safer practice in the case of a floating cap.

Touching the speaker terminals with your hands??? ....you might get a tingle with very high output levels, but not life threatening.
 
avionic said:
Put a multimeter across the terminals and measure there DC voltage.Safest way to discharge is with a 2 - 5 watt resistor clipped across the terminals 150 ohm or so.
Click to expand...

I've always used the following method, which has worked well for me with tube amps where the voltages are high enough to hurt you.

If

P(watts) = I(amps) x V(volts) and
V(volts)=I(amps) x R(ohms)

then solving for I in the first equation and substituting into the second equation yields

P=V^2/R

I have mainly 1/4w resistors on hand, so I base my calcs on 1/4w.

The power rails on a SS amp will be around 50V, so that's the max the caps should be charged to.

1/4w = 50^2 / R

or

1/4 x R = 2500

R = 2500 x 4

R = 10000 or 10K ohms

So in order to discharge 50V, dissipating no more than 1/4w, you need to use a resistor value of 10K ohms.

Using a 150 ohm resistor you get:

P = 50^2 / 150

P = 2500 / 150

P = 16.67 watts

Is my math wrong, or is 150 ohms on the low side for discharging a cap with potential of 50V with a 2 or 5-watt resistor? I get 500 ohms using a 5-watt.

Dave
 
Its all good this time, it had zero volts at all angles and I touched it with my, uh....fingers yeah! thats it my fingers.
 
dspear99ca said:
I've always used the following method, which has worked well for me with tube amps where the voltages are high enough to hurt you.

If

P(watts) = I(amps) x V(volts) and
V(volts)=I(amps) x R(ohms)

then solving for I in the first equation and substituting into the second equation yields

P=V^2/R

I have mainly 1/4w resistors on hand, so I base my calcs on 1/4w.

The power rails on a SS amp will be around 50V, so that's the max the caps should be charged to.

1/4w = 50^2 / R

or

1/4 x R = 2500

R = 2500 x 4

R = 10000 or 10K ohms

So in order to discharge 50V, dissipating no more than 1/4w, you need to use a resistor value of 10K ohms.

Using a 150 ohm resistor you get:

P = 50^2 / 150

P = 2500 / 150

P = 16.67 watts

Is my math wrong, or is 150 ohms on the low side for discharging a cap with potential of 50V with a 2 or 5-watt resistor? I get 500 ohms using a 5-watt.

Dave
Click to expand...
OK professor..Now what length of time would it take to discharge at 16.67 watts..
 
avionic said:
OK professor..Now what length of time would it take to discharge at 16.67 watts..
Click to expand...

It depends on what the capacitance value is that must be discharged. Let's say you have a 10,000uF filter cap. The time constant using a 150 ohm resistor is:

10,000 x 10^-6 x 150 = 1.5s

Hence, it'll take you 1.5s to discharge to 37% of the initial value or 18.5V; 3s to discharge to 6.8V; and 4.5s to discharge to 2.5V and so on. I'd guess your way works good as the majority of discharge occurs during the first (short) time constant. Can a 5W resistor handle 16.67W for a second or so? I have no idea but my gut feeling is yes as I don't think enough heat will build up in a second to fry it. I'll wear my safety glasses next time I try it.

Using my method and the 10k resistor, it'd take a LONG TIME to discharge.

10,000 x 10^-6 x 10,000 = 100s.

It'd take 300s or 5 minutes to get down to 2.5V. This doesn't make tons of sense to me, as I've discharged high-voltage 450V caps using this method and they're generally down to a very low voltage in about 10 seconds using very high value resistors such as:

1/4w = 400^2 / R
1/4 x R = 160,000
R = 640K

40uF x 10^-6 x 640K = 25s...

Using 1/2w resistors I drop to 148V in 12s, 55V in 24s and 20V in 36s.

I don't generally worry about getting below 50V or so as the current reserve at that point is low, at least with tube amp caps.
 
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dspear99ca said:
It depends on what the capacitance value is that must be discharged. Let's say you have a 10,000uF filter cap. The time constant using a 150 ohm resistor is:

10,000 x 10^-6 x 150 = 1.5s

Hence, it'll take you 1.5s to discharge to 37% of the initial value or 18.5V; 3s to discharge to 6.8V; and 4.5s to discharge to 2.5V and so on. I'd guess your way works good as the majority of discharge occurs during the first (short) time constant. Can a 5W resistor handle 16.67W for a second or so? I have no idea but my gut feeling is yes as I don't think enough heat will build up in a second to fry it. I'll wear my safety glasses next time I try it.

Using my method and the 10k resistor, it'd take a LONG TIME to discharge.

10,000 x 10^-6 x 10,000 = 100s.

It'd take 300s or 5 minutes to get down to 2.5V. This doesn't make tons of sense to me, as I've discharged high-voltage 450V caps using this method and they're generally down to a very low voltage in about 10 seconds using very high value resistors such as:

1/4w = 400^2 / R
1/4 x R = 160,000
R = 640K

40uF x 10^-6 x 640K = 25s...

Using 1/2w resistors I drop to 148V in 12s, 55V in 24s and 20V in 36s.

I don't generally worry about getting below 50V or so as the current reserve at that point is low, at least with tube amp caps.
Click to expand...

I was gonna say that. :p:
 
Sure. A 40W or 60W bulb is a great way to get it done, unless you're working with hundreds of volts from a tube amp...in which case the bulb isn't going to last very long.
 
MR_B said:
Pardon my ignorance, but can you do this with leads from a light bulb?
Click to expand...

A 40w bulb has a resistance (when hot) of:

P = V^2 / R or
R = V^2 / P

120v x 120v / 100w = 144 ohms, so, pretty close to Avionic's 150 ohm resistor, and the right power rating to boot. Would work great for discharging a solid state 50 or 63v cap. The resistance is much higher when cold so the time constant calc is more difficult without the use of calculus and some coefficients relating to the filament material.

For a tube amp though, with caps charged to 400V that'd put

V = I x R or

I = V / R

I = 400 / 144 = 2.8a of current through the bulb.

P = I x V or

P = 2.8 x 400 = 1,120W

The 40w bulb filament wouldn't last long.
 
badbadbad said:
:para:

once again, I start a thread that turns into a intellectual shootout:no:
Click to expand...
Its all good though, since most of us don't remember some of the calculations since we haven't used them since we studied electronics. Which in my case was in the late 70's.
 
I use a 60W construction service coated bulb with a rubber socket and leads with alligator clips...wedding ring off, one hand in pocket, safety glasses on...caps discharged with a soft light-show.
 
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