Discharge circuit options?

Folsom · « **on:** 19 Oct 2012, 06:12 pm »

Hi, I am looking at making a discharge circuit for AC capacitance. I am going to be using a 50w resistor to bleed, but I don't want the bleed on while the device is in use. My initial thought is to use a relay that turns on when the device is plugged in, and when it is off the normally closed position will activate the resistor circuit to bleed the capacitors. What I am not sure of is since it is the overall circuit that the AC is turned on with, that the capacitors are parallel on, is if the capacitors will keep the relay on until it they have discharged. That would be ok if one, it could discharge them fast (compared to 50w resistor at 2k, not that fast), and if just shorting the capacitors wasn't bad for them (and it isn't good for them). Kind of comical.

Thoughts?

*Scotty* · « **Reply #1 on:** 19 Oct 2012, 08:28 pm »

There are a large number of online bleeder resistor calculators available to help you figure what value you need use in order bleed off the charge in the amount of time you desire.
http://www.welwyn-tt.com/products/resistors/calculation-tools.asp
Use the capacitor discharge calculator.
Scotty

Folsom · « **Reply #2 on:** 19 Oct 2012, 09:21 pm »

That is actually why I chose 2k, it happens basically instantly but will produce a fair amount of heat, hence 50w resistors.

The problem is getting it to happen across the resistor and not the relay. What I am trying to get around is the fact that in order for the normally closed (no power going to relay from the wall socket) operation to discharge the capacitor across the load of the 50w resistor, the relay has to turn off. What I don't think it will do is turn off while the capacitors are bridging (parallel) the relay as the power from them will just be shorted across the relay; which is a problem because it is zero load so it shortens the capacitor life greatly. I need the burden to be on the resistor in heat, instead of chemically in the capacitor (which doesn't short as fast as the resistor will bleed).

cheap-Jack · « **Reply #3 on:** 19 Oct 2012, 10:07 pm »

Hi.

Quote from: Destroyer of Smiles. on 19 Oct 2012, 06:12 pm

Hi, I am looking at making a discharge circuit for AC capacitance. I am going to be using a 50w resistor to bleed, but I don't want the bleed on while the device is in use. Thoughts?

Is it an AC circuit with an AC cap across the hot line & neutral line?
Why need 50W such a huge wattage breeding resistor? What resistance would it be? Bleeding current should never be TOO large or it will only shorten the life of the cap.

c-J

Steve · « **Reply #4 on:** 19 Oct 2012, 10:19 pm »

Deleted

Folsom · « **Reply #5 on:** 19 Oct 2012, 10:32 pm »

Quote from: cheap-Jack on 19 Oct 2012, 10:07 pm

Hi.
Is it an AC circuit with an AC cap across the hot line & neutral line?
Why need 50W such a huge wattage breeding resistor? What resistance would it be? Bleeding current should never be TOO large or it will only shorten the life of the cap.

c-J

I have intended to use 2kohm resistors. I could change that, but figured they would work fine, and relatively instant as far as the human conception is concerned (or rather faster than anyone could possibly plug in something or unplug). The reason for going 50w is just overkill. The intention of this is to be full-proof, idiot proof, because it will live a hard life! $6 isn't bad either for such a beast.

I have been wanting to avoid a switch but it may be my only option. The relay is like an automatic switch but proving difficult to use in this application. Too bad you can make them phase sensitive!

*Scotty* · « **Reply #6 on:** 19 Oct 2012, 11:00 pm »

Why do the capacitors need discharge so quickly?
Many times I seen a 100k ohm or 50k ohm resistor used across the caps to bleed the charge off.
I am not understanding your motivation for reducing the voltage on the caps within a very short time frame.
Scotty

Folsom · « **Reply #7 on:** 19 Oct 2012, 11:28 pm »

Because I don't want any shorting of them as people touch the prongs on the AC jack, or extra arching to occur as people plug stuff in.

Think drunk people, they are not going to be careful where there AC jack goes after unplugging it from the receptacle.

*Scotty* · « **Reply #8 on:** 19 Oct 2012, 11:47 pm »

The drunk people could be a problem alright. It has been my experience that drunk people have impaired cognitive functions and no longer have the most acute hearing abilities, in otherwords their golden ears turn to tin. Perhaps you put your Audio Sector amp in the system and save yourself and the drunk people shocking experiences. Alternatively, you buy one of those Octoplug AC outlet multipliers and avoid unplugging the power amp entirely.

Scotty

Folsom · « **Reply #9 on:** 19 Oct 2012, 11:49 pm »

Unfortunately I won't even be present. The intent is portability. Bands tour.

Speedskater · « **Reply #10 on:** 20 Oct 2012, 12:28 am »

I just noticed the "AC" part. It's only capacitors in "DC" circuits that have a safety concern.

*Scotty* · « **Reply #11 on:** 20 Oct 2012, 01:27 am »

If the amplifier has a tube rectifier and the amplifier is turned off it will virtually impossible to discharge the caps through a non-conducting tube rectifier. It might be possible to discharge the caps by shorting the AC plug if you have a solid state rectifier.
Have you measured a voltage present at the prongs of the AC plug.
Scotty

Davey · « **Reply #12 on:** 20 Oct 2012, 03:24 am »

What is this AC line capacitance about? Are you trying to achieve power factor correction or something?
Get rid of the capacitance is my suggestion. Then you no longer have a "bleed" issue.

Whatever you do, be damn careful.

Dave.

Folsom · « **Reply #13 on:** 20 Oct 2012, 05:15 pm »

Quote from: Speedskater on 20 Oct 2012, 12:28 am

I just noticed the "AC" part. It's only capacitors in "DC" circuits that have a safety concern.

Really? That is good to know if you are referring to the life of the capacitors. They still need to discharge at least to not scare the people using them.

This particular circuit with AC capacitance (not amplifier) is not going to change for any reason what so ever.

Speedskater · « **Reply #14 on:** 20 Oct 2012, 06:47 pm »

The capacitors in the AC power input circuit discharge 120 (or 100) times a second. These capacitors (often marked type "X" or "Y") are designed to filter high frequency noise and interference. They are connected directly across the power transformers input windings, so these windings become a short circuit when the power is turned off.
It is the capacitors in the high voltage DC secondary circuit that are dangerous. After a long exposure to high voltage DC these capacitors can remember this voltage even after being removed from the circuit. These capacitors are sometimes stored with a shorting clip across their terminals.
But in circuit capacitors in a closed enclosure do not ever need to be discharged. The danger only starts when you open the cover to do maintenance.

Folsom · « **Reply #15 on:** 20 Oct 2012, 07:32 pm »

There is no transformer. Currently it takes 1-2 seconds for them to discharge across the relay.

Speedskater · « **Reply #16 on:** 20 Oct 2012, 10:54 pm »

Then they are not AC circuit capacitors.
What are the capacitors value?
What is the pre-discharge voltage? (AC or DC?)

Folsom · « **Reply #17 on:** 20 Oct 2012, 11:45 pm »

100uf, 120v.

cheap-Jack · « **Reply #18 on:** 22 Oct 2012, 01:15 pm »

Hi.

Quote from: Destroyer of Smiles. on 20 Oct 2012, 07:32 pm

There is no transformer. Currently it takes 1-2 seconds for them to discharge across the relay.

Quote from: Destroyer of Smiles. on 20 Oct 2012, 11:45 pm

100uf, 120v.

What you said so far is so confusing. Don't you mind post the schematc of yr device or amp or whatever you are talking about?

c-J

mgalusha · « **Reply #19 on:** 22 Oct 2012, 05:08 pm »

My guess is this is part of his AC conditioner that was posted some time back and I suspect he's talking about ensuring there is no residual charge in the caps if someone pulls the plug while they are charged and then gets their fingers (or other body part) across the prongs. IMO if I'm correct, you need to have the resistor across the line at all times, you cannot count on a relay and it's circuitry always working. Food for thought..