Bleeder resistors for cap bank

  • Thread starter Rhaged
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In summary, when calculating the bleeder resistors for a capacitor bank rated for 1000V at 1620 microfarads, it is important to consider the desired bleed down time and the amount of current that can be wasted on the resistor. A high value resistor can be used for a longer bleed down time, but it is important to use a high voltage resistor from a reputable company. The wattage of the resistor can be calculated by the peak current through the resistor at 1000V, and it is recommended to use a resistor that is two to three times the rated power. Light bulbs can also be used as bleeder resistors, but they must be rated for the correct voltage and precautions should be taken to ensure safety
  • #1
Rhaged
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Just wondering how my Ohms/Watts should my bleeder resistors be on a cap bank ratted for 1000v at 1620 microfarads (Each cap is 200v at 2700 microfarads) so a total of 810 joules if I'm not mistaken.

(5 caps in series and 3 strings)

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Thnx in advance,
 
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  • #2
It all depend on how fast you want to bleed off and how much current you can affort to waste on the resistor. If you can affort to have a bleed down time of many seconds, you use a high value resistor. The time constant is [itex]\;\tau=RC\;[/itex], so if you calculate using 5 time constant, you can find the value of the resistor. Make sure you get a high voltage resistor from companies like Caddock or Vershey that rated for this voltage, you don't get it from Radio Shack!

You calculate the wattage by the peak current through the resistor at 1000V since it has to be on the circuit when it is power up. Make sure you use a resistor that is two to three time the rated power.
 
  • #3
Ok, So let's see if i have this correct if i want to bleed off completely in 30 seconds i just need one 54 mega ohm resistor rated for 18.519m Watts?

Its been a very long time since I've done this kind of math
 
  • #4
This is my calculation, as I said 5 time constant. You said 30 second to bleed down, so the time constant is 1/5 of it.

[tex] R=\frac {6sec} {1620\times 10^{-6}\mu F}= 3.7037\times 10^3[/tex]

That is a very low value resistor. I think you need to give it a lot more time! Or you are going to need to use relay to kick the resistor in when you power down so while the power is on, the resistor is not there to draw current unnecessary.
 
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  • #5
Just wondering couldn't i dump the bank into some light bulbs or would that result in glass every were,

And for the r=3.7037*10^3 that's in ohms right? Oh and kinda funny thing, i went to check the voltage on the bank and my work bench is a mess so i grabbed two wires with alligator clips on it. Well it wasn't hooked up to the meter was pretty much a dead short, Ears are still ringing. Thnx for the help so far.
 
  • #6
Rhaged said:
Just wondering couldn't i dump the bank into some light bulbs or would that result in glass every were,

And for the r=3.7037*10^3 that's in ohms right? Oh and kinda funny thing, i went to check the voltage on the bank and my work bench is a mess so i grabbed two wires with alligator clips on it. Well it wasn't hooked up to the meter was pretty much a dead short, Ears are still ringing. Thnx for the help so far.

As long as the bulbs were rated at the correct voltage, no problem. Your meter, being a dead short, would cause all the energy to dissipate v. quickly. I guess you won't do that again in a hurry ( or with the same meter!).
 
  • #7
Lightbulbs in series are good. Lightbulb resistance drops when they cool down, and they also provide some indication of danger. Keep in mind that lightbulbs are rated for the RMS voltage while capacitor bank reaches the top voltage.

The power dissipated is v^2/r which, for 1000v and 4KOhm gives you 250 watts.

Speaking of time constants required, if we assume that the 'safe' voltage is 20 volts, then you need voltage to drop down by factor of 50 , so you need ln(50) time constants, which is about 4.

You can use 20W 230v lightbulbs in series-parallel (two strings of five) , and a relay switch that is open when you are charging capacitors. E.g. you can use some wall wart to power the relay off the same cord that plugs this into wall. Two strings so that it still works if one lightbulb burns out.

edit: ohh, there's one other thing. Keep in mind that the values of capacitors differ a bit, so after discharging some capacitors may end up charged backwards! And during charging, some capacitors may end up under over-voltage.
I'd recommend cross-linking your chains (so that it is parallel-series) and perhaps installing diodes and resistors across capacitors.
edit: other thing. I would also use capacitance meter to arrange the capacitors so that all parallel triplets of capacitors have the capacitances as close to equal as possible.
If you substitute lightbulbs for resistors there, make sure light bulbs are very under-volted as failure of a lightbulb will lead to 1000v being applied to a single capacitor!
 
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  • #8
If this is a product, you need to be very careful using the light bulb. The socket and the wires are not rated for the voltage and you have to worry about the creepage distance, dielectric strength. Even you use a relay, be careful of the contact rating. Anything of 1000V has to be look at in a totally different light. Even if it is a garage project, you still have to be careful for your own safety and other's safety as you can see yourself already. With that kind of energy, it is very very dangerous.

Light bulb might sound very easy to do, but if it is for a product, you are going to have to justify to UL or CE how you can make things safe. If the socket are not made to take the voltage, you have to justify how to make a casing to keep people from sticking their finger in, the creepage distance and it can end up to be a lot more complicate than you thing. I design a lot of HV supply and have to pass CE test and I am very familiar with the requirement.

I was going to suggest using a 1500V NPN transistor, using some simple control circuit to turn it off during operation but turn on when you switch off the supply. THe NPN can drain the cap through a few 2W carbon resistor in series that can hold up to 1000V to drain the cap down. It might sound more complicated. But in reality, the parts are rated correctly, a good pcb layout or careful building would make it work. At that, you still have to have transient suppressor for protection for component failure to prevent HV from entering into the low voltage circuits.

Just a thought.
 
  • #9
Rhaged said:
Oh and kinda funny thing, i went to check the voltage on the bank and my work bench is a mess so i grabbed two wires with alligator clips on it. Well it wasn't hooked up to the meter was pretty much a dead short, Ears are still ringing. Thnx for the help so far.

Another thing, you better be very careful working on this project. This is highly dangerous. What you have is like the photo studio flash lighting. Even it is not very high voltage, it is high enough to penetrate the top skin. With all the stored energy, you might not know what hit you! That is the reason there is so much warning on the studio lighting. It is not the voltage that kill, it's the current. Voltage help penetration and the current kill.

When we work on high voltage, we gave it a lot of respect. You definitely do not want to have a messy bench, things can arc over. When voltage get higher and higher, many people start to treat it like black magic. You don't know where is the path or least resistance, it go to places you don't expect. 1000V might not up to that level, but if you accidentally have something close enough, you can create unexpected path and with the stored energy, it can be a killer.

What happened to you last night is not funny, take it as a big warning to you. Be careful.
 

1. What are bleeder resistors for cap bank?

Bleeder resistors for cap bank are resistors used to discharge the energy stored in capacitors. They are typically connected in parallel with the capacitors and allow a controlled discharge of the stored energy to prevent electrical shocks and damage to the capacitors.

2. Why are bleeder resistors necessary for cap bank?

Bleeder resistors are necessary for cap bank because capacitors can hold a charge even when they are not connected to a power source. This stored energy can be dangerous if not discharged properly. Bleeder resistors allow for a safe and controlled discharge of the stored energy.

3. How do bleeder resistors work?

Bleeder resistors work by creating a parallel path for the capacitor's stored energy to flow through. They have a high enough resistance to limit the flow of current and prevent damage to the capacitor while also allowing for a gradual discharge of the stored energy.

4. What are the different types of bleeder resistors for cap bank?

The two main types of bleeder resistors for cap bank are fixed resistors and variable resistors. Fixed resistors have a set resistance value, while variable resistors can be adjusted to vary the amount of resistance. Other types include wirewound, carbon composition, and metal oxide film resistors.

5. How do I choose the right bleeder resistor for my cap bank?

The right bleeder resistor for your cap bank depends on the capacitance of your capacitors and the desired discharge time. A general rule of thumb is to choose a bleeder resistor with a resistance value that is at least 10 times higher than the impedance of the capacitor. It is best to consult a professional or refer to manufacturer specifications for guidance in selecting the appropriate bleeder resistor.

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