Faster capacitor bank bleeder - Page 1
Forum section: electronics / projects

author: grythumn
author_rank: Contributor
timestamp: March 06, 2019, 04:55:32 pm
content: I have built a 400V nominal, 500V maximum, 4800uF capacitor bank for a linear accelerator experiment. (Capacitor bank and charger are in a containment vessel, rails and feed wires are shielded; standard precautions.) Right now I'm using a pair of 1M Ohm half-watt resistors as bleeders, but it takes about 3 hours to reach safe handling voltages without shorting the rails, which will really slow down the tuning process.I've thought about using switches to short out sections of a series resistor discharge string as the bank voltage falls to approximate constant power. A string of 4 incandescent light bulbs would also work, I think, but are fragile and fairly bulky.A constant power electronic load would be ideal, but most of the examples I've found are designed for less than 200V, or are expensive desktop instruments.Any suggestions?Thanks!-Bob

author: DBecker
author_rank: Frequent Contributor
timestamp: March 06, 2019, 05:13:25 pm
content: Have you considered using a standard non-isolated power supply chip, something like a LinkSwitch, to run an incandescent indicator lamp? This will give a constant power discharge -- increasing current as the voltage drops -- to under 50V.Many are rated at 700-800V, since margin is needed for reliability in the face of line voltage surges and inductive kick during plug-in. You'll be a bit close to the margin for production-level reliability, but it should be reliable enough for a one-off build.Edit: There are now industrial-rated LinkSwitch chips with 900V MOSFETs. The examples are all for isolated supplies using transformers, but look in the datasheets for the cheaper chips to see how you can use them in the "white goods" non-isolated configuration with an inexpensive stock inductor.

author: nctnico
author_rank: Super Contributor
timestamp: March 06, 2019, 05:42:20 pm
content: It shouldn't be difficult to use a standard DC load schematic and replace the MOSFET with one which can handle 600V.

author: Sylvi
author_rank: Regular Contributor
timestamp: March 06, 2019, 05:48:13 pm
content: HiWhy would use 1M as a bleeder? In tube amps it is more typical to use about 220k to 330k. Use four Rs in series-parallel to have sufficient power rating without violating the voltage rating of standard 1W metal-oxide resistors.If you have caps in series, use the bleeders to assure voltage sharing of the caps. The bleeder Rs should be selected to carry more than the leakage current of the caps - which is what you rely on when no bleeders are present, so a total crap shootYou do not need active discharge if the power source is straight-forward. It is easy to come with active discharge circuits using an n-channel mosfet and a pulse circuit for the gate, BUT...It will damage the caps if you try to discharge them too quickly - even poly caps have a maximum current rise/fall rate.

author: coromonadalix
author_rank: Super Contributor
timestamp: March 06, 2019, 06:03:16 pm
content: Same as Sylvi, just use a good resistor value and proper wattage, no need of complicated gimmickTo be safe maybe add 25% more wattage than needed.Done that, for an 450v 36000uf capacitor bank, 220k ohms at 20 watts, thats what i had on hand.

author: grythumn
author_rank: Contributor
timestamp: March 06, 2019, 06:34:57 pm
content: I used 1M Ohm as that's what I had in my parts bin with at half watt; I would've needed a bunch of 1/8 watt resistors for an equivalent.Capacitors are in parallel, PCB mounted, 500V CDE photoflash electrolytics. Power supply is a cheap ZVS boost module, switched out of circuit before a test.For the actual acceleration tests, we're using a hot rail system; no switch in the main discharge path, which is why I need to be able to discharge the bank before making adjustments to the rail separation or to remove them for inspection.I want to get the discharge time under 5 minutes. 220k works out to ~40 minutes to 50V, an hour to 20V. With a simple resistor circuit... I'd need something like a 12W, ~27k resistor. Which are... way cheaper than I thought.Okay, thanks. Off to Mouser!-Bob

author: David Hess
author_rank: Super Contributor
timestamp: March 06, 2019, 11:02:59 pm
content: You need some of those negative value resistors which draw current inversely proportional to the voltage across them. The same store which sells prop wash and blinker fluid carries them.What you are asking for makes sense given your application but is significantly more complicated than any single circuit element and seldom needed. Essentially you need a constant power load circuit and these do exist but usually make use of a multiplier while something simpler would be more appropriate. I would probably design a "reverse fold-back" or "fold-forward" current limiter which is not constant power.

author: jmelson
author_rank: Super Contributor
timestamp: March 06, 2019, 11:33:07 pm
content: Quote from: grythumn on March 06, 2019, 04:55:32 pmUse a much lower value wire-wound resistor, and a relay that keeps the resistor out of circuit until the mains supply shuts off. You do need to find a relay with wide-spaced normally-open *** ACK*** I mean normally CLOSED, so the bleed resistor is connected when the coil is NOT energized -- contacts. DC would cause arcing of the contacts when breaking under load, but this should not see that condition.A 100 Ohm vitreous enamel resistor of 50 W capacity will only draw 5 A peak at the instant the contacts close, and the cap bank will be drained in a fraction of a second, and not get warmed at all.Jon

author: Circlotron
author_rank: Super Contributor
timestamp: March 06, 2019, 11:39:27 pm
content: A 1000W 240v radiator bar is 60 ohms so two in series will have a time constant of 0.576 sec. All done in under 3 seconds. An SCR might be a better choice than a relay because of no contact bounce and consequent arcing.

author: langwadt
author_rank: Super Contributor
timestamp: March 07, 2019, 12:11:31 am
content: Quote from: David Hess on March 06, 2019, 11:02:59 pma light bulb sorta does that

author: OM222O
author_rank: Frequent Contributor
timestamp: March 07, 2019, 12:11:58 am
content: simplest and safest way: use a constant current DC load, fairly cheap and simple to build.Use an IGBT with suitable high voltage rating, any basic rail to rail op amp (I've had good results with MCP6002) and high power resistors.I strongly recommend you use a heat sink for the IGBT and have a fan blow over the whole setup. you can increase the load by using dual/quad op amps and copying the schematic over. in your case precision doesn't seem to matter which makes it extremely easy and cheap to build.Instead of a fet use the IGBT. the load current would be Vref/Rshunt and that should allow you to pick suitably rated resistors for the task.This way you should be able to discharge the caps in 3 seconds if your load can handle about 200 watts! for a more moderate 100 watts, that'll still be only 6 seconds. not too shabby if you ask me.Edit: why are you looking for constant power? your main target is to discharge the bank ... constant current would be even better

author: Circlotron
author_rank: Super Contributor
timestamp: March 07, 2019, 12:36:57 am
content: For those suggesting mosfets and igbts as part of an electronic load, beware that igbts and recent hv mosfets don’t always have a very good safe operating area. A mosfet might have a dissipation rating of several hundred watts but this rating goes right down the toilet at high voltages. Seems more of an issue with newer types than for ones 20 years or so ago.

author: David Hess
author_rank: Super Contributor
timestamp: March 07, 2019, 12:55:52 am
content: Quote from: langwadt on March 07, 2019, 12:11:31 amIt does but over a limited range.Quote from: Circlotron on March 07, 2019, 12:36:57 amThat would need to be taken into account which a constant power load sort of does by making the current inversely proportional to voltage. Of course the secondary breakdown curve is nonlinear so a nonlinear slope would be used unless the part is derated. It is easier just to derate the part for the worst case part of the curve; power semiconductors at this voltage level are not that expensive.

author: OM222O
author_rank: Frequent Contributor
timestamp: March 07, 2019, 01:10:23 am
content: Quote from: Circlotron on March 07, 2019, 12:36:57 amThat is exactly why I suggested high power resistorsyou let them take the brute force while keeping the IGBT/FET almost fully on. let's say for each 100 watt dissipated by the resistor, the IGBT dissipates 10 wattsthat seems pretty coolable and SOA won't be an issue. again this is not a matter of accuracy as the resistance of the the Rshunt will change! but who cares if the load is drawing a bit more or a bit less current? the fan will keep the resistors safe.

author: DBecker
author_rank: Frequent Contributor
timestamp: March 07, 2019, 06:26:23 pm
content: Quote from: Circlotron on March 07, 2019, 12:36:57 amThat's why I suggested a power supply chip such as a LinkSwitch powering a low voltage incandescent indicator lamp. They use a hard-switching MOSFET well within in the SOA. The bulk of the power is dissipated in the lamp, which serves as an indicator as well as a power sink.

author: jmelson
author_rank: Super Contributor
timestamp: March 07, 2019, 07:50:17 pm
content: Quote from: Circlotron on March 06, 2019, 11:39:27 pmIf power is just blipped, the SCR could stay on, and set things on fire. Also, you need a circuit to detect power loss and turn on the SCR. The arcing issue should be limited on contact closure, and the cap bank should be at near zero voltage when the relay is energized and the contacts open.Jon

author: jmelson
author_rank: Super Contributor
timestamp: March 07, 2019, 07:53:03 pm
content: Quote from: OM222O on March 07, 2019, 12:11:58 amNEVER use an IGBT in the linear mode, they are specifically built to not tolerate that. In the linear mode, the current hogs to the warmest part of the die, due to the negative temperature coefficient of Vce. Even very small currents relative to the max collector current rating will destroy them, when not saturated hard.Jon