Discharging large capacitors in linear supply - Page 1
Forum section: electronics / general

author: akis
author_rank: Frequent Contributor
timestamp: June 07, 2014, 06:33:32 pm
content: In a typical linear supply with transformer, bridge rectifier and large electrolytic caps.How would you go about discharging the very large filter caps as soon as power is lost ?A compromise would be to bleed it with a fixed resistor, but you cannot take too much current because that is a waste, and too little bleed current will have no discernible effect.As a better solution, I was thinking of using an P-Channel FET (or PNP transistor) that is held closed by a voltage obtained by a separate bridge rectifier and a very small filter capacitor, say 10uF. As soon as power is lost, the small capacitor will discharge very quickly into a 10K resistor, and the FET will open and short the mains cap, through a 10R resistor. I attach a simple schematic. The problem is component count (extra bridge rectifier) and ideally we require a slightly higher than rail voltage to keep the PNP/FET properly closed, implying perhaps a separate secondary coil with a few more turns.If you have any ideas please let me know.

author: mariush
author_rank: Super Contributor
timestamp: June 07, 2014, 07:43:37 pm
content: A very simple way would be a relay that decouples a resistor when power supply is turned on. When power supply turns off, relay disconnects and resistor is put in contact with the large capacitor. Or use latching relay, but then your logic has to send signal to switch relay right before psu is turned off (will use less power but more complicated).Power supplies use ICs like Capzero from powerint to increase their efficiency :http://www.digikey.com/product-search/en?s=24287&FV=fff40027%2Cfff800d2&mnonly=0&newproducts=0&ColumnSort=0&page=1&quantity=0&ptm=0&fid=0&pageSize=25There's also TEA1708 from NXP which seems to be for X caps discharging though, not sure if it works for large caps :http://www.digikey.com/product-detail/en/TEA1708T%2F1J/568-11372-1-ND/4747821(here's some pdf about it:http://www.nxp.com/documents/leaflet/75017464.pdf)

author: nctnico
author_rank: Super Contributor
timestamp: June 07, 2014, 07:56:03 pm
content: Why would you want to discharge those capacitors? Some powersupplies discharge the output capacitors to have an orderly shutdown of the load.

author: akis
author_rank: Frequent Contributor
timestamp: June 07, 2014, 08:46:26 pm
content: Need to be able to discharge the caps quickly so that when I flick the mains switch the whole device powers down as quickly as possible. Although there is a relay that disconnects the load, there are other elements and components that are powered from those large caps and can take a while to shut down (many seconds). I would like the caps to be discharged in less than a second.

author: IanB
author_rank: Super Contributor
timestamp: June 07, 2014, 08:50:56 pm
content: Why not put a relay between the filter caps and the regulation circuitry, and power the relay from the transformer secondary? That way when you turn off the mains the relay will immediately isolate all the circuits downstream of the filtered supply.

author: akis
author_rank: Frequent Contributor
timestamp: June 07, 2014, 09:26:57 pm
content: That is an excellent idea - if I had any AC relays - need to buy some.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 07, 2014, 09:40:42 pm
content: If you've got an extra winding, you can use that too..Note that crowbarring circuits aren't necessarily sufficient for safety purposes. Like big old transmitter equipment with 10kV+ inside, safety switches all over the place shorting out the supply -- but they can often be defeated (manual bypass) as well. (These examples are for static safety in maintenance, but the same idea applies to a circuit that's been powered off.)Tim

author: AintBigAintClever
author_rank: Regular Contributor
timestamp: June 08, 2014, 10:54:45 am
content: Looking at it from a different angle (but by the way I'm not a PSU designer):-As it's a low voltage circuit and you're after rapid shutdown as opposed to a discharge to safe voltage levels (as opposed to discharging the HV cap in a switch-mode PSU for example) why not put the relay contacts in series with the cap? Contacts will need to be able to handle whatever current is flowing in and out of the cap. Relay can be fed by its own rectifier circuit off the same transformer secondary, this will stop the switched cap keeping the relay on.Power goes off, relay drops, cap stays charged but disconnects. Supply rail goes straight to zero.If the switch-ON surge (due to the cap already being charged from last time) is undesirable, use an SPDT relay. When on, cap is connected to the supply rail. When off, cap is connected to a low value resistor instead. The resistor never sits on the supply rail, all it ever gets is the "leftovers" in the cap.Also, what's with R1? Can you guarantee that that side of the supply is always going to be neutral, i.e. standard UK mains plug?Edit: perhaps 10R is a touch too low in my diagram, use whatever gets the job done quickly enough for you.

author: rob77
author_rank: Super Contributor
timestamp: June 08, 2014, 12:01:34 pm
content: Quote from: AintBigAintClever on June 08, 2014, 10:54:45 amit's a interesting idea, but there is a flaw in your design - the relay takes some time to switch on - typical relays need few milliseconds to switch. so you will have UNFILTERED voltage on the output for few milliseconds followed by a significant drop when the capacitor is connected. if the load is so sensitive that a quick shutdown is needed, then i doubt it'll survive the unfiltered voltage.

author: akis
author_rank: Frequent Contributor
timestamp: June 08, 2014, 12:04:59 pm
content: Quote from: rob77 on June 08, 2014, 12:01:34 pmYes a relay to disconnect the filter cap(s) from the rest of the circuit - rather than discharge, disconnect. The relay will be placed after the filter cap between cap and load.R1 is only there to allow the simulation to run.

author: AintBigAintClever
author_rank: Regular Contributor
timestamp: June 08, 2014, 12:30:48 pm
content: Quote from: rob77 on June 08, 2014, 12:01:34 pmIn that case move relay to between supply and load, like Akis said. May need a small cap (C3) to debounce the relay?

author: DJ
author_rank: Unknown
timestamp: June 08, 2014, 12:51:26 pm
content: Why not have an optocoupler kicking the fet?Power the opto straight from the ac side.

author: saturation
author_rank: Super Contributor
timestamp: June 08, 2014, 02:32:27 pm
content: A common term for what you need is a bleeder circuit, and the simplest form is a bleeder resistor.http://en.wikipedia.org/wiki/Bleeder_resistorThe principle is in the schematic. Naturally add support or protective circuitry as needed depending on the actual voltage or power involved in your PSU.In bleeder1, RLY1 is powered before the regulator circuits. AtX, R1C1 may be before the regulator circuits, thus shorting out the main filter capacitors, or it can be after the regulator circuits, discharging regulated DC output capacitors.In circuit bleeder2, an AC relay is used instead.Its a good ideal to have an LED or any annunciator reflect the status of the DC side to insure its discharged in case of failure of R1 or the relay.One advantage of old style analog meters in PSUs is the meter is powered by the output port. So unlike a digital panel meter, you know the status of your output without question. Analog meters are very reliable, and if they did fail, its easier to detect than a simple off/on LED and nothing can be gleaned once a digital panel meter is powered off even if the PSU outputs remained energized.

author: nctnico
author_rank: Super Contributor
timestamp: June 09, 2014, 06:26:35 pm
content: Quote from: akis on June 08, 2014, 12:04:59 pmDon't use relays. Relays and switching (!) large currents don't go well. Why not use an SCR? Push a small pulse into it and it will conduct by itself until the capacitors are discharged.

author: vk6zgo
author_rank: Super Contributor
timestamp: June 10, 2014, 04:55:10 am
content: You only have to worry about the external stuff.A properly designed linear supply should be capable of being turned off without any adverse effects to any of the regulator circuitry.Fit a honking great manual switch on the front of the PSU ,& use it to turn off the load,then,& only then,turn off the Mains switchA linear supply will fairly happily ride oververybrief power interruptions,due to the charge in those big capacitors.If you install circuitry which shorts those caps on loss of Mains,you may haveoutputdropouts every time you get one of these events.

author: vk6zgo
author_rank: Super Contributor
timestamp: June 10, 2014, 05:07:15 am
content: If you are really determined,you can do exactly what you want by fitting aMainson/off switch to the PSU,mechanicallyganged to another switch which either switches in a discharge path for the capacitors,or opens the DC circuit after the caps.This will not have the problems which are inherent with the automatic approach.You will,of course,have to get out of the habit of turning the thing off at the wall switch.

author: jlmoon
author_rank: Unknown
timestamp: June 10, 2014, 06:11:26 pm
content: Quoter/c time constants and laws of physics apply here.. how big are these electrolytic's, the operating voltage applied across them? I would assume the instantaneous power is going to be quite large if you're wanting to completely discharge them in < 1 sec.JLM

author: jlmoon
author_rank: Unknown
timestamp: June 10, 2014, 06:16:33 pm
content: I kinda like the soft load/short concept using a one shot pulse firing a SCR and gently discharging the caps then removing the SCR circuit if power is reapplied or when the voltage drops below a prescribed level.