How to safely discharge this capacitor faster in this application? - Page 1
Forum section: electronics / projects

author: XaviPacheco
author_rank: Regular Contributor
timestamp: October 09, 2018, 10:57:05 pm
content: Attached is the AC input circuit. When I remove the input voltage, the capacitor discharges way too slow. I can see the led (used to indicate any charge) lit for a very long time. I think I need to calculate a discharge resistor and put it across the bridge output, right? What would be the right approach?Edit: This happens when I remove the load and the ac input voltage, of course, as the capacitor doesn't have where to waste the stored charge.Additional data:C1: 2200uF/250VInput voltage: 120VAC/60HzR7: 100K

author: coppercone2
author_rank: Super Contributor
timestamp: October 09, 2018, 11:35:23 pm
content: I was told basically before a good way to size up the drain on the capacitor is to determine how long it takes to open up the enclosure and make sure the capacitor is drained to a safe voltage by the time someone opens up all the screws, and that theoretically you should build the design so that you cannot open it while it is connected, but obviously this turns into a idiocy prevention jigsaw mechanical puzzle.Some circuits will even have a blinking light that changes in blink frequency based on the voltage (often built from things like neon bulb oscillators). I have seen power supplies that stop blinking at around 70 volts or so. The idea behind them is that a flashing light is more of a warning then a simple bright LED because it really gets your attention.You need to make sure the resistor can withstand line overload conditions and such. Quite a bit of engineering to do this right.But you would time with a stopwatch how long it takes you to access the high voltage section and see if its reasonable to put a resistor that can discharge it in time.You would calculate the discharge rate by a simple RC circuit discharge formula. The main thing is to choose the right wattage (and over-rate significantly, since its an important part for safety's sake. ) and to choose a resistance that will drain it in time.Obviously more drain is safer but at some point it is considered wasted power, so there are enviromental concerns (i.e. if there is billions of these things wasting power).You could make a circuit that immediately drains it fast when unplugged too, but this should not be used unless your goal is to increase safety. Especially for an amateur device (not many people make this stuff anyway so your fine, but I don't know if there are modern safe enough equivalents for mass productions safe.. eventually there might be anyway.... but it will hard to beat a curled up wire.The other thing is that you will want this resistor to be a durable one, like carbon film, or pulse withstanding rating, or wire wound, all of a reputable brand (more so then the other design compromises you might end up making).And, even though you have a MOV, you still want the resistor sized up so it works without the MOV's protection just as well, because the MOV and Fuse are components that will likely cause the user to open up the box, followed by the capacitors, then the diodes, then the LED. You want that drain resistor to be the most reliable thing in that circuit you drew there.

author: coppercone2
author_rank: Super Contributor
timestamp: October 10, 2018, 12:05:26 am
content: http://www.vishaypg.com/docs/63501/how_to.pdfAlso, don't pick parts that are trimmed for this. Better to use lower accuracy classes since they won't fuck with it (like abrading or laser etching that makes a hot spot which decreases reliability in case of overload). You might wanna ask the manufacturer specifically what they do to the part you chose and how their bin sorting works etc.Also if forced to go SMD, use a MELF package or better.https://www.electronicdesign.com/power/carbon-film-melf-pulse-load-championThis however offers some insight as well:https://prod.sandia.gov/techlib-noauth/access-control.cgi/1981/810569.pdfI think you might wanna stick with untrimmed wire wound.Since the goal of this resistor is to discharge DC currents, you also might consider (but I am not sure) its inductance, which would be of benefit.

author: NiHaoMike
author_rank: Super Contributor
timestamp: October 10, 2018, 12:49:12 am
content: If you don't care about standby power usage, you can use a 10k, 10W resistor (actual dissipation ~3.5W). If that standby power usage would not be acceptable, add a MOSFET between the LED and ground, with a 1M or so resistor between gate and drain, a parallel 12V zener and 100nF cap from gate to ground, then a series 4007 and 1k resistor to one side of the mains such that if there's AC present, the cap will be kept discharged. Disconnect the incoming AC and the MOSFET will turn on, enabling the discharge circuit.

author: intabits
author_rank: Frequent Contributor
timestamp: October 10, 2018, 01:03:35 am
content: Use an active bleeder circuit based on a constant current sink.The transistor needs to be rated for the highest capacitor voltage. Adjust the emitter resistor to set the LED current.With values adjusted to give 10mA LED current while power is a applied, Capacitor voltage drops from 300V down to 30v 2.5 times faster.I would also add a higher value "backup" bleeder resistor in case of circuit/LED failure.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 10, 2018, 01:43:10 am
content: Industry often uses a relay and much lower-value bleeder resistor.The dumb and easy way: add a 120VAC coil relay to the input, and use the NC contact to control a 100 ohm 100W resistor connected across the capacitor.That way, when power is applied, the resistor is disconnected and no power is drawn.Eventually, this will fail, the contacts will remain closed and the resistor will melt itself into a blob. A safer circuit might use an optoisolator to detect presence of AC line, and use that to hold a transistor off. When AC line drops, the transistor turns itself on (powered by the high voltage), which turns on the resistor.Transistors are more reliable than contacts, but are more vulnerable to transients. The opto should be an AC type, or a DC type after a bridge rectifier; the current limiting resistor needs to be rated for the AC voltage and power dissipation. Often, two or more resistors are used in series, so if one fails shorted, the other can still bear the burden.One more catch: the transistor will not switch instantly. A couple transistors should be added, to implement a schmitt trigger circuit and make the switching transistor switch sharply. With fast switching, there is no danger of linear-mode power dissipation in the transistor. Finally, with faster switching comes the risk of transient voltages/currents; either slew-rate-limiting the switching transistor, or adding an R+C across it to dampen the load resistor's inductance, is all that is needed here.The inverse function is also handy on start-up, where a precharge resistor (R28, but passive, not a thermistor) is shorted out after the DC bus voltage comes up. In this case, protection needs to ensure that 1. the shorting contact never closes before the DC bus is nearly fully charged (lest it become welded from inrush current), and 2. the resistor is disconnected entirely (open circuited) if it remains on for too long and carries full load current. Again, contacts can be used, or semiconductors with benefits in reliability but also more considerations towards protection.A typical circuit uses a TRIAC based SSR on the AC line side, with a MOV at the input to prevent destruction of the semiconductors. It can be done on the DC side as well.Tim

author: Mechatrommer
author_rank: Super Contributor
timestamp: October 10, 2018, 07:24:23 am
content: Quote from: XaviPacheco on October 09, 2018, 10:57:05 pm1) understand your load requirement during operation, replace with equivalent R, and find C value for acceptable voltage sag. this way you wont put oversized capacitor. there is formula in the net but you can always SIMed it.2) during power off, set acceptable time constant and find suitable R bleed from formula t = 1/RC, R is R bleed.the shorter the t requirement, the higher the quiescent current and wasted power and the bigger the bleed resistor will be. if this is an issue, you may sort out with fancy stuff such as active bleeder. some example above, mechanical and electrical. this is another option (just for the fun of me) discharge 3.3mF power cap to a safe level in ~4sec @ 1.2A, quiescent current during operation 2.5mA or 0.3W wastage, give or take, the drawback is you have one more diode drop. size and pick your elements appropriately ymmv.

author: Henrik_V
author_rank: Regular Contributor
timestamp: October 10, 2018, 09:37:16 am
content: Or use a DPDT switchhttps://www.edn.com/design/components-and-packaging/4336763/Quickly-discharge-power-supply-capacitors

author: Zero999
author_rank: Super Contributor
timestamp: October 10, 2018, 10:11:29 am
content: Quote from: Henrik_V on October 10, 2018, 09:37:16 amYou beat me to it. Yes do that, it's the simplest option, just make sure the switch is break before make, i.e. not shorting, which 99.99% of switches are.