cap bleeder resistor - Page 1
Forum section: electronics / beginners

author: glinjik
author_rank: Regular Contributor
timestamp: December 27, 2020, 08:53:43 pm
content: Hi can anyone please advise where to put a bleeder resistor in this circuit I don't have I suitable cap for c4, so I'm using a bank of six 820mf at 200v and I don't want them charged when the circuit is not in use

author: civrays
author_rank: Newbie
timestamp: December 27, 2020, 09:25:16 pm
content: I wouldn't have thought that you would need one. There seems to be quite a few resistors in your circuit already providing a path from the +30v down to ground, so any charge remaining when you disconnect the power will discharge through those. I suppose you might want to calculate how fast it would discharge, but I wouldn't have thought it would take very long.You could add an LED on the +30v line with a suitable current limiting resistor, then you would have something visual to show you there is no remaining charge as well as providing another path for remaining charge to bleed through.Did you really mean 8x 820mF capacitors, or did you mean 8x 820uF ?

author: glinjik
author_rank: Regular Contributor
timestamp: December 27, 2020, 09:30:20 pm
content: 6 x 820mf

author: glinjik
author_rank: Regular Contributor
timestamp: December 27, 2020, 09:32:07 pm
content: 6x 820uf sorry my mistake

author: Doctorandus_P
author_rank: Super Contributor
timestamp: December 27, 2020, 10:08:55 pm
content: How fast do you want to discharge your cap's ? (And why?)If you want a fast discharge, then put an extra diode next to the main rectify diodes and charge a separate small capacitor from that, and put a resistor parralel to that, so it has a relatively quick discharge rate (A current sink will keep on discharging quicker through low voltages). After that timing circuit you can either add a simple PNP transistor to discharge the main caps, or a bit more elaborate circuit with a schmitt trigger that discharges the main capacitors quickly once the timing circuit goes below some voltage threshold.Ceramic power resistors have a relatively high thermal mass, and they can be overloaded significantly for short periods. I.e. you can put 100Watt into a 2W power transistor for a few seconds.Another option with such a timing circuit is to simply turn the output transistor of the power supply off once a voltage sag is detected. This can work relatively quickly, and then the main reservoir caps can be discharged at a more leisurely rate.Many variants are possible, and it is a good way to experiment with some simple analog circuitry to expand your knowledge.

author: glinjik
author_rank: Regular Contributor
timestamp: December 27, 2020, 10:13:28 pm
content: Speed is not an issue, but it will be connected to a set of open terminals so just want to cautious

author: innkeeper
author_rank: Unknown
timestamp: December 27, 2020, 11:38:42 pm
content: i'm with civrays, looks like that circuit will bleed itself. as someone else posted, it's more a matter of how long you want it to take.typically we talk about discharge time being 5 x RCso rough back of the matchbook math makes the 4700uf cap discharge in 162 seconds if we include the 47uf cap and its resistance in parallel with that your closer to 134 seconds...though it's not as simple as that as it is on the other side of a semiconductor and the all the other components that also bleed some current so the time might be a bit less or more. but my expectations would be this:63% discharged at 26.8 seconds86.3% discharge at 53.6 seconds95.0% discharge at 80.4 seconds98.2% discharge at 107.2 seconds99.3% discharge at 134 secondsof course, this is not just a simple resistive circuit, there are variables here i don't care to calculate for, like at what voltage level will various semiconductor components stop conducting. so that's why this is just a rough calculation.it will be 95% discharged (~1.5v) in a about min and a half, personally id be happy with that.rc discharge is a curve whereby 63% of the charge is discharged every 1 RC (resistance times capacitance) so in the first rc, you discharge 63% and in the next RC you discharge 63% of what remaining an so on. after 5 rc time constance your at 99.3 percent discharged.

author: srb1954
author_rank: Super Contributor
timestamp: December 29, 2020, 08:19:51 am
content: Quote from: innkeeper on December 27, 2020, 11:38:42 pmThe discharge time will be somewhat shorter than predicted by the RC time constants due to the action of the regulator feedback loop.Once the voltage on the input reservoir capacitors drops below the set output voltage the regulator control circuit will fully turn on the driver transistor VT2, which in turn will fully turn on the series pass transistor VT2. A larger current will be drawn through the emitter-base junction of VT1, through resistors R9 and R10 and transistor VT2 thus quickly discharging the reservoir capacitors.

author: Zero999
author_rank: Super Contributor
timestamp: December 29, 2020, 12:38:53 pm
content: The capacitor holding a charge is the least of the problems, with that circuit. It will oscillate and the voltage feedback is taken after current sense resistor, so it will have an output relatively high impedance of 0.1 Ohm.

author: innkeeper
author_rank: Unknown
timestamp: December 30, 2020, 03:14:07 am
content: Quote from: srb1954 on December 29, 2020, 08:19:51 amFair enough.. as I said I ignored all the semiconductor side of things...the explanation was more to give him/her the basic concept of RC discharge and to see if a bleeder was needed at all for his/her needs.it does get the OP in the ballpark though so they can tell it is not going to take hours or days to discharge. Next time if he needs to add a bleeder resistor or wants to know how long a cap will take to discharge, he/she can calculate it on thier own.. teaching him/her to fish.I would is curious about what he actually measures for time and hope they measure and post.It is pretty easy to poke holes in how accurate it is down to the second as I consciously ignored all the semiconductors in my replay so things like how much would actually be bleed through the pass transistor, as we have no idea the voltage setting so calculated as if it was fully conducting for simplicity sake and that's a variable based on the setting along with the tl431 .. and so on.. plus I used 4700uf instead of what he/she actually has which I realized after I posted which is 4920uf so overall it might be closer to a wash in time .... so it is totally a quick calculation for estemation to see if a bleeder was really needed and a teaching moment. Anyway like I said far more calculations and explanations than I cared to get into or felt would be understood by the OP if I got into all that, and frajly would be too far in the weeds for what they wanted to know.I mean to fully even explain how I even came with the numbers id have to explain how to calculate series and parallel resistances etc. .. so I do hope someday he/she tries to back into my numbers and can tell me why my numbers are not 100% accurate and that it has a variable aspect to it. that would make me smile.

author: innkeeper
author_rank: Unknown
timestamp: December 30, 2020, 03:33:57 am
content: Quote from: Zero999 on December 29, 2020, 12:38:53 pmI am sure the OP has a reason for using that design. maybe they will get into it. it looks like it's copied from some commercial design.personly id went for something like what s in the ti application notes for the lm317https://www.ti.com/lit/ds/symlink/lm317.pdf?ts=1609275533158there s a few simpler variations on that posted around the internet for using an lm317 to drive high currents.

author: Zero999
author_rank: Super Contributor
timestamp: January 01, 2021, 10:00:57 am
content: Quote from: innkeeper on December 30, 2020, 03:33:57 amI don't see how it can work, as drawn. It looks like several circuits stitched together. Not a single attempt is made at frequency compensation. A common emitter amplifier and op-amp in the voltage control loop is unstable and two common emitter amplifiers, plus three op-amps in the current loop will definitely cause oscillation.QuoteIt's not just the high currents which is the tricky thing. It's going down to 0V and adjustable current limiting from 0 to 3A, which makes it difficult.