Simple capacitor charging circuit problem - Page 1
Forum section: electronics / beginners

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: July 14, 2017, 08:20:40 am
content: I have the following circuit for charging a capacitor.Basically to back to back 230Vac / 12V ac transformers so that I can check the output on a scope without earth problems.Now at the output capacitor I am expecting 230V x 1.414 = 325V DCInstead when I check with a multimeter across the capacitor I get around 25V DC.What am I doing wrong here?

author: Benta
author_rank: Super Contributor
timestamp: July 14, 2017, 10:36:46 am
content: QuoteYour schematic looks OK. The only problem could arise if your cap is very small. What size/type/voltage is it?

author: Rbastler
author_rank: Frequent Contributor
timestamp: July 14, 2017, 10:44:12 am
content: Quote from: ZeroResistance on July 14, 2017, 08:20:40 amFor single wave rectification the output voltage is more like times 1.2I would post a picture page from of the book were I read it, but with a lot of my stuff atm I don't have it here. It's some older german electronics book.

author: AndyC_772
author_rank: Super Contributor
timestamp: July 14, 2017, 10:58:34 am
content: Reverse leakage through your diode, perhaps. At negative peaks you have about 325V plus the capacitor voltage across it. If its reverse leakage means its effective resistance is comparable to your 220K resistor, then it won't be rectifying.

author: Zero999
author_rank: Super Contributor
timestamp: July 14, 2017, 11:03:41 am
content: Quote from: Rbastler on July 14, 2017, 10:44:12 amWhere did you get 1.2 from?The original poster was correct. The capacitor will charge up to the peak mains voltage which is equal to the RMS voltage multiplied by root(2) or roughly 1.414. There will be a small 1V to 2V drop for the rectifier, depending on the load current.

author: alm
author_rank: Super Contributor
timestamp: July 14, 2017, 11:08:52 am
content: I would second the question about the type of capacitor. If you unplug it, how fast does the voltage across the cap decrease? The cap might be leaky at that voltage. Try putting it across a stable DC voltage source, ideally of at least 25 V, and measure the current with a DMM (or put a 1kOhm resistor in series and measure the voltage across it). Also, what AC voltage do you measure at the secondary winding of the second transformer (i.e. before the diode)?Quote from: Rbastler on July 14, 2017, 10:44:12 amThese values are usually based on some expected ratio of capacitance to load current, and take into account the ripple that will remain at that current draw. This is pretty much a peak detect circuit, so I would expect it to get very close to the peak voltage, which would be about sqrt(2) times the nominal voltage if not heavily distorted.

author: Rbastler
author_rank: Frequent Contributor
timestamp: July 14, 2017, 11:20:31 am
content: Quote from: Hero999 on July 14, 2017, 11:03:41 amFrom and quite old book about electronics

author: madires
author_rank: Super Contributor
timestamp: July 14, 2017, 11:38:42 am
content: 325V / 220kOhms = 1.5mA (peak current)

author: alm
author_rank: Super Contributor
timestamp: July 14, 2017, 11:45:02 am
content: What can you do with that figure unless you now the capacitance and leakage of the capacitor?

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: July 14, 2017, 11:56:39 am
content: Quote from: Benta on July 14, 2017, 10:36:46 amThe cap is 2u2, 400V its seems to be either a polyester or polypropylene cap is is epoxy coated and has MPP marked on it.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: July 14, 2017, 11:59:57 am
content: Quote from: AndyC_772 on July 14, 2017, 10:58:34 amThe cap discharges quite fast when I switch of the mains. I am checking this with a multimeter, so its discharging thru the multimeter resistance and the reverse leakage of the diode perhaps??I changed the diode from 1N4007 to 1N5408 and now the voltage has moved up to 270VDC, I have checked the 1N4007 with a multimeter and nothing seems to be wrong with it.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: July 14, 2017, 12:01:26 pm
content: Quote from: alm on July 14, 2017, 11:08:52 amI measure around 220Vac

author: madires
author_rank: Super Contributor
timestamp: July 14, 2017, 12:33:44 pm
content: Quote from: alm on July 14, 2017, 11:45:02 amIt gives an idea about the order of magnitude we're talking about, and also where to look for explanations. I agree, that the knowledge about the capacitor used is important, but we can start without that from another angle of view. Experience teaches us that sometimes you don't have all the information you need, and that you have to work with what you have.

author: AndyC_772
author_rank: Super Contributor
timestamp: July 14, 2017, 02:57:50 pm
content: Quote from: ZeroResistance on July 14, 2017, 11:59:57 amExcellent!QuoteYour multimeter isn't applying hundreds of volts of reverse bias. If you have a high voltage supply, you could test the reverse leakage current under more representative conditions.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: July 14, 2017, 05:01:39 pm
content: Quote from: AndyC_772 on July 14, 2017, 02:57:50 pm270V is better that 25V but I was expecting something around 320V any idea what could be the issue?I am checking this with a multimeter probably a scope would give a better idea of what the signal looks like?

author: MrAl
author_rank: Super Contributor
timestamp: July 14, 2017, 05:06:45 pm
content: Hello,270vdc would imply an AC voltage of about 191 volts rms.If you have a scope you can (carefully) check the waveform output of the transformer(s) as sometimes the waveform wont be perfect because of the properties of the core material and the excitation current.If you suspect the diode leakage, try two in series. With 230vac rms input the half wave diode will see a reverse voltage of about 650 volts.It goes without saying these are dangerous voltages here.

author: AndyC_772
author_rank: Super Contributor
timestamp: July 14, 2017, 07:26:51 pm
content: Quote from: ZeroResistance on July 14, 2017, 05:01:39 pmIf the problem is diode leakage, then it may well be that your new diode is much better than the original one, but still not ideal.Your transformer turns ratio may not be exactly what it says on the tin.Why such a high value resistor? If you use a lower value, what happens to the capacitor voltage?

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: July 14, 2017, 07:35:45 pm
content: Quote from: AndyC_772 on July 14, 2017, 07:26:51 pmI wanted to limit the peak power dissipation in the resistor. I tried a lower resistor with 1N4007 there was no difference. I think I tried 4 100k's in paralllel.

author: alm
author_rank: Super Contributor
timestamp: July 15, 2017, 02:17:31 am
content: The latter suggests a diode that is breaking down to me. If the cap was leaky, then I would expect increasing the current to increase the voltage across it.Since the circuit is floating, you could do measurements with a scope. The signal from the second transformer could be interesting, to see how sinusoidal that 220 VAC is and what the amplitude is. Another interesting measurement would the voltage across the 220 kOhm resistor (ground clip between resistor and cap, probe tip between resistor and diode). This should be proportional to the current through the cap, and might show the diode breaking down. You would expect a positive voltage during for a few milliseconds near the maximum value of the AC signal, and possibly a negative voltage during the negative cycle if the diode is breaking down.Obviously these are potentially dangerous (hands off) measurements and you should not be using more than one scope probe at a time.

author: viperidae
author_rank: Frequent Contributor
timestamp: July 15, 2017, 02:45:12 am
content: Are you sure it was a 1n4007 and not the 50v rated 1n4001?Wouldn't that explain the 25v across the cap? Half way between +325v and -275v

author: Damianos
author_rank: Frequent Contributor
timestamp: July 15, 2017, 12:03:30 pm
content: Its a simple circuit that is a good exercise, to show us that we must not overlook the parameters!Simulating the circuit, to evade from the math calculations, we have a result that shows a maximum voltage of around 261V.This is not strange, if we take in account that the capacitor, in each period of the input signal, has a short time to charge, from a high resistance source and a long time to discharge via the load plus the leakage current of the diode.In the simulation we have to examine what happens many seconds after applying the input voltage, because the time constant of the circuit is very high relatively to the period of the signal.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: July 15, 2017, 12:43:06 pm
content: Quote from: viperidae on July 15, 2017, 02:45:12 amI just inspected the diode and its turns out to be1N5819 schottkyDon't know how it landed in the 1N4007 box.The peak reverse voltage of this diode seems to be 40V, so there was probably no rectification happening, I would have probably got some distorted signal which the multimeter was showing as 25V on the DC rangeI'm so sorry I set you'll on a wild goose chaseHaving said that I would still like to investigate why I'm getting only 270V instead of the agreed upon 325V

author: StillTrying
author_rank: Super Contributor
timestamp: July 15, 2017, 12:52:04 pm
content: Quote from: ZeroResistance on July 15, 2017, 12:43:06 pmThe 10M ohm DVM accounts for most of it, and probably the diode capacitance for a bit more, plus maybe a flat topped mains waveform.You could try a smaller resistance, ~56K, or 2 diodes in series.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: July 15, 2017, 12:56:38 pm
content: Quote from: StillTrying on July 15, 2017, 12:52:04 pmCurrently I have used 4 100K in parallel (effective resistance 25K) and diode is 1N5408, will try with 2 of these diodes in series.

author: MrAl
author_rank: Super Contributor
timestamp: July 16, 2017, 01:36:16 pm
content: Hi,Where did i hear that before, oh yeah now i remember :-)Quote from: MrAl on July 14, 2017, 05:06:45 pm