Capacitance multiplier video follow-up questions for a bi-polar power supply - Page 1
Forum section: electronics / beginners

author: lightnb
author_rank: Newbie
timestamp: May 18, 2021, 09:33:53 pm
content: Hi everyone,I'm working on a simple circuit to convert 12VAC from a wallwart transformer into bi-polar DC with a +12VCD, 0VDC "ground", and a -12VDC rail. This is dual half-wave rectification.I tried building a version of a dual half-wave rectifier circuit that I found online that did not have a capacitance multiplier (it had 3x 4700uF caps on each rail), but, while working, there's still a lot of ripple on the rail even with a light load when viewed on an oscilloscope.Then I found and watched the video on the Capacitance Multiplier here:. It was very helpful, and I think this is the answer to smoothing the ripple.But there's a few followup questions I have on this, though, and I thought the best place to ask was on the forum. I've done some research on how to use this Capacitance Multiplier pattern on the negative rail, and I believe the answer is to use a Darlington PNP transistor instead of a Darlington NPN. (An NPN still goes on the positive rail).First Question: Now putting the caps backwards on the negative rail I understand, because the polarity flows from a higher potential to lower potential, and the "ground" is a higher polarity than the negative rail, so the "positive" side of the cap goes to "ground" because it's potential is higher than negative.Now, with a PNP transistor, I understand the current goes from the emitter to collector. So is it accurate to say that on the negative rail the power is flowingfromthe load device backtothe rail?Next on component selection: I understand that the idea is to get enough current to the base to get the transistor to operate, but the less current going in to the cap in the RC filter, the smoother the output. And as resistor size decreases (to let more current in), the required cap size increases. So use the biggest practical cap and the biggest resistor that still lets enough current in? But how much is enough? I guess the question is, how do you go about sizing and selecting the resistor, cap, and transistor to work as a set? Which part do you start with?In this case, after the smoothing caps and before the voltage regulator, the voltage is about 20 volts DC(ish). The transformer is rated for 1 amp at 12VAC on the output, and the DC output should be designed for 1 amp @+/-12VDC. But it's about 20VDC(ish) on the rail at the transistor placement point measured on a DMM. So are we using 20 volts to size the transistor because that's where it sits in the circuit? Or 12 volts because that's the desired in and out? Or 35 or 50 volts because more overhead is better and I don't care if a part costs ten cents more because I'm not manufacturing them.In otherwords, once I've drawn what I want on the schematic, how do I figure out the precise part that I need to buy? Resistors and caps are easy, but transistors have a lot of parameters. I was going to get a BJT PNP Darlington pair in a single package, and an BJT NPN Darlington pair in a single package. But there's still a lot of options to narrow down. There's 78 choices left in Digikey, after filtering by in-stock and showing only surface mount components. What spec is the most important one to start with?The next question is, how do you know when a heat-sink is required and when a through-hole package vs a surface mount part is required for thermal reasons? If we're flowing 1 amp @ 20VDC (or 12VDC at the output), how do we know how much heat is generated and if we need a physically larger package and/or a heatsink?Here's a schematic that I made in KiCAD for reference:Thanks,Nick

author: gnuarm
author_rank: Super Contributor
timestamp: May 18, 2021, 11:13:58 pm
content: I didn't watch the video because... well, for reasons, but I've never found a use for a capacitor multiplier. I'm pretty sure your circuit is not a good application for a capacitor multiplier.The capacitors on the input just after the rectifier are for storing the power that is needed when the input voltage drops below the voltage on the capacitor. The action of the cap multiplier is not a true capacitor, rather it uses the capacitor as a voltage reference and adjusts the transistor to drop voltage to maintain a constant output. But it does not source any current which is what is needed to supply power to the remainder of the circuit. So in this application the cap multiplier is doomed to fail.You will be better off using a center tapped transformer and a pair of full wave bridge rectifiers or use capacitors sized to the job. Have you done the job of calculating the capacitance required for holding the voltage required during half cycles there is no input power in each half of the circuit?I would also point out that you will never get rid of the ripple. The goal is simply to have the minimum voltage high enough that it can properly drive the output voltage regulators with their drop out voltages.

author: ledtester
author_rank: Super Contributor
timestamp: May 19, 2021, 12:28:23 am
content: QuoteWhat's your rectifier topology?http://www.hammondmfg.com/pdf/5c007.pdf

author: lightnb
author_rank: Newbie
timestamp: May 19, 2021, 11:17:07 pm
content: Quote from: gnuarm on May 18, 2021, 11:13:58 pmThe type of circuit that I'm building to rectify AC to DC is the example case that Dave gave in the video on when to use this pattern.The difference here is that I'm trying to build a bi-polar power supply, that has three outputs; a positive, a ground and a negative. His video only covers what would be the positive rail, so I'm trying to adapt it to use on a negative rail as well.Quote from: gnuarm on May 18, 2021, 11:13:58 pmSo if I understand correctly, we still want to use buffering/smoothing capacitors to store the current on the off-cycle of the AC wave, and need enough capacitors and large enough capacitors to hold this current. So we can't use a Capacitance Multiplierinsteadof real capacitors. But, I think the idea here is that when you need to get the voltage (or current?) as stable as possible, you would use the normal buffering/filter caps first to create a current buffer, but then, when there's still ripple on the supply, you use the "Capacitance Multiplier" not as a capacitor for storing current, but as a smoothing regulator to attenuate ripple before the voltage regulator, which only regulates voltage but doesn't deal with ripple at all.So my intention is to use a Capacitance Multiplier after the buffering caps for their smoothing effect, to get as clean and stable of a 12 volt signal as possible. So if I understand correctly, what you get from this pattern, which is an RC filter into an emitter follower, is the ripple smoothing effect of a giant capacitor, without any of the current buffering that a giant capacitor would give you. So all three parts, the real buffering caps (current storage), the transistor (ripple smoothing) and voltage regulator (voltage regulation) all work together to create a clean and stable power supply.You really should watch the video. When Dave demonstrates it, the ripple completely flat lines. He also demonstrates that is doesn't store current and is only for ripple reduction.Quote from: gnuarm on May 18, 2021, 11:13:58 pmI'm actually experimenting with both dual half-wave and full wave with a center tap. 3x4700uF seems to buffer enough current for a 1amp transformer with dual half wave.I used a calculator for full wave and it came up with 656uF or something odd like that. I rounded up to the next size that cap that wasn't special order, which is 1000uF.The schematic I posted is for a dual half wave design. I wanted to discuss that first to keep the scope of this question/thread small, rather than trying to discuss two different patterns at once. Second, the dual-half wave has more ripple than full wave with a center tap, so if I can size a capacitance multiplier to smooth the ripple for dual half-wave, that same part should work just fine for full wave. Third, if I'm experimenting on the bench, I want to use a circuit that has a 12v sealed wallwart as a power supply first.QuoteIt's not on that list in the PDF.It's a Dual Half Wave Capacitor Input Load, intended to produce a bi-polar output. ie. the line "hot" connects to two rectifier diodes, one backwards, to create a positive and negative rail. The line "neutral" becomes the circuit ground.The other option mentioned above is Full Wave Rectification with a center tap transformer. This uses a 24VCT (center tap) transformer into a bridge rectifier where the center tap is not connected to the rectifier, but used as the circuit ground.What I'm specifically trying to figure out is which PNP and NPN transistors to order to build an appropriate ripple smoothing circuit, and how you go about sizing them to work with each other.

author: penfold
author_rank: Frequent Contributor
timestamp: May 20, 2021, 12:00:49 am
content: A simple linear voltage regulator (a 7812, or LM317 for instance (negative versions also available), or even using a Zener diode and NPN) would produce the same desired effect, if not better as the resulting voltage would actually be 12V. There'll be plenty of videos about voltage regulators. Most linear regulators (sometimes also called low-drop-out regulators (LDOs)), will produce a suitable low ripple. Capacitance multipliers are best when you need even lower noise but have a regulated voltage to begin with.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: May 20, 2021, 12:37:06 am
content: Some hints for dimensioning:The maximum load current is an important design factor.You divide that by the Hfe of the transistor to get the minimum base current.(Note that Hfe may have a pretty big spread among a batch of transistors).You have a Base-emitter voltage of around 600mV, and on top of that some (DC) voltage drop to pull the base current through the transistor. So from that you can calculate / estimate the total DC voltage drop of your capacitance multiplier.Next step is to add the capacitor, which gives you the RC constant and the filtering.As an extra you can put a zener diode over the capacitor to make it a simple voltage regulator.But don't thing too much about theory. Get a breadboard and build the thing. Experiment with some values and different components. Observe the results and the differences when you change things.

author: lightnb
author_rank: Newbie
timestamp: May 20, 2021, 12:37:42 am
content: Quote from: penfold on May 20, 2021, 12:00:49 amThere's a 7812 and 7912 on the schematic already for voltage regulation. But Dave said in the video that voltage regulators are not suitable replacements for ripple reduction circuits. They are meant to be used together.Quote from: penfold on May 20, 2021, 12:00:49 amThat is what I'm trying to accomplish. I have capacitor buffered current and regulated voltage with a voltage regulator, but need to further reduce ripple as much as practical. The intended application is an analog audio synthesizer which uses very precise voltages to control oscillators. A stable and smooth power supply is necessary to properly tune the instrument.

author: gnuarm
author_rank: Super Contributor
timestamp: May 20, 2021, 12:48:54 am
content: Quote from: lightnb on May 19, 2021, 11:17:07 pmI'm not going to watch a half hour video just to explain why this is not a good use for the cap multiplier. If you understand what was said in the video you can reason through it to understand yourself.The video starts with a fallacy that a linear regulator is insufficient for the application. The data he presents at 4 minutes into the video only shows it performing poorly with a VERY low drop out voltage. With such a low drop out voltage nearly every spec of the regulator goes out of whack. Adding the cap multiplier will further reduce the input voltage to the regulator exacerbating that problem.He shows a real circuit at about 5 minutes but never says what the input DC level is. So how can he know what is causing the poor line regulation?QuoteThere is literally no way around this requirement. When the voltage from the AC input drops below the voltage required to drive the regulator the current has to come from somewhere.I repeat my question. Have you calculated the capacitance required to support the current and voltage to the regulators?QuoteThat has possibility, but why a capacitor multiplier? Why not use the same transistor with a fixed voltage reference? No matter how large the cap used in the multiplier, it will still put ripple on the emitter. So why not a voltage reference that has no inherent ripple?QuoteI don't think you have the voltage head room if using a 12VAC input. 17V peak before the rectifier, ~16V after. Some amount of ripple takes you down to about the voltage the regulator requires. Do you know how much is needed for the regulators to work right? Then how much does the cap multiplier need? Using a Darlington drops another ~1.5V minimum. Because of this loss the input caps need to be even larger to provide a sufficiently low ripple.Try simulating this design and you will see what's going on.QuoteExcept you don't need the cap multiplier if you pick a decent regulator and give it enough head room to operate.QuoteThat's fine, but it's not needed.QuoteFor the center tapped you will need twice the AC voltage.QuoteWhy don't you show your work here and we can take a look. What ripple amount did you assume, what current, what time?QuoteI suggest you simplify to a single positive supply irrespective of the rectifiers and transformer. Why make the same mistake on both polarities until you have a working circuit for one? In fact, I suggest you build the supplywithoutthe cap multiplier to determine there is even a need. Better yet, learn how to use LTspice and simulate the whole thing without spending a dime.A few months ago someone was trying to over design a regulator with a switcher in front to reduce the working voltage without too much heat, then cap multiplier to reduce the ripple. He changed that to a home brew linear regulator and an op amp but would not let go of the idea the cap was needed on the emitter, resulting in a mess. I never found out if he completed his project. He had something like 100 boards built with the crappy circuit he was so convinced it would work.

author: gnuarm
author_rank: Super Contributor
timestamp: May 20, 2021, 12:53:56 am
content: Quote from: lightnb on May 20, 2021, 12:37:42 amI gotta call BS on that one. I've never in my life as an electrical engineer seen any design with a cap multiplier other than a ham radio home brew circuit for a microphone power source. That was AFTER the regulator.QuoteHere is the difference between engineering and home brew. "As much as possible" means you never stop designing because there's always something else you can try.Figure out what level of noise you can live with and design to that. So start by coming up with a power supply noise spec that will let your oscillators work within their spec. Until you have that there is no use in trying to design a power circuit that may or may not meet the spec you didn't figure out.

author: penfold
author_rank: Frequent Contributor
timestamp: May 20, 2021, 01:01:03 am
content: Very sorry, that attachment only appeared as a particularly small icon, I completely missed it!The sentiment remains, however. The capacitance multiplier requires that the voltage at the collector of the NPN always be greater than the emitter. Good sizing of the rectifier capacitors should allow for a few volt's (I'll use 2V as an example) ripple (otherwise they would just be huge). So the output of the rectifier+filter capacitance would be something like 18VDC average with 2V ripple (so cycles between 17V and 19V). A low pass filtered version of this for the capacitance multiplier's base would be 18V with pretty much no ripple. The output of the capacitance multiplier would be 18V-0.7V (for the NPN's base-emitter voltage)... but the input ripple will cause the NPN to cut out when it gets below 17.3V (will happen at a slightly higher voltage, but I'm not going to work that out right now), and thus still causing a significant ripple to the regulator. Do you see the problem I'm trying to indicate?One solution around this would be to add a resistor in parallel with your C408 and C403 to reduce the voltage at the base slightly and keep the output sufficiently below the lowest voltage that's provided by the rectifier+filter cap stage.Dave's approach may work for switch-mode supplies or other supplies where the ripple is relatively small, but it is far less suited to line-frequency rectifiers without additional measures.EDIT: What's the target ripple, noise and drift, because the 78xx series aren'tbrilliant

author: lightnb
author_rank: Newbie
timestamp: May 20, 2021, 01:39:33 am
content: Quote from: Doctorandus_P on May 20, 2021, 12:37:06 amI'd like to do this, and I have a breadboard model working now without the capacitance multiplier. But I'm not sure what transistor(s) to buy to experiment with, so I feel like I'm stuck because I'm not sure what part(s) to order. Is there a good beginner focused resource on understanding the specs for ordering transistors? I get what it does, but not which specs are critical for getting the right one.Quote from: Doctorandus_P on May 20, 2021, 12:37:06 amThanks for this. I'm looking through some spec sheets now.Some spec sheets show hFE as "Static Forward Current Transfer Ratio" but Digikey calls it "DC Current Gain (hFE) (Min) @ Ic, Vce" in it's filter menu. Is it the same?So looking at a specific transistor, in this case, the FZTA14TA by Diodes Incorporated:https://www.diodes.com/assets/Datasheets/FZTA14.pdfI can see that it can handle one amp of continuous current. IC = Continuous Collector Current. Which should be good for our 1 amp circuit.It also says it can handle 30 volts of "Collector-Emitter Voltage" and "Collector-Base Voltage". That should be OK too.Now when we look at "Static Forward Current Transfer Ratio" which is labeled as the hFE mentioned above, it looks like it has three values, based on conditions.10K when the Continuous Collector Current is 10mA and the Collector-Emitter Voltage is 5V.20K when the Continuous Collector Current is 100mA and the Collector-Emitter Voltage is 5V.5K when the Continuous Collector Current is 1A and the Collector-Emitter Voltage is 5V.So the first question is, what units are we talking about when it says "10k" or "20k" or "5k"? It's just a multiplier? And second, all of the values on this sheet are based on 5 volts, but we probably have about 20 volts after the buffering caps and before the regulator. So we are talking about 20 volts going through it. Do we first need to convert the HFE spec to the correct voltage?Quote from: Doctorandus_P on May 20, 2021, 12:37:06 amSo in this case, we have one amp as the max load, and if we divide 1 by 5k, you get 0.0002 amps or 0.2mA. So does this mean we need at least .2mA on the base?Or do we need to alter that because we are dealing with 20 volts and not five volts flowing through the transistor?Thanks again,Nick

author: Doctorandus_P
author_rank: Super Contributor
timestamp: May 20, 2021, 02:28:24 am
content: If you like to tinker with breadboards, then buy a box with 600 or so transistors from your nearest china outlet. These have about 20 different transistors and are fun to play with for experiments, and compare the behavior of some different transistors. (Different vendors put different transistors in them).If you're on a tight budget, you can scavenge simple BJT's from lots of "recycled" PCB's. TO92 is easiest for breadboarding and experiments, and those have become less ubiquitous in the last 20 years..Your FZTA14 in SOT223 can only dissipate 2W, and that is quite low. It's usually not good to go anywhere near the maximum current rating of a transistor. If you need 1A, I would more be looking for a transistor that can handle 5A or so.SOT223 is very easy for breadboards, if you solder a piece of 3-pin header to it. I usually also solder a coin to the tab for both easier handling and as an ad-hoc heat sink.During normal operation the transistor has no knowledge of your input and output voltage. It only "knows" about it's three input terminals and the differential voltages between them. For a capacitance multiplier those would all be below 5V. Startup condition is different. The capacitor is empty, so there is no base voltage and the collector will (briefly) see the full input voltage.Hfe is indeed the ratio between collector and base currents. Wikipedia explains it as:https://en.wikipedia.org/wiki/Bipolar_junction_transistor#Etymology_of_hFEIf Hfe has a minimum of 5k, then your circuit probably works with only 200uA base current. Although I do not trust (or want to rely on) Hfe values, and would be more comfortable with something around 1mA. Your Collector-Emitter voltage is also probably lower (2 to 3 volts?), and then Hfe is also a bit smaller.For experiments though. Make your base current to small on purpose and observe how it fails.

author: lightnb
author_rank: Newbie
timestamp: May 20, 2021, 03:47:52 am
content: Quote from: Doctorandus_P on May 20, 2021, 02:28:24 amIs it "Current - Collector (Ic) (Max)" that I should be looking for above 5 amps? It looks like there's 4, 8, 10, 15 amp options.How about the NJVMJD122T4G-VF0?https://www.onsemi.com/pdf/datasheet/mjd122-d.pdfIt says it's rated for eight amps. So that's eight times the expected current.For heat, it says,Collector Dissipation (TC = 25 C) is 20 wattsCollector Dissipation (TA = 25C) is 1.75 wattsI looked it up and I found this:https://forum.digikey.com/t/ta-vs-tc-vs-tj-ratings/762And it seems like TA is ambient air temperature where TC is the case temperature. If I understand, that means, how many watts can you put through it until the air (TA) or case (TC) temperature rises by 25 degrees C.Now the other transistor said 2W and it was ambient (TA). So comparing ambient to ambient, this 8 amp transistor seems worse. But the other one doesn't have the case temperature rating at all. Which is more important, and how do you know when you need a heat sink?If I understand right, then this NJVMJD122T4G-VF0, when 20 watts goes through it, will have it's case temperature rise by 25 degrees C, which is an acceptable heat gain without a heat-sink. Is that correct?So I'm also trying to figure out which voltages and amperages to use where.So, if your final output, after the regulator, is 12 volts at 1 amp, then we have 12 watts as the final output. Before the regulator, the voltage is about 20v (measured on a DMM) due to the capacitors. So if it's still 12 watts being drawn through the 20 volt part of the circuit (is it?), then ohms law would say that we have 0.6 amps, 20 volts, and 12 watts on the input side of the voltage regulator, and 12 volts, 1 amp and 12 watts on the output side. Is that correct?If so, we would want to size the transistor for sustained .6 amps @ 20 VDC, which is 12 watts, and make sure the case temperature doesn't rise more than 25 degrees C at that power rating. Is that correct?Thanks,Nick

author: Doctorandus_P
author_rank: Super Contributor
timestamp: May 20, 2021, 06:41:31 pm
content: So you've learned Ohms law, and some other basic formula's such as basic power dissipation.Now learn to apply them properly.I repeat here what I wrote before:The transistor only has "knowledge" of it's own three terminals. It does not know that it's getting 20V at it's input, because it is not connected to the thing you have defined as "ground".This is a very basic concept that many beginners have some difficulty to fully understand.Look at your multimeter. It has two measurement probes.You can't measure a voltage with your multimeter when using only one probe.The multimeter measures only the voltage differential between those two probes.So when calculating the power dissipation in the transistor, then only use the currents through the transistor, and the differential voltages directly at the transistor terminals itself. A maximum DC output current of 600mA sounds about right for your 12VA transformer. If you've properly designed your capacitance multiplier, then there will only be a voltage drop of one to 3 volts over the transistor. So it's dissipation will then be 600mV * 3V = 2.1W.Simple iron core transformers are not ideal components. If you have a 12VA transformer which is rated for 1A 12V output, then those 12V are under full load. All transformers sag under load, which means that under a light (or no) load, the output voltage is higher. This is typically 3 to 6 percent for big transformers, but can be 30% or more for small transformers. (And of course output voltage also changes with input voltage, which has a pretty rough tolerance). But this is a whole different subject.Those temperatures....Nope, you've interpreted them wrong.Tc is case temperature and Ta is ambient temperature.Transistors can get quite hot when used. Your JMONDJDVJTATMDJ122-D is rated for a junction temperature of 150c, but that temperature is on the chip (junction) itself. So it means that if your ambient temperature is 25c, and the transistor dissipates 1.75Watts, then it's junction temperature will be 150c.The Tc value is for when the case has perfect cooling and is kept at 25c under all circumstances.So even if the outside of the case is kept at 25c, then with a dissipation of 25Watt the junction will still reach 150c.Dimensioning a heatsink is easiest done by trial and error. You can use the numbers above to calculate heatsink size, but the problems with all those formulas is that they omit a lot of real-life things. For example how much the air is moving. Putting a fan on a heatsink can easily improve it's ability to sink heat by a factor of 5. If you put your gadget (and heatsink) in an enclosure, then the heatsink won't work properly and it can easily overheat (after some time, a big heatsink can easily take half an hour to overheat)Dimensioning a heatsink with your fingers is very simple. If you build a circuit and you can touch the heatsink, then you can be sure the heatsink is big enough (or bigger than it needs to be). If you touch the heatsink (briefly!) with a wet finger (or drip a drop of water on it) and it sizzles, then your heatsink is too hot and you need to use a bigger one, or add a fan.Dimensioning a heatsink by trial and error is not very scientific. You can look up thermal resistances and throw in some math, and if you have all that info (including from your scavenged heatsink out of that 3rd hand gadget) then you still have big uncertainties because of the mounting method on the heatsink and the amount of air movement around it.