What is going on? Strange component placement in PSU - Page 1
Forum section: electronics / beginners

author: FriedMule
author_rank: Frequent Contributor
timestamp: October 19, 2020, 06:35:06 am
content: I have stumbled upon this strange power supply for a 5GHz unit I found in an old magazine. So I can't copy more of the circuit since it may still be under copyright.On the image have I marked the "strange" components, so you know what I am asking about.1) the PSU contains two Voltage controllers, 7808 and 7805, why?2) The Zener diode (D17) is in front of a LED, why and what does it do there?3) There are two inductors (SD12K100) is they the same as SD12-102-R :https://www.mouser.dk/ProductDetail/Bussmann-Eaton/SD12-102-R?qs=%2Fha2pyFadug4QZjN0cSDOUFn6uFsp6ssRidD%2F6taA6M%3D4) At the 9V is there two protection diodes (DD214AA), what are they doing there, are they not to protect against lightning and ESD?Hope you'll please explain it to me. :-)EDIT: Sorry but U7 is reversed, please ignore that.

author: ataradov
author_rank: Super Contributor
timestamp: October 19, 2020, 07:21:40 am
content: Dual stage voltage regulation is for filtering. You first drops the voltage to 8 V, but 7808 is not great at noise rejection. Then you see a whole lot of filtering and then it drops to 5 V for the final voltage. The inductors are there for the same filtering reason.D21 / D22 logically prevent back feeding the first stage voltage regulators, but I'm not sure why it would be a concern here. They may know something about their load that would make them necessary.No idea about the Zener diode. Are you sure there is connection to the ground? R16 makes no sense here.

author: FriedMule
author_rank: Frequent Contributor
timestamp: October 19, 2020, 08:47:05 am
content: Oh, thanks for your great reply. I have checked again and yes the zener is going to ground as shown, besides it is written 8V2.Could the R16 be to make the same resistance / voltage drop as the zener and LED combined?Do you know what data the DD214AA should have to work best?

author: ataradov
author_rank: Super Contributor
timestamp: October 19, 2020, 09:08:50 am
content: Equalizing the two parts should not make any difference, since the voltage levels will be further regulated later on. I have no idea what that resistor is doing. It is possible it is some load so that 7808 is stable, but that should not be necessary.If Zener is 8.2 V, then this is some sort of over voltage indicator. I feel like it could be safely omitted.Diodes should not matter too much, but low forward voltage Schottky would be best, since 7805 needs at least 2 V headroom to regulate properly.It is also not clear why it says 9V. 7808 will produce 8V.How old is that magazine? There are probably better ways to get clean +/- 5V supply.

author: Vovk_Z
author_rank: Super Contributor
timestamp: October 19, 2020, 09:21:30 am
content: Quote from: ataradov on October 19, 2020, 09:08:50 amYes, it looks like an over-voltage indicator, but it will work no longer then LED can withstand short-circuit current.And in general - this circuit looks a bit like its designer not really know well what he do (so I don't see a reason to investigate it very deep). Possibly I am wrong because I haven't worked with 5GHz devices (but...).

author: digsys
author_rank: Unknown
timestamp: October 19, 2020, 09:24:47 am
content: Agreed with above comments, something isn't shown maybe? A tap off the center ? a second feed in there? Definitely ODD with a capital D :-)

author: Zero999
author_rank: Super Contributor
timestamp: October 19, 2020, 09:30:39 am
content: Presumably the transformer is 7.5V-0-7.5V? If so, the LM7808 and LM7908 won't do anything, because the input voltage is too low to give 8V out, with decent supply rejection.

author: ataradov
author_rank: Super Contributor
timestamp: October 19, 2020, 09:33:17 am
content: 7.5 VAC will be rectified into 10 VDC, so the 8V regulators will work. Probably not very well.The whole thing makes no sense.

author: Vovk_Z
author_rank: Super Contributor
timestamp: October 19, 2020, 09:35:58 am
content: Quote from: ataradov on October 19, 2020, 09:33:17 amIt depends.

author: FriedMule
author_rank: Frequent Contributor
timestamp: October 19, 2020, 10:04:06 am
content: The design is from the end of 2015, and made someone I understand is considered as a bit of a guru who should know a lot of what he is talking about and people are listening to.As I understand your comments should the 7.5VAC transformer, give 10V and maybe via some voltage drop, give about 8V/9V?If so should the zenor-LED maybe glow if voltage is ok? If Voltage is too high?I think it may be smartest to just follow his drawing, but yes a lot of it does not make sense. I.e. the 8.2V zener, I simply did not understand what it meant for the circuit.Also, I can not find out the impedance on the zener, the values needed on the two safety diodes or if the LED's needs to have some exact specification / color.

author: tunk
author_rank: Super Contributor
timestamp: October 19, 2020, 10:41:08 am
content: Do you know how much current that is drawn from the PSU?

author: srb1954
author_rank: Super Contributor
timestamp: October 19, 2020, 10:47:17 am
content: Quote from: FriedMule on October 19, 2020, 08:47:05 amIs it DD214AA or DO214AA as shown on your schematic?DO214AA is not a diode type number but a diode package style. It is more commonly known as an SMB package.

author: srb1954
author_rank: Super Contributor
timestamp: October 19, 2020, 11:10:17 am
content: Quote from: Zero999 on October 19, 2020, 09:30:39 amEven if the LM7808 and LM7908 somehow barely manage to get enough voltage to regulate their outputs the ripple rejection performance will be terrible. The LM7805/LM7905 won't fare much better in the ripple rejection numbers as they will have barely enough input-output differential to work with.In my experience the ripple rejection of this type of regulator performs well below spec unless there is at least 3V input-output differential to work with. With such a low AC input voltage you would probably get better overall ripple rejection from this circuit by dispensing with the LM7808 and LM7908 pre-regulators altogether and just running the smoothed DC supplies straight into the the LM7805 and LM7905.

author: Gyro
author_rank: Super Contributor
timestamp: October 19, 2020, 11:42:04 am
content: R16, the Zener and the LED make sense if there is no intermediate connection to 0V, maybe you mistook a crossover for a join (especially if it's a print from an old magazine).Maybe the circuit also makes use of the intermediate +/- 8V rails as well as the +/- 5V. Impossible to say without seeing the whole thing.

author: FriedMule
author_rank: Frequent Contributor
timestamp: October 19, 2020, 01:52:19 pm
content: You are all great! Thanks!!The schematic you can see is one I have drawn from a copy of a copy of the original article.I have checked and rechecked and I have not forgotten any part or lines, but by looking again, can it maybe look like I have mistaken a smudge for a connection. Therefor have I deleted the ground connection between R16 and D17.I have looked again and luckily is "DO214AA" very clear, it's not "DD" but on the original schematics are they shown as diodes.As I can read it, is the PSU a 3VA6 7.5-0-0-7.5 with the two zero's connected.

author: tunk
author_rank: Super Contributor
timestamp: October 19, 2020, 02:23:50 pm
content: I think Zero999 and srb1954 are right.7.5Vac has a peak of 10.6V and about 9.5V after rectification.The dropout voltage of the L7808 is ~1.6V, so it won't regulateat all (it's even worse if you add the capacitor ripple). Are yousure it says 7.5V?Edit:It could possibly work if it draws 2x0.1A and that the transformervoltage at that current is 8.5V: ~10.8V after rectification andsubtracting ~0.5V ripple gives 10.3V, which is sufficent for regulation.

author: FriedMule
author_rank: Frequent Contributor
timestamp: October 19, 2020, 03:33:45 pm
content: I'll try to find a better copy of the article, about 90% are missing, maybe there are some clues in there.To me does it all look strange and a bit like something an amateur would have drawn, but this person is apparently looked at as someone who knows his stuff. Maybe a bit like if Dave had drawn the schematic, nothing would be on the schematic without a perfect reason:-)

author: Mechatrommer
author_rank: Super Contributor
timestamp: October 19, 2020, 03:35:30 pm
content: Quote from: srb1954 on October 19, 2020, 11:10:17 amthats why inductors and diodes (D21/D22) are there. as Alex said, its input 20Vpp rectified.Quote from: srb1954 on October 19, 2020, 11:10:17 amLM7x05 will get hot! you dont know how much load they are supplying. D21/D22 can also help reducing heat. i believe that circuit had been tested, including R16/D17 (equalizing load seen by +/-, clamping -ve rail who knows? only the one who put there can tell), i believe some thought had been put into that the designer awared of the load behaviour. if its me i wont remove R16/D17 from pcb, i can "do not populate" later no problem if proved useless or unecessary. ymmv.

author: srb1954
author_rank: Super Contributor
timestamp: October 19, 2020, 08:39:41 pm
content: Quote from: Mechatrommer on October 19, 2020, 03:35:30 pmThe inductors are only 100uH so they will provide negligible filtering of any 100Hz/120Hz ripple on the supply rails. With this low value they only provide useful filtering above 10KHz. To get significant filtering of low-frequency ripple with an inductive filter you need an inductor in the range of hundreds of milliHenries.The diodes themselves provide no useful filtering of noise at low frequencies. In fact they have a negative effect on the ripple rejection because the voltage drop across them leaves less input-output voltage differential across the LM78Lxx/LM79Lxx to perform their function as regulators.Since the output regulators are 78Lxx/79Lxx the load can't be more than 100mA. At this load current, and with the low input-output voltage differential, power dissipation in the regulators is not going to be a major problem.

author: tkamiya
author_rank: Super Contributor
timestamp: October 19, 2020, 10:24:58 pm
content: I think a lot of stuff in this circuit is neither necessary nor beneficial.I'd simplify it a lot and use more modern LDO type regulators. They tend to have better noise/ripple suppression. Also, with 3 terminal regulators, capacitors across input and output to ground needs to be carefully selected. Otherwise, they may oscillate. These things likes to oscillate in VHF range. My favorite way is to use 0.1uF and 10uF in parallel.

author: Mechatrommer
author_rank: Super Contributor
timestamp: October 20, 2020, 12:23:53 am
content: Quote from: srb1954 on October 19, 2020, 08:39:41 pmditto! lets do some exercise.. PSRR @ 100Hz is maxed at -80dB, that make like what? 50uV out from 0.5V in ripple? does that really matter? who want to design a 0.5V input ripple (retarded small caps) he must be bad, no? PSRR starting from 10KHz onward is degrading bad, so i guess... thats why the inductors are there... btw i guess the design is old when LDOs are not common thing but i'll still use these jellybeans "HDOs" any days for some non-demanding tasks. ymmv cheers.

author: Zero999
author_rank: Super Contributor
timestamp: October 20, 2020, 03:09:50 pm
content: Quote from: Mechatrommer on October 20, 2020, 12:23:53 amBut that graph is probably when the difference between the input and output voltages is between 3V and 15V, not when it's operating close to drop-out, which will have poor ripple rejection.

author: Mechatrommer
author_rank: Super Contributor
timestamp: October 20, 2020, 06:01:54 pm
content: the dropout in the OP circuit is 2V i wouldnt say that is close, that is the max dropout at 1A at room temperature. edit: well before anyone spot it, there's 2 diodes dropout in rectifier so we can continue argue, putting a little bit higher output transformer will be the easy fix, or as anyone suggested except that original design. ymmv cheers.

author: FriedMule
author_rank: Frequent Contributor
timestamp: October 20, 2020, 09:40:16 pm
content: So I have finally found out a bit but if it's plausible, makes sense or is understood correctly by me, is another matter. :-)1) The two inductors should be to stop hf signal from the unit to go back out in the outlet, to keep it silent in the lab.2) The 9V should apparently be an indicator for MAX Voltage, normally will the Voltage be 8V3) the two LED's D15 and D16 is to make an "on board" indicate that there is power on both rails.4) The LED on the zener diode is to show that there are sufficient power, a sort of "power on" led meant to be mounted visible.But I could still not get information about the value of the SD12K100, it's a SD12 type but K100 is that a 1000uH like the SD12-102-R :https://www.mouser.dk/ProductDetail/Bussmann-Eaton/SD12-102-R?qs=%2Fha2pyFadug4QZjN0cSDOUFn6uFsp6ssRidD%2F6taA6M%3D

author: ataradov
author_rank: Super Contributor
timestamp: October 20, 2020, 09:49:37 pm
content: 1. It is dubious whether they will actually do anything to reject the HF noise. But I guess they don't hurt, so it is fine.2. Ok, sure.3. That makes sense. Along with D25/D26. I would say it is an overkill, but whatever.4. With 8.2 V Zener it will never be on if things are fine. And it will burn out almost immediately when voltage gets higher. I bet there is no connection between the lower part of R16 and the ground. In that case that LED will indeed show that voltage between +8 and -8 V rails is bigger than 8.2 V. And I guess the idea is that if either of the rails fails, then LED will be off.

author: FriedMule
author_rank: Frequent Contributor
timestamp: October 20, 2020, 10:08:15 pm
content: Okay thanks, it looks like "overkill" is the idea here?Yes the R16 is not connected to ground. If the circuit is hoping for 8V or below 9V, can the LED also be constructed to "burn off" voltage above the 8.2V? A sort of voltage / current drop if too much power?

author: ataradov
author_rank: Super Contributor
timestamp: October 20, 2020, 10:13:53 pm
content: Quote from: FriedMule on October 20, 2020, 10:08:15 pmWell, I don't know what the author was going for. If you showed me this schematic and asked to guess its age, I would have guessed mid to late 1980s. It has that look and feel.In modern days, just use proper LDOs and you will get way better results without that much effort.Quote from: FriedMule on October 20, 2020, 10:08:15 pmR16+D17+D18 are purely and indicator that both voltage rails are present (at least somewhat).It is not really clear why they needed so much indication. Again, feels like 1980 where you cold expect those 78xx/79xx devices to actually fail. This does not really happen with modern components anymore.

author: temperance
author_rank: Super Contributor
timestamp: October 20, 2020, 10:30:04 pm
content: This circuit has been designed by a guru. It can mean two things:Guru type 1: an audio fool.Guru type 2: a shaman, a healer of some kind, who cures people by blasting low frequency waves at peoples brains generated by circuits with 555 timers for which they charge you at least 300 bucks. The power supply must be very will filtered to keep out dangerous radiation by-products generated by nuclear plants which somehow manage to circulate on the power lines.This circuit is plain nonsense.Any other gurus with brilliant ideas about simple circuits?

author: FriedMule
author_rank: Frequent Contributor
timestamp: October 20, 2020, 10:48:10 pm
content: I think it's a 2015 design but maybe an old fellow who just love old components / designs?The Guru part was my word, I think he did work as some sort of lead tech engineer at Tektronix, not sure of his complete title, did not find out of much.

author: Mechatrommer
author_rank: Super Contributor
timestamp: October 20, 2020, 10:57:16 pm
content: Quote from: temperance on October 20, 2020, 10:30:04 pma "good" way to learn and grow..

author: srb1954
author_rank: Super Contributor
timestamp: October 20, 2020, 10:59:23 pm
content: Quote from: FriedMule on October 20, 2020, 09:40:16 pmIt is more normal for suppression inductors to be placed on the primary side of the power transformer, as close as possible to the mains inlet, to control all forms of HF emissions escaping from the unit.Without considering the emissions from any of the loads connected to this PSU the main source of emissions from this PSU is likely to be from the reverse recovery snap-off of the rectifier diodes. The position of the inductors in this circuit does nothing to control this source of HF emission.

author: FriedMule
author_rank: Frequent Contributor
timestamp: October 20, 2020, 11:09:09 pm
content: Hmm okay, do I understand correctly that the two inductors could be to reduce noise in other circuits on the transformer? I.e. control circuit, front panel or something else? I have not been able to find the complete schematic of the entire unit, but if it did contain several circuits, could the inductors then make sense?

author: srb1954
author_rank: Super Contributor
timestamp: October 20, 2020, 11:49:50 pm
content: Quote from: FriedMule on October 20, 2020, 11:09:09 pmWithout having the full circuit details it is impossible to be sure what the reasoning was behind some of the considerations that went into the design of this PSU.However, if there were other circuits connected off the +8/-8V supplies and those circuits generated significant HF noise on the supply lines then the inductors would provide useful filtering of that noise and prevent most of it reappearing on the +5V/-5V outputs.

author: FriedMule
author_rank: Frequent Contributor
timestamp: October 21, 2020, 03:31:18 am
content: Thanks, too bad I haven't the whole article but only a pert of the schematic.Anyone any idea on what safety-diodes I should use? And the value of the inductors? I don't care about the 5GHz noise, as long 2GHz-1.5GHz + downward is taken a bit care of.What about shielding? I am thinking of maybe shielding the HF circuit + an outer shield including the whole board and just let the PSU board be without shielding but kept at a distance, all inside a case.

author: ataradov
author_rank: Super Contributor
timestamp: October 21, 2020, 03:35:32 am
content: What are you powering? What makes you think it will actually generate that noise on the power supply rails?Any diodes will work, they don't do anything here. You can just remove them altogether.

author: FriedMule
author_rank: Frequent Contributor
timestamp: October 21, 2020, 04:50:35 am
content: I am building a sort of impedance tester for finished circuit boards. By adjusting the frequency, it goes up to 5GHz but I do in no way think that I will need anything above 2GHz, so as long it is okay for that, am I happy:-)

author: Zero999
author_rank: Super Contributor
timestamp: October 21, 2020, 09:58:24 am
content: Quote from: Mechatrommer on October 20, 2020, 06:01:54 pmTo get the characteristics specified on that graph, the input voltage will need to be a good volt or so higher, than the drop-out voltage. The regulator will work, at lower voltages, but its performance will gradually deteriourate, as the voltage falls. 2V pretty darn close to the drop-out specification of the regulator. Yes, the diodes are a factor, but when the mains voltage passes through zero, the filtering capacitors will discharge, causing the input voltage to the regulator to fall even more.To get decent performance from an LM78xx regulator, the peak mains voltage needs to be > output voltage + rectifier losses + ripple + 2V drop out + a volt, just to be sure.The rectifier losses are around 2V, for a full bridge rectifier.The ripple depends on the load current, filter capactor value and mains frequency. The higher the filter capacitor value, the lower the ripple and the higher the load current, the higher the ripple, with the relationships being roughty inversly proportional. At 50 to 60Hz mains frequencies, a filter capacitor is 10µF/mA of load current, gives roughtly 1V.Example:Output voltage = 8VLoad current = 200mACapacitor = 2200µFMinimum mains peak voltage = 8V output voltage + 2V rectifier loss + 2V drop out + 1V extra + 1V ripple = 14VOn top of this, the mains voltage in most countries has a wide tolerance, often +/-10%, so we need to ensure good charateristices, even when the voltage is on the low side.Add an extra 10%Minimum mains peak voltage = 14*1.1 = 15.4VThe RMS voltage rating of the transformer should be at least 15.4∕√2 = 15.4/1.414 = 10.9V.The nearest standard transfomer voltage is 12V.Note that the voltage rating of the output capacitor will need to be greater than the maximum unloaded peak secondary voltage of the transformer. If the transformer has a load regulation factor of 15% and the mains voltage is on the upper end of the tolerance:V peak = 12×√2×1.1×1.15 = 21.5V, so use a 25V capacitor.

author: FriedMule
author_rank: Frequent Contributor
timestamp: October 21, 2020, 11:36:47 am
content: The transformer should deliver 7.5VAC-0-0-7.5VAC, if that is RMS do we end with about -10.5VDC / 10.5VDC, or am I wrong?

author: Mechatrommer
author_rank: Super Contributor
timestamp: October 21, 2020, 12:01:33 pm
content: well it depends on what name printed on it... i've met transformers rated for ±12V that outputting significantly more than that.

author: tunk
author_rank: Super Contributor
timestamp: October 21, 2020, 12:13:06 pm
content: The transformer specification is at full load, at noload or light load it will be (significantly) higher.

author: FriedMule
author_rank: Frequent Contributor
timestamp: October 21, 2020, 01:23:03 pm
content: I think it's a 200mA fuse and the transformer is a 3VA6 7.5-0-0-7.5hmm is it even a transformer I can get?

author: Mechatrommer
author_rank: Super Contributor
timestamp: October 21, 2020, 01:39:35 pm
content: Quote from: tunk on October 21, 2020, 12:13:06 pmunderstood... the secret recipe is, at what percentage from full load the actual load is?

author: tunk
author_rank: Super Contributor
timestamp: October 21, 2020, 02:38:56 pm
content: In the schematics there's a L78L05 which is max 0.1A.And if 3VA6 means 3.6VA, then the max load is a bit above 40%.

author: FriedMule
author_rank: Frequent Contributor
timestamp: October 21, 2020, 09:01:39 pm
content: Thanks:-) Yes I assume 3VA6 does mean 3.6VA but chose to write it as it is on the paper.Have I maybe solved the value of the SD12K100 inductor? I found this articlehttps://how-to.fandom.com/wiki/How_to_identify_inductor_markingsSD12K100 = SD12 10% 10uHseriesPresisionvaluemultiplierSD12K100How doe it sound?

author: Mechatrommer
author_rank: Super Contributor
timestamp: October 21, 2020, 09:50:47 pm
content: Quote from: FriedMule on October 21, 2020, 09:01:39 pmit sounds like you need to work out at what cutoff frequency you want the filtering to work, and field test them with frequency sweep or... entirely throw them away you probably wont need them, the inductor convention that i have here tells differently that SD12K100 is the packaging type with inductance value comes later after the dash, but you could be right you could be wrong, anyway wild guess is not the proper way to go, make your own value there is no such recommendation in LM7800's datasheet.. other ref:https://neosid.de/en/products/inductors/inductors/tht-inductors/52093/sd-12k/100-h

author: FriedMule
author_rank: Frequent Contributor
timestamp: October 21, 2020, 11:41:55 pm
content: That sounds as a great idea!! Right now do I only own a 350MHz scope, a 70MHz sig gen, so not much possibility to generate or test around the frequency I will need.Any advice on gear I'll need? I mean something that works but is cheap-ish to come by?What do I mean by cheap-ish, I don't know, but I am willing to save up, so maybe used gear that has a good reputation.

author: Mechatrommer
author_rank: Super Contributor
timestamp: October 22, 2020, 02:39:29 am
content: thats quite obvious the bandwidth you require. iirc 5GHz? not so long ago, the common thing to get is used spectrum analyzer with track generator. if no track gen, one need separate rf signal generator, but tedious work to plot bode respond (PSRR or load regulation). even if used and old, they are still pricey. but today we have $50 NanoVNA V2 that theoritically should do the same job (currently max to 4.4GHz i just got one to accompany my KC901V) if you really need up to 5GHz, there is development for 6GHz NanoVNA taking place today you may wait for a while.

author: FriedMule
author_rank: Frequent Contributor
timestamp: October 22, 2020, 05:27:09 am
content: I think I may max have use for 2-2,5 GHz the only reason for me talking about 5GHz is that that impedance tester could go to 5GHz, but I do in no way think I'll ever need that. I understand you can get cheaper gear when below 2,5GHz but that also a lot of bad gear is made in the "low" range?