Old and basic linear power supply - Page 1
Forum section: electronics / repair

author: tony359
author_rank: Frequent Contributor
timestamp: August 22, 2020, 05:21:13 pm
content: Greetings all,I have acquired a VERY old signal generator - it's a LEVELL and it's from 1977 or whereabouts I suppose as a note inside the PSU said 1977! It can generate sinewaves and square waves up to 1Khz and up tp 20V p/p.It was working intermittently and then it stopped working entirely.It turned out the PSU was gone so I opened it up and found a capacitor and a resistor (R2) gone. I did not have a suitable large 1.2K resistor so I just botched something together for a quick test and replaced both caps: it worked but I could not keep the PSU on for more than a few seconds as the resistor was getting very hot.I ordered a 2W 1.2Kohm resistor which arrived today. I have soldered it in place in but it still gets very hot. I also found that R1 is also getting very hot, even with no load.As you can see from the schematics I am attaching, this is a very simple power supply. And probably it would be worth replacing it entirely. However, I'd like to understand what is wrong with it before I just throw it away. The PSU says "33V - 20mA" on the case.A few notes:1. The positive rail draws 11mA when the unit is running.2. if I check the current flowing through the Zener diodes (betweenAandB, opening the circuit of course), it reads 45mA, which I feel it's quite high considering the whole unit draws 12mA from positive when running. I believe this explains why the 1.2K and 33Ohm resistors are getting so hot - current had to be flowing somwhere!3. The original capacitors were 100uF or so, I did not have those so I went up in capacitance hoping that that could only make things better?Across the R2 resistor I read 50V - which is puzzling but it may be just be not fully understanding the circuit? Between 0V andCI read 68V which I suppose it's a byproduct of the fact that half of the circuit is not ground referenced?As it seems to be such a basic circuit, can anybody help me fixing this thing?Thank you!

author: SilverSolder
author_rank: Super Contributor
timestamp: August 22, 2020, 05:58:04 pm
content: Is there a switch to select between 110V and 220V mains? What is the AC voltage coming out of the secondary side? (i.e. from the bottom of the two capacitors to the point where the two diodes meet)

author: bob91343
author_rank: Super Contributor
timestamp: August 22, 2020, 06:05:47 pm
content: It does appear that the source voltage is far too high. Half voltage would be better. It just has to supply a little more than the zener diode voltage to function. Run it from 120V and all should be well. Or make sure the primaries are in series and not parallel.As it stands it is a poor design. But not so with half voltage.

author: tony359
author_rank: Frequent Contributor
timestamp: August 22, 2020, 06:45:52 pm
content: SilverSolderNo, there isn't and the box says 200-250V (I am in England). It is also "Made in England".The voltage I measure at the secondary is 36VAC. Each of the two winding is labelled 15V so that makes sense to me. Though I previously measured 46VAC - probably when the 1.2K resistor and one of the caps were open.BobThat's what I thought: voltage too high at the Zener, and they drain way too much current to keep the voltage "regulated" which in turns burns everything.I am happy to test amending the primary but I did not change it. And They are currently in series, did you mean I should change it to parallel?Some pics attached - thanks for your help!!

author: MarkF
author_rank: Super Contributor
timestamp: August 22, 2020, 07:12:41 pm
content: Why not just build a better supply with simple 3-terminal regulators?

author: tony359
author_rank: Frequent Contributor
timestamp: August 22, 2020, 07:28:36 pm
content: MarkFIndeed that’s an option.But before going that direction I’d love to understand what’s wrong with this one and ‘fix it’.The only thing is: there is a slight chance that this PSU has always been a poor design (or maybe modified by others) and that I’m chasing ghosts here. Maybe it’s always been running hot as the sun? I do not know to be honest.

author: Gyro
author_rank: Super Contributor
timestamp: August 22, 2020, 07:46:07 pm
content: Itlooksoriginal. It was obviously an add-on PSU option to replace the original battery option of four PP9 batteries.If it only has to supply 20mA max (in line with PP9 battery life), then you probably have the option of just increasing the resistor value and/or replacing it with a higher wattage one, maybe attached to the Aluminium chassis.

author: bdunham7
author_rank: Super Contributor
timestamp: August 22, 2020, 07:48:34 pm
content: What voltage do you measure between the top ends of C1 and C2 (the sides that aren't connected to each other) ?How sure are you that it was all connected this way in the beginning? Can you try connecting both transformers in parallel (watch the polarity--the sides that are connected now will need to be disconnected and crossed) and just putting the 33R resistor in between one side of the transformers and the diode junction?

author: tony359
author_rank: Frequent Contributor
timestamp: August 22, 2020, 07:51:44 pm
content: Yes I suspected it was a replacement for some batteries indeed.What resistance value should I choose? I suppose you mean R2? Would it be connected between the positive rail and ground?Just increasing the wattage of R2 may not work: R1 is burning too after a few seconds of power being on.BdunhamSorry, you posted while I was replying.I read 85.6VDC.I am 100% sure, I did only swap the capacitors and R2. I did not change how the windings are wired.Let me make a drawing of your suggestion so that I confirm I understand correctly!

author: MarkF
author_rank: Super Contributor
timestamp: August 22, 2020, 08:00:18 pm
content: Don't you need a resistor on the negative side as well?And tie the COM (i.e. 0V) to the center tap to keep it balanced.

author: tony359
author_rank: Frequent Contributor
timestamp: August 22, 2020, 08:10:00 pm
content: MarkFI was wondering 🙂The 0V is in fact connected to mains ground. Do you suggest to remove the 33ohm resistor between secondaries and ground it?

author: Kleinstein
author_rank: Super Contributor
timestamp: August 22, 2020, 08:30:15 pm
content: The 33 Ohms resistor in between the 2 transformers should not be removed in the original circuit. It reduces the current peaks when charging the filter caps and this way has a little less loss compared to using only the 1.2 K to limit the current. It would be more like making the 33 Ohms larger (e.g. 47 Ohms, maybe 100 Ohms - than with higher power rating).The power loss in the circuit is quite high as the rectified voltage is so high with the voltage doubling rectifier. When using a normal bridge rectifier and only 1 filter cap the voltage would be about half. So some 40-45 V DC. This would still be just enough to get the 2x16 V from the zeners. The 1.2 K resistor would than need to get a little smaller (e.g. some 300 Ohms), but would get less hot.The 3 Pin regulators would probably also be OK, though likely with 7815 / 7915 and no need for excessively larger filter caps. However it depends on the circuit, as the current circuit also provides current limiting. So it depends on the rest of the circuit.

author: bdunham7
author_rank: Super Contributor
timestamp: August 22, 2020, 08:32:01 pm
content: Quote from: tony359 on August 22, 2020, 07:51:44 pmIf it is very old, it may have been both marginally designed and designed for lower ranges of line voltage. The math works--you have ~30VAC which is 42.4V peak, a voltage doubling circuit so 85 volts seems right. This would give plenty of margin for the capacitors discharge cycle. If you run the xformers in parallel, you'll have half the voltage, but still more than 10 volts over the sums of the zeners. You might have to lower the value of R2 as it will drop more than 10 volts at the 12mA you said the device drew. Try 500R, you'll see 8mA thru the zeners and 12mA to the device.Or, you can bodge together a better supply as MarkF suggests.

author: SilverSolder
author_rank: Super Contributor
timestamp: August 22, 2020, 09:46:00 pm
content: Quote from: MarkF on August 22, 2020, 08:00:18 pmYou only need one resistor, because the same bias current flows through both zener diode sets. I.e. with no load connected, the current flows from the positive side of the voltage doubler (point C in the diagram), through R2, then through D6+D4, then up again through D9+D7, and back into the negative side of the voltage doubler.The zener diodes act as a kind of floating "rail splitter" so COM / 0V should NOT be tied to the transformer center tap.

author: SilverSolder
author_rank: Super Contributor
timestamp: August 22, 2020, 09:49:07 pm
content: The circuit appears to work as designed. Back in the day, using a couple of watts this way was probably not something anyone would lose any sleep over, even though it seems very inefficient by today's standards. Back then, it was probably often more important to use less components than it was to get high efficiency. That's why this doesn't use a bridge rectifier - after all, that's two more diodes!Using a 2W resistor for R2 is marginal. Probably better to use at least a 3W type, if not a 5 watter, so it can lead a more relaxed life!

author: tony359
author_rank: Frequent Contributor
timestamp: August 22, 2020, 10:18:31 pm
content: Uhm... Now I am thinking about that I just replaced R2 based on the 20mA figure printed on the case. I went back to the bin and fished out the original resistor and maybe that's a bit more than the 2W I used as a replacement - picture below.Even assuming that a 5W resistor would do the trick, what about R1 which is getting over 80 degrees seconds after I power up the PSU? Granted, I did not try WITH a load.

author: bdunham7
author_rank: Super Contributor
timestamp: August 22, 2020, 11:00:15 pm
content: The thing about power resistors is that even if you get one big enough to easily handle the load, it just cooks everything around it anyway.

author: SilverSolder
author_rank: Super Contributor
timestamp: August 23, 2020, 02:46:17 am
content: Quote from: tony359 on August 22, 2020, 10:18:31 pmBasically, these resistors will be loaded with the full 45mA at all times, no matter what load you connect to the supply (the current either flows through the zeners or the load, that's how it regulates the voltage).So, the 300 ohm resistor will burn about 0.6 watts constantly, and the 1.2K one will burn about 2.5 watts constantly. Just make sure both are over-rated and they will last a long time.

author: SilverSolder
author_rank: Super Contributor
timestamp: August 23, 2020, 02:48:22 am
content: Quote from: bdunham7 on August 22, 2020, 11:00:15 pmIn this case, there is no PCB - so as long as there is air flow around the resistor, and it is rated for the heat produced, everything will be fine for another several decades!

author: tony359
author_rank: Frequent Contributor
timestamp: August 23, 2020, 11:51:08 am
content: SilverSolderThank you.I realised I never checked the currents flowing through those resistors. Indeed I measured 43mA on R2 (Voltage drop of 50V) which makes 2.08W. And on R1 I measured 122mA and a voltage drop of 4V which makes a 0.49W dissipation.I suppose that when a resistor is rated, say, 2 Watts, it is supposed to be 100 degrees when running at 2 Watts?I botched together a 4W 1.2Kohm resistor and indeed it gets warm but ok. I suppose I will order a 5 or 6W resistor and be happy.R1 however is quickly going to 90 degrees after a few seconds. I suppose this is the way it's been running since day one?? I can replace it with a 2W one and it should be ok.Thanks so far.Now, part of this whole exercise for me is also to learn something new. So I hope you don't mind if I ask a few very basic questions to understand this circuit.1. What is the purpose of R1? Current limiting?2. I measure 50VDC voltage drop across R2 but the transformer is outputting 36V AC - which is 50.76V p/p. If I understand correctly, the diode and the capacitor will rectify the waves and make something close to (36 x 1.41 = 50.76) 50.76VDC - as this is DC, I do not consider the RMS value, that is for alternate current only, correct? So how I do end up with the whole 50V dropped by the 1.2KOhm resistor then? Is that because I had no load attached?3. How did you come up with the power values from the diagram?I appreciate your - or anybody's - help on this, thank you!!

author: aqibi2000
author_rank: Regular Contributor
timestamp: August 23, 2020, 12:37:30 pm
content: Great now that you have it up and running with low wattage resistors, get a 7912 and 7812 regulator like MarkF suggested.

author: Kleinstein
author_rank: Super Contributor
timestamp: August 23, 2020, 01:02:32 pm
content: The rectifier is build as a voltage doubling rectifier. One gets nearly the peak value at the positive and negative side. So close to the peak to peak voltage. So 2.8 time 36 V or some 100 V. Due to R1 and finite capacitor values the voltage at the caps is a little lower (e.g. some 80 V).Both resistors are there to limit the current. R1 does this before the filter cap and R2 does it behind the filter caps. R1 is slightly (with the very large caps the difference is not large) more effective (less heat) here, as it takes more of the current when the AC voltage is lower. R1 sees the RMS current and thus more current, so it takes less resistance at R1.Chances are one could than get a large value for R1 and increase it's power rating correspondingly. One may than get away with the original 1.2 K resistor. With twice the caps twice the value for R1 should be OK.

author: SilverSolder
author_rank: Super Contributor
timestamp: August 23, 2020, 01:39:39 pm
content: The circuit consists of two main parts: a voltage rectifier/doubler part, and a zener regulator part. Let's look at each individually.During the positive part of the sine wave from the transformer, C2 is charged via D2.During the negative part of the sine wave, C1 is charged via D3.The DC output + and - are made from C1 and C2 in series, hence the voltage is double what it is on either C1 or C2 alone.You could try measuring the DC value with your voltmeter and see if it hovers around 100V as @Kleinstein predicted. Use alligator clips or something, keep fingers away, 100V is enough to kill in the "right" circumstances!

author: tony359
author_rank: Frequent Contributor
timestamp: August 23, 2020, 02:27:21 pm
content: Oh don't worry, I am very careful. I do use alligator clips and I always discharge the caps before fiddling. I also have a incandescent bulb current limiter on the line. I appreciate the warning though!Yes, I think understand that part. I have 36VAC at the transformer which are then rectified by the diodes - D2 will only keep the positive part and D3 the negative part. The capacitors will charge and then smooth the "missing wave" in between each one. I can see that with my scope, there is about 1V of ripple. Hence if 36AC is my RMS value, the actual sinewave is around 36 x 1.41 = 50.76V peak to peak.Hence my Positive rails becomes around +50VDC and my negative rail becomes around -50VDC - which measure around 100VDC with a DMM. I made a very professional sketch below.

author: SilverSolder
author_rank: Super Contributor
timestamp: August 23, 2020, 02:54:09 pm
content: OK so I re-drew the zener part of the circuit to make it more obvious what is happening.The zener diodes always drop a total of 32 volts (nominally 8 volts each, but maybe a little more in reality). So the resistor R2 has to drop the remaining voltage, for example 100V - 32V = 68VWhen you connect a load that uses current from the +17 or -17 volt rail, what happens is that the load "steals" current that the zener would otherwise have taken.The way this happens is that as the load uses current, the voltage across the zener drops ever so slightly, causing the current through the zener to drop to make up for what the load is using. The drop in voltage required to change the current is tiny - this is "secret" to how the zener maintains a near constant voltage across it.This way, the current through R2 is essentially constant no matter what load you put on the supply... the current either flows through the zeners or through the load, the sum is always a constant (give or take a tiny little bit).

author: bdunham7
author_rank: Super Contributor
timestamp: August 23, 2020, 03:48:20 pm
content: Quote from: SilverSolder on August 23, 2020, 02:48:22 amIt's still in a closed box, along with a transformer that is constantly at or near full power. Also, we have no idea if it has lasted 'several decades', as the total runtime might actually be quite low. It may have failed years ago, then sat in a closet for 20 years at which time it was sold on eBay as 'pulled from a working environment'.I think it would be a perfectly interesting and rewarding exercise to see how cheaply and easily it could be modified using 'modern' components. I think the design objectives are a well-regulated +/- 17VDC supply up to 20mA with short circuit tolerance. I think it can be done with a BOM under $2, but perhaps we should make it a contest!I would:a) retain point-to-point wiring, no need for PCBsb) parallel the secondaries with a 75R current 3W resistor in series with each winding to limit short circuit current to 200mA (0.2A x 15V = 3VA, the rating of the xformer) and to balance the secondary currents to some extent, as the primaries are in series.c) retain the half-wave voltage doubler configuration.d) replace R2 and the zeners, which appear to be externally mounted, with an LM317/LM337 combo + 4 resistors to set the voltage. I don't think there would be any need for heat sinks in any case, but if you really wanted to make it bulletproof (impervious to sustained overload or reverse transient currents) you could add small heatsinks and regulator bypass diodes.

author: tony359
author_rank: Frequent Contributor
timestamp: August 23, 2020, 04:46:04 pm
content: Bdunham,I am not ruling out to improve the PSU - My approach here is to understand what's happening, why it happened, fix the PSU and then maybe improve it. I am in stage 2 right now, let me have it fully up and running and then I may move to phase 3The LM317 is an interesting item. I was wondering how I would be able to get 17V when regulators are usually 5-12V.SilverSolderMany thanks for taking the time to draw the schematic for me. Let me study it and experiment with the PSU and I'll get back to you

author: bdunham7
author_rank: Super Contributor
timestamp: August 23, 2020, 05:06:04 pm
content: Quote from: tony359 on August 23, 2020, 04:46:04 pmOK! It's just regulation by dissipation, relatively common in old low-power designs. It's not just cheap stuff, even fairly expensive audio equipment used this method for secondary power supplies. It's quite common for these things to burn up, especially as they get older and line voltages go up, encroaching on their thermal margins. I'm fixing some Fluke calibrators right now (very expensive units) that use zeners as dropping resistors before LM-series regulators to get a +/- 15V supply from +/-40V rails. It isn't pretty--the power supplies look like they've been on fire. Your current design will work with the proper size components, and as you said, it will run as hot as the sun inside that box. Perhaps it really wasn't intended for continuous use if it was a battery replacement device.And, just in case you missed it, the LM317 is not (specifically) a 17 volt regulator!

author: Wallace Gasiewicz
author_rank: Super Contributor
timestamp: August 23, 2020, 08:37:29 pm
content: What happens if you ground the link between the primary windings of the two transformers (to the 240V third wire)Essentially making them two 120 V transformers.Or linking the secondaries together in parallel??

author: MarkF
author_rank: Super Contributor
timestamp: August 23, 2020, 09:36:31 pm
content: Quote from: Wallace Gasiewicz on August 23, 2020, 08:37:29 pmYou NEVER ground the primaries.You either put them in parallel or series (noting the winding direction).

author: bdunham7
author_rank: Super Contributor
timestamp: August 23, 2020, 09:50:25 pm
content: Quote from: MarkF on August 23, 2020, 09:36:31 pmI had to go back and look because your drawing of the transformer(s) shows a single core where the OP shows two transformers. In fact, the OP's photo shows a single xformer, as you have drawn. That's an important difference!No worrying about balancing primary winding voltages and no choice in how you connect the windings.

author: tony359
author_rank: Frequent Contributor
timestamp: August 23, 2020, 10:23:17 pm
content: apologies for that, I could not find a way to draw a single transformer with two windings and I forgot to mention that detail.

author: Wallace Gasiewicz
author_rank: Super Contributor
timestamp: August 23, 2020, 10:31:13 pm
content: Yea I think I meant wiring the transformer primaries for 120, Since there is no center tap so cannot be done.Just wondering why so much electricity is wasted and the resistor has such a load.Does not seem right.

author: bdunham7
author_rank: Super Contributor
timestamp: August 24, 2020, 12:50:08 am
content: Quote from: tony359 on August 23, 2020, 10:23:17 pmNo worries. But if you want to configure the new supply with dual full-wave rectifiers, that will work nicely as well. The primaries can't become unbalance if they share a core.

author: MarkF
author_rank: Super Contributor
timestamp: August 24, 2020, 01:27:41 am
content: Quote from: Wallace Gasiewicz on August 23, 2020, 10:31:13 pmCan be done.You wire the two primary coils in parallel to double the current for 115V.For 240V the coils are in series which requires half the current for the same power rating.I don't understand your question.The photo of his transformer is exactly the same as the wiring diagram I posted.You can wire it for either voltage.If he builds a +/- supply with a full wave bridge, he will also tie the secondary windings in series and create the center tap.

author: Kleinstein
author_rank: Super Contributor
timestamp: August 24, 2020, 06:59:45 am
content: In the old days designers were not that afraid of more heat as they were used to power hungry tubes.Electrolytic capacitors were more expensive so they used lower capacitance. With more heat wasted at the resistors the regulation from the zeners gets a little better. It is still a slightly odd circuit.Another point is that in the old times the mains voltage was more variable - so they may have designed for a 20% drop in mains voltage.2 x 15 V AC is normally just enough to get some +-16 V DC.Something like LM317 would however not include a current limit to some 60 mA like the original circuit. For most circuit it would still be OK.The 33 Ohms resistor may also be a fusible one and serve as a fuse.

author: SilverSolder
author_rank: Super Contributor
timestamp: August 24, 2020, 01:36:41 pm
content: This circuit would never be designed this way today, but it is kind of cute and probably gets the best out of the handful of parts that are used.In the 1970's, the decisions might have looked like this:1) Using just 2 diodes for rectification is a cost saving over a bridge rectifier.2) Using a voltage doubler allows use of smaller (cheaper) capacitors as @Kleinstein said - makes more ripple but is dealt with by the drop resistor and the zeners which we have to have anyway, so this advantage comes essentially for free.3) Voltage doubler may allow getting away with less copper windings on the transformer, so we can use a cheaper model4) The circuit gets away with just one expensive power resistor to drop the voltage for both positive and negative rails.The guy that designed this circuit could only dream of what we have today... where it would be cheaply possible to use a switch mode power supply having dozens of advanced components, including a microprocessor if we wanted, with a total BOM cost possibly lower than he had!

author: bdunham7
author_rank: Super Contributor
timestamp: August 24, 2020, 01:40:07 pm
content: 5) The power factor is probably pretty good compared to most 'better' designs!

author: tony359
author_rank: Frequent Contributor
timestamp: August 28, 2020, 08:29:48 pm
content: @silversolderI studied the diagram and did some tests on the actual unit. I do not measure 100V but 85V - I believe @Kleinstein mentioned that because of R1 and the capacitors the value was not going to be exactly VAC x 2.8 but more in the 80 range - which is exactly what I am measuring.I now understand why R2 is reading 50V across it - the way you draw the circuit is very clear, thank you a lot!Indeed I see the voltage slightly dropping when I connect a static load. Coincidentally, I played with Zeners just a few days ago on a breadboard and I know what you mean that a tiny change in voltage causes a large change in current - which is how Zeners work indeed. I have also tested with my thermal camera and indeed I can see the Zeners getting much hotter when I switch the unit off from the selector (which only disconnects the main PCB but not the PSU).After replacing the resistors with a 5W and 2W versions and verified that the whole thing still burns like a star, I think I am happy to move to phase 3 and improve on the PSU as many here recommendedI understand the LM317 (and I guess LM337 for the negative rail) are the variable regulators of choice - Wouldn't a voltage regulator get very hot when dropping lots of volts though? . I do not have the skills to design something like that by myself so I would appreciate your help.Based on the typical drawings found on the datasheets, it could be something like that - but this is the first time I am doing something like this so I am 100% sure that it'll make many of you cringe!Please be kind! (Yes, I know 3.024Kohm resistors do not exist, but I am struggling a bit finding the correct values so for now I put the ones given by the calculator!)https://crcit.net/c/d71559e0902e4b2d84d78f4fad22bbb7Thank you!

author: bdunham7
author_rank: Super Contributor
timestamp: August 28, 2020, 09:49:34 pm
content: Connect the point where the two secondaries are common to the point where the two input capacitor are common. Add 75R (or something) resistors to the non-common sides of the secondary if you want short circuit current limiting.Also, I would put the larger capacitors on the input side, you shouldn't need so much on the output side. You need reservoir caps to supply power when the transformer can't, but the regulators only need small output caps to be stable. And the term 'mF' is confusing, nowadays it refers to a really, really big capacitor. Years ago, when capacitors that size were all but impossible, this term was used to mean 'microfarad', but nowadays 'uF' is needed because 100 millifarad capacitors are perfectly common--and there are even 'kFkilofarad units!The regulators will not get hot because the current through them is very small--just the load and the small current through the divider resistors. They will be dissipating less than 1 watt and likely won't need heat sinks at all.

author: MarkF
author_rank: Super Contributor
timestamp: August 28, 2020, 10:25:10 pm
content: A summary of changes.Use the LM337 on the negative side instead of two LM317'sThe power(P) dissipated by the regulators is the (voltage across regulator) * (current)It's the combination of both voltage and current, not just the voltage.I would use 5K multi-turn pots instead of the 3K resistors.https://www.mouser.com/ProductDetail/Vishay-Sfernice/T93YA502KT20?qs=%2Fha2pyFaduiha0Qd7DX8mLYS%252BFwYHADuGWgWC1%2F3SdC2ktWgCdFBcA%3D%3DEdit-Your LM317 on the negative side will NOT work as you have it.

author: SilverSolder
author_rank: Super Contributor
timestamp: August 28, 2020, 11:11:40 pm
content: Quote from: tony359 on August 28, 2020, 08:29:48 pmLearning by doing is the best way, well done!For your improved circuit, the suggestions from @bdunham7 and @MarkF appear to be sound, although I have never actually used the LM337 or LM317 in a project - but these things tend to be quite docile.The way these regulators work isn't like a zener diode. These things actually pass the entire current through a large pass transistor, which is controlled by a built-in op amp. This way, pretty much all the current goes to the load. Only a very small current flows in the "voltage setting" part of the circuit. So the power dissipated by them is equal to V*I, where V is the voltage drop across the regulator, and I is the current going to the load (I guess, 22mA max).My penny's worth on the design discussion so far is that I would stick with fixed resistors instead of putting a pot or a trimmer in the circuit. The old design that is being replaced was not ultra precise, so using fixed 1% resistors will be more than good enough to set the voltage reliably.The less parts used, the more reliable the result will be (all else being equal). The reason for disliking trimmers and pots is that over time, they have a tendency to go bad (usually by the wiper arm going open circuit or intermittent against the track). Depending on the circuit configuration, this can cause the voltage to either max out or drop to zero... there are commercial power supplies where it maxes out! - needless to say, that usually isn't good for the load connected to the supply...The other thing to consider is that the improved circuit as it stands will be able to supply much more current than the 40 odd mA that the old one was capable of. It is probably wise to add resistors on the transformer outputs to the bridge as @bdunham7 suggested, in order to put a limit to this - making sure they are high enough wattage to handle a short circuit of infinite duration if necessary. Also, it would be wise to calculate how much power will be dissipated by the regulator(s) during a short circuit condition - this might influence your decision as to whether the regulators should be bolted onto the case for cooling, or if it is OK with a small heat sink internally, or even no heat sink at all!

author: MarkF
author_rank: Super Contributor
timestamp: August 28, 2020, 11:42:51 pm
content: I'm just going to toss this out there.If you do want the pots to be able to adjust the voltage (your preference), I saw a configuration where by the outputoutput voltage drops to 0V if the wiper goes open. I never wired this up and would be interested in how well the voltage adjust works.I wouldn't be surprised though if you couldn't just get away with +/-15V fixed regulators.

author: SilverSolder
author_rank: Super Contributor
timestamp: August 29, 2020, 01:13:19 am
content: You can get 18V fixed regulators as well - a couple of those might be a good solution?

author: ArthurDent
author_rank: Super Contributor
timestamp: August 29, 2020, 02:39:59 am
content: I think the easiest way to fix what you have is to and not have 100 VDC across a string of 4 zeners is to rewire the circuit as shown in the redrawn schematic below. The resistors in series with the rectifiers can limit inrush and can also be selected so too much current doesn't flow through the zeners. If the secondaries are 15 VAC each, the rectified peak would be about 23 VDC. By adjusting the values of the resistors before and after the rectifiers you can adjust the current and spread the dissipation between the resistors.

author: SilverSolder
author_rank: Super Contributor
timestamp: August 29, 2020, 02:53:05 am
content: Quote from: ArthurDent on August 29, 2020, 02:39:59 amThis would only be a half wave rectifier for each rail, which may lead to too much ripple?If the two diodes were replaced with two bridge rectifiers...

author: ArthurDent
author_rank: Super Contributor
timestamp: August 29, 2020, 03:02:39 am
content: Quote from: SilverSolder on August 29, 2020, 02:53:05 amThe best repair would be to throw it all away and start over! With the low current drawn I think the half wave rectifiers should work. Without knowing more about the load (audio preamp or battery charger) we can't be sure. Maybe increasing the value of the caps may be needed. This is pretty much guess work without the power supply and known load to test. The original wiring didn't indicate that much filtering was needed so I went with the simplest modifications. The caps are probably dried out by now as well.

author: tony359
author_rank: Frequent Contributor
timestamp: August 29, 2020, 01:50:28 pm
content: it's a vintage Signal generator. Generates up to 1Khz sines and square. The system is already a half wave setup and I have 1V ripple on the rails already.As mentioned before, it seems the PSU is replacing 4 batteries which I am not familiar with - PP9? which seem to be 9V each so 18V both. So maybe an 18V regulator could work. I see rechargeable PP9s are 8.4Vx2 = 16.8. To be honest I feel that 18V could be ok.yesterday after I posted my sketch I realised that I don't think it's going to work. The way it's configured won't give me a positive and negative output from the bridge, am I right? If I am not mistaken I would need to ground the "central" ends of each 15V winding?Or, as suggested by someone, I could just keep the half wave configuration - which is what I have now.@MarkFThanks. Are the TIP3055 and 2N2905 in your drawing for dropping the voltage to zero in case the regulation opens?

author: vk6zgo
author_rank: Super Contributor
timestamp: August 29, 2020, 02:45:21 pm
content: Quote from: SilverSolder on August 29, 2020, 01:13:19 amEspecially as, if you fit new batteries, 18v is what you get!The OP's Levell Signal Generator sounds like the one I have.Mine didn't come with a power supply, but with four horribly corroded Brit 276 batteries, (the whole thing being in a very nice leather carry case, which was the original reason for my purchase).The Brit 276 batteries are different from the 276P batt which were sold in Oz, in that the former uses "press stud" type connectors, but both types are quite large.The battery compartment is, hence, also large, & my reading of the available information was that the PSU fitted in there, in lieu of the batteries.Even so, the PSU described by the OP seems like it would be even larger, so if it is separate, I would suggest it may be a home brew effort, hence the "sneaky" way of getting a split supply.In my case, I stuck with batteries, using four 6 cell AA battery holders.

author: tony359
author_rank: Frequent Contributor
timestamp: August 29, 2020, 02:53:16 pm
content: Here is the signal generator!

author: MarkF
author_rank: Super Contributor
timestamp: August 29, 2020, 02:54:34 pm
content: Quote from: tony359 on August 29, 2020, 01:50:28 pmNo. The transistors are to bypass the regulator and provide higher current.I was only pointing out the pot connections.If the wiper opens, the pot will pull the ADJ pin low and lower the output voltage since there is no path to the output pin.

author: vk6zgo
author_rank: Super Contributor
timestamp: August 29, 2020, 03:02:24 pm
content: Oh! You've got the "Luxury" model!Mine doesn't have a meter.

author: vk6zgo
author_rank: Super Contributor
timestamp: August 29, 2020, 03:24:06 pm
content: By the way, PP9 is just another number for the same thing as a 276.It seems "Ever Ready" in the UK stuck to the same numbers as "Eveready" in Aust (& the USA?), although it was no longer the same company.in Oz, Eveready generally used a totally numeric system, though they "fiddled"with it a bit.The zinc carbon "D" cells were "950s" if zinc-carbon, & "D50" if alkaline."AAs" could be "915s" for penlights, or "1015s" for use in Electronics".Their competitor, "Diamond" used the same numbers, but prefixed with "3".Japanese companies used another numbering system & Taiwanese ones, yet another.

author: bdunham7
author_rank: Super Contributor
timestamp: August 29, 2020, 03:28:26 pm
content: Quote from: tony359 on August 29, 2020, 01:50:28 pmThe circuit as drawn by MarkF, in the post where he points out that you have put LM317 in the negative rail instead of LM337, is your best option--and should be dirt cheap to implement, adding the current limiting resistors I mentioned if you want. One big advantage of the LM317/LM337 combo is that if your voltage choice isn't perfect, you can easily change your mind later. In fact, for sensitive split-rail systems that rely in one way or another on exactly equal positive and negative voltages, they can easily be 'shimmed' up or down by bodging on higher value resistors in parallel to one or the other divider resistor.I had suggested retaining the half wave configuration under the mistaken assumption that you had two separate transformers. Since that's not the case, there's no reason to retain that system. I would point out that the capacitor values might actually be a bit excessive at your current levels, which is OK if you have them on hand. Otherwise you can likely decrease them each by a factor of 5 and still be OK.

author: SilverSolder
author_rank: Super Contributor
timestamp: August 29, 2020, 04:50:49 pm
content: Here are a couple of ideas for fixed regulators:Fixed positive 18V regulator, 2V dropout voltage. ~$0.50 qty 1www.digikey.com/product-detail/en/MC78M18CTG/MC78M18CTG-ND/1481494Fixed negative 18V, 1.3V dropout, ~$0.50 qty 1https://www.digikey.com/product-detail/en/MC7918CTG/MC7918CTGOS-ND/1481547These are just examples of what appears to be available. You'd need to read through the specs to see if there are any "gotchas"!Using fixed regulators would keep the "spirit of simplicity" of the original design, but using somewhat more modern components.

author: MarkF
author_rank: Super Contributor
timestamp: August 29, 2020, 05:43:02 pm
content: One thing that caught me was the tab gauge.Carefully read the datasheets when picking the TO-220 style regulators.The single gauge tabs are very thin and you need to be careful when mounting them.In your case with a low current draw, the heatsink isn't a real big deal.

author: vk6zgo
author_rank: Super Contributor
timestamp: August 30, 2020, 12:11:09 am
content: Sorry, I missed your original pix of the PSU, so forget my comments about being "home brew".I think it is an earlier design than the 1970s, maybe early to mid 1960s, as single device regulators were mfairly widely available by the beginning of the 1970s.I wonder if they used the rather "bodged" system because, just using the transformer windings for "standalone" pos & neg supplies did not allow enougn "headroom" for those older regulators.You may need to keep that in mind if specifying a regulator.

author: vk6zgo
author_rank: Super Contributor
timestamp: August 30, 2020, 04:15:17 am
content: But wait! There's more!https://worldradiohistory.com/hd2/IDX-Site-Early-Radio/Archive-Wireless-World-IDX/70s/Wireless-World-1971-08-OCR-Page-0003.pdfThis gives you a bit of info about the capabilities of your generator.It doesn't "top out" at 1kHz as you supposed, but goes out to 1MHz.Over on the "vintage-radio.net forum", there are several interesting threads on the Levell TG200D series, probably the secnd one being the most useful, as it is about the PSU.https://www.vintage-radio.net/forum/showthread.php?t=159533https://www.vintage-radio.net/forum/showthread.php?t=143263They are all "mad as hatters" on that forum, but are a good (& very knowledgable) crowd, & I stronglyrecommend it.Googling also brought up the "Radio Museum" site which, apart from the one we are looking for has a listof other Levell devices.One, in particular, took my eye, which was an RF voltmeter which went up to 410MHz.That's HP, Marconi & Rohde & Schwarz territory!It seems the little Levell instruments, although their appearance seems to shout "hobbyist level", definitely"punch above their weight" when it comes to performance!

author: tony359
author_rank: Frequent Contributor
timestamp: August 30, 2020, 10:13:40 am
content: 1Mhz! I'll check again but I maxed out the frequency selector and I could read 1Khz - maybe I just missed a step!Interesting, to be honest I picked that up at my workplace some time ago and never thought it would be usable or useful in any way. I'll check out the other thread, thanks!Edit: I think I just mis-read the frequency value when I tested last time! Yes, 1Mhz! Wow!

author: tony359
author_rank: Frequent Contributor
timestamp: August 30, 2020, 08:36:09 pm
content: @MarkFI missed your message when you pointed out the 317 on the negative rail: indeed that was just a typo. I intended to change the label but I must have missed it! Thanks for mentioning.I guess that 2x18V regulators will be perfectly fine so I don't have to worry about the regulation. I am actually impressed that my schematic was not just a bunch of nonsenseand it only needed minor changes (besides the 317 in the wrong place but that again was not intentional).Now a question before I struggle more with drawing a circuit: what do you use to make yours? I have tried a couple online but the first would not allow me to save, the second is missing some basic components!While I find a better software, I guess I can go with what @MarkF posted based on my first sketch (thanks) and1. Replace the LM317 with 7818 and 79182. Add 75R on both positive and negative end of transformers as current limiters as @bdunam7 suggested - May I ask what is the math behind this selection? If I consider 18VDC and 75R I get 240mA and 4.32W.Regarding the regulators, I was let to believe many years ago by an engineer friend of mine that the heat dissipated by a voltage regulator depends on the supply voltage. That is, if I supply a 7805 with 40V I have a much higher dissipation than if I supply it with 10V. That does not seem to be correct and the dissipation is only linked to the current that passes through it - I grabbed a 7805 from my storage and connected it to my bench PSU: indeed with no load the 7805 would just stay cold up to the max voltage I can supply of 15V.(Hence the 7815 and 7915 that are screaming hot in my Behringer active speakers are like that because of the current drawn by the circuit and not because the supply voltage is 41V! I have been misled!!)Thanks again all!

author: bdunham7
author_rank: Super Contributor
timestamp: August 30, 2020, 09:55:23 pm
content: Quote from: tony359 on August 30, 2020, 08:36:09 pm...but if they aren't, you can't adjust them!Quote15VAC and 75R = 200mA, 15VAC x 200mA = 3VA and 3W into a purely resistive load. Thus your worst case scenario, a dead short, puts the transformer and resistor right at their max rating.QuoteBoth the supply voltage and the current matter. The power dissipation of the regulator will be the product of the voltage drop and the output current, including the voltage divider section whether it is internal or external, plus whatever the regulator dissipates due to internal operation--which is quite small. So a 7805 delivering 1A from a 10V supply will dissipate approximately 5 watts, while a similar device delivering 500mA from a 15V supply will also dissipate approximately 5 watts.

author: tony359
author_rank: Frequent Contributor
timestamp: August 30, 2020, 11:16:02 pm
content: Quote from: bdunham7 on August 30, 2020, 09:55:23 pmHonestly I feel it's ok - that unit used to run on 9V batteries.QuoteRight, I had forgotten the PSU is 15V!QuoteAh, ok. Then I was not misled, thanks for clarifying!Now, before I attempt to calculate the power dissipation for the regulators, as @silversolder suggested, am I correct in assuming that the output of the bridge is going to be +/-19.8VDC?

author: bdunham7
author_rank: Super Contributor
timestamp: August 30, 2020, 11:56:59 pm
content: Quote from: tony359 on August 30, 2020, 11:16:02 pmMeasure the open circuit AC voltage of one primary, it will probably be higher than 15VAC by a surprising amount. Take that number multiplied by 1.4 and you'll have your approximate OC voltage. Yes, it's 1.4142818... and there's a diode drop in the bridge, etc, but that is all nothing to worry about. Then subtract your output voltage (18) from that number and multiply by your current, which if IIRC was .012A, plus the quiescent current of the regulator due to its internal voltage divider if the datasheet shows that it is significant enough to bother with.

author: tony359
author_rank: Frequent Contributor
timestamp: August 31, 2020, 03:02:14 pm
content: thanks @bdunham7I've experimented a bit with a bridge rectifier, a load and a capacitor. Very interesting session!I have found that with a load and a capacitor the actual voltage out of the bridge is +/- 25.07VDC. Which looks perfect for a 7x18 - I thought I would end up with not enough voltage to be honest.The voltage out of the transformer with no load is 23.2VAC which after the bridge turns into 31.2VDC (I measured with just a capacitor in) and it's consistent with the math which gives me 32.7VDC.If my regulated output voltage is 18VDC, that gives me 13.2V for the regulator to drop. Multiplied for the max current of... 200mA I guess (this is the short circuit current) plus 3.5mA as quiescent current, that makes for 2.68W to dissipate.I've found this:https://www.ti.com/lit/an/slva118a/slva118a.pdfAnd the 7xxx datasheet:http://www.farnell.com/datasheets/1878412.pdfThe datasheet I found does not give a PDmax - as it says the component is internally limited. But it gives a TETA-JAmax of 65 for the TO-221 package. My calculation (page 12 of the document I found) says that my Teta-JAmax is = (Tj-Ta)/PDmax = (150-20)/2.6862 = 48.39 so it looks like the regulator won't reach the 150 degrees temperature even in a short circuit without a heatsink.THAT SAID, I may attach them to the case just in case, nobody wants a regulator to go up to 100 degrees I guess!Is the above correct?If so, can @MarkF please confirm the values of the capacitors he recommended? I have 100mF and 1000mF - is that millifarad?

author: bdunham7
author_rank: Super Contributor
timestamp: August 31, 2020, 10:29:53 pm
content: Quote from: tony359 on August 31, 2020, 03:02:14 pmYou're overestimating the power because the voltage will drop as the load increases. You didn't say what load you were using, but if it is close to your real-life 12mA load, look at how much the voltage already dropped. At 200mA, it will likely drop into the 15VAC range. In addition, this is why I was recommending a 75R resistor on each primary--they'll both limit the short circuit current and dissipate most of the power.As for capacitors, IIRC you had those in your original drawing and both MarkF and I recommended you swap them, putting the larger in front of the regulators. I also pointed out that they were significantly larger than needed--in microfarads. Millifarads would be ludicrously huge for this. You can use 220uF for the ones right after the bridge and 22uF after the regulators and that's still likely more than needed.

author: tony359
author_rank: Frequent Contributor
timestamp: August 31, 2020, 11:10:37 pm
content: Quote from: bdunham7 on August 31, 2020, 10:29:53 pmI thought what I was calculating was the short circuit power dissipation - which is the worst case scenario. If I don't need a heatsink in that case, then I am safe. The test with the load was just to have an idea of what voltage I was getting out of the bridge. That was a 170 Ohm resistor if I remember correctly (7x1.2K in parallel). At 25VDC that gives me 147mA which is way more than I would normally use. I did not check what the AC voltage was then but I can do that.Am I following your trail of thought or am I completely off track?QuoteThat's why I was confused!I thought "mF" stood for millifarad! I have a selection of capacitors, I'm sure I can find something suitable, thanks for clarifying.

author: MarkF
author_rank: Super Contributor
timestamp: August 31, 2020, 11:38:59 pm
content: Since you arenotbuilding a 'generic use' power supply and your current requirements are very low, I would take a minimalist approach. You don't need a bunch of reverse polarity connection protection, etc. In addition, I would not worry about inrush current limiting (unless you are going to useverylarge capacitors. You needed to worry about current limiting in the original design because of the zener diodes. Here the regulators do have some limited overcurrent protection builtin.I redrew the from an earlier post of my with what I would do.Since you already have some capacitors, I would use 1000μF-2200μF on the regulator inputs and 10μF-47μF on the output.One thing that has not been mentioned is that your current requirement is close to the minimum for the MC78xx and MC79xx series regulators. Therefore, I would add power ON indicator LEDs on the output set for a 15mA-20mA current draw to insure you always have the minimum.You ask about software tools. I useDipTracefor schematics and PCB layouts.

author: tony359
author_rank: Frequent Contributor
timestamp: August 31, 2020, 11:43:53 pm
content: Thanks @MarkFI did not notice the 7x18 had a minimum current requirement - good idea for the LEDs, thanks!And really thank you for the software suggestion, I'll give that a go asap!

author: MarkF
author_rank: Super Contributor
timestamp: August 31, 2020, 11:55:49 pm
content: Quote from: tony359 on August 31, 2020, 11:43:53 pmThey don't come right out and specify a minimum current.They hit at it when specifying the conditions under which their data is given.But, I've had problems with ripple in no-load or small load cases.Some of it depends on manufacturer too. You may not have a manufacturer choice for 18V regulators.The 5V, 12V and 15V regulators are more common.And both anyone asks... I only buy parts from Mouser and Digikey.

author: tony359
author_rank: Frequent Contributor
timestamp: September 01, 2020, 10:05:15 am
content: I used to buy from CPC or RS but RS has quite high minimum quantities to purchase for each item. So I recently switched to Farnell where you can buy a single component - for a slight higher price. But I still need to buy at least £20 to have free shipping. So for less important projects I rely on - cough cough - EbayI try to stick to good sellers when I can. When the project is really not very important I purchase from China and cross my fingers!I have to say that so far I've been pleased!I've ordered regulators and resistors. I will fish a suitable capacitor within my stock. I shall keep you posted with the outcome.For now, thank you very much for being so patient with me, I will try to retain what you taught me!

author: SilverSolder
author_rank: Super Contributor
timestamp: September 01, 2020, 02:52:46 pm
content: Quote from: MarkF on August 31, 2020, 11:38:59 pmNot wanting to divert the thread too much, but DipTrace looks pretty cool for hobby use. I use an ancient unsupported product and will sooner or later have to move to something else... DipTrace seems pretty flexible with what it lets you do?

author: SilverSolder
author_rank: Super Contributor
timestamp: September 01, 2020, 02:55:01 pm
content: Quote from: tony359 on September 01, 2020, 10:05:15 amI use eBay a lot for small quantities as well. - Digikey is flexible enough to use even for small quantities (they will even mail it as a USPS small package for not much $) and I think there must be a depot near where I live, because it usually arrives in 1-2 days.

author: tony359
author_rank: Frequent Contributor
timestamp: September 01, 2020, 04:04:34 pm
content: If I am over £20 I am going with Farnell hands down. I get everything via DHL for free (*) next working day and I know what I am buying. But if I need 2 capacitors and I am not in a rush, then Ebay is the key(*) well, everything is more expensive there in small quantities hence the "free" shipping

author: tony359
author_rank: Frequent Contributor
timestamp: September 07, 2020, 08:15:29 pm
content: Hello allUpdate: I have received all the components and have rebuilt the PSU! It works like a charm. The regulators get lukewarm with 200mA load so with normal use I am very safe. Under short circuit the 75 Ohm resistors get very hot as expected but regulators stay cool.Not the most tidy circuit I guess - I have used a board I had already which features "linked" rows and that did not help in making things tidy. But I didn't want to invest any more money on that thing!The circuit didn't work at the beginning, -15V would start ok and then drop to zero after a few seconds. It turned out I had forgotten to connect the common wire to the central section of the secondary windings! After that everything worked ok. I loaded the rails with 200mA to test it and all went well. I was planning to screw the regulators on the chassis but then realised that the 7918 has INPUT on the tab so rather than insulating the tab I opted for a couple of small heatsink I had around - not really needed anyways! Outputs are stable and clean. To add a bit more protection I swapped the fuse in the mains plug with a 1A fuse.Once connected, everything worked ok and after fixing a couple of snagging (loose knobs etc) the unit now works like a charm. When the dial is set to the "battery" symbol I didn't realise the needle moves to a symbol that I guess means the battery is ok. Indeed now the needle stays towards the top of the range but within it so happy days. (the picture is just a crop of the larger one, it does not mean the needle is over the maximum when checking the supply - I just didn't take a picture of that)Thank you all for your help and for being patient with me, I have really appreciated that and I have thoroughly enjoyed this small project which has taught me lots!!