Simple DIY Linear PSU - Page 1
Forum section: electronics / projects

author: panoss
author_rank: Frequent Contributor
timestamp: March 05, 2020, 10:27:45 am
content: I want to build arelatively simpleLinear PSU having the ability of transformer taps' switching (I have a transformer with output of 30V with a center tap, so it's 0-15 and 15-30).I 'm in the process of buildingthis power supply. Relatively simple, has tap switching and works ok (according to the writer).What do you think?Can you see anything wrong with the schematic?Maybe someone has built it?

author: Zero999
author_rank: Super Contributor
timestamp: March 05, 2020, 11:19:18 am
content: It looks reasonable to me, although the transformer will multiple secondary windings, with odd voltages might be difficult to source.I would probably have used a comparator such as the LM393, rather than the LM324 op-amp (seems odd they're using a quad, rather than the dual LM358, as only two are used) and some PNP transistors to drive the relays.

author: panoss
author_rank: Frequent Contributor
timestamp: March 05, 2020, 11:52:12 am
content: Quote from: Zero999 on March 05, 2020, 11:19:18 amWhat do you mean by 'odd voltages'?Should I stick to the exact voltages (of the transformer) shown in the schematic?I 'm planning to use the transformer I mentioned, a 0V-15V-30V. (that is a 30V with center tap)And a second small transformer for the 21.7V that feeds the 7818.

author: kallek
author_rank: Regular Contributor
timestamp: March 05, 2020, 02:59:20 pm
content: I am not very familiar with lm723, but I still want to mention what I think. I think that even with center tap transformer, these transistors have too much heat to dissipate.You have few options.- Add thermal protection, the simplest solution is mechanical thermal switch.- Use big heatsink and more transistors, and maybe transistors with lower junction-to-case thermal resistance.- Reduce maximum current.I would myself add a fuse to the secondary side also. Often with big transformers your primary fuse have to be quite high value so it does not blow from inrush current.That 0.18 ohm 5W resistor have to be current sensing resistor. It will be very hot with max current and therefore not very accurate. For example, three 0.56 ohm 5W resistors in parallel would be better solution.Edit. I would also add protection diode over pass transistors.

author: coromonadalix
author_rank: Super Contributor
timestamp: March 05, 2020, 03:29:17 pm
content: This psu design is good, not perfect, many companies use switched xformer secondaries to limit thermal dissipation at low voltage output ,,, you have some models with 5 secondaries taps ...The lm 723 design(s) are proven, you can add some trick(s) to get it down to zero volts etc .... this psu as a floating supply design for the regulation section, it give better stability.You could help the cooling with an temperature controlled fan on top of the heatsink, or add an small temperature controlled fan driver board ...

author: panoss
author_rank: Frequent Contributor
timestamp: March 05, 2020, 07:42:23 pm
content: I 'm trying to simulate it in Proteus 8.I have ommited the part with LM324, for simplicity.I 've replaced some parts that I didn't find in Proteus:Instead of KSA940 I used ZTX753.Instead of 2SD667 I used MPSA20.Instead of 2SC9014 I used 2SC2547.But it doesn't go lower than 7V.Can you check if I do the simulation correctly? I must have done something wrong with the groundings...I have attached the Proteus project, just change it from .pdf to .pdsprj and it will become again a Proteus project (the platform does not allow .pdsprj files to be uploaded).

author: panoss
author_rank: Frequent Contributor
timestamp: March 05, 2020, 08:48:52 pm
content: I tested it again, it looks like working ok...!I increased the primary inductance of main transformer (TR2) to 50H, and secondary inductance to 9H and Secondary DC resistance to 15Ohm and looks better now.It goes down to 0V!i don't know if this is very realistic...

author: Zero999
author_rank: Super Contributor
timestamp: March 05, 2020, 09:50:32 pm
content: Quote from: panoss on March 05, 2020, 11:52:12 amThat seems reasonable.Quote from: panoss on March 05, 2020, 08:48:52 pmDon't bother with the transformer. No one simulates mains transformers. Just use an AC voltage source, with a series resistor, equal to the impedance of the transformer.Look at the data sheets for the transformers you plan to use. There will be a regulation specification, which tells you how much higher the unloaded voltage will be and can help you work out the impedance. Suppose it's 10% and is rated for 3A and the voltage is 30V. The unloaded voltage will be 10% than the rating, so 30*1.1 = 33V, a difference of 3V from the unloaded voltage, therefore R = V/I = 3/3 = 1Ohm.

author: mariush
author_rank: Super Contributor
timestamp: March 05, 2020, 10:54:13 pm
content: You may want to look into transformers that use 12v secondary windings ... a 24v AC transformer with 2 x 12v windings should be easy to source.You could get two of those and switch between 0 - 12 - 24 - 36-48v or maybe have two independent outputs up to 24v rectifiedLooking at tme.eu in the 80..160va range, the most common windings are 11.5v, 12v, 18v and 24vtoroidal :really long tme.eu linkregular transformers :another long tme.eu linkexample100va 2 x 12v toroidal for 17$ :https://www.tme.eu/en/details/tst100w_2x12v/toroidal-transformers/indel/tst-100-004/160va 2 x 12v for 24$ :https://www.tme.eu/en/details/tst150w_2x12v/toroidal-transformers/indel/tst-160-002/I've posted some schematics in the past, which may give you some ideas :https://www.eevblog.com/forum/beginners/secondary-current-rating-of-a-center-tap-transformer/msg1958257/#msg1958257https://www.eevblog.com/forum/projects/design-power-supply-30v-5a/msg1789325/#msg1789325

author: panoss
author_rank: Frequent Contributor
timestamp: March 06, 2020, 08:02:26 am
content: Has anybody tried the simulation?Unfortunately, it doesn't seem to work very well.The output voltage is not stable, for example it fluctuates from 14V to 16.4V.I have replaced the main transformer with a Voltage Source, see the attachment (don't forget to replace the .pdf with .pdsprj)

author: exe
author_rank: Unknown
timestamp: March 06, 2020, 08:41:58 am
content: Do you really need a 30V supply? I mostly use voltages below 5V. Occasionally I use 9V, 12V and rarely 15V to provide supply for devices.I'd say a two-channel 15V psu is much more useful. If I ever need 30V I'll just connect outputs in series.I suggest have two or four isolated low-voltage channels, rather than one beefy channel. This also helps with heat dissipation.In terms of heat dissipation, a better approach could be to use a pre-regulator like this one:http://www.bramcam.nl/NA/NA-01-PSU/NA-PSU-SCH-21.png. I successfully build it, I set the drop out to ~3V, and it works really well. The downside is increased complexity.Finally, make sure your transformer is rated properly. It is commonly suggested to buy a beefier transformer as VA != W.

author: panoss
author_rank: Frequent Contributor
timestamp: March 06, 2020, 08:49:29 am
content: I need a 30V supply. Ok, most of the times I need 5V, but not rarely I need voltages up to 25V.I 'll consider your suggestions, of course.I changed the values of C4 and C3 to 1n and 100n respectively, and looks very stable now!

author: xavier60
author_rank: Super Contributor
timestamp: March 06, 2020, 09:10:15 am
content: Quote from: panoss on March 06, 2020, 08:02:26 amNot saying it's a bad design, the whole auxiliary supply hanging off the LM723's reference pin, sort of, scares me.This design doesn't need a separate winding for its control supply and can be further simplified,https://www.eevblog.com/forum/beginners/lm324-power-supply-with-variable-voltage-and-current/725/

author: Zero999
author_rank: Super Contributor
timestamp: March 06, 2020, 09:18:52 am
content: Quote from: panoss on March 06, 2020, 08:02:26 amI think it should work. Unfortunately, few people here use Proteus and changing the file extension will just confuse people. Zipping it to a .zip file is a more acceptable way to upload file types which are not supported by the forum.

author: panoss
author_rank: Frequent Contributor
timestamp: March 06, 2020, 09:21:07 am
content: Quote from: xavier60 on March 06, 2020, 09:10:15 amHave you built it? Does it work ok?Could you please send me the Eagle files (I have Eagle 7.7.0) and the LTSpice files?I 'll send you a pm with my email.EDIT: I tried but you have blocked PMs?

author: panoss
author_rank: Frequent Contributor
timestamp: March 06, 2020, 09:28:30 am
content: Quote from: Zero999 on March 06, 2020, 09:18:52 amWhen I had zipped a file I was told it 'd be better not to.Because people avoid zipped files because of viruses.So, here it is zipped, in the attachment.

author: xavier60
author_rank: Super Contributor
timestamp: March 06, 2020, 09:37:50 am
content: Quote from: panoss on March 06, 2020, 09:21:07 amSorry, I don't use a simulator. The design has been thoroughly tested by the thread's opening poster.I had built a bench supply with almost identical design which works perfectly well,https://www.eevblog.com/forum/projects/linear-lab-power-supply/375/.But I prefer the other design as it has better Under-Voltage-Lock-Out behavior.If you decide to go down this path, you need to be prepared to do bread-board testing. I want some changes made.

author: panoss
author_rank: Frequent Contributor
timestamp: March 06, 2020, 12:27:36 pm
content: Quote from: xavier60 on March 06, 2020, 09:37:50 amBy 'this path' you mean mike_mike 's design?Why do I need bread-board testing? mike_mike has tested it already

author: xavier60
author_rank: Super Contributor
timestamp: March 06, 2020, 12:48:56 pm
content: Quote from: panoss on March 06, 2020, 12:27:36 pmThat's right, the design that mike_mike settled on. I would like to make some minor changes. There are some parts that aren't needed like the op-amp for the CC LED and a few other bits. I like to test a design on a bread-board before making a PCB.Testing on a simulator is ok also I guess, I have never tried it.I can draw a schematic.

author: panoss
author_rank: Frequent Contributor
timestamp: March 06, 2020, 01:11:59 pm
content: Draw it and I 'll try it on Proteus as it 's a lot easier than the breadboard.Also add the transistors (for 5A output current) in the 'Power block' as in mike_mike 's design they have been ommited.

author: xavier60
author_rank: Super Contributor
timestamp: March 06, 2020, 02:56:03 pm
content: It will work with an LM358.

author: panoss
author_rank: Frequent Contributor
timestamp: March 06, 2020, 03:16:56 pm
content: Quote from: xavier60 on March 06, 2020, 02:56:03 pmIt looks really nice! I will simulate it and let you know.I suppose there is some advantage in using TLC072 instead of LM358?

author: xavier60
author_rank: Super Contributor
timestamp: March 06, 2020, 03:27:14 pm
content: Quote from: panoss on March 06, 2020, 03:16:56 pmAlthough the TLC072 is much faster, it only makes a worth while improvement to the speed of the fast current limiting.

author: panoss
author_rank: Frequent Contributor
timestamp: March 06, 2020, 03:43:09 pm
content: Proteus has the TL072 but not TLC072.I 'll simulate it with LM358 as I already have it (and this is what I 'll use), while the TLC072 will take some time to come if I order it.

author: xavier60
author_rank: Super Contributor
timestamp: March 06, 2020, 03:55:38 pm
content: Quote from: panoss on March 06, 2020, 03:43:09 pmThe TLC072 mainly has a much higher output current which might have some affect on the fast current limiting only.

author: panoss
author_rank: Frequent Contributor
timestamp: March 06, 2020, 04:29:54 pm
content: Also it has no BAT85, so I put BAT42. It 's the D5.

author: panoss
author_rank: Frequent Contributor
timestamp: March 06, 2020, 07:43:41 pm
content: Ok, I managed to simulate it!I use TL072 as for LM358 Proteus has no simulation model.I 've also replaced D45H11(Q2) with a BD912 because D45H11 has no simulation model.I power it with a voltage source, 30V.I put the CV Pot in the middle, CC Pot on the top (max current).These are the results:RL=30 Ohm, 0.47A, 14.2VRL=25 Ohm, 0.57A, 14.2VRL=20 Ohm, 0.71A, 14.2VRL=15 Ohm, 0.94A, 14.2VRL=10 Ohm, 1.40A, 14.0VRL=5 Ohm, 1.56A 7.8VConclusion: It cannot give enough current.I have uploaded in .pdf (change it to .pdsprj) and in zip format.

author: xavier60
author_rank: Super Contributor
timestamp: March 06, 2020, 10:48:05 pm
content: The BD912 likely has low current gain. Try a BD138-16 for now. Most of the BJTs I tested with had gains of 200 and more.The D45H11 has unusually high gain for a large BJT.The voltage at the Base of Q1 should be checked during testing to get an idea of how much total gain the output path has.

author: panoss
author_rank: Frequent Contributor
timestamp: March 07, 2020, 04:39:33 am
content: Quote from: xavier60 on March 06, 2020, 10:48:05 pmIt 's not included in Proteus' libraries.

author: xavier60
author_rank: Super Contributor
timestamp: March 07, 2020, 04:51:24 am
content: The design on mike_mike's thread uses a BD244a. Check what the modeled current gain is.

author: panoss
author_rank: Frequent Contributor
timestamp: March 07, 2020, 07:48:57 am
content: Quote from: xavier60 on March 07, 2020, 04:51:24 amI don't have the simulation file of mike_mike's design.I 'm expecting someone to send it to me.

author: xavier60
author_rank: Super Contributor
timestamp: March 07, 2020, 07:56:56 am
content: Quote from: panoss on March 07, 2020, 07:48:57 amYou mainly need to find a driver transistor with reasonable gain. What did you use in the simulation of LM723 design?Keep in mind to regularly check Q1's Base voltage.The .pdf you posted seems corrupted.

author: panoss
author_rank: Frequent Contributor
timestamp: March 07, 2020, 08:06:20 am
content: Quote from: xavier60 on March 07, 2020, 07:56:56 amOk, I imported the D45H11 's model from a PSpice model for this transistor. (it 's doable)We do have an improvement but it's far from proper working:RL=10 Ohm, 1.42A, 14.2VRL=5 Ohm, 11.5V, 2.31AI reuploaded both .zip and .pdf files.

author: xavier60
author_rank: Super Contributor
timestamp: March 07, 2020, 08:16:15 am
content: Did you note Q1's Base voltage? Can you obtain Q1's Collector current from the simulation?

author: panoss
author_rank: Frequent Contributor
timestamp: March 07, 2020, 08:32:51 am
content: RL=10 Ohm, Q1 Vb= 7.12V Ic=0ARL=5 Ohm, Q1 Vb= 7.60V Ic=0ANo current through Q1??

author: xavier60
author_rank: Super Contributor
timestamp: March 07, 2020, 08:47:54 am
content: Quote from: panoss on March 07, 2020, 08:32:51 amQ1's Collector current should be the same as the current through R10 witch can be calculated from its voltage drop.I would also expect the op-amps to be allowing Q1's Base voltage to go up much closer to 9V. Check this.Without knowing the current gain of the output transistors, I can't work out the expected maximum output current.You can try lower values for R10 as a temporary fix.Watch for stability problems.

author: kallek
author_rank: Regular Contributor
timestamp: March 07, 2020, 08:49:59 am
content: That .pdf file does not open. At least base current go through because voltage drop. Are you sure you are measuring dc current?

author: panoss
author_rank: Frequent Contributor
timestamp: March 07, 2020, 08:54:22 am
content: kallek, the pdf file is not actually a pdf file but a pdsprj file: you have to change the .pdf to .pdsprj.Yes, I'm measuring DC.I changed the Amp Meter and now it shows current flow! (the other amp meter I used has a low resollution!!!):RL=10 Ohm, Q1 Vb= 7.12V Ic=0.00143A Ve=6.45VRL=5 Ohm, Q1 Vb= 7.60V Ic=0.00189A Ve=6.9V

author: panoss
author_rank: Frequent Contributor
timestamp: March 07, 2020, 09:01:20 am
content: Quote from: xavier60 on March 07, 2020, 08:47:54 amRL=5 Ohm, Q1: Ic=0A 0.00189A IR10=0.00190A

author: kallek
author_rank: Regular Contributor
timestamp: March 07, 2020, 09:15:16 am
content: No I did not. I assume that then I need Proteus too. Currents looks like ok, the difference is probably base current. Next measure Q2 current to check if pass transistors get enough base current.

author: panoss
author_rank: Frequent Contributor
timestamp: March 07, 2020, 09:25:04 am
content: Quote from: xavier60 on March 07, 2020, 08:47:54 amWith R10=1k:RL=5 Ohm, Q1: Vb=7.60VRL=10 Ohm, Q1: Vb=7.15VWith R10=300 Ohm:RL=5 Ohm, Q1: Vb=6.43VRL=10 Ohm, Q1: Vb=6.13VWith R10=300 Ohm it seems to work a lot better (it can give more current and voltage remains stable).For knowing the current gain of the output transistors I should simulate my first design? (the kakopa design from my first post?)

author: panoss
author_rank: Frequent Contributor
timestamp: March 07, 2020, 09:33:40 am
content: Quote from: kallek on March 07, 2020, 09:15:16 amWith RL=5 Ohms and R10=300 Ohms:Q2 Ie=0.147AQ3 Ib=0.0691AQ4 Ib=0.0691A

author: xavier60
author_rank: Super Contributor
timestamp: March 07, 2020, 09:43:21 am
content: Quote from: panoss on March 07, 2020, 09:33:40 amThat works out to 20 for the output transistors. The simulator must be using the minimum spec from the data sheet.

author: xavier60
author_rank: Super Contributor
timestamp: March 07, 2020, 09:50:16 am
content: 150 for the driver transistor which is reasonable.A suitable value for R10 can be chosen after the design is built with real parts.

author: panoss
author_rank: Frequent Contributor
timestamp: March 07, 2020, 11:10:29 am
content: So you 're saying the circuit is ok?I should built it and after this experiment with the value of R10?

author: xavier60
author_rank: Super Contributor
timestamp: March 07, 2020, 11:37:00 am
content: Quote from: panoss on March 07, 2020, 11:10:29 amThat's right. Aim for a gm of 10 to 15. That is the amount of output current change to change of Q1's Base voltage.A gm of 10 would mean that the output current changes by 1A for a change of 0.1V at Q1's Base.

author: panoss
author_rank: Frequent Contributor
timestamp: March 07, 2020, 11:50:38 am
content: Output current changegm = -----------------------------------Q1's Base voltage changeIs this correct?e.g if Q1's Base voltage increases by 10% and Output current increases by 100%, this is a gm of 10?

author: xavier60
author_rank: Super Contributor
timestamp: March 07, 2020, 12:07:31 pm
content: Quote from: panoss on March 07, 2020, 11:50:38 amYou have the right idea but use absolute changes, not percentages.Use 2 resistors in parallel for R17 so that the LM317 output can be easily set to the desired voltage.It is important that the unreg voltage stays high enough to prevent the LM317 from dropping out of regulation.I don't expect a problem with a 15VAC secondary.The regulation stability can only be as good as that of the LM317 which I think is reasonably good.

author: panoss
author_rank: Frequent Contributor
timestamp: March 07, 2020, 12:16:07 pm
content: Quote from: xavier60 on March 07, 2020, 12:07:31 pmI 'll try with 15VAC and 30VAC as these are my transformer 's output voltages.Another question:Which resistors should be power resistors?I think R13, R14 and R18 should be 5W rated?

author: xavier60
author_rank: Super Contributor
timestamp: March 07, 2020, 12:44:58 pm
content: At 5A the dissipation is 0.625W for each of the 100mΩ resistors and 1.25W for the 50mΩ resistor which can be 2x 100mΩ resistors in parallel.So 2 or 3 watt resistors will be fine. 5W resistor on ebay are likely to be cheaper.I like to place larger areas of copper print under power resistors.`

author: panoss
author_rank: Frequent Contributor
timestamp: March 07, 2020, 02:24:45 pm
content: Quote from: xavier60 on March 07, 2020, 12:44:58 pmAre you referring to SMD resistors?I 'm asking because through hole resistors are on the top side of the PCB, while the copper on the bottom.

author: xavier60
author_rank: Super Contributor
timestamp: March 07, 2020, 02:32:52 pm
content: Quote from: panoss on March 07, 2020, 02:24:45 pmNo, the axial resistors for R13, R14 and R18. As well as keeping resistance low, the extra wide print helps take the heat away.

author: xavier60
author_rank: Super Contributor
timestamp: March 08, 2020, 12:46:45 am
content: With the early version PCBs for my bench supply I was inputting unreg on one end and outputting on the opposite end of the PCB.With later versions I put the input and output close together at one end of the PCB to avoid having long tracks carrying high current across the PCB.

author: xavier60
author_rank: Super Contributor
timestamp: March 08, 2020, 07:28:04 am
content: About R15. I picked its value out of the air. Some consideration needs to be given to its possible high dissipation.1 watt would be a safe choice and try to keep it away from electrolytic capacitors to avoid heat degradation.

author: xavier60
author_rank: Super Contributor
timestamp: March 09, 2020, 08:50:27 am
content: My bench supply's PCB illustrates my layout suggestions.C6 does not strictly need to be 1000uF.

author: panoss
author_rank: Frequent Contributor
timestamp: March 10, 2020, 05:54:02 pm
content: I 'm in the process of building the circuit and I thought of trying something in the simulation.BUT I found out that I had made a mistake:OnReply #32I had written that I made the simulation with D45H11.Well...that 's not true...I had not connected the D45H11 model to the transistor.So the simulator used a 'Generic PNP transistor' and not a D45H11...So, I connected the model to the transistor and run it, but it wouldn't run because of errors.So, for Q2 I thought of using a Darlington transistor (BDX54), because of it 's high gain.I changed the value of R10 (from 300 Ohms) back to it's original value, 1k.And looks like working fine!

author: xavier60
author_rank: Super Contributor
timestamp: March 11, 2020, 08:17:44 am
content: If you are adding one of those cheap Volt Amp panel meters to the project, it should not be powered directly from the 9V rail. A shunt regulator should be added to prevent the LEDs from causing current fluctuations through the CS resistor.

author: panoss
author_rank: Frequent Contributor
timestamp: March 14, 2020, 09:49:32 am
content: Ok, I built it!But, I 've made some modifications because these are the parts I had in hand.I avoided getting out of the house (corona virus, you know).So, I replaced Q2 with a DIY darlington pair, with BC557 & S8550.Q3=Q4=D209L.D5=BAT41 (instead of BAT85)I did some tests, with no current limit:RL= no load, Vout=16.79VRL= 20 Ohms, Vout=14.93V, Iout=0.71ARL= 10 Ohms, Vout=13.59V, Iout=1.26A

author: xavier60
author_rank: Super Contributor
timestamp: March 14, 2020, 10:02:31 am
content: Quote from: panoss on March 14, 2020, 09:49:32 amBoth of those PNP transistors have fairly high gain.The CV loop is likely oscillating, can you check?Use just one PNP transistor.Extra; The D209L has very low gain so a lot of load will be placed on the undersized PNP driver. So don't power anything valuable with it yet.

author: xavier60
author_rank: Super Contributor
timestamp: March 14, 2020, 10:14:16 am
content: Do you have any scrapped audio power amplifiers? They might have suitable transistors.

author: panoss
author_rank: Frequent Contributor
timestamp: March 14, 2020, 11:42:45 am
content: I have no oscilloscope so I can't check for oscillation.I made a list of the Power NPN transistors I have (these without any number of pieces are 1 piece):E13009L 2PCSE13009 2PCS (they are a lot smaller than E13009L)BU426VBU508AON4046 (PHILIPS)BU508DFC5027-R 2PCSTT2146 2PCSD13007K 2PCSBD241C 5PCSC5763D1408 2PCSBD139CORRECTION in my previous post:I did some tests, with no current limit:RL= no load, Vout=16.79VRL= 20 Ohms, Vout=14.93V, Iout=0.71ARL=10 Ohms, Vout=13.59V, Iout=1.26A

author: xavier60
author_rank: Super Contributor
timestamp: March 14, 2020, 12:06:36 pm
content: Try the two D1408 with a BC557.Also measure the voltage drop across R10 for each load.Extra: If the BC557 fails, try the S8550 although it has a low voltage rating and is likely to fail also.

author: panoss
author_rank: Frequent Contributor
timestamp: March 14, 2020, 12:14:30 pm
content: I have already removed the BC557 (as it was more convenient) so I will firstly try with the S8550.

author: panoss
author_rank: Frequent Contributor
timestamp: March 14, 2020, 12:58:03 pm
content: Unfortunately one of the pins of one of the D1408s was half-broken and I broke it totally...Maybe I can use one D1408 with some other transistor?

author: xavier60
author_rank: Super Contributor
timestamp: March 14, 2020, 01:02:52 pm
content: Quote from: panoss on March 14, 2020, 12:58:03 pmUse one for now and don't go much more than 1A.I notice that the voltage has been dropping by about 2.5V per amp of load. There could be a wiring mistake.

author: Kleinstein
author_rank: Super Contributor
timestamp: March 14, 2020, 01:12:00 pm
content: Combining different power transistors in parallel is not that great and would need more emitter resistance.The BD241C should be OK, at least for lower power (about 20 W per transistor).Less current and only 1 transistor should be OK too.

author: panoss
author_rank: Frequent Contributor
timestamp: March 14, 2020, 01:14:50 pm
content: Quote from: xavier60 on March 14, 2020, 01:02:52 pmI 've split it in two PCBs, in the attachment you will find my Eagle (7.7.0) files:1. main pcb.2. power section, contains the power transistors' section

author: panoss
author_rank: Frequent Contributor
timestamp: March 14, 2020, 02:49:20 pm
content: -I removed the BC557 so now Q2 = S8550-I removed the D209Ls and I put one D1408I get the same measurements.I use an LM358. It's voltage at pin 8(in respect to pin 4) is 9.69V.With or without load.

author: iMo
author_rank: Super Contributor
timestamp: March 14, 2020, 03:12:37 pm
content: OT: It has been demonstrated many times here the LTspice gives pretty good results with DIY linear PSU simulations..

author: xavier60
author_rank: Super Contributor
timestamp: March 14, 2020, 03:59:14 pm
content: Quote from: panoss on March 14, 2020, 02:49:20 pmWhat are the CV op-amp's input voltages for the 3 loads? Also Q1's Base voltage.

author: panoss
author_rank: Frequent Contributor
timestamp: March 14, 2020, 04:30:46 pm
content: By 'input voltages from LM358' you mean voltage on pin 5 (and maybe 6), right?Here they are:RL=no load, Vpin5=4.63V Vpin6=3.18, Q1's Vb=8.97VRL=20 Ohm, Vpin5=4.67V Vpin6=2.26, Q1's Vb=8.55VRL=10 Ohm, Vpin5=4.70V Vpin6=1.72, Q1's Vb=8.30V

author: xavier60
author_rank: Super Contributor
timestamp: March 14, 2020, 04:45:04 pm
content: Quote from: panoss on March 14, 2020, 04:30:46 pmSo the voltage across R10 is about 4V?What is the unregulated input voltage?Try doing the test at a lower output voltage setting.When the Inverting input voltage is less than the Non-inverting input voltage, the output of the op-amp should go to its maximum voltage. About 1.4V less than supply voltage for the LM358.

author: panoss
author_rank: Frequent Contributor
timestamp: March 14, 2020, 04:52:56 pm
content: Quote from: xavier60 on March 14, 2020, 04:45:04 pmIt 's 3.26V, no matter the load.Quote from: xavier60 on March 14, 2020, 04:45:04 pm18.50VQuote from: xavier60 on March 14, 2020, 04:45:04 pmA regulated output from an SMPS would not be a good power supply, right?

author: panoss
author_rank: Frequent Contributor
timestamp: March 14, 2020, 05:00:41 pm
content: Quote from: xavier60 on March 14, 2020, 04:45:04 pmRL=no load, Vpin7=8.52V Vpin8=9.69VRL=20 Ohm, Vpin5=8.13 Vpin8=9.30VRL=10 Ohm, Vpin5=7.87 Vpin8=9.04V

author: xavier60
author_rank: Super Contributor
timestamp: March 14, 2020, 05:05:51 pm
content: No problem using a SMPS so long as the voltage is correct and holds up when loaded.The Collector of Q1 should be close to 18V?It seems that the output stage has too little gain or a fault.

author: xavier60
author_rank: Super Contributor
timestamp: March 14, 2020, 05:11:17 pm
content: Quote from: panoss on March 14, 2020, 05:00:41 pmVpin8 going from 9.69V to 9.04V means that the input voltage is dropping by a lot.

author: panoss
author_rank: Frequent Contributor
timestamp: March 14, 2020, 05:46:22 pm
content: Quote from: xavier60 on March 14, 2020, 05:05:51 pmRL=no load, Q1 Vc = 17.33VRL=20 Ohm, Q1 Vc = 15.22VRL=10 Ohm, Q1 Vc = 14.04VI also tried with an SMPS (12V):RL=no load, Vout=11.4VRL=20 Ohm, Vout=10.3VRL=10 Ohm, Vout=9.9VEDIT: I tried with a second SMPS (a PC power supply). The results are the same.

author: xavier60
author_rank: Super Contributor
timestamp: March 14, 2020, 05:55:54 pm
content: Quote from: panoss on March 14, 2020, 05:46:22 pmRL=10 Ohm, Q1 Vc = 14.04V, means that 4.5V is being applied to B-E of Q2 which means that it's damaged. This is assuming that the supply is holding up at 18.5V.But why was the 9V rail dropping before? Are you using an LM317?

author: panoss
author_rank: Frequent Contributor
timestamp: March 14, 2020, 06:00:13 pm
content: Quote from: xavier60 on March 14, 2020, 05:55:54 pmYes, I 'm using an LM317. (I just took one more look to make sure)I 'll replace it to be sure, maybe it 's damaged.

author: panoss
author_rank: Frequent Contributor
timestamp: March 14, 2020, 09:12:03 pm
content: Well, I replaced:-the S8550 with the BC557-the LM317 with another LM317Results are the same.

author: xavier60
author_rank: Super Contributor
timestamp: March 14, 2020, 09:15:35 pm
content: Quote from: panoss on March 14, 2020, 09:12:03 pmJust with the 20 ohm load, measure all of the voltages in the output stage,Q2 EmitterQ2 BaseQ2 CollectorQ3 EmitterQ3 BaseQ3 CollectorWith respect to the top end of the CS resistor.

author: panoss
author_rank: Frequent Contributor
timestamp: March 14, 2020, 09:27:14 pm
content: Quote from: xavier60 on March 14, 2020, 09:15:35 pmWhich is the CS resistor?

author: xavier60
author_rank: Super Contributor
timestamp: March 14, 2020, 09:30:23 pm
content: Quote from: panoss on March 14, 2020, 09:27:14 pmThe Current Sense resistor R18. Always take measurements with respect to the end that connects to the negative output terminal,What value is it?

author: panoss
author_rank: Frequent Contributor
timestamp: March 14, 2020, 09:35:13 pm
content: R18 is two 100m resistors in parallel, so it's 50milliohm.

author: panoss
author_rank: Frequent Contributor
timestamp: March 14, 2020, 09:48:48 pm
content: Input voltage: 12V.RL=20 Ohms.Q2:Ve=11.67VVb=10.83VVc=11.58VQ3:Ve=11.66VVb=11.59VVc=10.30V

author: xavier60
author_rank: Super Contributor
timestamp: March 14, 2020, 10:14:10 pm
content: Quote from: panoss on March 14, 2020, 09:48:48 pmThe voltages for Q2 look about right. 0.84V B-E voltage means it is being driven hard for its size which is expected.Voltages for Q3 only make sense if the Vc and Ve are swapped? B-E voltage of 1.29V means that it's being stressed also.Looks like output voltage setting is higher than what's available. Try decreasing the setting.You need to get the proper transistors for it. Don't test at more than 500mA for now.When it's happily regulating, the Base of Q1 should drop to a lower voltage and the CV op-amp's input voltages will be the same as each other.

author: xavier60
author_rank: Super Contributor
timestamp: March 15, 2020, 12:20:10 am
content: The D45H11 and 2x TIP35C combination in my bench supply project have a total current gain of about 10,000.Q1 only needs to sink about 1mA for 5 amps of output.There needs to be another 3 or so milliamps reserve to drive Q2 deep into saturation as the regulator approaches its dropout state of about 0.9V.I have some minor changes in mind that will allow the regulator to work better with low input voltages. I would like to see it working properly first of all.

author: panoss
author_rank: Frequent Contributor
timestamp: March 15, 2020, 08:13:24 am
content: Quote from: xavier60 on March 14, 2020, 10:14:10 pmVb=11.59V Ve=11.66VVb-e=-0.07VThat is, b is lower than e for 0.07V. (So Q3 is in cutoff?)You made a mistake: you swapped Ve with Vc (Vb=11.59V Vc=10.30V , this is a difference of 1.29V)

author: xavier60
author_rank: Super Contributor
timestamp: March 15, 2020, 08:41:40 am
content: Q3 is NPN. Its Collector should connect to the input rail and should always be more positive than its Emitter which should connect to the output rail.

author: panoss
author_rank: Frequent Contributor
timestamp: March 15, 2020, 09:33:43 am
content: Indeed...you 're right Xavier...I had swapped C & E in Q3 and Q4... (omg..... I 'm gonna hide in some cave...In Eagle, I had connected Q3 and Q4 to a PNP package...)Ok, I corrected it:RL=no load, Vout=11.39VRL=20 Ohms, Vout=10.82VRL=10 Ohms, Vout=10.55V

author: xavier60
author_rank: Super Contributor
timestamp: March 15, 2020, 09:41:36 am
content: That's holding up much better. You still need to reduce the CV setting down to something like 9V to allow it to regulate.If the voltage measurement is taken at the correct points on the PCB, the voltage should only drop by millivolts with increasing load.

author: panoss
author_rank: Frequent Contributor
timestamp: March 15, 2020, 09:46:45 am
content: Quote from: xavier60 on March 15, 2020, 09:41:36 amHow will I do this?

author: xavier60
author_rank: Super Contributor
timestamp: March 15, 2020, 09:51:39 am
content: Quote from: panoss on March 15, 2020, 09:46:45 amReduce the setting on the CV Pot. If the the output voltage doesn't reduce, there could be damage to the output transistors.

author: panoss
author_rank: Frequent Contributor
timestamp: March 15, 2020, 10:12:08 am
content: The output voltage changes when I turn the CV Pot.But not linearly:starting from max (Vout=11.37V), I turn the pot and for almost the half of the route voltage does not change.After this it starts lowering till 0V!The potentiometer is a Piher 10k 426M.

author: panoss
author_rank: Frequent Contributor
timestamp: March 15, 2020, 10:40:03 am
content: Ok, I got it!I lowered Vout to 8.89V.RL=no load, Vout=8.89VRL=20 Ohms, Vout=8.86VRL=10 Ohms, Vout=8.83VWow! This is rock solid!

author: xavier60
author_rank: Super Contributor
timestamp: March 15, 2020, 11:04:44 am
content: Quote from: panoss on March 15, 2020, 10:40:03 amWell done for now.Depending on how much more experimenting you want to do.I would like D3 replaced by 2 forward biased signal diodes in series and the control rail changed to 7V.The zener diode does not need to be removed.

author: panoss
author_rank: Frequent Contributor
timestamp: March 15, 2020, 11:15:11 am
content: Quote from: xavier60 on March 15, 2020, 11:04:44 amWhat are the signal diodes? Zener? (ok, I found ithere. A 1N4148 is a signal diode) What shall I achieve with the signal diodes? Greater stability in Zener 's Vout?Which is the 'control rail'? The output of the LM317? And why should I change it to 7V? I'm gonna use it with Vin=15V or Vin=30V.If D3 (which is the zener diode) gets replaced, how the zener diode will not be removed?

author: xavier60
author_rank: Super Contributor
timestamp: March 15, 2020, 11:36:19 am
content: Quote from: panoss on March 15, 2020, 11:15:11 am"Signal diodes" usually refers to small general purpose silicon diodes like 1N4148 or 1N914.The 2 diodes will drop the voltage across D3's location to about 1.4V. The original 5.1V zener will have no effect and can be left in circuit in case the mod has to be undone.I'm trying to achieve a control range at the Base of Q1 of about 4V. The output of the LM317 needs to be dropped to about 7V for it to work out.This can be achieved by going to a lower voltage zener but low voltage zeners don't regulate very well.

author: panoss
author_rank: Frequent Contributor
timestamp: March 15, 2020, 11:44:44 am
content: You 're saying I should do this?:

author: xavier60
author_rank: Super Contributor
timestamp: March 15, 2020, 11:47:22 am
content: Yes, its a very simple mod including the LM317 voltage change. I would normally try it here.

author: panoss
author_rank: Frequent Contributor
timestamp: March 15, 2020, 11:50:25 am
content: To change LM317 's Vout I must change R16? Or R17? Or something else?

author: xavier60
author_rank: Super Contributor
timestamp: March 15, 2020, 11:58:36 am
content: Quote from: panoss on March 15, 2020, 11:50:25 amTry R17 first. ll try to work it out here.

author: xavier60
author_rank: Super Contributor
timestamp: March 15, 2020, 12:03:34 pm
content: 680Ω for R17 will set it to a bit under 7V.

author: panoss
author_rank: Frequent Contributor
timestamp: March 15, 2020, 01:02:22 pm
content: Ok, I did the mods:R17 = 680ΩTwo 1N4148 in place of D3, voltage across them is 1.22V.Also I changed Vin to 18V.RL=no load, Vout=16.93VRL=20Ω, Vout=14.61VRL=10Ω, Vout=13.34VI lowered it to 12V.RL=no load, Vout=12.04VRL=20Ω, Vout=11.99VRL=10Ω, Vout=11.96VSo at lower voltages it 's more stable.Maybe if I put a second transistor it 'll be more stable at higher voltages?

author: xavier60
author_rank: Super Contributor
timestamp: March 15, 2020, 01:20:04 pm
content: Quote from: panoss on March 15, 2020, 01:02:22 pmAs the amount of current approaches a transistors maximum rating, the gain usually drops a lot.What maximum current do you need?The TIP35C transistors in my bench supply have a combined current rating of 50A to handle 5A. This isn't unusual.What is Q1's Base voltage with the output voltage set to 9V unloaded?

author: panoss
author_rank: Frequent Contributor
timestamp: March 15, 2020, 03:25:46 pm
content: Quote from: xavier60 on March 15, 2020, 01:20:04 pmI 'll make it for 5A too.So two TIP35Cs (they have maximum Collector current 25A) will be ok, right?Quote from: xavier60 on March 15, 2020, 01:20:04 pmIt's 2.55V.

author: Kleinstein
author_rank: Super Contributor
timestamp: March 15, 2020, 04:54:16 pm
content: With higher current the limiting factor is usually the SOA (at higher voltage) and maybe the heat sink. So the current rating of the output transistors is often not that important. The other point to watch for with higher power is the heat dissipation in the shunt.

author: panoss
author_rank: Frequent Contributor
timestamp: March 15, 2020, 05:48:44 pm
content: Quote from: Kleinstein on March 15, 2020, 04:54:16 pm??Which is the shunt? R13 & R14?

author: kallek
author_rank: Regular Contributor
timestamp: March 15, 2020, 06:56:32 pm
content: Quote from: panoss on March 15, 2020, 05:48:44 pmShunt means low value resistor what is used to measure current by its voltage drop. So it is R18.

author: panoss
author_rank: Frequent Contributor
timestamp: March 15, 2020, 08:02:39 pm
content: R18 is actually two resistors, R18 and R18-B in parallel, both rated at 2W.I hope this will be proved enough for heat dissipation.

author: kallek
author_rank: Regular Contributor
timestamp: March 15, 2020, 08:42:21 pm
content: Sure it is enough. R18 is total 0.05 ohm so you can calculate power dissipation with 5 amp current by formula P=RI^2 = 1.25W. But as mentioned earlier, make sure that traces next to these resistors are quite wide. As well as other high current traces too.

author: Kleinstein
author_rank: Super Contributor
timestamp: March 15, 2020, 08:46:33 pm
content: With the usual voltage drop of some 200-500 mV one already has a power of 400 mW to 1 W at 2 A.To get a stable resistance the temperature rise should be low. So the 2x2 W may be just OK for 2 A if the voltage drop is low (some limitations when the set current is low, may want a good OP for the current limit than).For the resistor the permissible temperature rise is also different, depending on the type - some of the power wire wounds are allowed to go very high, so there rating is high, but the useful power as a shunt is still relatively low.

author: xavier60
author_rank: Super Contributor
timestamp: March 15, 2020, 09:15:58 pm
content: An important test for a power supply is how the output voltage responds to mains power loss.Without an oscilloscope, this test will need to be done with a multimeter while input voltage(18V) is gradually decreased to zero.One way is by temporarily connecting a very large capacitor across the 18V so that the voltage falls gradually when the mains is switched off.The output should be set to 1V, then monitored as the 18V falls.

author: panoss
author_rank: Frequent Contributor
timestamp: March 16, 2020, 10:23:29 am
content: Which value is 'very large'?4700uF?10000uF?More?And I should measure the time that it takes for Vout to go from 1V to 0V?

author: xavier60
author_rank: Super Contributor
timestamp: March 16, 2020, 10:38:43 am
content: Quote from: panoss on March 16, 2020, 10:23:29 am10,000uF will make it about 1v/s. You are looking for any increase of output voltage which is what we don't want to see.

author: panoss
author_rank: Frequent Contributor
timestamp: March 16, 2020, 07:44:40 pm
content: I did the test and everything seems normal.I saw no increase of the output voltage.I plug it in, set the Vout=1V, unplug it. It stays for a few seconds at 1V and then gradually, in a few seconds, decreases till 0.Till now (I have tried a few things) it works VERY WELL!

author: xavier60
author_rank: Super Contributor
timestamp: March 16, 2020, 08:00:05 pm
content: That's good to know.

author: panoss
author_rank: Frequent Contributor
timestamp: March 16, 2020, 08:24:02 pm
content: This is thanks to you Xavier, thank you very much.Have you build it or I 'm the first?

author: xavier60
author_rank: Super Contributor
timestamp: March 16, 2020, 08:39:24 pm
content: I built the design into a working bench supply some time ago. The circuitry around Q1(Q2) is a little different and it relies on a micro-controller to shut off the output when the mains is switched off. It uses a higher performance TLC072 op-amp which does not have much benefit.Actually, using a TLC072 or other types in the current design might cause a mild voltage spike during power down. It would need testing.https://www.eevblog.com/forum/projects/linear-lab-power-supply/msg2388873/#msg2388873As far as I can tell, mike_mike 's implementation of the design worked out well also.

author: panoss
author_rank: Frequent Contributor
timestamp: April 11, 2020, 06:22:16 pm
content: Hi again.I have a problem with the PSU: I short circuited the output and something was burnt (I saw some (very little) smoke).Since then it only outputs the maximum voltage (20V).I checked every component one by one: I desoldered one pin from every diode and checked them, I desoldered Q1 and checked it, I desoldered Q2, Q3, Q4 (they are in a separate schematic & pcb, not seen in the attached schematic) and checked them.I measured every resistor, they all are ok...I even checked the LM358 as perthese suggestions, it 's ok.These are the voltage measurements, input voltage is 20V, any help is more than welcome:

author: xavier60
author_rank: Super Contributor
timestamp: April 11, 2020, 07:17:51 pm
content: Because there is no drive current flowing through Q1, there must be a problem in the output block.If a BJT seems ok on diode test, also check for junction leakage on resistance range, mainly C-B.Are you still using undersized transistors?

author: panoss
author_rank: Frequent Contributor
timestamp: April 11, 2020, 07:25:28 pm
content: Quote from: xavier60 on April 11, 2020, 07:17:51 pmSo I measure resistance between C-B? Which measurement is ok and which implies a damaged transistor?Quote from: xavier60 on April 11, 2020, 07:17:51 pmI went back to D209L(for Q3 and Q4).Q2 is a BC557.They worked fine (before short circuiting the PSU 's output).

author: xavier60
author_rank: Super Contributor
timestamp: April 11, 2020, 07:35:23 pm
content: Quote from: panoss on April 11, 2020, 07:25:28 pmAny good silicon PN junction should measure totally open circuit, no leakage at all with the probes connected in the direction that will reverse bias the junction.You need to get suitable parts some day.

author: panoss
author_rank: Frequent Contributor
timestamp: April 11, 2020, 08:00:34 pm
content: I measured Q2, Q3 & Q3.Q2: C-E, resistance: 32 ohms! Both directions...So it 's like a resistor of 32Ω, it 's bypassing Q3 & Q4, so this is why it 's outputing the input voltage!Btw, in diode mode it looks fine!Quote from: xavier60 on April 11, 2020, 07:35:23 pmThis will happen some day...

author: WattsThat
author_rank: Frequent Contributor
timestamp: April 12, 2020, 04:54:17 am
content: Quote from: panoss on April 11, 2020, 08:00:34 pmI suspect meter operator error. Chances are about 99.999% that Q2 transistor is shorted collector to emitter. Checking on DMM diode setting should show less than 0.1 volt C to E regardless of lead polarity. A good transistor should show INF.Measure a 27 ohm resistor on your meters diode check function - what does it show?

author: panoss
author_rank: Frequent Contributor
timestamp: April 12, 2020, 08:20:47 am
content: Following measurements are all in DMM 's Diode Mode.C-E(Negative lead to C, positive to E): 0.027.E-C:(Negative lead to E, positive to C): 0.033.I measured a 30 Ohm resistor (as I don't have a 27 Ohm resistor, even with combinations) : 0.027.

author: WattsThat
author_rank: Frequent Contributor
timestamp: April 12, 2020, 05:19:42 pm
content: Those are not not resistance values, they are voltage values as the meter is displaying the voltage drop of the device’s PN junction. You check both forward and reverse biased directions, forward bias with red probe on P (anode), black on N (cathode) show proper voltage drop for the material type, 0.6-0.7 volts for silicon, 0.2-0.3 volts for germanium. When you reverse the probes and reverse bias the part, the meter should show INF with no current flowing.When you see a very low voltage in both the forward and reverse directions, like you identified on the failed transistor, it is because the charge junction between the P and N sections has been destroyed and it is shorted/fused together.

author: panoss
author_rank: Frequent Contributor
timestamp: April 12, 2020, 05:48:05 pm
content: Quote from: WattsThat on April 12, 2020, 05:19:42 pmBut I did not see a very low voltage in both directions. (we 're always talking about diode mode measurement)As I mentioned in reply #123, 'in diode mode it looks fine'.So the transistor looked fine when I measured it in diode mode, this is why I was thinking it 's ok. It was proved to be damaged only when I measured it in resistor measurement mode after Xavier 's suggestion.