How to get 5V from a 12V SMPS without using Buck and not heat up LDO or 7805 - Page 1
Forum section: electronics / projects

author: getfast_kiran
author_rank: Regular Contributor
timestamp: December 23, 2017, 10:32:23 am
content: I have build a SMPS of 12V, 0.6A rating using Tiny switch TNY286. I want to get a 5V from this 12V without making another buck because it causing problem in the measurement circuit that I am running using this smps. When I had a 9V SMPS this could have been easily done with helpof the LDO and use for weeks continously without turning it off. Now since I had changed it to 12V suddenly LDO is heating and SMPS is failing after use for few hours for no apparent reason that I could else than replacing the transformer.Can anyone suggest me how to get 5V using LDO properly from this SMPS. Cause I have read somewhere the SMPS and LDO operated like this may result in problems in compensation circuit or feedback loop components of the smps to fail or the LDO to fail when operated like it.Would anyone suggest me how to go about the same.

author: German_EE
author_rank: Super Contributor
timestamp: December 23, 2017, 11:03:47 am
content: How much current are you drawing at 5V?The problem is that you want to drop 7V at an unknown current, this lost power is converted to heat and that heat has to go somewhere. If it's a small amount then you can drop 7V using a series zener diode, this technique is good up to about 5W or 1A before things start to get complicated.

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: December 23, 2017, 11:16:23 am
content: If you're going from 12 to 5 you do not need an LDO, just a normal linear reg like a 7805. LDOs are more fussy about output capacitance etc.Of course this is only viable for low currents - 50-100mA depending on how much heatsinking you can provide.

author: David Hess
author_rank: Super Contributor
timestamp: December 23, 2017, 05:05:02 pm
content: 12 volts could be charge pumped down to 6 volts and then regulated with a low dropout linear regulator.An inverter could convert 12 volts directly to 5 volts or to a slightly higher voltage and then a linear voltage regulator used.Quote from: getfast_kiran on December 23, 2017, 10:32:23 amWill you expound on this? Was switching noise interfering with the measurement circuit?

author: getfast_kiran
author_rank: Regular Contributor
timestamp: December 27, 2017, 11:01:57 am
content: Quote from: German_EE on December 23, 2017, 11:03:47 amI am drawing 0.434A from the 5V output while keeping the 12V pin open.Quote from: German_EE on December 23, 2017, 11:03:47 amI plan to draw only 0.6A max at 12V. Do you advice using zener to drop this to 5V and use 0.45A max at 5V cause I have seenzener getting heated up and reduce in lifetime when regulating such high differences in voltage or have I got it wrong.

author: getfast_kiran
author_rank: Regular Contributor
timestamp: December 27, 2017, 11:07:39 am
content: Quote from: mikeselectricstuff on December 23, 2017, 11:16:23 amMy circuit will consume a max of 0.45A 5V and 0.6A at 12V. I used the LDO because right before increasing my smps voltage to 12V I use to run the same circuit at 9V and a LDO to get 5V. I was able to run the circuit with just ground plane to dissipate the heat for weeks without any heating.Right when I switched to 12V the problem with the smps producing audible noise and LDO heating started.Thats when I read somewhere in a arduino forum that a SMPS and LDO when coupled could push each other out of stability eventually destroying one of the two. Can you or anyone please confirm the same.And I tried putting 7809 before LDO to bring 12V to 9V and still the 7809 and LDO heat pretty good.

author: getfast_kiran
author_rank: Regular Contributor
timestamp: December 27, 2017, 11:18:11 am
content: Quote from: David Hess on December 23, 2017, 05:05:02 pmI first used to use a 5V SMPS and boost converter to get 12V and run my circuit. I found that when using this my analog IC was giving me wrong reading no matter what I do with the filter. Then I removed the boost circuit and powered it externally and got the analog IC to work perfectly. And thats when I designed a 9V SMPS and 5V LDO output from it to power my circuit. It is working fine and was under testing for last two months. It stills performs good. But the company from which I bought the relay to be switched using 12V asked me to change the voltage to 12V for stable operation with ageing. So I redesigned the SMPS for 12V and 0.6A and tested it in electronic load for full load and more and worked fine.Then when I made the circuit for LDO to get me my 5V the LDO started getting heated and my transformer started making audible noise. And after two days run. My TVS diode to supress my spikes from the input of flyback transfomer in smps got burned.Can you help me diagnose the reason for the same.

author: David Hess
author_rank: Super Contributor
timestamp: December 28, 2017, 05:01:26 am
content: Quote from: getfast_kiran on December 27, 2017, 11:18:11 amSo the input noise from the boost converter got into your circuit. That is not really surprising and could probably be solved with a better grounding arrangement and input filtering. If the inductor was unshielded, then it may have been leaking magnetic flux.QuoteBut running on the output through a 5 volt linear regulator was not a problem. Linear regulators are overrated for filtering switching power supply noise. Just their effect as an additional passive filter has more effect.QuoteThe TVS diode problem should be solvable. Some other resistor/capacitor/diode snubber configuration for suppressing flyback leakage inductance spikes might be better. That does not do anything for the linear regulator overheating of course. The power capability of the linear regulator could be upgraded. One easy way if there is a minimum load current is to bypass some of the current around the regulator with a power resistor. Or a low voltage zener diode can be placed in series with the input to the linear regulator.A second winding or tap could be added to the flyback transformer to produce a lower voltage output.Is the input SMPS isolated? There might be an easier to use solution which yields two outputs without any custom parts.The charge pump idea I suggested is more complicated than other solutions.

author: Peabody
author_rank: Super Contributor
timestamp: December 28, 2017, 06:11:20 am
content: Quote from: David Hess on December 28, 2017, 05:01:26 amIndeed. Back when we were all much younger, I discovered that the Commodore C16 computer was designed that way. Most of the current went around the regulator through a big honking 10W resistor, and of course most of the heat was dissipated by it. Then the regulator just kinda topped it off at 5V and regulated the result. Worked like a charm. But you'd have to sure of the absolute minimum current the 5V circuit would draw, and make sure the resistor would supply a bit less than that so some current would always flow through the regulator.A more conservative approach might be to just put that resistor in series coming into the 5V regulator so that a maximum current draw there's still enough dropout room for the regulator to operate.But none of the 5V solutions is going to help with the noisy transformer. Is it rated high enough for what you're asking of it?

author: getfast_kiran
author_rank: Regular Contributor
timestamp: December 28, 2017, 08:42:41 am
content: Quote from: Peabody on December 28, 2017, 06:11:20 amI tried including a diode too in series with the resistor but the resistor and the diode where both heating up so badly that I am sure would not have lasted long in that heat.But none of the 5V solutions is going to help with the noisy transformer. Is it rated high enough for what you're asking of it?Yes the transfomer was specifically made for the SMPS even though there was no varnish used. But if properly working most of the time transformer use to work for mewithout audible noise.Can you suggest me what could go wrong in my circuit that could cause this noise in transformer else than reasons mentioned below.http://www.electronicdesign.com/power/troubleshoot-flyback-supply-generates-audible-noise

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: December 28, 2017, 09:44:20 am
content: Quote from: getfast_kiran on December 27, 2017, 11:01:57 amYou need to use a buck reg. Whatever you use to drop voltage without us going to burn it off as heat. Makes no difference if it's a zener, resistor or regulator. If you need it quiet, add an LC filter. Unless you have a heatsink, linear is simply not going to work

author: David Hess
author_rank: Super Contributor
timestamp: December 28, 2017, 11:49:53 am
content: QuoteThe resistor or diode has to be sized for the power dissipation and the power dissipation has to be acceptable in the design. This just moves the power dissipation away from the regulator.QuoteThe article pretty much covers it. Capacitors and transformers and inductors may be noisy due to construction. Improper operation including pulse skipping or subharmonic oscillation can result in noise.

author: getfast_kiran
author_rank: Regular Contributor
timestamp: December 29, 2017, 07:51:23 am
content: Quote from: mikeselectricstuff on December 28, 2017, 09:44:20 amThe buck reg is not usable cause the radiated emi will intefere with my analog IC which is the main function of my circuit. And it will require complete reshaping of circuit to continue on it and I want to keep it as last step. And yes there is a LC filter. 100uf and 3.3uH.

author: getfast_kiran
author_rank: Regular Contributor
timestamp: December 29, 2017, 07:59:28 am
content: Quote from: David Hess on December 28, 2017, 11:49:53 amYes the power dissipation can be diverted like that as you said but it may increase the temperature inside the casing of the board and it is a very compact box with little room to handle airflow to get rid of this heat. Do you think the audible noise in transformer due to low switching speed might be behind this...IF so how can I correct it. I am suspecting the clamp circuit failing with the TVS diode of 180 V heating might also be behind this problem of heating at the output.Can you please look into it.I have attached the image of the smps circuit.QuoteQuoteAnd yes as you told I tried changing all the capacitors and inductors in the circuit. I also rewound the transformer only thing improved was the load carrying capability of the transformer.

author: mariush
author_rank: Super Contributor
timestamp: December 29, 2017, 08:17:28 am
content: A linear regulator or a LDO will take just as much current from the source, as it goes on its output.So if you want up to 5v at 0.45A at the output , then you're going to take up to 12v at 0.45A from the input, leaving you with just 0.15A out of the 0.6A budget on the 12v. You'll also produce (12v - 5v ) x 0.45A = 3.15 watts of heat, which means you absolutely need a heatsink on your linear regulator.If you insist on using the linear regulator, you could easily reduce the input voltage the linear regulator sees by placing several diodes in series before the linear regulator. Each diode will have around 0.5-0.7v drop at around 0.45A and if you use diodes rated for at least 0.5 watts then it would perfectly fine to dissipate 0.7v x 0.45A = 0.315 watts on them.So basically something as generic as IN5817 / 1N5818 / 1N5819, these have forward voltage between 0.45v and 0.6v at 1A , or something like 1N4007 that would have around 0.8v drop at 0.45vPlace 3-4 of these in series with a bit of leads to lift them off the circuit board and you'll have around 3 volts drop on them and around 1w dissipated on the diodes, leaving a more manageable 2 watts to dissipate on the regulator.I actually suspect your power supply is not really capable of even 12v at 0.6A and you actually need more than 0.15A on 12v, so you're overloading the power supply you have.If you'll change to a buck regulator for the 5v which will be more efficient, you're probably going to have more reserves for 12v and the power supply will probably behave better.For example, get a buck regulator and configure it to output 5.2v at 0.45A with an efficiency of around 90% - that means the regulator will only take 5.2v x 0.45A x 100/90 = 2 .6 watts / 12v = ~ 0.217 A from 12v, less than half the amount you'd normally need.Then, you can use a LDO to further smooth your 5.2v down to 5v dissipating only 0.2v x 0.45A = 0.09 watts on the linear regulator.Example of cheap and simple to use buck regulatorAP65111 : 0.45$ :https://www.digikey.com/product-detail/en/diodes-incorporated/AP65111AWU-7/AP65111AWU-7DICT-ND/7652826up to 97% efficient, requires only a small inductor and a bunch of ceramic capacitors and resistors easy to get

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: December 29, 2017, 10:46:30 am
content: Quote from: getfast_kiran on December 29, 2017, 07:51:23 amIt is quite possible to get switching regulators to be quiet with good design of both the regulator and analogue parts, and of course the layout.However a linear reg will always burn off the voltage as heat.The only choice you have is where that heat gets dissipated, and design the load for minimum power draw.And don't forget the effect heating may have on the stability of your analogue stuff.Sometimes the only option is to tear it up and start again, and learn from the experience.

author: chris_leyson
author_rank: Super Contributor
timestamp: December 29, 2017, 12:13:21 pm
content: You could always center tap the secondary or use two 6V windings bifilar, something like thishttps://www.dialog-semiconductor.com/sites/default/files/iw1830-eval_summary.pdf.

author: matseng
author_rank: Frequent Contributor
timestamp: December 29, 2017, 12:56:19 pm
content: The bulk 100u and 1000u caps you're using in the LC filter issnt really ideal for filtering out any high frequency hash. Sprinkle some 1u, 100n and 10n ceramic caps there as well and see if you're getting a cleaner rail.

author: Peabody
author_rank: Super Contributor
timestamp: December 29, 2017, 03:05:24 pm
content: If it was working ok at 9V, did anything else change when you switched to 12V? Still drawing the same current? It just seems from your description that the heat situation has changed more than the added 3V drop would predict.Perhaps you could go back to the customer and have a discussion about their 12V requirement. Talk about added heat and noise, reduced performance and reliability. More is not better.

author: getfast_kiran
author_rank: Regular Contributor
timestamp: December 30, 2017, 12:09:34 pm
content: Quote from: mikeselectricstuff on December 29, 2017, 10:46:30 am

author: getfast_kiran
author_rank: Regular Contributor
timestamp: December 30, 2017, 12:12:47 pm
content: Quote from: matseng on December 29, 2017, 12:56:19 pmGot it. I will look into it. do you think it could help in the heating up issue.

author: rstofer
author_rank: Super Contributor
timestamp: December 30, 2017, 03:51:17 pm
content: Quote from: getfast_kiran on December 27, 2017, 11:01:57 amIt doesn't matter much how you do it, you are passing a current while dropping a voltage - you will dissipate watts.If I had to do it, I would pass the 12V through a series of 1N4001 diodes. Each will drop about 0.6V and dissipate 0.27 Watts. The 0.6V drop per diode isn't exact (it varies with current) so measure the results. Once the voltage is within range of a 7805, just use one.A SMPS is the last thing I would use if I were at all concerned about noise and sensitive measurements.

author: getfast_kiran
author_rank: Regular Contributor
timestamp: January 01, 2018, 09:02:41 am
content: Quote from: mariush on December 29, 2017, 08:17:28 amFirst off big thanks for such a comprehensive calculation and analysis of the circuit. It really helped much.As you said I will try the method you told to dissipate heat. Could you please comment on the effect of heat generation change from 12v to 9v. Is it big enough to change the intensity of temperatureto bearable after one full day of operation at 9v to finger burning hot temperature just after operating for 3 minutes OR could it be problem with my control loop working properly.Quote from: mariush on December 29, 2017, 08:17:28 amI had problem when using boost before when I was using smps of 5V. The radiated EMI was creating problem with Analog IC I was using in my circuit. Hence why I was trying toget it working using linear. But if this my only option I will switch to buck regulator as you told. And if you could forward me to info on how to get away with radiated EMI it wouldbe great once I start with it if it ends as only option.

author: getfast_kiran
author_rank: Regular Contributor
timestamp: January 01, 2018, 09:11:43 am
content: Quote from: rstofer on December 30, 2017, 03:51:17 pmCan you tell me how would you have done it if you have space constraint of accommodating a low frequency transformer. I would like to know what alternatives to this problem.

author: getfast_kiran
author_rank: Regular Contributor
timestamp: January 01, 2018, 10:24:49 am
content: Following is the signal I got across SB1100 Schottky diode at the rectifier stage. Can anyone please tell me if it has any problems. This waveform is at no load.

author: nctnico
author_rank: Super Contributor
timestamp: January 01, 2018, 07:33:37 pm
content: I'd use a multi-tap transformer. A couple of zeners at the output can take car of no-load condition. Since the SMPS basically is a buck converter the Vin/Vout ratio is determined by the pulse width and transformer ratio.

author: jbb
author_rank: Super Contributor
timestamp: January 01, 2018, 08:46:39 pm
content: Like nctnico and chris_leyson, I suggest a dual output; say 6V and 12V out. The 6V out can go through an LDO to make 5V with little loss.Quote from: getfast_kiran on January 01, 2018, 10:24:49 amOK, here's what I see on that waveform:The trigger might give a less messy response if set to 20VThe pulse width might be a bit uneven - I suggest you check with trigger @ 20VThe fast ringing when the diode turns off around t = -2us suggests oscillation between diode capacitance and tx leakage inductance. It's around 10MHz and probably more prone to sneaking into other stuff. Maybe you could damp it better than this by adjusting R26. Also, I expect that it will get worse as the load increases.The slow ringing off to the right (circa 1MHz) is, IMHO, between transformermagnetisinginductance and assorted capacitances (TNY286, D1/D2, D6, C20 etc.). The R26/C20 snubbers won't be very effective against it. However, this may change a lot at higher loads (e.g. you might go into continuous conduction mode), so try it at operating current.In general, try it at a few load points to see what changes.On the schematic:I suggest leaving room for an extra RC snubber across D1/D2; it might be helpfulThe feedback loop formed using R18, D4 and OK2 won't be very well controlled (but could be fine in many applications)The output filter L4, (C18+whatever output load) could resonate at an inconvenient frequency. You should check out what happens with the real system capacitance and assorted load steps. For extra fun, it could interact with the main control loop too.HOWEVER, the hard switched nature of the flyback converter will always generate some high frequency energy (could be from 10MHz - 200MHz+) due to the fast dV/dt and dI/dt of the switching circuits. With careful design (and good PCB layout), you can try to attenuate as much of this as possible before it leaks out of the switching supply and into the analogue stuff.But a better way might be to use a soft-switching converter and generate less noise in the first place. These converters use resonance or auxiliary switches to help with the switching, and inherently produce less noise than a more violent hard switching converter. They normally wouldn't be used for a low power supply, but might help with your particular requirements. Possible contenders:Active clamp flyback converter. (TI has just started sampling the UCC28780 which could be interesting.)Active clamp forward converter. (LT has several controllers. Probably too many parts for this application...)LLC converter. This one is nice because it uses the transformer leakage inductance; you can use a split winding bobbin to get really good primary-secondary isolation. Power Innovations makes the LCS700 which is a bit overpowered but might do the trick.As a final note: doing your own off-line supply does give you a lot more safety worries than just buying a qualified one. Be careful, and prepare to spend extra time reviewing everything to check creepage, clearance, component ratings etc. And be very thorough with your transformer design and winding...Edit: fixed format boo-boos

author: chris_leyson
author_rank: Super Contributor
timestamp: January 01, 2018, 09:12:56 pm
content: One example of a tapped secondaryhttp://katalog.we-online.de/ctm/datasheet/750871630.pdfor separate windingskatalog.we-online.de/ctm/datasheet/750316585.pdf

author: getfast_kiran
author_rank: Regular Contributor
timestamp: January 03, 2018, 07:01:34 am
content: Quote from: nctnico on January 01, 2018, 07:33:37 pmHi, Yes I was thinking of this but I am trying to stay away from it since my loads on the two taps are so different so regulation will take me a good amount of time to test again and settle on. If everthing else fail I will definitely go as you said.

author: getfast_kiran
author_rank: Regular Contributor
timestamp: January 03, 2018, 07:32:56 am
content: Quote from: jbb on January 01, 2018, 08:46:39 pm

author: getfast_kiran
author_rank: Regular Contributor
timestamp: January 03, 2018, 07:36:22 am
content: Quote from: chris_leyson on January 01, 2018, 09:12:56 pmThanks for giving the reference.I will look into multi tap once this solution seems unsolvable cause it will take more time for me to build the circuit and test it again.

author: getfast_kiran
author_rank: Regular Contributor
timestamp: January 09, 2018, 11:58:35 am
content: Solved one of the problem.The heating in the clamp circuit was reduced by reducing the number of turns in the primary and seconday of the transformer keeping the turns ratio the same. I think my circuit since it was designed for universal input supply. The duty ratio of Tiny switch at 230 V was at the other end of the spectrum which means it was off for most of the time. So add this to problems in leakage inductance...(btw which Ido not know yet how to link them perfectly) resulted in heating my clamp diode.Now reducing the turns in both the winding might have reduced the spikes coming at low duty ratio( hence helping in minimizing the effect of leakage inductance - according to friend) hence causing less voltage spikes and relaxation for the clamp diode.Now my only problem is the heat being dissipated in the LDO (which also has come down drastically) as per calculation given to me here before. Now I will try to design the heatsink as per the specs and get back to you with the result of the cooling effectiveness of the same.

author: james_s
author_rank: Super Contributor
timestamp: January 09, 2018, 10:00:58 pm
content: There seems to be a common misconception around LDOs. They are linear regulators, the *only* thing special about the LDO type is that the dropout voltage is lower than a standard linear regulator. That is you need less overhead, the input voltage to the regulator can be lower for a given output voltage, reducing the voltage that the regulator has to drop and that is what makes the regulator run cooler. Under equal circumstances a LDO will get exactly as hot as a standard regulator. A resistor, diode(s), or any other linear dropping device will dissipate the same power and produce the same heat. Dropping a voltage with linear methods always converts all that excess voltage into heat. If you want to avoid making this heat you either need to use a lower voltage secondary on the transformer or use a switching regulator to drop the voltage. There is no magic here, those are your options.

author: getfast_kiran
author_rank: Regular Contributor
timestamp: January 10, 2018, 05:15:49 am
content: Quote from: james_s on January 09, 2018, 10:00:58 pmYes james. Thanks for the info.I was not able to drop the voltage using switching voltage regulator due to issues with my following analog IC and former cannot be used cause of my circuit needs. I understand now. LDO was heating up unusually for a 3.14W power dissipation before the turns where reduced. Now its heating up reasonably. As you said I will try to dissipate this heat off with standard technique. Let me verify it for few weeks and I will mark this thread as solved.