Ethernet transformer based DCDC - Page 1
Forum section: electronics / general

author: Miha
author_rank: Contributor
timestamp: December 13, 2023, 10:03:54 am
content: Hello.I need to design isolaing DCDC.Output 12..15 V, min 2 mA. Input voltage any from fixed from 3.3 to 12 VThe project has very little space and requires very little power, so the idea came up to use a cheap and tinyEthernet transformer and a microcontroller to generate the frequency. The terminals of the primary winding of the transformer can be connected between the power supply and the collector of the transistor, which is controlled by a microcontroller. The terminals of the secondary windings can be connected to get a voltage boost.Transformer like TDK ALT3232M-151-T001:https://www.mouser.fi/ProductDetail/810-ALT3232M151T001https://product.tdk.com/system/files/dam/doc/product/transformer/transformer/lan/catalog/trans_alt_en.pdfIs this a workable idea, and what frequency is best to use?

author: retiredfeline
author_rank: Frequent Contributor
timestamp: December 13, 2023, 10:37:12 am
content: You know you can buy DC/DC isolated conveters of various combinations of input and output voltages off the shelf for a buck or two? Search for DC isolated power module on AliExpress for example.

author: Miha
author_rank: Contributor
timestamp: December 13, 2023, 10:42:10 am
content: Quote from: retiredfeline on December 13, 2023, 10:37:12 amOf course, but not 5x5x5 mm...Spase is critical.

author: Berni
author_rank: Super Contributor
timestamp: December 13, 2023, 10:46:27 am
content: You should be able to use it for that, but at very low powers. Datasheet gives a spec for inductance at 8mA bias, so it should be safe to use it up to 8mAId assume your reasoning for using this is having a very tiny transformer. But you can still get similar things that are actually designed to be used with power. If not actual power transformers, you can even use coupled inductors (typicaly for use in SEPIC switchers) as a transformer.But if size is not too big of an issue, then off the shelf isolated DC/DC modules are indeed the easiest solution.

author: Kleinstein
author_rank: Super Contributor
timestamp: December 13, 2023, 10:50:14 am
content: This should be viable. With a 1:1 type transformer one would not get the 12-15 V directly, but could use some voltage doubling rectifier or diode+capacitor charge pump to increase the voltage.One way may be a full bridge drive between 2 PWM pins of the µC. It depends on the transformer details how high the frequency must be without needing to much magentizing current or reaching saturation. The datasheet does not give all the details. Chances are one would need some 100 kHz, maybe 200 kHz. A lower frequency would give less EMI, but more loss in the driver(s).If cost is not such a issue, there are also small (SOT23-6) drivers for push pull DCDC converters, like SN6501. They may work with the given transformer or want a similar one made for that purpose.

author: Miha
author_rank: Contributor
timestamp: December 13, 2023, 11:04:25 am
content: Quote from: Kleinstein on December 13, 2023, 10:50:14 amPictures connection of wire wounds will give voltage doubling.Cost and spase are problems.

author: Berni
author_rank: Super Contributor
timestamp: December 13, 2023, 11:26:16 am
content: No this will potentially produce >100V on the output because you have created a flyback DC/DC topology.

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: December 13, 2023, 12:02:25 pm
content: May also be worth looking at using a common mode choke as a transformer if high isolation voltage isnt needed

author: nctnico
author_rank: Super Contributor
timestamp: December 13, 2023, 01:22:56 pm
content: Quote from: Berni on December 13, 2023, 11:26:16 amNot a problem at all. A tiny bit of load current will take care of that. A resistor of ballpark 10k Ohm will do the trick for a small transformer as the excess voltage is mostly due to the leakage inductance (IIRC). The advantage of flyback is that you can create whatever voltage you want by changing the duty cycle.I like the suggestion of a common mode choke. But there are also coupled inductors available. Like these:https://www.we-online.com/en/components/products/WE-MTCIBut likely you can find similar inductors from other suppliers as well.

author: Miha
author_rank: Contributor
timestamp: December 13, 2023, 01:57:05 pm
content: Quote from: mikeselectricstuff on December 13, 2023, 12:02:25 pmCan you recommend a similar small common mode choke?

author: nctnico
author_rank: Super Contributor
timestamp: December 13, 2023, 02:53:50 pm
content: Not similarly small but smaller:https://www.we-online.com/en/components/products/WE-CNSW-HF. Again, Wurth is just one of many manufacturers.

author: DavidAlfa
author_rank: Super Contributor
timestamp: December 13, 2023, 03:23:38 pm
content: For 10mA or so, probably can be driven by the pin itself without problem?This will make about 4x the input voltage, but very low currents!.

author: Berni
author_rank: Super Contributor
timestamp: December 13, 2023, 03:48:50 pm
content: Quote from: nctnico on December 13, 2023, 01:22:56 pmNo this is an actual flyback topology that requires feedback to run at a stable output voltage. However much current your duty cycle pushes into the transformer determines however much current gets pushed out of the transformer on the other side once it it swings back, since that is the only available path for the core to demagnetize itself after a cycle. So you need feedback to actually hold a stable 2x the input voltage.To use the transformer ratio to determine the voltage then the secondary needs to be flipped around so that it operates in a forward converter topology. In that case you also want to drive the transformer in push pull fashion (or have a catch diode) to aid id demagnetizing after a cycle.Some care also has to be taken when using common mode chokes as transformers since the ferrite material in those tends to be a different blend that is very lossy on purpose to aid in dissipating power rather than storing it. Tho if ran at low enough currents it might not be an issue, and might actually provide a electrical noise advantage if the core itself eats up all the high frequency components of your square wave drive waveform.

author: Miha
author_rank: Contributor
timestamp: December 13, 2023, 03:57:55 pm
content: Quote from: nctnico on December 13, 2023, 02:53:50 pmYes, I see these and like these, but Ethernet transformer have inductance 200 uH, but little common mode filters, like WE in 0603/0805 packages just 100..200 nH. How can to understand the dependence of transformer transfer power fron his induction?I need 50..100 mW.

author: Miha
author_rank: Contributor
timestamp: December 13, 2023, 04:08:17 pm
content: Quote from: Berni on December 13, 2023, 03:48:50 pmIt is very important notes.So. I find 4-wonds Ethernet transformer (about $0.1).https://datasheet.lcsc.com/lcsc/2112021030_Dongguan-Mentech-Optical---Magnetic-CL4532A-201_C2904734.pdfAnd the scheme could be like this?

author: nctnico
author_rank: Super Contributor
timestamp: December 13, 2023, 04:20:05 pm
content: Quote from: Berni on December 13, 2023, 03:48:50 pmNO. Hard no! Please refresh your buck / boost converter theory! The voltage conversion ratio depends solely on the PWM duty cycle. If you don't need precise regulation and have a regulated supply voltage, it is perfectly fine to use a flyback topology open loop. Only effects you'll have is variation due to diode forward voltage and inductor DC resistance versus current.

author: Kleinstein
author_rank: Super Contributor
timestamp: December 13, 2023, 06:14:10 pm
content: Quote from: nctnico on December 13, 2023, 04:20:05 pmOpen loop is working with Buck, but not with boost or fly-back. When using open loop the flyback is approximately fixed power and with no load can get rather high voltage. In this application one may get away with a zener to clamp the voltage.

author: DavidAlfa
author_rank: Super Contributor
timestamp: December 13, 2023, 07:24:46 pm
content: I'd just go with a cheap module like the B0312S, B0315S, B0512S, B0515S...About $1 at Aliexpress:https://www.aliexpress.com/item/1005005644608026.htmlhttps://www.aliexpress.com/item/1005005582106287.htmlAbout 60 cents at LCSC:https://www.lcsc.com/product-detail/Power-Modules_EVISUN-B0512S-1WR3_C7465188.htmlhttps://www.lcsc.com/product-detail/Power-Modules_EVISUN-B0515S-1WR3_C7500907.html

author: Miha
author_rank: Contributor
timestamp: December 14, 2023, 08:27:31 am
content: After all, here is a flyback. Attached the diagram. Please rate the selection of components.

author: Kleinstein
author_rank: Super Contributor
timestamp: December 14, 2023, 09:23:32 am
content: The TL431 is not that suitable for clamping as it needs quite some current to work at all and may be a bit slow to react. Unless one really needs a well regulated output voltage a simple zener would be more suitable. For the clamping there is no need to have the extra series resistor R35. If at all the resistor would make sense when used with an extra filter capacitor.The earlier version with the diode on the input side is a bit more elegant: here the excess power is largely send back to the input supply. As a downside the voltage is set by the turns ratio and not as free to choose.Quote from: Miha on December 13, 2023, 03:57:55 pmThere is no simple way to calculate the possible power from the inductance. The more relevant factors are the coil resistance and saturation current. It also depends on the topology used. With a forward converter the power can usually be higher than with a flyback type. The inductance can set a minimum working frequency.

author: Miha
author_rank: Contributor
timestamp: December 14, 2023, 09:55:09 am
content: Quote from: Kleinstein on December 14, 2023, 09:23:32 amForward converter that uses a diode in primary side of a transformator will take little more spase, becourse of diode. If i finde extra spase, so will try to use forward topologi.It is required to set the voltage quite accurately to 10V. A regular zener diode also requires current to control. The selected TL432 begins to operate at a current of 100 μA.Thank you for mentioning the low operating speed of TIL432.Does this mean that I will get a "saw" instead of constant voltage?If I delete the resistor R35, so can the "saw" be not large?

author: Berni
author_rank: Super Contributor
timestamp: December 14, 2023, 10:17:57 am
content: Quote from: nctnico on December 13, 2023, 04:20:05 pmThis is true for Buck topology but not for Boost. The flyback is basically the same as a boost topology except that the power is tapped off with a separate winding to give isolation and more flexibility in the output voltage range by using the turns ratio to the advantage.The isolated brother of the Buck is the Forward converter topology, there you can drive it with a fixed duty cycle and get a stable output while changing the load.Quote from: Miha on December 13, 2023, 04:08:17 pmForward converters are a bit more complicated than just that. The part count for those is fairly high, so not worth it.In your case just remove VD12 and instead place a zenner diode across the output capacitor to provide regulation. The zenner will just burn excess power into heat, but at such low powers it shouldn't be a problem.

author: iMo
author_rank: Super Contributor
timestamp: December 14, 2023, 10:18:30 am
content: The sim (for the sim purposes is the secondary grounded)..

author: magic
author_rank: Super Contributor
timestamp: December 14, 2023, 10:27:00 am
content: R2 seems unnecessary hereI believe this only works thanks to the zener burning excess power, I think Vout would go much higher without it.That being said, you can as well put the zener on the pramiray, and instead of a zener maybe use MOSFET switch with suitable avalanche rating(For the sake of minimizing component count, as requested by the OP. A suboptimal solution in every other regard.)

author: Kleinstein
author_rank: Super Contributor
timestamp: December 14, 2023, 10:36:13 am
content: I would not remove R35 completely, but use a smaller value (e.g. more 100 ohm) and maybe add a capacitor after that. For stability reasons with the TL431 may need a small series resistor though. Not sure if the C-R-C filter type is already enough damping - maybe with a relatively small resistor. The rest of the circuit connected to the 10 V may also play a role.There would be a little ripple at the output, but not very much as with the high frequency the filter capacitors are quite effective. The first fitler cap may want a bit more capacitance for less ripple.If one plans to build more units, maybe do a simulation to check the stablity and ripply.

author: iMo
author_rank: Super Contributor
timestamp: December 14, 2023, 10:42:06 am
content: Yep, the resistor there was used for sim to work, but it needs grounded secondary to show results, anyhow..The voltage without the zener is rather high, sure..

author: iMo
author_rank: Super Contributor
timestamp: December 14, 2023, 11:12:05 am
content: And with the TL431 and a single secondary winding..

author: iMo
author_rank: Super Contributor
timestamp: December 14, 2023, 12:28:17 pm
content: And a more realistic sim with a Wuerth 10u common mode choke (from LTSpice library).I think it has sense to try, provided the size fits.I would go with something like 47-100uH.

author: Miha
author_rank: Contributor
timestamp: December 14, 2023, 12:52:59 pm
content: Quote from: iMo on December 14, 2023, 12:28:17 pmThanks! Great work!Best is to move C1 to U1.

author: nctnico
author_rank: Super Contributor
timestamp: December 14, 2023, 12:53:15 pm
content: I wouldn't try this circuit. Look at the collector current versus the output current (where does all the energy go?). Also look at the voltage across the primary winding. Looks like the inductor used is satured quickly.

author: magic
author_rank: Super Contributor
timestamp: December 14, 2023, 01:06:21 pm
content: I suspect there is a chip somewhere which could do it all, efficiently, with built-in switching transistor and output regulation by sensing peak primary voltage. Add a diode and capacitor on the secondary and it's done.Such technique is routinely used in 5V USB chargers.

author: Kleinstein
author_rank: Super Contributor
timestamp: December 14, 2023, 02:11:07 pm
content: For a split primary there are chips liek SN6501 / SN6505 to drive a transformer. There are also some cheaper Chinese equivalents. However no regulation but straight square wave drive to the transformer and simple forward converter. The only added feature may be some current limit. As the frequency is usually fixed one need a suitable transformer to run at the voltage, so not to run into saturation.

author: jonpaul
author_rank: Super Contributor
timestamp: December 14, 2023, 02:17:36 pm
content: Ethernet trsf are $0.10 chinese toroids of cheapest material.Not suitable for any power supply use.Easier to just wind your own with a stock toroid or other ferrite corej

author: MarkL
author_rank: Unknown
timestamp: December 14, 2023, 07:02:04 pm
content: There is a chip that could do it all: 3.3V or 5V in, and multiple isolated voltages output, including 12.5V @ up to 20mA:https://www.analog.com/en/products/ltm2883.html15mm x 12mm x 3.4mm BGA. It doesn't meet your 5mm^3 requirement, but I think you were referring to just the transformer. No other components are needed for this chip, including decoupling caps.It's not cheap, but you didn't say much about requirements for the rest of the design. Maybe you need data too across the isolation barrier too.

author: Miha
author_rank: Contributor
timestamp: December 15, 2023, 10:37:20 am
content: Quote from: Kleinstein on December 14, 2023, 02:11:07 pmThanks, good idea. But MCU, that already on the board, can generate to.

author: temperance
author_rank: Super Contributor
timestamp: December 17, 2023, 01:54:09 pm
content: @ ntcnicoQuoteIn a flybck converter, energy stored during the on time is delivered to the output during the off time. As such, changing the load requires a change in output voltage. Only buck derived converters can work open loop.

author: magic
author_rank: Super Contributor
timestamp: December 17, 2023, 03:33:55 pm
content: Quote from: temperance on December 17, 2023, 01:54:09 pmYes, ntcnico needs to refresh his flyback theoryThat being said, stored energy may also be dissipated by the primary - shunt it with a zener of value target output voltage × turns ratio (edit: plus a diode in anti-series).Stupid and inefficient but it could work.

author: langwadt
author_rank: Super Contributor
timestamp: December 17, 2023, 03:44:52 pm
content: Quote from: magic on December 17, 2023, 03:33:55 pmOP said 2mA so it probably doesn't make much difference

author: nctnico
author_rank: Super Contributor
timestamp: December 17, 2023, 05:29:27 pm
content: Quote from: temperance on December 17, 2023, 01:54:09 pmNo. If what you where saying is true, then the output voltage of any mains transformer would be infinitely high with no load attached. In reality it is better to mind-model a transformer as input current = output current * winding ratio. No output current, means no input current (in an ideal transformer).The same goes for a flyback converter. The input / output ratio is governed by the winding ratio and PWM duty cycle. Just like with a buck converter, the actual load doesn't matter (for as long as the core of the transformer doesn't saturate and the converter stays in continuous operating mode). You get tripped up because there is a transformer in the middle but keep in mind that there still is a single core in there and all the windings see the same magnetic flux (variations). Just look at the formulas for a flyback converter; the output voltage is determined by duty cycle and winding ratio. The math doesn't lie. Run some simulations with various loads (in continuous operating mode) if you want and you'll see for yourself.Another thing to keep in mind is that the energy in higher power flyback / buck /boost converters is stored in the air gap in the core. Core material itself can't store much energy; it only serves to channel the magnetic field lines through the windings. For very low power applications (several mA), you might get away by using a small signal transformer but it is important to check for core saturation.

author: magic
author_rank: Super Contributor
timestamp: December 17, 2023, 07:40:20 pm
content: Quote from: nctnico on December 17, 2023, 05:29:27 pmThat's exactly what will happen if you open-circuit the primary near voltage zero-crossing, at peak magnetizing current. Not only the secondary but even the primary voltage will be infinite.Quote from: nctnico on December 17, 2023, 05:29:27 pmYes, and there is always a current in the primary of a flyback, that's the problemWhat you think would stop it from being there when the switch turns on and full input voltage appears across the primary?

author: temperance
author_rank: Super Contributor
timestamp: December 17, 2023, 08:21:24 pm
content: Sorry, nctnico, but almost everything you've stated is not correct.Buck derived converters: the transformer operates as a transformer. The transformer doesn't store energy except for the magnetizing current and the transformer doesn't require an air gap (ok, that's not correct because some implement an air gap but not to store energy which is delivered to the output). It's the output inductor which stores the energy.Flybck derived converters: the transformer if you can call it that does store energy. Ferrite will quickly saturate and because of that, they use an air gap to store energy.For any flyback converter without feedback, the output voltage depends on the load current, maintaining volts second equality.QuoteA "transformer" in a flyback converter operates very different from a mains transformer.If you're an LT spice user, I will post a flyback converter for you to play around to demonstrate the difference.

author: nctnico
author_rank: Super Contributor
timestamp: December 17, 2023, 10:52:01 pm
content: Yes, please do the simulations (as I already did). You'll see it works as I described (be sure though the converter is in the continuous mode region).

author: magic
author_rank: Super Contributor
timestamp: December 17, 2023, 11:07:55 pm
content: Maintaining CCM or BCM requires a suitable combination of minimum average load current, minimum frequency and minimum inductance. As load current approaches zero, one of the others must approach infinity.Practical BCMs (I haven't seen a true CCM) limit switching frequency and thus degenerate to DCM at some point in the name of efficiency. Then "effective" duty cycle is different than switch duty cycle due to idle periods and feedback is used to keep the output in check.By attempting "unlimited frequency" perhaps one could burn enough energy in commutation losses to make it work, I don't know. I think it could be considered an exotic design.

author: nctnico
author_rank: Super Contributor
timestamp: December 17, 2023, 11:36:38 pm
content: For the OP's purpose having an open loop design is the simplest and very doable as the load current is low and predictable. I have used these kind of designs myself as well. For example to create a power supply with 8 (or even 10; I forgot) isolated outputs to create bias voltages for various circuits. A simple NE555 to create a fixed PWM wave, two transistors and a bunch of small signal transformers.

author: temperance
author_rank: Super Contributor
timestamp: December 17, 2023, 11:47:13 pm
content: QuoteThat condition can only be met with a minimum load attached. With no load attached the converter will try to maintain volt/sec equality and because of that the output voltage will increase.

author: magic
author_rank: Super Contributor
timestamp: December 17, 2023, 11:51:55 pm
content: It is, however, a valid point that once a flyback enters CCM, its output impedance becomes low.So perhaps the output zener could be replaced with a suitable resistor

author: nctnico
author_rank: Super Contributor
timestamp: December 18, 2023, 12:15:09 am
content: Quote from: temperance on December 17, 2023, 11:47:13 pmAnd how is that a problem? You can design the combination power supply + circuit so that the minimum load requirement is always met. If you design a power supply for a specific circuit, you can tailor the design of the power supply to match the operating conditions. We're not talking about designing a lab PSU which has to operate over a wide variety of operating conditions.

author: Kleinstein
author_rank: Super Contributor
timestamp: December 18, 2023, 09:02:14 am
content: Reaching the CCM mode depends on having a predictable load and the right choice of the transformer / inductor. It may need a rather high frequency to reach this and this may come with excessive losses with the transformer not specially made for this use. The problem is anyway possible saturation.

author: Berni
author_rank: Super Contributor
timestamp: December 18, 2023, 12:36:03 pm
content: Yep flyback converters do get a more predictable output voltage once under enough load to become CCM, but this is usually a fair bit of load, to the point where increasing the load might actually start causing the output to drop some too.As already said before, for a low power converter like this, just put a zenner diode on it and it will act as a "dynamic load" that eats up exactly enough current to keep the voltage from going too high. The power dissipation is perfectly reasonable for it on such low powers (but completely impractical for higher power designs)You can even have multiple outputs out of a flyback by placing the zenner at the primary side and then using winding ratios, not quite a pure flyback at that point, but it works (mostly a creative way of driving a transformer). Another good trick is also having an extra windings for use as feedback, that way you don't need to sense the isolated side.But since the OP is trying to keep it simple, small and low part count at low powers just wacking a zenner directly across the output is easiest.