Compact power supply - Page 1
Forum section: electronics / beginners

author: vava
author_rank: Newbie
timestamp: May 02, 2019, 11:33:41 am
content: I'm a complete noob in electronics. However, for my current project I need to do something very tricky - create a very compact power supply. I need it to power ESP8266 and measured the required power to be 125mA (3.3V). Power input would be mains (not grounded).Requirements:- small size, 10mm height and 25mm width are most restrictive requirements, length could be up to 60mm- powerful, while I've measured 125mA, I'd like to have a bit of leeway and targeting 300mA of output current- safe, fire safety is a major concern. It's going to operate in a very tight sealed plastic box. I don't care much about electronics being destroyed as long as it doesn't burn my home down- as little components as possible, being a noob it's hard for me to keep track of more than a dozen or so components on a PCBWhat I've tried so far:- capacitive dropper: more experienced friend of mine suggested using this method but calculation shows that for my power requirement the mains capacitor has to be huge in size- integrated solution: Myrra 48021 seems like an almost perfect fit but a bit too tall unfortunately- MPS MP103: a tricky IC that charges capacitor without an inductor. This almost works but it looks like I can't get high enough current. Even though I can get ~1W of power out of it, I can only do so at 15V and 66mA- AC-to-DC supply: can't find small enough off-the-shelf transformer that is capable of handling mains voltage- SMPS: this seems like a way to go but it is soooo complicatedOn topic of SMPS, I've tried Webench power designer and some of the designs seem to fit the bill. I can use Wurth EFD15 core and have 8mm tall transformer, which is really nice. Unfortunately, it seems that I have to make such transformer myself. Quick research showed that all of my usual sources (digikey and rs components) don't have either the needed core or magnetic wire of needed thickness.So at this point I'm not sure how to proceed further. I will try to research more on how to source necessary parts for making a transformer but to be honest, I'd rather buy sufficiently close one pre-made. I am however have no idea which parameters of the transformer Webench shows me are important and how close they have to be to stated values. It seems I have no one around knowledgeable enough on the topic of transformers but it was suggested to me that this forum might be helpful. So here I am.

author: MarkF
author_rank: Super Contributor
timestamp: May 02, 2019, 11:54:27 am
content: ThisAMS1117-3.3V 800ma Power Supply Module Voltage Regulator+ a 5VDC Wall Wart ?ORhttps://www.ebay.com/itm/AC-DC-Power-Supply-Buck-Converter-Step-Down-Module-3-3V-5V-9V-12V-15V-24V-36V-US/122832954547?hash=item1c996a24b3:m:mNop_RHwBUxr99MJ7MTU6Cg

author: Psi
author_rank: Super Contributor
timestamp: May 02, 2019, 12:02:54 pm
content: Quote from: vava on May 02, 2019, 11:33:41 amThere is a trick you can do with that.With capacitive droppers higher currents costs you space to fit a bigger cap but you get voltage for free.So, Have a capacitive dropper and then a little low voltage switchmode DCDC PSU to get your 3.3v.Set the capacitive dropper output voltage quite high, like 48V so the current needed into the DCDC is very low.That way you only need a small capacitive dropper cap.etc, if you need 3.3V at 300mA that's 1W. So at 48V you only need 20mA to get your 1W into the DCDC,Ideally use as high of a voltage as your DCDC switchmode chip can handle to keep the current and cap size small.Another option is you can get small DCDC ICs that will run from the full 350V rectified mains,They are more expensive than other approaches but if space is at a premium and you're only making small volumes then maybe they would work for you.

author: mariush
author_rank: Super Contributor
timestamp: May 02, 2019, 12:23:52 pm
content: An idea.... really must make it in real world within your constraints and see how hot it gets.So basically you want 3.3v at max 300mA ... or 1wtiny bridge rectifier to get a peak voltage of 1.414 x Vac ex.. peak of ~ 340v for 240v AC or ~150v DC for 110v ACcould use a single diode and do half wave rectification, but then you may need a bigger input capacitor.LR8K4 linear regulator: 1.2V to 440V/20 mA out, TO-252AA-3 :https://uk.farnell.com/microchip/lr8k4-g/linear-volt-reg-0-02a-440v-to/dp/2448524Use this to bring the voltage down to something that a cheap (relatively) and small switching regulator can handle, ex 60v ... 60v x 0.020 A = 1.2w ... you'll need only around 1.1w to get 3.3v @ 300mA with 0.1w for losses.So the linear regulator needs at least 12v above the output voltage to regulate, so for 60v out, you need 72v min input, let's go with 75v for safety.You have 60v 0.02A output, and with linear regulators that's also the input current.So you can estimate the capacitor size required to have a minimum of 75v DC at the input of the regulator.C = Current / [ 2 x AC freq. x (Vdc peak - Vdc min) ]For 110v AC ( go with 100v AC to account for variations in grid power) you have 100 x 1.414 = ~ 140v peak DC soC = 0.02 / [ 2 x 60 Hz x (140 - 75) ] = 0.02 / 120 x 65 = 0.02/7800 = 2.564 uF ... so a 2.7uF or 3.3uF or 4.7uF 250v+ capacitor would be enoughFor 230v AC (go with 210v AC to account for variations ) you have 210x1.414 = ~300v DC soC = 0.02 / [ 2 x 50Hz x (300 - 75) ] = 0.02 / 22500 = ~0.89uF so a 1uF 420v+ would be enoughnow you have min 75v DC in, 60v out at 20mA ... you can use a switching regulator to convert 60v to 3.3vIf budget is not an issue, a 9$ (in 1qty) LTC3630 would fit the requirements : chip, inductor, couple ceramic resistors and capacitors and you're good to go. over 80% efficient with 65v in , 3.3v out at 100mA+Link :https://www.digikey.com/product-detail/en/linear-technology-analog-devices/LTC3630IMSE-TRPBF/LTC3630IMSE-TRPBFCT-ND/8566932Something cheaper at 4$ would be LM5166 :https://www.digikey.com/product-detail/en/texas-instruments/LM5166YDRCR/296-47663-6-ND/8133067As the switching frequency is lower (~200kHz with such high voltage) , you'd need bigger inductor... also a bit less efficient , a bit over 70% efficient at 65v in - > 3.3v outand others you can easily find with some filters on a distributor like Digikey...The big issue you're gonna have is HEAT ... the microchip part being a linear regulator, it will throw away the excess as heat.For example, if you design with an input capacitor big enough to have minimum 75v at 20mA but your product actually consumes much less than 1w, the input capacitor will go up close to your high voltage because it never discharges completely.Let's say your product uses 150mA at 3.3v (0.33 watts) and these are converted with 80% efficiency by the switching regulator ... that means 0.41 watts were taken from LDO.0.41w / 60 = 0.006875 A ... let's just round to 7mASo you may have an average of 120v DC at the input, and 60v on the output ... that means the regulator will dissipate (120v - 60v) x 0.007A = 0.42 watts ... this will raise the temperature of the LDO by around 50 degrees, so the chip will be around 70-80c hot ... datasheet says 80c/w thermal resistance... if placed on pcb with plenty of copper that drops a bit but still...Still, you'll cook that 1uF..4.7uF electrolytic on the input...Ideally you really want to decide on a maximum power that's lower, like let's say absolute maximum 200mA at 3.3v

author: vava
author_rank: Newbie
timestamp: May 05, 2019, 09:39:21 am
content: Ok, so if I understand it correctly, the proposed solution is to drop the input voltage before converting power to voltage/current I would need using some kind of DCDC converter.The main problem seems to be that the first stage that drops power is dumb and can only provide constant amount of power, which can either be used or dissipated as heat.So based on those I can see two ways to deal with this:- have first step to deliver less power, so less of it has to be dissipated. I still want to be able to handle short spikes though, not sure how to deal with this.- have first step to be smarter. MPS MP103 would be almost good as the first step, potentially delivering 15V/60mA in a very smart way but that's only 0.9W, meaning after DCDC conversion I'll get slightly more than 200mA or so. It should be enough to be honest but then option 1 (less powerful first step) would probably be as good.If there's a simple way of somehow storing 150mA of current after power supply, I can deal away with supply that delivers slightly more than 125mA and use that storage to cover short spikes. Is it as easy as just putting it a capacitor? How would I choose the capacitance knowing voltage and current?

author: mariush
author_rank: Super Contributor
timestamp: May 05, 2019, 10:52:48 am
content: If you have a low average power consumption but for a brief moment you need a lot more current, you could consider super capacitors.ex your circuit averages 50mA but every 30s or so you need a burst of 200mA for 2s ... then maybe design psu for 75-100mA ... and rely on 25..50mA to charge a supercapacitor during those 30smajority of super capacitors are typically rated for 2.7v (but you can also find 5.5v rated) so you'd step-down to ~2.6v..2.65v and then boost to 3.3v as needed.

author: Simon
author_rank: Unknown
timestamp: May 05, 2019, 01:26:13 pm
content: Silly question but why must it be so small? You are just starting out, why not learn how to do it first before adding space constraints.

author: vava
author_rank: Newbie
timestamp: May 05, 2019, 02:02:20 pm
content: Not so sillyBut I don't plan to learn how to build power supplies, I'm just trying to finish a hobby project of mine that requires one. Project has size restrictions, purely for aesthetic reasons. I'd buy something already built if there was anything available, but I couldn't find any and in a pure hobby style decided to build my own. I hope it answers the question

author: Simon
author_rank: Unknown
timestamp: May 05, 2019, 02:07:25 pm
content: Maybe there is a reason you can't buy what you want? As you yourself pointed out power density and heat production are a concern. Thermal stuff is not the sort of thing that can be estimated on an unknown design.So, get ready to burn your house down........

author: mariush
author_rank: Super Contributor
timestamp: May 05, 2019, 02:52:09 pm
content: I did some research for another thread, so I figured maybe it would help you.If you can stretch the height to 15mm plus the pcb, TME.EU has classic transformers that are quite flat, 33.4 x 28.9 x 14.9mm :1.https://www.tme.eu/en/details/bvui210012/pcb-transformers/hahn/bv-ui-21-0012/230v -> 6v ac 1VA ( ~7v..8v peak dc voltage ~ 175mA of DC current)2.https://www.tme.eu/en/details/bvui210002/pcb-transformers/hahn/bv-ui-21-0002/115v + 115v -> 6v AC + 6v AC 1VAIt's a bit bigger than 60 x 25 x 10 but maybe close enough.If you want to risk it, maybe you could remove the plastic cover around the transformer to shave a few mm off from each dimension.You have on the length size enough room for a tiny bridge rectifier, a capacitor and a 3.3v LDO or a switching regulator.Also, another option ... but it's expensive.... but no transformers requiredcapacitive dropper / resistive dropper power supply - > optional dc-dc converter (higher voltage -> 5v) -> adum5000 or other digital isolator -> 3.3v or 5v -> optional dc-dc converter 5v -> 3.3vcapacitive dropper power supplies and resistive power supplies (a very simple application note, not quite correct but good enough) :http://ww1.microchip.com/downloads/en/appnotes/00954a.pdfSee attached PDF for a more thorough document about transformerless power supplies.You could use a capacitive dropper power supply + dc-dc converter (optional) to get 5v and then you could use a digital isolator like ADUM5000 to create the isolation :https://www.analog.com/media/en/technical-documentation/data-sheets/ADuM5000.pdfThe Adum5000 can take in 5v or 3.3v and produce 5v or 3.3v :Input Voltage (V) Output Voltage (V) Output Power (mW)5 5 5005 3.3 3303.3 3.3 200So basically you could get ~ 0.1 A worth of 5v from one of these digital isolators.... but note that the efficiency is lousy, let's say 34% at maximum output.This means you more or less need to create a ~ 5v 0.4A (~1500-2000 mW) power supply just to get 5v 0.1A isolated.Or you could use a capacitive dropper psu to get 24-48v ... let's say a 48v 50mA with capacitive dropper or resistive power supply, then convert that using a dc-dc converter to 5v 0.3..0.5A , then use the digital isolator to get 5v 0.1AIf you need more than 5v @ 0.1A , you could parallel two digital isolators... the chips are more or less 10x10 mm so you could solder each one on a tiny circuit board with some pins on one side and then solder them on an vertical extension board kinda like this ___|E where each horizontal line in the E would be one tiny pcb with a digital isolator.With capacitive dropper power supply, the problem would be the big capacitor. With careful calculations, you could probably parallel several capacitors to keep the height less than 10mm and taking advantage of your 60mm of length.

author: KrudyZ
author_rank: Frequent Contributor
timestamp: May 05, 2019, 03:21:53 pm
content: Since this is the beginner's section and you declared yourself to be a noob, maybe it should be mentioned that working on circuits directly connected to mains can easily kill you.Even if the final product is completely (double) insulated, you would still need to test it while you bring it up.Without the right knowledge, experience and equipment, you put yourself in danger.That being said, your requirements are very close to what USB or phone chargers provide.It is extremely unlikely that you will be able to do better than that, considering they are being designed by specialist and being produced by the billions every year.I would recommend that you find yourself the smallest USB adapter and take it apart. Remove the AC plug and USB connector and see what's left. If that doesn't fit, get another one. If none of them fit, change your size requirement so that it does.