Electronics > Beginners
Compact power supply
vava:
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 PCB
What 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 complicated
On 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.
MarkF:
This AMS1117-3.3V 800ma Power Supply Module Voltage Regulator + a 5VDC Wall Wart ?
OR
https://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
Psi:
--- Quote from: vava on May 02, 2019, 11:33:41 am ---What 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
--- End quote ---
There 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.
mariush:
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 1w
tiny bridge rectifier to get a peak voltage of 1.414 x Vac ex.. peak of ~ 340v for 240v AC or ~150v DC for 110v AC
could 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/2448524
Use 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 so
C = 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 enough
For 230v AC (go with 210v AC to account for variations ) you have 210x1.414 = ~300v DC so
C = 0.02 / [ 2 x 50Hz x (300 - 75) ] = 0.02 / 22500 = ~0.89uF so a 1uF 420v+ would be enough
now you have min 75v DC in, 60v out at 20mA ... you can use a switching regulator to convert 60v to 3.3v
If 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/8566932
Something cheaper at 4$ would be LM5166 : https://www.digikey.com/product-detail/en/texas-instruments/LM5166YDRCR/296-47663-6-ND/8133067
As 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 out
and 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 7mA
So 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
vava:
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?
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