Power Supply requirements calculation - Page 1
Forum section: electronics / beginners

author: advark
author_rank: Contributor
timestamp: April 20, 2017, 05:37:59 am
content: Hi everyone,I'm actually designing a home made DYI kind of Electronic DC Load. I have 4 power rails in the project: a 15V @ 100 mA for the opamps, a 12V @ 1A for the cooling fan and a 5V @ 3A which also feed a 3.3V @ 300 mA for the LCD display, so 2.3A for the other stuff. According to Ohm's law, as always, this makes a total of 28.5W (1.5W + 12W +15W). These power rails are sourced by an 18V transformer. The question is: How much VA does my transformer needs to be?An 18V transformer would actually output around 24.5V (18 /1.414 = 24.45). My understanding is that 35W divided by 24(.5) would give me roughly 1.5A, times 18V would give me 27VA. So far, how am I doing?This would be in a perfect world, but... I should account for power loss (rectifer bridge, regulators, etc). Is there a rule of thumb for this, like adding an extra 25% (or whatever) to account for? My preliminary tests for the regulator circuitry showed an efficiency around 80% for the 5V and 12V rails. That was on a breadboard with cheap ass 1 ohm and 10 ohms resistors as the load. I'm guessing the result will be much better on a PCB... right?Thanks for any hint.

author: Ian.M
author_rank: Super Contributor
timestamp: April 20, 2017, 06:05:03 am
content: If you are using switching regulators you can divide the output power by the efficiency to get the input power, then divide that by the input voltage to get the input current, but for linear regulators, the input current is the output current (neglecting the quiescent current.) Once you have the total average DC current out of the bridge, for a quick & dirty solution, apply the formulae from the Hammond Transformer division'sDesign Guide for Rectifier Use.It seems over-complex - cant the OPAMPs be run from 12V?

author: advark
author_rank: Contributor
timestamp: April 20, 2017, 01:08:55 pm
content: Quote from: Ian.M on April 20, 2017, 06:05:03 amExcellent! That's pretty much what I was looking for. Thanks.Quote from: Ian.M on April 20, 2017, 06:05:03 amYes this circuit is kind of over complex... on purpose. I'm building this because, well, I need one and mostly because I want to use this as a base to experimient all kind of stuff (DAC, ADC, Op-Amps, RAM, USB, etc).The op-amps can be run from 12V. The 12V rail uses a switching regulator while the 15V uses a linear one. I've read somewhere that switching regulators are more noiser than the linear ones and I don't know (yet) what would be the impact on the voltage reading since the output goes to an ADC. I've never used switching regulators, Op-Amps nor DAC (and a bunch of other stuff!) before, so don't have much "real life" clue what the results would/might/will be.

author: Ian.M
author_rank: Super Contributor
timestamp: April 20, 2017, 01:26:19 pm
content: You've got precious little headroom for a clean 15V rail using a linear regulator. Also is there any chance you'll need a negative rail for the OPAMPs?You also have to consider protecting the ADC against out of range inputs - if you run the OPAMP and ADC from different supply rails, you'll probably need clamping at the ADC input, especially during development.

author: advark
author_rank: Contributor
timestamp: April 20, 2017, 02:03:11 pm
content: Quote from: Ian.M on April 20, 2017, 01:26:19 pmEventually probably. For now, I'm focusing on getting my hands dirty with positive input voltage only.Quote from: Ian.M on April 20, 2017, 01:26:19 pmNot in detail yet. But since you mentionned it, I thought using a Zener diode, something like 4.3V. The ADC (MCP3204) is 5V power with a 4.096V reference for the conversion. The thing is that the Zener usually need more current than what an Op-Amp can source to be stable. The LM324 can source ~40mA while a 1N4731 needs 58mA to be stable. Or maybe, just a regular diode will do? Again, I have not done a lot of research on this topic yet. Any leads to point me to?

author: Zero999
author_rank: Super Contributor
timestamp: April 20, 2017, 07:12:39 pm
content: Get rid of the transformer and replace it with a 2A +24VDC switch mode power supply and change the 12V for a 24V model. Having a bit of extra capacity on the main power supply always helps, as it enables you to prepare of unexpected additions to the design.There are plenty of switch mode voltage regulator ICs/modules which can convert 24V to the other voltages you require.Yes it's true that switch mode power supplies are more noisy than linear but filtering can virtually eliminate the noise. For example a decent sound card in a PC runs from a switched mode power supply, yet doesn't pass the noise through to the speakers because the power supply will be well filtered.What analogue voltages are you amplifying? Is there anything with a very high impedance or a very low voltage?

author: advark
author_rank: Contributor
timestamp: April 20, 2017, 09:45:31 pm
content: Quote from: Hero999 on April 20, 2017, 07:12:39 pmI thought about that but the main goal being a learning experience, I want to build as much as possible from the ground up.Quote from: Hero999 on April 20, 2017, 07:12:39 pmThe input will be 0-30VDC @ 4-5 amps max. The Op-Amps are mainly use to isolate the input from the circuit and to reduce voltage down to a 0-4V range for the ADC (Vref=4.096V) to pick it up. Still have to work on some protection circuitry around it (mostly over and negative voltage).

author: Zero999
author_rank: Super Contributor
timestamp: April 21, 2017, 12:42:16 pm
content: Quote from: advark on April 20, 2017, 09:45:31 pmThere's still plenty to learn, when designing circuits to convert the 24V to the lower voltages used in the rest of the circuit. Connecting a transformer to a bridge rectifier and filter capacitor is very basic and I question how much you'll really learn in doing that.Quote from: Hero999 on April 20, 2017, 07:12:39 pmA switched mode power supply shouldn't cause any problems with proper filtering. Why use +/-15V? Go for +/-5V, then the op-amp circuitry can share a positive rail.

author: advark
author_rank: Contributor
timestamp: April 21, 2017, 02:54:27 pm
content: Quote from: Hero999 on April 21, 2017, 12:42:16 pmPower requirement (the topic of this thread), voltage ripple calculation (see it on the scope, playing around with different cap and see the effect). I like to know how things work/react in the inside. The best ways that I know of are either to tear them down or build them from scratch. I tend to favor the second option because you can go through the design problems one may encounter when designing something and that makes you understand better why people choose one design/component/path instead of another. Just my opinion.Quote from: Hero999 on April 21, 2017, 12:42:16 pmI thought using a differential amplifier (INA117) and I read somewhere (can't remember where) that it performs better with ±15V power rails (better common-mode rejection, I think). I dropped that option and decided to go with a jelly bean op-amp (LM324) because I have a few handy, but the 15V rail... well... it just stuck in the landscapeI sure wil give it a try at 5V and compare the differences (if any).

author: mariush
author_rank: Super Contributor
timestamp: April 21, 2017, 04:18:22 pm
content: Do you actually need 12v at 1A for fans? You can buy computer fans (120mm or 140mm with lots of airflow) that only need around 0.15-0.35A to work just fine.Alternatively, if you're going to use a 18v AC transformer why not just use a 24v fan? Just like 12v fans which are designed to tolerate up to 13v, the 24v fans should work at up to 26-27v just fine.Your 18v AC transformer will be rectified to 18v x 1.414 - 2 x ~ 0.7v = ~24v ( PEAK voltage, the minimum voltage will depend on how much capacitance you're willing to add) and even with +/- 2v due to mains voltage variation, the fan (or fans) should be fine.Digikey has plenty of them, for example here's a 24v @ 0.08A max 92mm fan for 7$ :https://www.digikey.com/product-detail/en/nmb-technologies-corporation/3610SB-05W-B50-B00/P15567-ND/3594465It feels like it's expensive at 7$, but if it saves you 1-2$ for the 12v switching regulator, your cost is only around 5$ ... still more than a cheap 120mm 12v fan, but a quality 12v fan with so much air flow will be close to 5$ as well.So the fan is 12v to 27.6v DC , 0.08A (1.8 watts), up to 3050 rpm and produces up to 54 CFM ... very efficient fan moving lots of air and you can run it straight from input voltage. Connected straight to the output of the transformer (after bridge rectifier and capacitors) it will see a input voltage around 18v to 26v depending on how much capacitance you have and how much the other rails take... either way you'd still have plenty of air flow over the heatsink.Oh.. and keep in mind that you could also use this fan and everything in your "product" with a laptop adapter style power supply which has a default 16.5v to 19v .. usually 18.5v. The fan would still run quite fast from this voltage, and the voltage is still high enough that a ldo would output a cleaner 15v for the opamps and so on...You'll have a peak voltage of 24-26v and you say you need 15v at 100mA and 5v at 3a (15w) and 3.3v at 300mA (1w). Assuming 90% efficiency that means you're going to need about 1.2w for 3.3v and about 18w for 5v, so overall about 20 watts.So let's say we want a minimum input voltage of 18v and let's work with peak voltage of 22v (for situations when your mains voltage will be slightly less than 110v AC so your output is less than 18v ac) .. then for 20 watts you'd need 20w/18v = about 1.15 A so you should maybe put enough capacitance for around 1.5A of current..C = Current / [ 2 x mains frequency x (vpeak -vdc min)] = 1.5 A / [2 x 60 hz x (22v-18v) ] = 1.5A / 480 = 0.003125 Farads or minimum 3125 uF ... so I'd probably suggest going with a 3300uF or maybe a 4700uF or 2 x 2200uF 35v DC rated capacitors, at least (use more than 3125uF if you want higher minimum voltage all the time).Now that you know your minimum dc voltage will be at least 18v and at most around 28v, for just 100mA at 15v you won't get much higher efficiency with a switching regulator compared to just using a linear regulator.So for the 15v 100mA, i'd just use a ldo like this one for example :https://www.digikey.com/product-detail/en/on-semiconductor/MC78M15CDTG/MC78M15CDTGOS-ND/1481487DPAK makes it nice heat dissipation wise and at 0.4$ it's cheap. It has a 2v dropdown voltage at 0.5A, much less at just 100mA ...So 18v in , 15v out at 100mA is an efficiency of around 84%, with 22v in it would be around 70% - a switching regulator may cost 2-3$ and give you 85-90% efficiency but at only 100mA you'd only save something like 0.1w to 1 watt (in worst case scenario with 22-24v in and 15v out). Rather than spending 2-3$ on a switching regulator, spend an extra dollar on a higher VA rating transformer if needed.For the high current 5v it makes sense to use switching regulators to get the better efficiency.You could buy ready made POL dc-dc converters (for a one off or just a few pieces it makes sense) or you could make your own switching regulator.So yeah.. you'd probably need a 35VA rated transformer to get your voltages at those currents you want. A 35VA 18v AC transformer has about 2A ac current, or about 0.62xIac = 1.25A dc current, which is slightly more than what you need.ps. Thinking about it, you probably wouldn't need a switching regulator for 3.3v at 300mA ... that's only 3.3v x 0.3 = 1 watt. You can just use a linear regulator to get 3.3v from 5v, so instead of designing a 5v 3A switching regulator, design one for 5v at 3.5 A or slightly more.5v in at 0.3A , 3.3v at 0.3A .. that's 66% efficiency and about 0.5w loss. A switching regulator will give you 95% efficiency and only about 0.1-0.2w loss but is it really worth spending 1-2$ more to save 0.5 watts? Not really...//edit : i don't know why i read 3.3v @ 3A so i did the numbers assuming 10w for 3.3v ... edited to correct it.

author: advark
author_rank: Contributor
timestamp: April 21, 2017, 11:09:30 pm
content: Quote from: mariush on April 21, 2017, 04:18:22 pmActually "the fan" is an old CPU heatsink block that I salvage. The power mosfet(s) will be bolted on as a heat sink and so will be the temperature sensor. And no, I don't need 1A right now. Between 150mA and 450mA, depending on speed. However, I want to keep some room for future expansion where I "might" need extra airflow for other components. Ambient temperature in my lab can get very hot during summer time (~30°C). Tight room, 2 computers, bunch of equipements and direct sunlight through the windows!The same rule applies to the 5V rail. I actually need around ~2A worst case, at the moment. But it also sources the 3.3V LDO (less power loss, no heatsink required).Quote from: mariush on April 21, 2017, 04:18:22 pmYep! Actually, it's 24.45V. The diode bridge I'm using only drops 1V.*edit: The correct CPU cooler current is between 150mA and 450mA. These numbers come from a similar FoxConn cooler. I cannot find the exact specs for the FoxConn 2ZQ99-096 that I'm using. If anyone knows where I can get it, let me know.