How to set up this project/experiment? - Page 1
Forum section: electronics / projects

author: FenderBender
author_rank: Super Contributor
timestamp: January 29, 2012, 08:27:00 pm
content: Greetings,I am in high school and I have been assigned a project for the electronics course I am taking. My teacher is pretty lax but I thought I would go all out since I have the ability to.Anyway, my topic is "power supplies". Half of my project is going to be video and half a voiced over PowerPoint. I am going to start off with some really basic AC and DC theory since I am supposed to be explaining my topic.For the video part, I plan on showing different types of power supplies in action. Self-built of course.One of the main things I want to show is the difference in efficiency of SMPS vs linear power supplies. So my main question here is what are the parameters I should set up for a fair and accurate test?My thought is to have a high power transformer and a fairly high powered linear regulator as a prestage to the actual power supply circuits. So maybe a 12volt trafo and a 10v regulator. Dont want to waste too much power. Any other ideas? And how should I keep it fair?I pretty much have the equipment. Though I'm planning on whipping up a FET constant current dummy load for testing. I currently have 4 multimeters, a function generator, and an oscilloscope. Of course I dont have any Flukes or Gossens or $50k Tek MSOs but I have decent enough equipment I believe.Does anyone have any other ideas for tests or demonstrations? I'm totally open!Thanks!

author: IanB
author_rank: Super Contributor
timestamp: January 29, 2012, 10:10:49 pm
content: I think you could just use a DC power adapter is your upstream supply, a wall wart or a power brick, something of that sort. It's small, safe, convenient and portable.For your test you could set up a simple DC-DC buck converter based around one of the standard chips made for the purpose. You just need to add an inductor and a few other components according to the datasheet and you get a nice compact circuit. Maybe buck down from 12 V to 6 V (5 V?) and drive a 6 V bulb as a load? (A bulb "does something" so it makes for more visual impact than a bunch of electronics. An incandescent bulb also approximates a constant current load.)You could then compare your buck converter with an LM317 (7805?) regulator circuit. Put a nice heat sink on the LM317--again for visual impact. A heat sink won't be needed on the buck converter.Then you're all set up to measure and compare efficiency by measuring current and voltage on input and output sides.

author: Bored@Work
author_rank: Super Contributor
timestamp: January 29, 2012, 10:41:28 pm
content: This is one of the things I would plan from the end. What do you want to demonstrate? That switching-mode power supplies can be more efficient than linear power supplies?If yes, then I would start with assuming that a mediocre SMPS can be build with 70% or better efficiency. Conversely I would set up a linear power supply that is worse than 70%, e.g. 50%. Doing that is easy. For 50% efficiency of a linear power supply you need to have twice the output voltage as input voltage, e.g. 12V in, 6V out. For comparison, build a 6V DC converter.The rest is selling that to the audience.

author: FenderBender
author_rank: Super Contributor
timestamp: January 30, 2012, 01:10:14 am
content: Thank you both Ian and Bored at Work.I guess my dilemma is that there really isn't any real point to prove according to the projects guidlines. The guidlines are pretty much "Tell me how power supplies work and what they do". I mean linear supplies do have positives, though a good switcher has very few downfalls.So I was thinking pros and cons, but it's hard not to sound biased.Yikes.I think I need some time with myself to lay this project out.Thank you.

author: bfritz
author_rank: Regular Contributor
timestamp: January 30, 2012, 02:13:39 am
content: Quote from: FenderBender on January 30, 2012, 01:10:14 amI think there are pros and cons to each. There are always trade-offs in design, that is what engineering is about. The trade-offs include factors like: Design time (which is really design cost, ie: If you are making a million you can spend more time optimizing design, than if you are making 10.), Component Cost (If you are making 10 you care less about individual component cost, than if you are making millions.)Switcher:Pros: Efficiency, Size, Weight. Cons: Complexity (More difficult design), Noise (Both Radiated & Conducted)Linear:Pros: Simplicity (Ease of design), Noise (Both Radiated & Conducted). Cons: Efficiency, Size, Weight.Now the above attributes are typical. For example, if the power levels are low, efficiency of a switcher can become a "don't care", and the size becomes almost fixed. For the linear, the simplicity at very low power levels can mean cheaper and simpler. This makes for some interesting topics in your paper. For example, at high power levels the losses in a linear become very important, and overweigh the complexity of design of a switcher, as the linear becomes difficult to design well at such high power levels. Looking at the various points at which these swap places could be interesting.It might be interesting to examine a typical power supply scenario, For example a charger for a cell phone, and look at a solution that uses a linear supply, and another that uses a switching supply. If they both are designed by professionals they will both be good examples of what can be done. This also guarantees that you aren't putting in your own bias by overdesigning one or the other, to fit your bias.If you wish to do something like the above, I might suggest a trip to a hotel. Tell them what you are doing, and ask if business travelers have left chargers that are unclaimed. They'll probably give you a box of them to take home for free! Pick one that gives you 5V out, probably at 500mA, as it will likely be a linear design, tear it apart to make sure it is a linear, and test it for efficiency. I would then suggest an Apple supply, as all that I've seen are switchers with good efficiency. Comparing those will give you something to talk about.

author: mstevens
author_rank: Regular Contributor
timestamp: January 30, 2012, 02:25:19 am
content: Quote from: BoredAtWork on January 29, 2012, 10:41:28 pmI believe, the general definition of efficiency of a Power Supply is Pout/Pin. So the OP should be looking at overall Power Out versus Power In to get the correct efficiency of the power supplies.Melvin

author: Psi
author_rank: Super Contributor
timestamp: January 30, 2012, 03:01:56 am
content: Burn up a resistor until it catches fire to show why heat is an issueEveryone will love that.And you can make a comment on the smell of burning electronics

author: FenderBender
author_rank: Super Contributor
timestamp: January 30, 2012, 04:04:46 am
content: Thank you everyone. I think I will try to do pros and cons. I have a number of chargers and wall warts. Some linear, some SMPS.So I guess I'll show a 5V 1A linear (if I have one). It's gonna be big. And thena 5V 1A SMPS, which can be pretty darn small. Smaller = better?So I guess my new questions are: What tests to run now to show the pros and cons?Thanks.

author: siliconmix
author_rank: Frequent Contributor
timestamp: January 30, 2012, 07:09:41 am
content: cool.electronics taught in schools.if they did that when i was in school i would have went more often ..meh.efficiency isn't everything .

author: Bored@Work
author_rank: Super Contributor
timestamp: January 30, 2012, 07:48:58 am
content: Quote from: mstevens on January 30, 2012, 02:25:19 amGuess what that boils down to for my linear regulator example?Pin = Vin * Iin; Pout = Vout * Iout;With Iout roughly equal Iin (ignoring the little bit of current used by the linear regulator itself) you getPout/Pin = (Vout * I) / (Vin * I) = Vout / VinTherefore 50% efficiency for a linear regulator means the output voltage is half the input voltage.

author: Rerouter
author_rank: Super Contributor
timestamp: January 30, 2012, 08:14:51 am
content: not that its in anyway easy to demonstrate, switchers also can get around the need for transformers, both for input and for step up,aslong as it's in the device limits, you can feed line voltage through a diode bridge, capacitors to filter it, and get away with no input transformer,and for the output, even if stepping it up further, for general increases, e.g 2x you usually only need an inductor, transformer only needed for isolation or inverting,now that there is a massive pro for a switcher, when you get to high power supplies, as high wattage transformers can easily quadruple the cost of a supply.as for linears, they are good as references, and low noise supplies, similar performance can be achieved with switchers, but you have to filter the output, generally quite heavily, if your school has a scope, that would be a good thing to hook up to compare them with,

author: Psi
author_rank: Super Contributor
timestamp: January 30, 2012, 08:31:35 am
content: If you get the right value coil and capacitor in series with a battery and add a push button switch in parallel with the capacitor you should be able to generate some high voltage sparks when you let go the button.Basically the same kind of back emf you get from a relay but the capacitor makes it start ringing and produce more voltage. Also the same thing used in cars for the ignition.You can then show that switching power through an inductor can generate a higher voltage than the source.And also explain that a switch mode supply does the button pushing automatically and adjusts itself to get a steady output voltage.It's a crude explanation but it will help people who have no idea about electronics understand a bit better.

author: FenderBender
author_rank: Super Contributor
timestamp: January 30, 2012, 09:14:42 pm
content: http://en.wikipedia.org/wiki/Switched-mode_power_supplyWikipedia has a rather nice list of pros and cons for linear vs smps. I think I'll take a look at that in some detail.

author: mstevens
author_rank: Regular Contributor
timestamp: January 31, 2012, 01:06:14 am
content: Quote from: BoredAtWork on January 30, 2012, 07:48:58 amI never said your example was wrong or even implied that it was. So I don't understand your necessity to post this response. But since you seem to be bothered by my statment let me be more specific.Specifically, I *KNOW* that the efficiency of a power suppy IS: eta=Pout/Pin. The OP should be using that equation to determine the efficiency of a power supply and not Vout/Vin; Vout/Vin is a voltage gain, which is not necessarily equal to efficiency of a power supply.This definition can be found in (I would guess) any first semester Power textbook or any text on Power Electronics.Melvin

author: IanB
author_rank: Super Contributor
timestamp: January 31, 2012, 02:45:14 am
content: Quote from: mstevens on January 31, 2012, 01:06:14 amI understand. I came very near to posting the same response myself, but I let it drop. Life is too short.QuoteWe know that you know, and so does everyone else know. This is called "stating the obvious". Hence the strange puzzlement about why you seem so compelled to state the obvious, when nobody in this thread has suggested anything different. What exactly is causing you issues here?

author: mstevens
author_rank: Regular Contributor
timestamp: January 31, 2012, 03:27:53 am
content: We know that you know, and so does everyone else know. This is called "stating the obvious". Hence the strange puzzlement about why you seem so compelled to state the obvious, when nobody in this thread has suggested anything different. What exactly is causing you issues here?While those with "experience", i.e. old people, may know this, FenderBender, the OP, stated that this was for a high school project. Thus, I make no assumption(s) that he would "know" the definition of efficiency. BoredAtWork's post can be interpreted that the efficiency of a power supply is Vout/Vin. I was merely trying to clarify for the OP. BoredAtWork's reply to my post indicated the he had the urge to emphasize that he was correct; which I acknowleded that he was.But when trying to educate, particularly those still in grade or high school I feel it is important to give them the correct information. And again, make the assumption that one does not have the formal education of the definition of efficiency then re-read BoredAtWork's post, can one not see how it is possible that someone may conclude that eta=Vout/Vin, which is wrong. Not trying to flame here, just trying to pass along the most precise info possible for the OP.So I guess succinctly I don't really have any issue(s); rather, I would urge the OP to use the correct definition of the efficiency in his paper/presentation.

author: bfritz
author_rank: Regular Contributor
timestamp: January 31, 2012, 04:50:40 am
content: Quote from: FenderBender on January 30, 2012, 04:04:46 amQuote from: FenderBender on January 30, 2012, 09:14:42 pmFenderBender,I think you are headed in the right direction. I could give you the answers up what to do, but I think you know how to proceed, and I'm more of a "teach them to fish" sort of guy. :-)Take the list of pros and cons from wikipedia, and decide how to show those, with what you have. For example, I think you know how to show what the efficiency is of each. Do that for as many of the items as you can in the Pros/Cons section of the wikipedia article.If there are certain ones that you can't figure out how to demonstrate, come back and ask for specific help on those items. I'll make sure to come back, and look for any questions you may have. Note that for some items you may not have the correct equipment to test the Pro/Con. For example, being able to tell the difference for line conducted emissions or radiated emissions will require access to a spectrum analyser, which you likely don't have. So, for some you may just have to state the theory of why one is better/the other less so, and state why you can't demonstrate it without designing test gear.You're doing great. Keep going!

author: Bored@Work
author_rank: Super Contributor
timestamp: January 31, 2012, 06:28:51 am
content: Quote from: mstevens on January 31, 2012, 03:27:53 amIf you manage to read it, which you apparently failed at, it can't.QuoteIt is right for linear regulators. And I stated nothing else. Your fault is that you apparently didn't know this and are now shitting your pants, making a fuss about your mistake.QuoteIf it would be really about giving the most precise info you wouldn't shit your pants about my second posting with more information.QuoteI think you have, but I leave it to you to figure it out.

author: FenderBender
author_rank: Super Contributor
timestamp: February 01, 2012, 10:12:51 pm
content: So here are some parameters and methodology I plan on testing--Output noise/ripple>Outputs will be monitored with an oscilloscope. Linearmayhave an advantage here. Since I am using the MC34063 as the switching regulator here, I may show the output with the optional filter and without. Output may be very reasonable with the optional filter, but I will have to make the point that it is now more complex and possible more expensive/larger.-Efficiency>Compare input power to output power via several multimeters. SMPS should be superior in this aspect.-Heat>Make a small enclosure. Put the power supply circuits in the box with some openings for wires + K-type thermocouple. I'll load the supply up for a while and record ambient temperatures over time. Should show that SMPS produce much less heat. To illustrate why heat is important, I could say that if you used a linear regulator in your phone, your pocket would be on fire or something.-Complexity->For smaller scale implementations, linear designs can be less of a hassle, and can cost less (depending on the situation).-Load Regulation->I don't know if this is the right term, but I am going to try to see what happens if I suddenly apply a load. How well does it maintain the output when this load is applied? How quick does it respond. Not sure whoshouldwin here.

author: erupter
author_rank: Contributor
timestamp: February 01, 2012, 11:17:08 pm
content: May I suggest a stupid simple setup?Get a couple of decent mosfets.One is part of the feedback loop of an opamp, use it to light a lamp.Put a temp probe on it.The other is pulsed by a NE555 (or similar) in a PWM mode.Same load: a lamp.Put a temp probe on it too.Crank the output voltages of both to the same level, show temp difference.

author: FenderBender
author_rank: Super Contributor
timestamp: February 01, 2012, 11:43:33 pm
content: To help show the difference in operation between SMPS and linear? Sounds good!Sounds like a good idea, though I'm running out of time.The problem I'm having with this is that the kids in my class frankly don't give a shit. Let's put it this way. I live in an urban area and my school has about a 20% dropout rate....not all of the kids in my class necessarily want to be there. They just got thrown in there because it was the only opening.So the problem is how to explain everything on a lower level.I guess the point would be to impress the teacher, but I want to also benefit the kids in someway as they are the ones that should be learning.The word PWM means nothing to anyone..maybe not even my teacher. I don't know.I'm going off on a tangent here, but the point is: I'm not sure how to go about any of this. How do I explain that the op amp is "always on" and the PWM switches off and on, which is why its more efficient. It's just tough.

author: ivan747
author_rank: Super Contributor
timestamp: February 02, 2012, 01:20:20 am
content: You could explain things this way:Just be aware that switching regulators are not just PWM. Actually, they don't change the duty cycle too much."The problem I'm having with this is that the kids in my class frankly don't give a shit."<--- At this point I would focus on impressing the teacher.You can also bring a linear power brick and a switching equivalent and the difference in size an weight will be obvious if the supplies deliver something like 12V at 700mA.Good luck!Ivan

author: FenderBender
author_rank: Super Contributor
timestamp: February 02, 2012, 01:59:54 am
content: Oh yeah I know SMPS isn't PWM though the SMPS might have PWM as part of it.

author: bfritz
author_rank: Regular Contributor
timestamp: February 02, 2012, 06:05:51 am
content: Quote from: FenderBender on February 01, 2012, 10:12:51 pmWhen doing this test, be sure to load the output at close to maximum current. You will almost always see more ripple and noise with a loaded output.QuoteIt would be informative to discuss that efficiency is proportional to heat produced. In a supply, there is nothing present that stores energy for long periods of time. Since physics tells us that "energy is conserved", any energy that doesn't exit the output is converted to some other energy, and the only other energy being produced is heat. So, the lower the efficiency, the greater the heat.QuoteI suggest skipping this for your presentation, as I don't think this fits exactly with what you are trying to show in your presentation, as each topology can be designed for good regulation, it is mostly a function of loop response, and output capacitor choice. So, I don't think this does much to highlight the differences. I think for your purposes, this may sort of "muddy the water" and you might want to just skip this. I'm guessing you have more than enough material with the rest.As far as load regulation, the items that will determine this:Size of output capacitor (uF, not physical). A larger output capacitor will tend to supply pulse currents to the load better, and as the load draws less current, absorb the energy that might cause overshoot because the loop response has not kicked in.Impedance of output capacitor. A capacitor with a low impedance will do a better job of supplying pulse currents, and absorbing energy that might cause overshoot.Feedback system loop response. For a given loop response, it should be about the same cost and complexity for both systems. You may find that one system has a better loop response, just because it happened to be easily possible for that IC manufacturer.