Problem calculating L in SMPS with variable Vout. - Page 1
Forum section: electronics / beginners

author: king.oslo
author_rank: Frequent Contributor
timestamp: January 21, 2012, 11:03:46 pm
content: Hello,On page 20 of the datasheet of my buck controller, Maxim lists formulas to help me calculate specs of my inductor.http://datasheets.maxim-ic.com/en/ds/MAX8792.pdfL = ((Vin - Vout) / (fsw*Iload(max)*LIR))(Vout/Vin)This is all fine, only problem is that I have variable Vin (swinging between 16.9 and 16.05). I will also have adjustable Vout.I am aiming for a LIR of 0.2.What should I chose to substitute Vin and Vout in the above formula?Thanks.Marius

author: IanB
author_rank: Super Contributor
timestamp: January 22, 2012, 12:08:41 am
content: On reading the data sheet I learn that bigger inductors will be better (within reason), so you need to choose an inductor that is big enough for your worst case scenario (make a conservative selection).From that formula, it is evident you should select the smallest value of Vin and the largest value of Vout to find the minimum inductor size. If the inductor is slightly too big everything will be fine. If the inductor is too small, everything won't be fine.Keep in mind that the inductor should have the lowest DC resistance practical. It won't help to choose a large L that also has an excessively large R.(As Richard notes in the post below, "big enough" includes magnetically big as well as inductively big.)

author: amspire
author_rank: Super Contributor
timestamp: January 22, 2012, 12:09:26 am
content: Using the maximum input voltage, you have to make sure that the inductor is equal or greater then the maximum L required over the range of output voltages. So plot L for different Vout and see what the maximum L is.The next step is really important. Calculate the Ipeak current. The inductor must be able to handle that current without saturation. I would probably add a 20% extra margin to make sure.One of the things you realize is that a switching power supply is all about finding the best compromise. You can have high ripple current and a large output filter capacitor, or a low ripple current and a smaller output filter capacitor. Each choice has its own different benefits or problems. A big output capacitor means that changing the output voltage will be slow as you have to charge or discharge the output capacitors. Also even if the regulator has a current limit set, there is no current limit on the charge stored in the capacitors. So with a large output capacitor, the current limit is more to protect the regulator then the output load.With small output capacitors, and small ripple current (ie a large inductor), the capacitors can change voltage more quickly, but the inductor cannot. You will get signifigant undershoot and overshoot for sudden load changes. The undershoot as you go from zero to full current will be worse at minimum input voltage and may happen at low output voltages or high output voltages - it all depends on how the numbers work out.The maximum overshoot as you go from full current to minimum current will happen at low output voltages and will be independent of the input voltage.The higher the switching frequency, the lower the inductor and the more you can minimize the problems above, but you get higher switching losses.So what I am saying is that if your needs are not to demanding, follow the data sheet calculations and you will end up with a supply that is stable and reliable. It is much easier designing a switching supply for one output voltage then a wide range of output voltages. If you need to minimize overshoot and undershoot, minimize ripple, and maximize output voltage range, then it can take a lot of trail and error to find the magical set of numbers that meet your minimum specs, and tht give the best overall result.If you need high performance, and you are not confident about the calculations, then a linear regulator is much easier. Some lab power supplies use a switching regulator followed by a linear regulator, so they can get the performance of the linear regulator, but they can limit the voltage drop across the linear regulator to improve efficiency and reduce heatsink size.Richard

author: IanB
author_rank: Super Contributor
timestamp: January 22, 2012, 01:39:07 am
content: For background information, a general note about saturation current in an inductor might be of value.Inductors are similar in many ways to capacitors. It can even be said that an inductor is the inverse of a capacitor.Capacitors and inductors are energy storage devices. A capacitor stores energy in an electric field proportional to the square of the voltage. If you double the voltage you quadruple the energy. But there is a limit to the voltage for any given capacitor. If you go above the maximum voltage the dielectric insulation will break down and the capacitor will not function properly. Capacitors have a maximum working voltage marked on them, and usually capacitors of the same construction and value with higher working voltages are bigger. In principle the bigger size correlates with being able to store more energy (higher voltage before breakdown).Analogous to a capacitor, an inductor stores energy in a magnetic field proportional to the square of the current. If you double the current you quadruple the energy. But as with a capacitor, there is a limit. If you increase the current too much the magnetic core will saturate and it will be unable to store any more energy. Inductors have a maximum current rating where the core saturates and they will no longer function as inductors if you increase the current beyond this limit. Usually inductors with the same inductance and core material but a higher current rating are bigger. The bigger size correlates with being able to store more energy in the magnetic field before the core saturates.With both capacitors and inductors the increased physical size and cost is the reason why you would not just choose really big values. To keep things reasonable it is necessary to choose the smallest size that will work, but no smaller.

author: king.oslo
author_rank: Frequent Contributor
timestamp: January 22, 2012, 05:12:03 am
content: Richard, Ian: Thank you. This information was worth it's weight in gold (unfortunately, it didn't weigh anything).Hehe!The datasheet only has typical setups for sub 2V supplies.I was thinking about Richard's advice regarding alterations in fsw. This is a 1.5-15V 7A supply. I understand that the next thing I will ask is dependent on a variety of factors. However, I am only asking for an approximation: Circa what kind of losses are we talking about when increasing fswfrom 200kHz to 600kHz? 1s or 10s of percents of losses?Thanks.M

author: amspire
author_rank: Super Contributor
timestamp: January 22, 2012, 05:50:25 am
content: Can't really say without seeing a circuit, pouring through the data sheets and doing the calculations.But I would advise you design and test at 200KHz first. If everything is running very cool, you can try and increase the frequency.600KHz means a much smaller inductor and smaller capacitors which would be important if you were designing a production board, but you have 7A current at a fairly high input voltage. Both can lead to high switching losses very easily.Richard

author: king.oslo
author_rank: Frequent Contributor
timestamp: January 22, 2012, 09:51:37 am
content: Richard, yes. I will do that. My goal was 5mV peak to peak out. I worked out I need ESR <= 5.4mOhm. That should be possible. But it does surprise me that the datasheet doesn't mention what kind of actual capacitance is needed? Surely this must be relevant. 1uF vs 1000uF is a big difference at 15V 7A. Right?Have I missed something?Thanks.M

author: jahonen
author_rank: Super Contributor
timestamp: January 22, 2012, 10:16:25 am
content: I think it might be beneficial to do a crude simulation of the power stage to get idea of what kind of currents and voltages etc. can be anticipated for different output inductors and capacitors. You can use pulse voltage source of SwitcherCAD to do that. Just set fixed duty cycle and see how it behaves.Regards,Janne

author: amspire
author_rank: Super Contributor
timestamp: January 22, 2012, 02:35:14 pm
content: Quote from: king.oslo on January 22, 2012, 09:51:37 amThe capacitance value comes out of your performance specs and the frequency and inductor choices.Lets say you design for 100mA p-p ripple in the inductor and you want 5mV p-p from the capacitor. The ripple will be set by two capacitor parameters - the capacitance and ESR. Lets just assume half the ripple is from ESR and half from capacitance.At 200KHz and 2.5mV capacitive ripple, you have a voltage slew rate (dv/dt) of 0.0025V * 200KHz = 500V/sec.C = I/(dV/dT) = 0.1/500 = 200uFFor 2.5mV of ESR ripple, you need an ESR of less thanRESR= .0025V/0.1A = 0.25 milliohmsSo at minimum, you want a single capacitor of combination that has a capacitance of 200uF or greater, and 25 milliohms or less.Now the next issue is the voltage overshoot or undershoot.I will just look at overshoot.At 1A current through the inductor, then energy stored in the inductor is E = 1/2 L x I2= 1/2L joules.If you suddenly drop the load down to zero, the only place left for this energy to go is the output capacitor.For a capacitor, E = 1/2 C x V2The biggest voltage overshoot will occur when the capacitor voltage starts at 0vIn this case the inductor energy all goes into the capacitor soE = 1/2 L = 1/2 C x V2So the capacitor output voltage V will be (L/C)1/2If the inductor is 10uH and the capacitor is 200uF, then the voltage overshoot is (10-5/2x10-4)0.5= 0.7VIn other words with a 10uH inductor and a 200uF capacitor, if you had set the power supply to near 0V out at a current of 1A, and suddenly you drop to 0A, the voltage on the capacitor will jump up to 0.7V, and the it will only drop if there is a load to discharge it back to the regulated voltage near 0V.I won't do the calculations here, but at 12V out, the same parts would have an overshoot of about 4mV if my quick calculation is right.To reduce this overshoot, you can reduce the inductor, or increase the capacitor. If you increase the output capacitor, then a lot of energy is stored in that capacitor - if you short the output, you get many amps flowing momentarily instead of the 1A current limit.Undershoot is caused by the fact that the bigger the inductor is, the lower the current ripple as you know, but the longer time it takes for the current to increase after a load current increase. This causes the output voltage to undershoot, which is again reduced with a big output capacitor.There are other subtleties, but I hope I have shown how the choice of C is based on your design specs and your other regulator design choices.The reason why they didn't spend much time on C, is that if you are designing a fixed voltage regulator, just whack in a big enough output capacitor.If you are designing a variable input and output voltage regulator, optimizing performance can be complex. In comparison, a linear regulator has no problem (other then heat) at handling widely varying input and output voltages, and you can often use a much small output capacitor, which means the output can change much faster, and there is no large capacitor to discharge into you load under fault conditions.Richard

author: king.oslo
author_rank: Frequent Contributor
timestamp: January 23, 2012, 09:41:47 am
content: Thanks guys.M

author: king.oslo
author_rank: Frequent Contributor
timestamp: January 23, 2012, 12:42:24 pm
content: Some good news: Whilst simulating, I was able to reach my goal of 5mV ripple with LIR .4. 6.8uH and 100uF!I also noticed that if I place several output capacitors in parallel, I get less ripple than one of the same value. Any guesses as to why?Are there any problems doing this? MEDIT: And must I use the forumla for parallel resistances when using multiple capacitors for further calculations using total ESR?

author: jahonen
author_rank: Super Contributor
timestamp: January 23, 2012, 03:27:35 pm
content: Quote from: king.oslo on January 23, 2012, 12:42:24 pmYou get less ripple because ESR is smaller when paralleling multiple capacitors. ESR-based ripple can easily dominate the ripple voltage if switching frequency is over 100 kHz or so. You can separate ESR and capacitive ripple components by noting that ESR ripple is a square wave, capacitive ripple is triangle wave.I think it is almost always better to put multiple pieces of same capacitors in parallel than just put one big one due because paralleling improves ESR and ESL consistently. BOM price might be higher with multiple pieces of capacitors, though.And yes, you can calculate combined ESR of parallel capacitor just like parallel resistors.Regards,Janne