Making a boost converter, which inductor value should I use? - Page 1
Forum section: electronics / projects

author: WarFreak131
author_rank: Contributor
timestamp: October 01, 2022, 08:30:41 pm
content: I want to create a boost converter for a project that takes an input of 5 V and outputs anywhere from 24 V to 40 V. I have decided to use the LM2578A (https://www.ti.com/product/LM2578A).If I power my load with 24 V output, it will take about 26 mA. If I power it from 40 V, it will take about 62 mA.Page 15 of the datasheet shows how to calculate the necessary inductor value based on the load current and other factors.IL, MAX DC= ILOAD* (VO/VIN)andE-TOP= (VO-VIN)*(VIN/VO)*(1000/F,kHz)Assuming I use a timing capacitor to choose a frequency of 100 kHz:If you use a 40 V output, IL, MAX DC= 0.5 A, and E-TOP= 44.When you locate the intersection of 40 V and 44 on the bottom chart, you get somewhere between 220 and 330 uH.If you use a 24 V output, IL, MAX DC= 0.12 A, and E-TOP= 40.That gives you somewhere between 1 and 1.2 mH.Which inductor value should I choose to make this work across all the output ranges, the larger one or the smaller one, or is that even possible?

author: tooki
author_rank: Super Contributor
timestamp: October 01, 2022, 09:14:33 pm
content: TI has a quite good online simulator that will take your input parameters, produce a schematic, with component values (and the ability to select other components within a recommended range), and simulate the circuit, and even a PCB layout.I’ve used it a few times and it’s worked well.https://www.ti.com/design-resources/design-tools-simulation/webench-power-designer.html

author: WarFreak131
author_rank: Contributor
timestamp: October 01, 2022, 09:32:29 pm
content: Thanks for the link. However, I think this is a more theoretical question about the function of a boost converter. Does the larger inductor cover the functionality of the smaller one or vice versa? My knowledge of how switching converters work isn't extensive, but if this were built with a smaller inductor, but was running at a low current which favored the larger inductor, doesn't the chip reduce its own duty cycle so that it will output the correct current?

author: Faranight
author_rank: Unknown
timestamp: October 03, 2022, 12:35:47 pm
content: I have done some buck research/designs, and I can perhaps share a few things.You should ideally select the inductor according to the worst expected case plus some margin. Higher inductance value will generally mean lesser voltage ripple.Please also pay special attention to the inductor saturation current rating, which must be higher than the maximum expected inductor current (average inductor current + voltage ripple / 2).You can also decrease the size of your components (i.e. if your inductor is too big) by choosing a regulator with higher switching frequency (i.e. 300kHz to 1MHz).

author: WarFreak131
author_rank: Contributor
timestamp: October 04, 2022, 12:30:42 am
content: Thanks for the information, so by "worst expected case," I'm assuming you mean the case where the maximum amount of current is being drawn, correct? I'm perfectly content using an MC34063A now that I know the true nature of my current requirements.Some background on the project, I am building a clock using a multiplexed VFD tube. To get the current requirements, I powered all grids and all segments simultaneously (something that would never occur since this will be multiplexed). That is how I got to the 62 mA figure. Ergo, I will never actually reach 62 mA, so my true current requirements are even lower. Originally I thought I was gonna need 200 mA+, which, with the voltage boost, brought me over the MC34's internal current limit of 1.5 A. But I can go for something cheaper now that I've measured it accurately.

author: TimNJ
author_rank: Super Contributor
timestamp: October 04, 2022, 01:38:11 am
content: Interesting set of charts in the LM2578A datasheet. I guess it's handy as quick reference, but seems a bit cumbersome beyond that, to be honest. It might be more useful to think in terms of %ΔI(L) instead of E-Top. At least, that's a more typical approach, in my experience.You can use a wide range of inductance values and have the converter "work". If you use an inductor lower than the prescribed value per the datasheet, at a given set of load conditions, this just means that the inductor ripple current ΔI(L) will be higher. Conversely, if you use a higher inductance, than ΔI(L) will be lower. For a number of practical reasons, %ΔI(L) between 20 - 30% is usually selected as a reasonable compromise. In an ideal world, we'd want %ΔI(L) to be low, to keep voltage ripple low, switch peak currents low, and output capacitor ripple current low. But, if we selected a (high value) inductance based off of, say, %ΔI(L) = 5%, then we'd have a monstrously large inductor - usually larger than the given application can tolerate. So, we back down %ΔI(L) to somewhere in the 20 - 30% range (typically), to use a physically smaller inductor.So, then you see that inductance value is proportional to physical size of the inductor, for a given application, if switching frequency and loading conditions are held constant. The reasons for this are probably beyond the scope of this discussion, but are related to saturation of the core material (most generally).

author: Faranight
author_rank: Unknown
timestamp: October 04, 2022, 06:04:50 am
content: Keep in mind we're talking about a boost converter here, not a buck. So, when picking an inductor for your application you should dimension it according to the input current, not the output current.From your measurements you can calculate the average inductor current (with assumed 85% efficiency):24V * 26 mA / (5V * 0.85) = 147 mA40V * 62 mA / (5V * 0.85) = 583 mAKeep in mind that this is only an average inductor current. You still have to add an appropriate amount of ripple current (ΔIL/ 2) to the numbers. Assuming 30% current ripple (ΔIL= IL* 30%) we get 169 mA and 670 mA respectively. The latter current is the worst case, so you would pick an inductor with its saturation current higher than this value (with some margin) i.e. at least 800 mA. One more thing to keep in mind are the capacitors. You would generally use two types of caps here in parallel. MLCC's with extremely low ESR are used to keep the voltage ripple to a minimum, and the bulk alu electrolytics should be used for proper transient response i.e. so that the output voltage doesn't fall too much when a large load suddenly appears at the output.

author: Terry Bites
author_rank: Super Contributor
timestamp: October 11, 2022, 06:16:19 pm
content: The inductor size is essentially dictated by the amount energy you need to move from input to output per cycle.You can doing the maths. I love maths in all its tedious forms. Read TIs SLVA797 and SNVA038BOr stop mucking about.www.coilcraft.com/en-us/tools/dc-dc-optimizer/#/search