Design an low power boost converter - Page 1
Forum section: electronics / beginners

author: hitech95
author_rank: Regular Contributor
timestamp: February 12, 2017, 09:09:30 pm
content: Hello,I just started a new project, nothing complicated: a simple board for an OLED display, or so it seemed. Having its little brother I saw that only needed extra component are some capacitor and resistors .... but reading the datasheet I've found that the bigger display that I have chosen, unlike the one in my possession needs a power supply to 12.5V.According to the datasheet the maximum current is about 50mA (44.5mA to be picky) and the minimum of 11mA (11.8mA).As supply i'm using it to 3.3V (comes from a AMS1117 in MCU onboard regulator), so I need a boost converter. I never built anything like that.I had been looking for some infos but I do not understand how I should proceed to the design this circuit.Doing a bit of research I found in available components the TPS61040: a little overkill for my needs but has a package that can be soldered.Reading of the applications notes I realized that the first step is to choose the inductor and consequently make various calculations on the peak current, load current, frequency convertiore and so on.Remaining on the chip I reported I doubt watching the first formula (Paragraph "7.3.1Peak Current Control"), the unit of measurement for inductance is in H? What is the result of that formula, what would be the unit of measure? I do not understand.The biggest problem is how to choose the components correctly, the datasheet provides some examples without explaining (of course) why.Can anyone make me clear?Yeah, I saw that in the datasheet there is already an example 3.3V -> 12V but when I have some time start projects to learn not to copyRegards,Nicolò.EDIT:OLED Datasheet:http://www.buydisplay.com/download/manual/ER-OLED015-2_Series_Datasheet.pdfTPS61040 Datasheet:http://www.ti.com/lit/ds/symlink/tps61040.pdf

author: tron9000
author_rank: Frequent Contributor
timestamp: February 13, 2017, 09:25:54 am
content: Quote from: hitech95 on February 12, 2017, 09:09:30 pmThe units of that formula is Amps (A)

author: hitech95
author_rank: Regular Contributor
timestamp: February 14, 2017, 10:50:19 am
content: Quote from: tron9000 on February 13, 2017, 09:25:54 amUhm... ok.I tried to use the "My Ti" suite to design a circuit but looks like that the TPS61040 cant handle the required current (Not sure why)Anyway after some research i switched to a LM2733x, same package similar simple circuit...

author: MagicSmoker
author_rank: Super Contributor
timestamp: February 14, 2017, 02:35:42 pm
content: Quote from: hitech95 on February 12, 2017, 09:09:30 pmThe key parameters in designing a boost converter are: switching frequency, duty cycle, peak and ripple (ie - peak to peak) current.In this case, the TPS61040 takes a slightly different approach that is likely the source of your confusion - this IC turns on the boost switch until the peak current rating is reached, or at approximately 400mA, though with some overshoot in that value due to an (appallingly long) 100ns propagation delay.The inductor equation is L * dI = V * dt, where dI is the change in current, dt is the change in time, and V is the voltage across the inductor.In this case, let's use nanoseconds for dt and nanoHenries for L, as that keeps the scale consistent on both sides of the equation.We want to find L, so we need to know or estimate the following parameters: 1) the voltage across the inductor (which is simply Vin during the switch on time in a boost converter); 2) the change in current (which is given as ~400mA in the TPS60140 datasheet); 3) the change in time (which is the switch on time).In a boost converter, the duty cycle, or percent time that the switch is on, is approximately given by the simplified equation, (Vout - Vin) / Vout, or ~73% here.Knowing that the maximum switching frequency for the TPS60140 is 1MHz, the "minimum" on time for the switch will need to be at least 730ns (1MHz = 1000ns period), so there needs to be sufficient inductance that it takes at least 730ns to hit 400mA with 3.3V across it. Plugging all that back into the inductor equation, we get a minimum value of 6000nH (6uH) for the inductance.Now, go to Digikey or Farnell or whatever you use over in Italy to select an inductor of at least 6uH and at least 400mA current rating - in actuality, you will want to select a much higher current rating, and a higher inductance value (which will cause the IC to lower the switching frequency). Let's pick 8.2uH for the inductance, as that is a standard value, and search for parts rated for at least 600mA.Making a semi-random selection, here is one part that should work just fine:https://www.digikey.com/product-detail/en/abracon-llc/ASPI-0316S-8R2N-T3/ASPI-0316S-8R2N-T3CT-ND/3060073But as you should infer from the above explanation, there will likely be hundreds of inductors in stock that will work. Just make sure the construction is shielded (e.g. - don't select a rod type) and that the DC resistance (DCR) value doesn't result in too much of a voltage drop at 400mA (the above part has a DCR of 0.34 ohms, so will drop about 0.14V at 400mA).

author: hitech95
author_rank: Regular Contributor
timestamp: February 14, 2017, 05:37:13 pm
content: Quote from: MagicSmoker on February 14, 2017, 02:35:42 pmThanks for the explanation, you have made the whole process much simpler and understandable to me.I only have one doubt, opening "My TI platform" I checked why it believes that this chip will not work with my parameters. The error is "Toff minimum constraint" not respected.If I understood the speech: if Tonis equal to 730ns the Toffat 1MHz frequency is 270ns which is less than the minimum value of 400ns.This value, however, is indifferent to which inductance I choose because it is calculated on the duty cycle.So that is why I can't use this chip...What happend if I change the inductance value, theoretically the value is equal to 6uH, but you said to use a standard value (of course). According to the formula it would change the peak current, right?

author: MagicSmoker
author_rank: Super Contributor
timestamp: February 14, 2017, 07:35:37 pm
content: Quote from: hitech95 on February 14, 2017, 05:37:13 pmThe inductance value is inversely proportional to frequency, and only frequency if all other parameters held constant.Duty cycle is only proportional to the ratio of input to output voltage, though see additional explanation below.This particular IC regulates output voltage by varying its switching frequency (up to a maximum of 1MHz) while the off time (~400ns) per cycle and the peak current in the switch (400mA) are fixed. The more frequently the switch turns on the more energy is delivered to the output and so this type of regulation is called, appropriately enough, pulse frequency modulation.This is different from pulse width modulation in which the duty cycle is varied to regulate the output against changes in supply voltage or load current, and so it would appear that we don't really need to know duty cycle to use this IC, but that is not correct.More specifically, the datasheet says the maximum on time per cycle is 6us, while the minimum off time can range from 250ns to 550ns (always look at the guaranteed min/max values of a spec, rather than rely on "typical"). This means the maximum guaranteed duty cycle is 6us / (6us + 0.55us), or ~92%, and would occur at a switching frequency of ~153kHz (period of 6.55us). However, at the maximum switching frequency of 1MHz (period of 1us, or 1000ns) and if the minimum off time is at the datasheet maximum of 550ns, then the maximum duty cycle will be limited to only 45%!So my previous recommendation of 8.2uH "just to be safe" would not actually be totally safe (I should have followed my own advice about looking at the max/min for each spec)! Bumping the inductance up to 12uH should guarantee operation even if the IC has the worst case 550ns of minimum off time, but going to the next standard value up of 15uH would better accommodate variation in the inductance value itself (most commercial inductors have a 20% tolerance) as well as the various voltage drops/losses involved. You still want the current rating to be substantially higher than 400mA, so an example part number that should work well is:https://www.digikey.com/product-detail/en/bourns-inc/SRN4018-150M/SRN4018-150MCT-ND/3821523

author: hitech95
author_rank: Regular Contributor
timestamp: February 15, 2017, 12:29:22 pm
content: Quote from: MagicSmoker on February 14, 2017, 07:35:37 pmStill not sure on how you have found that 12uH value....The formula should be, but I'm wrong....:l*0.400A=3.3V*450ns

author: MagicSmoker
author_rank: Super Contributor
timestamp: February 15, 2017, 02:15:33 pm
content: Quote from: hitech95 on February 15, 2017, 12:29:22 pmThe major* thing you seemed to have forgotten is that the duty cycle, D - which is the ratio of Ton / (Ton + Toff) - needs to be approximately 0.73 (ie - on for 73% of the switching period) to achieve the desired boost ratio.This means thetotal switching periodmust exceed Toff / (1 - D), or 2.037us in this case, and therefore the on time must exceed 2.037us - 0.55us, or 1.487usConsidering the worst case current limit value is 450mA (0.45A), then the proper numbers to plug into the inductor equation are:L > (3.3V * 1.487us) / 0.45AL > 10.9uHNote that I used thegreater thansign to emphasize that the inductance value can be higher, but note that it can't be arbitrarily large because this IC also has a maximum allowed on time of 6us, and if that value is plugged into the inductor equation instead of 1.487us you get a maximum allowed inductance of 44uH. So anything between ~11uH and 44uH will theoretically work, with the switching frequency automatically dropping linearly as you go up in inductance. Since higher switching frequency means less capacitance is required for filtering/ripple reduction you generally want to use the lowest value of inductance you can get away with, leaving some margin for tolerances and such, hence why I recommended 15uH before.* - you also did not use the worst case values for minimum off time and peak current.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: February 15, 2017, 03:25:22 pm
content: While this is all well and good for an educational excercise, for a cookie cutter application like this I suggest just using TI's Webench to design a circuit that will suit your needs.

author: hitech95
author_rank: Regular Contributor
timestamp: February 16, 2017, 08:16:55 am
content: Quote from: MagicSmoker on February 15, 2017, 02:15:33 pmThanks, now I got it.Quote from: suicidaleggroll on February 15, 2017, 03:25:22 pmWell, the purpose of this project is to learn something....