Need some advice for my first buck regulator circuit - Page 1
Forum section: electronics / projects

author: superKris
author_rank: Regular Contributor
timestamp: March 14, 2020, 02:12:15 pm
content: I'm designing some hardware using EasyEDA. I'm planning to do the PCB and SMD assembly at JLCPCB. I just got my first print last week, and i'm very happy with the price and quality.My current design features a Arduino mega on a PCB with a 3" OLED, some LED,s chips, and expansion ports. I need to power the circuit with 11~15V from a car battery, and am looking to use a more efficient regulator than the AMS1117 on the Arduino.So i want to convert 11 to 16V to 5V, and i'm targeting a current usage between 200mA and 1A (the 1A is for future expansions)JLCPCB SMT service has a very limited list of basic components, and i want to prevent extra charges from their extended library. I made quick evaluation of the available types below:So i wanted to go with the last one. I played around with their great design tool, but i can really get the efficiency above 88%. This means i hit the maximum dissipation of the SOT-23-6 package before i get to 1A. This may be acceptable, but i rather go up to 1A and dont love the idea of the component working at its maximum disipation.I'm kind of getting the idea that it is not realistic getting getting more than 1A out of SOT-23 or SOIC-8 package buck converter. Is this true, or am i doing something wrong here?Which regulator from the list would you pick?This will also be the first time i work with coils. How does choosing a higher or lower uH effect the efficiency of a buck converter? I understand the advantage of a higher current coil, and a lower DC resistance.But what is a smart SMD package size to choose? There are so many options here.By the way, i'm also trying to keep the BOM costs low, as i may want to bring this product to market. That is why i did not choose the LM2576.Also while on the subject; Is there a reason all Arduinos use linear regulators? Are there disadvantages to supplying the 5V pin with a buck regulator?

author: mariush
author_rank: Super Contributor
timestamp: March 14, 2020, 09:42:50 pm
content: Don't look at just the price of the switching regulator IC.Look also at the components they need to work.Some inductors are more expensive than others, some are cheaper. Often an IC may cost you a few cents more but it may run at higher frequency so you can use a smaller and cheaper surface mount inductor, or that regulator doesn't require some ceramic resistors and/or capacitors that others do need) so overall you're saving money.Also look into the total area used by the whole circuit.They use linear regulators because they're basically a single chip, only needs maybe an input and output capacitor. Switching regulators often require an inductor, an extra diode etc all these add up to the cost.There's something to think about in switching regulators may be more noisy so the ADC performance could be affected a bit I suppose, but cheap linear regulators don't filter well whatever crap psu most noobies use to power the board so I don't think this is the reason they use linear regs.// also pay attention to description and information and recommendations in datasheets when it comes to inductors and specifications for inductors and filtering capacitors ... they definitely affect efficiency and badly chosen components can even cause oscillations or issues from time to time.Linear regulators aren't excepted, especially the 1117 linear regulators. For example, most 1117 regulators require a capacitor on the output which ESR between 0.1 ohm and 1 ohm, otherwise they output can go bad. Some datasheets even go so far as to recommend 0.3 ..1 ohm .... so ceramic capacitors are not recommended, unless you add a resistor in series with the ceramic capacitor.

author: Siwastaja
author_rank: Super Contributor
timestamp: March 15, 2020, 08:52:58 am
content: I like your color-coded decision-making spreadsheet, I work the same way.Note that the efficiency curve is for a full "typical" system (funnily enough, often they don't specify the values of the components which it was measured at).This means, it's possible to getbetterefficiency than the datasheet curve, by using better external components; but also that the dissipation calculated from the efficiency number is not fully in the regulator IC, but also in the external components.At 5V, non-synchronous rectification is already the efficiency killer (at 3.3V, even more so, and below 3.3V, synchronous rectification becomes an absolute must). So the external diode takes big part of the total dissipation. Sure, due to the high f_sw (and related fast switching), the diode needs to be laid out very close to the regulator, so you can't spread the heat over a large area; but it's still way better than if the diode was integrated on the tiny package.Note that although you can find some schottky diodes with extremely low Vf around 0.2V, these tend to have reverse leakage currents so high that the losses may go up instead of going down, even causing destructive thermal runaway. So always look at the reverse current (at the actual reverse voltage,at the high temperature) when selecting the diode, as well.Typically such complete ICs are rated for a certain maximum current for a reason; they can actually deliver it (of course, add a bit of your own margin) without going to extreme measures; no need to cool it with liquid nitrogen to achieve that 1.2A spec, which is the case with some MOSFET "continuous current" ratings, for example.These 1-2MHz very small integrated switchers are usually the way to go because not only they themselves are of the cheapest price, the passives also end up cheap due to the high f_sw.If you want to up the efficiency and reduce losses further, try to see if a similarly integrated switcher with synchronous rectification is available - then you get rid of the external diode, as well, so one part less.The elephant in the room is the automotive environment; if you want to power the device while the engine is running, you should seriously consider better input protection. OTOH, if this is a hobby one-off or a cheap entertainment device, there is no absolute need to protect againstload dumps(Google it), which are relatively rare events - never seen one myself, for example.As a side note, do not use 1117 regulator. It totally sucks: it isn't even very low-drop, but comes with all the downsides of older generator low-drop regulators. Either use an older classic like the 78xx, LM317 series which have better stability and easier output capacitor selection, or use an actual modern regulator which have lower drop-outs than 1117andbetter stability. We don't have to deliberately add ESR anymore.

author: superKris
author_rank: Regular Contributor
timestamp: March 16, 2020, 03:09:55 pm
content: Quote from: mariush on March 14, 2020, 09:42:50 pmThanks for your reply! Its indeed the price of the whole cirquit i care about. In my case i have many expensive components like a 3" OLED, 8 pcs of $2,- pushbuttons, a Arduino Mega Pro, etc. That doesnt mean however that i dont care about other component prices. I if ever bring this product to market, every $ counts, but it will never become a sub $35,- board. So i dont care about the difference between a $ 0,20 and a $ 0,25 regulator, but do care about the difference between a $ 0,20 regulator and a $ 1,50 regulator. The cost of a very small coil vs a larger value coil ($0,20 vs $0,30) isnt effecting the BOM costs that much. Also i have plenty of board space because of spacing i need for the switches.As vintage auto enthusiast i have always avoided switching regulators because of their suspected noise levels, but i doubt this was very fair. In this case i'm only using 1 ADC on the arduino, and that is for a LDR only. I dont need it to be accurate at all.This board will however be combined with another board that has HALL current sensor IC's on it. These IC's also have a analog output in the 0-5V range that goes to a external ADC (ADS1015). Also i'm using some INA219 for external shunt measurements. I'm planning to add the same circuit to this PCB too. Can i expect noice problems? I know the right PCB design/placement of components will effect the amount of noise, but should i sick with the MP2359, or look for another type?Quote from: Siwastaja on March 15, 2020, 08:52:58 amAlso thank you very much for your reply.I'm currently usingSS34diodes in my design, and was planing to use this in the regulator too. Its a Schottky, but with a Vf of 0,55V which isnt the lowest. It was choosen because its one of the few schottkys available in the JLCPCB basic component library that has a decent current rating. Do you think this diode is suitable? I tried it in the the MPS design tool, and it did not effect efficiency in that tool.I'm nut really understanding the current rating of these smaller chips yet. The MP2359 is rated at 1.2A. I bieve this can be reached without surpassing the maximum package disipation, but the conditions would need to be perfect. For the 3A regulators form my list i dont see how they can ever do close to 3A with a 0,56W max package disipation.Also thank you very much for informing me about load dump protection. I was aware of this, but kinda forgot again while doing this design. I may want to bring this product to market in a later stage, so it needs to be solid. I understand there are many different solutions for this. I'm leaning towards a RVS diode with a poly fuse in front of it, or maybe a crowbar circuit with either a poly fuse, or a blade fuse. Do you have any advise in this?Anyway, for no i went with the MP2359. Please find a screenshot of the circuit below:Does anyone have any recommendations to this circuit?I Choose a 15uH coil instead of a 4.7uH coil hoping to boost the efficiency. I'm planning to use a 12x12mm sized coil with a DC resitance of 50mOhm and a current rating of 3,2A. Is this a good choice?

author: Siwastaja
author_rank: Super Contributor
timestamp: March 16, 2020, 04:56:56 pm
content: Increasing L may or may not increase efficiency; increase it too far likely decreases efficiency, especially if you are space-constrained.If you keep the package size the same, larger L means more loops of thinner wire inside the inductor, hence larger DC resistance and lower current rating.Excessively small L means higher magnetic flux ripple and possibly higher core loss, though.I'd use the value of L suggested by the manufacturer, and only increase it if low output ripple voltage is of utmost importance (then, increase C as well, of course). Or, if you run at lower loads.If you have the physical size and the budget to use a larger value L, and found one that is suitable per current rating, don't pick that; pick the lower value (the original suggested L) from the same series, and you'll get much lower DC resistance and extra thermal budget!Note though that at such high f_sw, inductor AC losses (core losses + winding AC losses) are absolutely paramount. Sadly, they are not in the datasheets; some manufacturers provide web-based tools to calculate them. Often, you just pick a small ferrite-cored (shielded type!) inductor for prototyping and hope you are lucky...What comes to your wondering of chip losses, switching losses are difficult to calculate in such integrated-FET controller, but let's calculate conduction losses for a simple example case:Vin=12VVout=5VDuty = 5/12 = 42%MOSFET on-time = DutyDiode on-time = (1-Duty)MOSFET I_rms during on-time = approx. I_L_avg = 1.2AMOSFET Rds(on) (from the datasheet) = 0.35 ohm typicalP_MOSFET_cond = Duty * I_rms_during_on^2 * 0.35 ohm = 0.42*(1.2A)^2*0.35ohm = 212 mWNow just assume conduction losses and switching losses are 2:1, so add 50%, and we are somewhere around 300mW, well below the absolute maximum dissipation 568mW (as it should be, because this is absolute maximum rating, and at Ta=25 degC.)From RthJ-A = 220 degC/W (note: 4-layer board), the temperature rise is around 66, say 70 degC. At T_amb = 60 degC, you are at Tj=130 degC, which is, IMHO, a barely acceptable margin. It's still 20 degC away from the typical thermal shutdown value. You could run into problems using it in a hot car on a sunny day in direct sunlight, at full 1.2A output all the time, though.But this is a bit hand-wavy, because we can't know for sure what the switching losses are. You also need to consider thelocalambient; you get roughly correct mental simulation by placing all the other components, like the diode and inductor, on the PCB, run power through them to heat them up the same amount they do in the actual circuit, and put it in the box like you finally do. This is the ambient the chip "sees". (Proper thermal simulation is much more complex, though).The schottky diode will dissipate approximately (1-Duty)*Vf*I_L = 58%*0.4V*1.2A = 0.28W, heating it up by just 0.28W*55degC/W = 15.4 degC over ambient. So note, while the diode dissipates as much as the chip, it has way better metal leads to heatsink it. Tj ends up well below 100degC, which means the reverse leakage likely isn't a problem, but let's calculate it (at Tj=100degC) just to be sure:P_leak = Duty * Vsupply* Ileak = 0.42*12V*6.5mA (from the curve) = 32.8mW. Insignificant here, but funnily enough, notthatfar from being a concern!