How to properly select inductor size for buck converter? - Page 1
Forum section: electronics / general

author: Jim from Chicago
author_rank: Regular Contributor
timestamp: February 08, 2023, 06:23:59 am
content: I am currently working on a buck converter which I know will supply 600mA to its load. The buck converter is rated to be able to supply up to 4A. If I select my ripple current to be 30% of load current, then ripple current will be roughly 200mA. If I then plug that value into the standard equation for inductor size (see attached image from TI article), I end up needing an inductor of about 29uH, which seems like a lot (and seems like it will result in terrible transient response).However in page 6 of this TI article:https://www.ti.com/seclit/ml/slup384/slup384.pdf, they say that if the buck IC has its FETs internal to the IC (true in my case), then you should just calculate your inductor size using the max rated current capability of the IC (see attached image for the exact statement). So in my case, that would be 30% of 4A which is 1.2A. Using 1.2A as my ripple current, I would need an inductor of about 4.8uH.There's a big difference between 4.8uH and 29uH, so which is correct/better to use? Is the TI article correct in its claim? I don't understand why it says that you should just use the nominal current rating of the IC. If I know the exact load current my regulator will supply, shouldn't I be using that value, especially if that value is well below the max current rating of the regulator?

author: Siwastaja
author_rank: Super Contributor
timestamp: February 08, 2023, 06:48:05 am
content: It is up to you, finally an arbitrary choice. Larger dI_L means also larger ripple current over the output capacitor, and therefore, larger ripple voltage (because output capacitor has >0 ESR and less than infinite capacitance). With less inductance, the converter runs in DCM even at higher loads, but even that might not be a problem, it's up to you to decide.30-40% offull load current(not the IC maximum; not the IC current limit) is some kind of sensible default. You can't have too much inductance per se. If the inductor is physically too large, the options to make it smaller is to increase dI_L or increase switching frequency.It makes absolutely no sense to use the maximum current capability of the IC. The "usual rule of thumb default" is a decent compromise between inductor size, output capacitor size/stress, and transition point from DCM to CCM, maybe by assuming that the load "usually" is more than 40-50% of maximum. Even if the IC is capable of supplying 99999999999A, it does not affect this decision.It seems someone has just mixed up inductance and saturation current. What you really should do is to look at the pulse-by-pulse current limitation feature of the IC, specifically the worst-case maximum value (which can be significantly more than the rated max I_out), and make sure the inductor does not saturate there, at least fully. If the inductor saturates, it loses inductance, and as a result, the current rise rate will increase, possibly so much that the current limit feature (which turns the FETs off) is too slow to react in time. Unfortunately, the reaction delay is almost never specified, but it tends to be fast enough that losing maybe 50% of inductance won't be a problem. Therefore, with the usual 70% saturation criteria as given in inductor datasheets, there is no need to leave extra margin. Just don't make the mistake to choose an inductor which saturates at 1A "because my load never is over 1A" and then the IC is capable of supplying 2A continuous and worst-case current limit value is at 4A.

author: Faringdon
author_rank: Super Contributor
timestamp: February 08, 2023, 09:42:03 am
content: Please give Vin, vout, Pout and Fsw.But yes, theres no right answer......just make sure you dont exceed conduction losses and core losses.If its HF you may need the core loss calculator of the vendor.Wuerth and vishay and coilcraft do these.When you tell the vout, vin, pout etc, i can help you...with things like slope compensation and if its needed or not.....if you need added slope, then pick a "shallow ramp " inductor.......if your power is low then you coudl go DCM even....remember the added slope is approx half the downslope......if you want i can send you a paper on this.Do you use current mode control? Or constant off time control?...or voltage mode control?These all have bearings on it.In this post there is an LTspice buck sim for you to download and you can play with it if you wanthttps://www.eevblog.com/forum/renewable-energy/massive-reduction-in-smps-undershoot-and-overshoot/

author: Wolfram
author_rank: Frequent Contributor
timestamp: February 08, 2023, 01:04:09 pm
content: Quote from: Siwastaja on February 08, 2023, 06:48:05 amFor peak current mode control chips, the inductance cannot be arbitrarily large. A certain ripple fraction is required for the correct operation of the control scheme. Also a lot of integrated buck converters have integrated slope compensation where the ramp is scaled for the recommended value of inductance.This is why it can be very disadvantageous to use a high current integrated buck for a low current application. The ripple current and and component size will be much worse than with a part with a more suitable current rating.

author: Jim from Chicago
author_rank: Regular Contributor
timestamp: February 08, 2023, 06:09:40 pm
content: Thanks for replies. More info is Vin = 12V, Vout = 5V, Fsw = 500kHz typical ±50kHz. The load draws about 600mA and never varies from that.

author: Siwastaja
author_rank: Super Contributor
timestamp: February 08, 2023, 07:01:44 pm
content: Then pick an IC with maximum current near 0.8 - 1A and thus, current limit at around maybe 1.5A. You can get the inductor size down significantly as you don't need to support ridiculously high saturation current.