Why Does No-one Place Inductance on Buck Converter Input? - Page 1
Forum section: electronics / beginners

author: Froff
author_rank: Newbie
timestamp: March 26, 2025, 09:58:07 am
content: Hi all! I'm working on a project where I have access to a 24V supply, but my board needs 3v3, so naturally I use a buck converter to lower the voltage. Buck converters are supposed to give more current out than they take in on average (power conservation). However, while the converter is in the ON state (charging the output inductance/capacitor) the instantaneous input current should be greater than the output current, as the current supplied at the input will be split between the load and the output cap. Can this be a problem if my 24V supply is supposed to supply other systems too? The 24V supply is only rated to supply pulses of 6W (so 250mA).To test this, I set up a circuit simulation in Falstad (see Basic circuit below), which indeed shows that the input current spikes to 450mA, while the output is at 250mA.So I thought I could add an inductance between the supply and the bypass capacitor. This nicely smooths the input current, so the maximum instantaneous input current is 73mA (see Circuit with inductance below). This seems like a great fix, but I am confused as to why I can find no references of this being done online. Is there something glaringly wrong with doing this?Basic circuit:https://tinyurl.com/22egz8v6Circuit with inductance:https://tinyurl.com/23ej2nv3

author: MrAl
author_rank: Super Contributor
timestamp: March 26, 2025, 10:08:08 am
content: Quote from: Froff on March 26, 2025, 09:58:07 amHi,First off, use a REAL simulator then make your judgements. Try the free LT Spice simulator.When the switch is on, the output inductor is already in series with the output cap. If the inductor is of the right value, the peak inductor current should be minimal. You can allow it to go higher or lower depending on the choice of inductor, and the inductor must be rated for a high enough saturation current level.Of course there are limits. If you are allowed 1 amp input current you should not be operating near that value so you may need a higher value inductor.We could look at some cases if you like.

author: golden_labels
author_rank: Super Contributor
timestamp: March 26, 2025, 12:00:44 pm
content: For most cases the right answer is: “because it works.” If it doesn’t require a bulky, expensive component, why use it?In more general sense, your Falstad simulation gives a good hint. You even has that visible in one of the scopes, in plain sight. Look at the current through the input capacitor. The current spikes aren’t removed; you just moved them to the capacitor.

author: Siwastaja
author_rank: Super Contributor
timestamp: March 26, 2025, 12:12:07 pm
content: Oh, such inductor is sometimes used. It's called differential mode filtering, and it does exactly what you see in simulation, reduce the current spikes seen by whatever is powering the buck converter. I have done it.It's just that if you want decent efficiency out of your converter, it needs really low impedance at the input of the MOSFET stage, i.e. the input capacitor. Meaning very low-ESR, low-ESL type needs to be used anyway. And once you have enough capacitance and low enough ESR and ESL to satisfy the rest of the circuit, you will often see that the ripple seen by the input supply is small enough already not to matter.But sometimes it does matter and then such extra LC stage for filtration is sensible thing to do.

author: coppercone2
author_rank: Super Contributor
timestamp: March 26, 2025, 12:15:55 pm
content: make a inductor kit and try it

author: PGPG
author_rank: Frequent Contributor
timestamp: March 26, 2025, 04:00:43 pm
content: Quote from: Froff on March 26, 2025, 09:58:07 amSo the pulse will be more taken from input capacitor than from this 24V supply.Quote from: Froff on March 26, 2025, 09:58:07 amNothing wrong in it.I think that any appnote circuits only show how supply works. In TI appnotes you can find equations for calculating Vpp at input capacitor. What you than will do with this is up to you. I don't want my circuit to emit disturbances through supply lines so I always have pi filter at input.When my supply is used to supply for example RS485 transceiver and I don't want to have disturbances from DCDC being emitted through RS485 lines (the one being at the moment connected by driver with VCC follows its disturbance) I also add pi filter at DCDC output.

author: Terry Bites
author_rank: Super Contributor
timestamp: March 26, 2025, 04:23:47 pm
content: Why do you say that?Its not uncomon to see an LC filter on the input.There also are current fed switcher circuits that rely on an input indcutor to operate

author: Froff
author_rank: Newbie
timestamp: March 26, 2025, 04:51:53 pm
content: Thanks for all the answers!Quote from: MrAl on March 26, 2025, 10:08:08 amWhat is the preferred way of sharing designs on this forum? Posting Spice netlists?Quote from: MrAl on March 26, 2025, 10:08:08 amI can't get this to make sense. When the switch is on and the load draws 250mA (for simplicity assume a resistive load), then the inductor current must be at least 250mA (Kirchhoff's current law), unless the output cap is discharging exactly whenever the switch is on, which is not my understanding of how buck converters work. And since the diode doesn't draw current when the cathode is at 24V, all that current must come from the 24V supply.In reality I suppose parasitic resistance and inductance between the 24V supply and the bypass capacitor will make the converter source some current from the capacitor and not the supply, but it seems like an unreliable solution.Quote from: Siwastaja on March 26, 2025, 12:12:07 pmI think I will go for a small LC filter just in case. From the (perhaps not accurate enough?) Falstad simulation 1uH should be plenty, even with 45mΩ ESR for the input cap. And a 1uH/1A inductance is just about 5 cents and in a 1207 package, so it's not exactly costly.The switching freq. for the XL1509 I'm using is only 150kHz, so I doubt the ESL will make much of a difference. I'm not running on battery, so efficiency isn't the highest of priorities.Final circuit (sorry, still Falstad):https://tinyurl.com/223j5d8d

author: Froff
author_rank: Newbie
timestamp: March 26, 2025, 04:59:01 pm
content: Quote from: Terry Bites on March 26, 2025, 04:23:47 pmBefore starting this thread, I didn't think to search for LC filters for some reason (just "inductor on input" and similar). But searching for "buck converter LC filter input" yields a lot more helpful info.

author: madires
author_rank: Super Contributor
timestamp: March 26, 2025, 06:29:56 pm
content: A very simplified explanation: an inductor opposes change of current which conflicts with the function of a buck converter

author: Vovk_Z
author_rank: Super Contributor
timestamp: March 26, 2025, 06:49:28 pm
content: Quote from: madires on March 26, 2025, 06:29:56 pmYes, it opposes to it. But if you need to lower EMI noise from your Buck converter then you have to use a small inductor at input (of cause together with good LOW-ESR capacitor). It is a typical noose-cancelling circuit. It is sometimes even showed at some datasheets from manufacturers. This additional inductor should have inductance value about an order or two smaller then main Buck inductor (but it depends and varies).

author: Whales
author_rank: Super Contributor
timestamp: March 26, 2025, 07:18:10 pm
content: Make sure to have L+C, not just L, at the input of your buck converter. Otherwise it will turn into a boost converter under some situations.You might also findCuk convertersinteresting.

author: John B
author_rank: Frequent Contributor
timestamp: March 26, 2025, 07:51:04 pm
content: Without LC filtering before the buck converter, as you see you get a discontinuous current. Not only does this create noise in the rest of your power rails, it can also interact with your input DC supply.I have had situations where a buck converter caused the AC/DC supply to oscillate. This is corrected more by C than L, but the L is useful for dealing with the high frequency content.

author: MrAl
author_rank: Super Contributor
timestamp: March 27, 2025, 10:38:10 am
content: Quote from: Froff on March 26, 2025, 04:51:53 pmHi,A lot of people use LT Spice and share using the .asc files. That means each person can quickly replicate what the circuit is doing.The situation you seem to be describing is when the inductor is at peak current. If the output current was really 250ma then the peak current would have to be higher such as 275ma, 300ma, 350ma, 400ma, 500ma, whatever. That value depends on how big the inductor value is. If it was (for example) 50uH it might be 400ma, and if it was 100uH it might be 300ma, etc.Also what is interesting is the input capacitor is supplying a lot of that current for a time. That's why in a lot of applications a low ESR capacitor is advised on the input.The math is not that hard to think about in simple terms. If the input cap ESR is 1 Ohm and the input voltage source impedance is 1 Ohm and output current is 300ma, then the current from the input supply would be around 150ma and the current from the input cap would be around 150ma. If the ESR is only 0.1 Ohms however, then the current from the input supply would be roughly 30ma and from the cap 270ma. If the ESR was actually higher though, like 10 Ohms, then the input supply would be providing roughly 270ma and the input cap just 30ma. So you see how important the ESR of the input cap can be. That's why a lot of applications require a low ESR input cap. It also has to be of a value that can support the average current over one cycle, but startup can be harder on it because the input current might be 5 times higher or more during startup if there is no slow start mechanism. That, and it's effects, would have to be investigated as well during simulation and of course during actual real life bench testing.I think it would be rare to find an input inductor but much more common to find a very low ESR input capacitor. An input inductor has the unwanted effect of lowering efficiency as well as the possibility of lowering the apparent input voltage level itself. It might be an option if you really need it I guess, but a low ESR capacitor should help a lot without it.A really, really, simple view on this is that in a lot of these circuits the capacitor is viewed as a voltage source with zero series impedance. Under this ideal condition, the voltage across that input cap would never decrease during the inductor peak current and thus it would not draw any current from the input voltage supply (until the second half cycle). This idea is used in some theory to get a quick idea what to expect, but of course in real life the presumption is that a low ESR capacitor would be used.

author: mtwieg
author_rank: Frequent Contributor
timestamp: March 27, 2025, 12:36:38 pm
content: Adding an LC filter on a buck converter (or any topology) input is fine, but should be done carefully. The input filter affects the transfer function of the overall circuit, and in some cases can cause an otherwise stable DCDC converter to become unstable.This white paper explains it in detail:https://www.venableinstruments.com/hubfs/White%20Papers/Documents%20/Source-Load%20Interactions%20In%20Multi-Unit%20Power%20Systems.pdfIn practice I often use LC input filters, but always make sure they are damped with an impedance which is guaranteed to meet themiddlebrook criteria.Even with an input filter, it's still important for the capacitor to have very low ESR and ESL, especially for buck and buckboost converters.

author: Froff
author_rank: Newbie
timestamp: March 28, 2025, 06:26:33 pm
content: Quote from: mtwieg on March 27, 2025, 12:36:38 pmI must admit this is getting quite a bit more complicated than I thought it would be when starting the thread. (Is adding another huge cap really necessary?) But I foundthis answerfrom Tim Williams to a similar problem on SE, and it seems the important part is making sure that Zin<< -Rin(e.g. one fifth or less), where -Rinis the small-signal negative resistance of the converter. According to my calculations, Rinis given byRin= -Vin^2 / (Vout* Iout)which gives me Rin= -700Ω. The impedance of my LC filter at 150kHz is Zin= 0.916Ω. This should be plenty alright, assuming that I understood his answer correctly.

author: mtwieg
author_rank: Frequent Contributor
timestamp: March 28, 2025, 07:19:31 pm
content: Quote from: Froff on March 28, 2025, 06:26:33 pmYeah that's basically the Middlebrook criteria.QuoteSounds right.QuoteThe Middlebrook criteria should be correct at all frequencies. The worst case is at the resonant frequency of your input LC filter, where the source impedance becomes infinite (if the LC filter is ideal). That's why some damping on the LC filter is necessary, it prevents the source impedance from peaking excessively.

author: MrAl
author_rank: Super Contributor
timestamp: March 30, 2025, 01:34:03 am
content: Quote from: Froff on March 28, 2025, 06:26:33 pmHello again,Maybe you are just letting this get complicated. This simplification might help a little.First, your ideal conversion factor is 3.3/24 which is k1=0.1375 and that means that you cannot get better than Iin=k1*Iout as the average input current. Because your factor is so low though, you can get significantly lower than Iout. With an average Iout of 250ma that comes to about 34.4ma in the ideal case.All we have to do then is apply a little analysis and see what we can come up with.An approximation to this average input current is:Iin=(Iout*sqrt(w^2*C^2*R2^2+1))/sqrt((w*C*R2+w*C*R1)^2+(1-w^2*C*L)^2)whereIin is the average input current,Iout is the average output current,C is the input filter capacitance,L is the input filter inductance,R1 is the inductor ESR,R2 is the capacitor ESR,w is the angular frequency equal to 2*pi*f with f=switch frequency in Hertz.Note we immediately approximate the frequency as the most significant harmonic of the inductor current frequency.This actual makes it kind of easy to approximate the filter components. In fact, since L and C kind of enter into this in an almost completely symmetric way, we can start the approximation by letting L=C and we end up with:Iin=(Iout*sqrt(w^2*C^2*R2^2+1))/sqrt((w*C*R2+w*C*R1)^2+(1-w^2*C^2)^2)which as you can see contains no L just C.The idea after that is since L*C is symmetric, once we calculate C we also have an idea what L is, because it can be initially 'said' to be the same as C.To actually pick the value for L though we just have to square C and then divide be the actual value of C we might want to use. This leads to:L=C^2/Cawhere C is the value we calculate from the formula, andCa is the actual value of the capacitor we might want to use or try first,and L is the actual value of the inductor.You should also probably make sure you choose the buck converter L to keep the converter in the continuous mode.Of course a simulation will show your final results, which should be very good.One of the assumptions in the above is that the input surge is not much greater than the average input current once the buck is running normally. If this is not the case, you probably have to add a soft start circuit or else we'd have to make some modifications to this such as assuming the output current is 5 times what it really is, and then choosing the L and C of the input filter accordingly.We can go through an example if you like.

author: Smokey
author_rank: Super Contributor
timestamp: March 30, 2025, 02:05:32 am
content: https://www.ti.com/tool/WEBENCH-CIRCUIT-DESIGNERCheck out TI's design tool. It's online and free. Gives you full schematics for most of their power supply chips, including optional input filters based on the design parameters. It's often a really good starting point.

author: MrAl
author_rank: Super Contributor
timestamp: March 31, 2025, 08:58:18 am
content: Hello again,I should have also pointed out that the success also depends on the quality of the components used. A good quality inductor that can handle the peak current without a problem would help to reduce peaks that go too far above the theoretical requirements. If the peak current does up higher than usual, it's going to cause a higher peak current at the input.

author: Froff
author_rank: Newbie
timestamp: March 31, 2025, 11:12:10 am
content: I ended up writing a little python script to calculate the output impedance of my filter for a range of frequencies, and with the ESR in the inductor and capacitor, I get a spike of around 300mΩ @ the resonance frequency f=27.7 kHz. I also see now why no dampening can be such a huge problem. As you say, with ideal components (or very very very low ESRs), the output impedance for my filter is huge! I think part of my confusion was mixing up the input and output impedance of my filter.Thanks again for all the help!Code:import matplotlib as mplimport matplotlib.pyplot as pltimport numpy as npfrom math import piL = 1e-6C = 33e-6RL = 60e-3RC = 45e-3def Z(f):w = f * 2 * piZL = w*L*1j + RLZC = 1/(w*C*1j) + RCreturn ZL * ZC / (ZL + ZC)Z = np.vectorize(Z)f = np.logspace(2, 6, 401)z = Z(f)fig, ax1 = plt.subplots()ax1.plot(f, np.abs(z), label='abs', color='black')ax2 = ax1.twinx()ax2.plot(f, np.angle(z) / pi, label='phase', color='orange')ax1.set_xscale('log')ax1.set_xlabel('f (Hz)')ax2.yaxis.set_major_formatter(mpl.ticker.FormatStrFormatter(r'%g $\pi$'))ax2.yaxis.set_major_locator(mpl.ticker.MultipleLocator(base=0.25))ax2.set_ylim((-1, 1))fig.legend(loc='upper right', bbox_to_anchor=(1, 1), bbox_transform=ax1.transAxes)plt.show()

author: Froff
author_rank: Newbie
timestamp: March 31, 2025, 12:11:36 pm
content: Quote from: MrAl on March 31, 2025, 08:58:18 amThank you for all your help in this thread!I actually calculated my first current draw number as the largest possible peak current (all components added up) with an added 100% margin. Just to be on the safe side with current spikes. That's where I got my 250mA from. In reality, the average current will be way, way, way lower, closer to 20-30mA.

author: Froff
author_rank: Newbie
timestamp: March 31, 2025, 12:12:39 pm
content: Quote from: Smokey on March 30, 2025, 02:05:32 amThis looks awesome, will definitely be using it next time!

author: MrAl
author_rank: Super Contributor
timestamp: March 31, 2025, 07:06:32 pm
content: Quote from: Froff on March 31, 2025, 11:12:10 amHello again,This turned out pretty interesting. I also did a Fourier analysis just to see what would happen.It looks like you chose values 1uH and 33uF, how did you decide on those values?

author: Froff
author_rank: Newbie
timestamp: April 01, 2025, 08:04:47 am
content: Quote from: MrAl on March 31, 2025, 07:06:32 pmMostly chose them based on practical concerns (price, pcb area). The datasheet of XL1509 states CIN should be a 470uF/35V cap, but those are quite bulky and I am space limited! I think I can get away with a smaller cap sinceI'm outputting maximum peaks of 250mA (and usually a lot less), while the XL1509 is max 3AThe input 24V does not have the craziest ripples (1V pp @ 10kHz)

author: MrAl
author_rank: Super Contributor
timestamp: April 01, 2025, 12:03:19 pm
content: Quote from: Froff on April 01, 2025, 08:04:47 amWould you happen to know if the converter is operating in the continuous mode?