Inductor saturation current for buck converter

[Silenus]

Hey Guys. I'm trying to design a buck converter that takes 12V in and outputs 5V. I am basing the circuit off http://www.farnell.com/datasheets/85220.pdf. For the inductor, I am aware that it has to be able to handle the switching currents so that core saturation is not reached, however I am unsure of how to determine the peak switching current. The data sheet says "current limit" is around 3.4A so should my inductor have at least a saturation current of 3.4 A?

[Simon]

Depends on the topology, for buck it's generally "Iout x 2"

[Zero999]

V_OUT/V_IN gives you the duty cycle, except it'll be higher than that due to the losses.

Here's some more information:
http://simonthenerd.com/files/smps/SMPSBuckDesign_031809.pdf

[Golana]

The link you provided doesn't work so I can't give an exact answer but...  If for example your load current is at 3.4A then this will be the average current through the inductor but the inductor current will go above and drop below 3.4A based on the duty cycle and switching frequency of the switch.  If you are familiar with the equation V = L (di/dit) where di/dt is the rate of change of current through the inductor then you can figure out the max current of the inductor.  The duty cycle of your device is going to be 5V/12V=0.417.  Now just figure out the period of your switching frequency and multiply it times  0.417 to get the on time of your switch.  Now if you re-arrange the inductor equation you can figure out how much above the load current the inductor current will rise.  So, (12V-5V)/L = rate of change of current.  So, max inductor current, I_Lmax = (12V-5V)/L*on time/2+I_load.  So, just find an inductor saturation current that exceeds I_Lmax and of course you probably want to oversize it a bit but I'll leave that to you to decide how much.

[Mad ID]

Depends on the topology, for buck it's generally "Iout x 2"

This is incorrect. The peak inductor current is equal to the output current plus half the ripple current. If the output current is say 2A, and the ripple is e.q. 0.6A (30% of the max output current) than the peak inductor current is 2.3A, a far cry from your 4A estimate.

For anyone trying to design a SMPS converter I highly recommend that you first go through the current waveforms during one switch cycle. Than it's easier to understand what are the component requirements.

Rule of thumb is to pick an inductor value which will result in the ripple current being 0.2-0.4 of the maximum output current.

[Simon]

I have read datasheets for SMPS controllers with formulas asking that the current capability of the inductor x2 the output, that is why I said "generally", you should refer to the controllers datasheets of whatever formulas are driving your design.

[johansen]

without cycle by cycle current limiting or current mode control, the inductor can saturate at turn on, and that is why you need more headroom than simply output current plus half the ripple current plus 10%.. but you can do that if you want.. just make sure you have no unknown unknowns..

[Zero999]

I was wrong before so edited my post.

The link posted by the original poster didn't work so I've fixed it:
http://www.farnell.com/datasheets/85220.pdf

[Simon]

without cycle by cycle current limiting or current mode control, the inductor can saturate at turn on, and that is why you need more headroom than simply output current plus half the ripple current plus 10%.. but you can do that if you want.. just make sure you have no unknown unknowns..

Often switch mode power supply controllers limit the peak current as a method of overcurrent protection. On a chip which I forget the number of but is now quite old but still very popular it uses but 2x output current factor. Obviously controller designs vary so there is no definitive answer.

[c4757p]

If the inductor saturates it becomes little more than a wire. di/dt will be massive and the overcurrent protection may not be fast enough to prevent damage. Not very many DC-DC regs handle saturation gracefully.

By the way, Silenus asked for the peak switch current, not a rule of thumb to estimate the same in an assumed properly designed converter. This is somewhat varying. One way you can think of a buck converter is as an LC low-pass filter, which averages a square wave of varying duty cycle.

- Approximating the output voltage as fixed, the inductor will act as an integrator, where di/dt is proportional to the voltage across the inductor and inversely proportional to the inductance:

V = L di/dt

- Working from the rising slope, V across the inductor will be equal to Vin-Vout; from this we get di/dt.

V = Vin - Vout

di/dt = (Vin - Vout) / L

- We also have the duty cycle (if the square wave's top and bottom are Vin and 0, and it averages to Vout, the duty cycle must be Vout/Vin), and the switching frequency (by design).

dc = Vout/Vin

- The positive width of the waveform is equal to the duty cycle divided by the switching frequency.

Wpos = dc/fs = Vout/(Vin * fs)

- The height of the waveform is equal to the positive slope (di/dt) multiplied by the positive width.

Delta I = di/dt Wpos = (Vin - Vout)/L * Vout/(Vin * fs)

This is the ripple current. The peak will be roughly half the ripple current plus the average current, as the inductor's current waveform is equal to the ripple with an offset of the average.

Example:
Vin = 12V
Vout = 5V
L = 33uH
fs = 100kHz

Delta I = (12V - 5V)/33uH * 5V/(12V * 100kHz) = 883mA

This assumes, of course, continuous mode (the inductor's current does not reach zero).

You'll note no dependence on the average output current! This just comes from the rule of thumb - assuming L will be chosen depending on the output current.

Now, given an output current of 2A, and our approximation of the peak as the average output plus half the ripple:

Peak I = Mean I + (Delta I)/2

Peak I = 2A + (883mA)/2 = 2.44A

This is all obvious to someone who actually understands how a buck converter works - and rules of thumb like that are rarely of much use to people who don't first understand the actual operation.

This assumes ideal everything; adding in inductor loss, FET resistance and diode V_F is an exercise left to the reader.

Where the rules of thumb do come in is in the less ideal cases, like (as johansen mentioned) startup. This is quite variable, and I'd highly recommend testing to find this - overspecify an inductor at first with significant excess capacity, and then measure the peak startup current.

Soft start is nice.

[T3sl4co1l]

Ripple varies by control method.  There are generally two preferred control schemes in use: peak current mode control (usually augmented by slope compensation), and average mode control (or something roughly similar).

Peak current mode control (for any topology, whether buck, flyback or other) is distinguished by its relation to the Logistic Map, which is an iterated chaotic system.  In particular, it becomes ill-behaved as current goes into continuous conduction mode (CCM).  Therefore, it is preferred for DCM to BCM (discontinuous to boundary) circuits, which are, in turn, preferred for low power levels, where the relatively high ripple fraction is tolerable, in exchange for the simplified circuit design.  High ripple is basically restricted to ferrite based inductors, which are relatively expensive; though composite molded inductors are looking pretty good these days.

Average current mode control is best for low ripple cases, which are, in turn, preferred for higher power levels, where an emphasis on smaller or cheaper capacitors, and cheaper inductor materials (though not necessarily smaller inductors), motivates design.  These are distinguished by relatively low ripple (often Ipp < 20% Idc), and accordingly, relatively slow response (at 20% ripple, it takes over 5 cycles to change output current from max to min or vice versa).

At low ripple, inductor design admits very cheap powdered iron materials, saving cost, at some expense to losses (#26 and #52 powdered irons are the cheapest materials, but make good resistors at most switching frequencies..), and usually size as well (it takes a lot of core, and a lot more turns, to keep ripple and losses low enough to survive).

Proprietary controller designs, which vary frequency as well as PWM, are commonly stable in the BCM (boundary conduction mode: 100% ripple) to 30% ripple range.  Many powder-composite and ferrite cored inductors are available for this sweet spot, which seems to be where the market is centered these days.

Ed: also, don't forget this lovely tool: http://schmidt-walter.eit.h-da.de/smps_e/smps_e.html
I do suggest playing with the suggested values of filter inductor and (to a lesser extent) turns ratio (if applicable).  Default calculation gives 20 or 30% ripple, which is good for some types of controllers, but deadly for others (namely, the peak current mode family).  As always, use your noggin -- it's only a tool, and blinding following its default setting is like using a hammer to pound eggs...
Tim

[Simon]

If the inductor saturates it becomes little more than a wire. di/dt will be massive and the overcurrent protection may not be fast enough to prevent damage. Not very many DC-DC regs handle saturation gracefully.

By the way, Silenus asked for the peak switch current, not a rule of thumb to estimate the same in an assumed properly designed converter. This is somewhat varying. One way you can think of a buck converter is as an LC low-pass filter, which averages a square wave of varying duty cycle.

- Approximating the output voltage as fixed, the inductor will act as an integrator, where di/dt is proportional to the voltage across the inductor and inversely proportional to the inductance:

V = L di/dt

- Working from the rising slope, V across the inductor will be equal to Vin-Vout; from this we get di/dt.

V = Vin - Vout

di/dt = (Vin - Vout) / L

- We also have the duty cycle (if the square wave's top and bottom are Vin and 0, and it averages to Vout, the duty cycle must be Vout/Vin), and the switching frequency (by design).

dc = Vout/Vin

- The positive width of the waveform is equal to the duty cycle divided by the switching frequency.

Wpos = dc/fs = Vout/(Vin * fs)

- The height of the waveform is equal to the positive slope (di/dt) multiplied by the positive width.

Delta I = di/dt Wpos = (Vin - Vout)/L * Vout/(Vin * fs)

This is the ripple current. The peak will be roughly half the ripple current plus the average current, as the inductor's current waveform is equal to the ripple with an offset of the average.

Example:
Vin = 12V
Vout = 5V
L = 33uH
fs = 100kHz

Delta I = (12V - 5V)/33uH * 5V/(12V * 100kHz) = 883mA

This assumes, of course, continuous mode (the inductor's current does not reach zero).

You'll note no dependence on the average output current! This just comes from the rule of thumb - assuming L will be chosen depending on the output current.

Now, given an output current of 2A, and our approximation of the peak as the average output plus half the ripple:

Peak I = Mean I + (Delta I)/2

Peak I = 2A + (883mA)/2 = 2.44A

This is all obvious to someone who actually understands how a buck converter works - and rules of thumb like that are rarely of much use to people who don't first understand the actual operation.

This assumes ideal everything; adding in inductor loss, FET resistance and diode V_F is an exercise left to the reader.

Where the rules of thumb do come in is in the less ideal cases, like (as johansen mentioned) startup. This is quite variable, and I'd highly recommend testing to find this - overspecify an inductor at first with significant excess capacity, and then measure the peak startup current.

Soft start is nice.

You mean you have explained in one post what many application notes that only tackle one area of design fail to explain.  I've given up reading app notes as they often promise much but then just talk about one little aspect and don't say anything definitive. Although I'll probably understand more as i learn the math through the maths module I'm on that is mandatory to commencing a HNC

[Silenus]

Thank you everyone for the replies! It's been very helpful in understanding this more. Thanks c4757p especially, the background theory makes a lot more sense now 

[planet12]

If you're actually using the LM22676 as linked in the data sheet, page 9, "External Components" - "Inductor" specifically covers the saturation current requirements.