[ §1 · modulator ]

Buck choke, switching frequency and the gate supply

Frequency looks like the cheapest way to a smooth ramp. Its second price is the gate supply, and at 30 kHz a two watt module is already past its limit.

CommonGnd4 August 202610 min read

Three things set how well a buck follows its PWM: the inductance, the switching frequency and the output capacitance. They do not cost the same, and the one that looks cheapest has a second price that nobody puts in the budget.

The ripple does not change; the average under it does. When the valley reaches zero the choke empties every cycle, and from there the converter has a different transfer function from the one its controller was tuned on.

The arithmetic

inductor ripple
  dI = V_in · D(1-D) / (L · f_pwm)

output ripple
  dV = dI / (8 · C · f_pwm)
  so dV goes as 1 / (L · C · f^2)

Frequency enters the ripple squared and inductance only linearly, so doubling the frequency does more for smoothness than quadrupling the choke. That is the obvious conclusion and it is a trap.

The gate supply nobody budgets for

The buck's own switch has to be driven, and the average gate current is Q_g · f, which grows linearly with frequency:

P = Q_g · f_pwm · rail-to-rail swing

For the buck's own switch, a brick at Q_g = 2.8 µC:

20 kHz: 56 mA. 0.84 W on +15/0, 1.33 W on +15/-8.7, 1.68 W on ±15.
30 kHz: 84 mA. 1.26 W, 1.99 W, 2.52 W.
40 kHz: 112 mA. 1.68 W, 2.65 W, 3.36 W.
60 kHz: 168 mA. 2.52 W, 3.98 W, 5.04 W.

There is a second trap in the same place. The buck's switch sits on the full bus, so its gate supply has to isolate the full bus. A module with 400 V of working isolation on a 650 V bus is over by a factor of 1.6, and it is a blocker that is easy to miss because everything else in the build has margin and this one line does not. Check it separately, alongside the device voltage.

What actually breaks the ramp

Ripple is usually tolerable. The real problem is discontinuous conduction.

When the inductor current reaches zero inside a chopping cycle, the gain rises above the duty cycle and starts depending on the load, and at nearly no load the filter output floats up towards the bus at any duty at all. The relationship between duty and voltage does not merely distort. It stops existing.

The threshold is dI/2, the load current below which it happens:

CJ: 100 µH, 40 µF, 25 kHz. 40 A of ripple, 5.0 V of output ripple, discontinuous below 20 A. The rest of that machine, for scale: 2.39 m of arc on a 420 V bus, a 23 ms ramp and an 11 nF tank.
gsch.labs: 210 µH, 13 µF, 60 kHz. 7.9 A, 1.3 V, discontinuous below 4 A.

The cure is inductance. davekni's own QCW buck is built with a high L for exactly this reason. Frequency moves the threshold too, and you pay for it twice.

When you do bench it on a resistive load, use a low enough resistance: ten ohms rather than thirty. On a high resistance the current goes discontinuous and you see a nonlinearity the coil will never show you, or miss one it will.

Cores

Micrometals T184-2, the red one, or T300-26D, often as a stack of two. Loneoceans has a calculator for it.

Sizing one

Which current. Not the primary's. The choke sits on the bus side and carries (2/π)·I_primary — the average the bridge draws, not the peak the tank circulates. Gao runs 100 to 200 A here against a primary of many hundreds.

Which material, and what that changes about the question. A ferrite saturates: push far enough and the domains run out of room, extra current buys no extra flux, and the inductance falls off a cliff. Every "stay under B_sat" rule exists because of that cliff. A powder core is iron particles in a binder — millions of tiny air gaps spread through the volume, a distributed gap — and it has no cliff at all. Its permeability starts falling from the first amp and keeps falling smoothly. Nothing dramatic ever happens, which sounds safer and moves the question: not whether you are under a limit, but how much inductance you have left at the current you actually run.

So the catalogue inductance is not the working inductance. A ring sold as 800 µH quotes A_L near zero current. Wound to 108 turns it measures 793 µH on an LCR meter and delivers 341 under 102 A of bias — under half. Design the ripple at 793 and you get twice what you planned.

And the sag is not a second effect. L = A_L·N² and A_L is proportional to permeability, so if the permeability at your operating point is 43 per cent of its initial value, so is the inductance:

793 uH catalogue
  x 0.43 permeability left
= 341 uH working

Turns — drag

Working current — at zero the two curves are the same line

The catalogue value, the permeability left, and the working value are not three numbers. They are one. L = A_L·N² and A_L ∝ µ, so whatever the permeability loses, the inductance loses with it.

Field strength

427 Oe

Permeability left

43 %

Window used, 7 AWG

30 %

Copper, peak

254 W

catalogue, measured at zero current — leaves the top of the plot what you actually get at working current the two published bias points

What to watch. Drag the working current down to zero and the two curves become one line — the catalogue number is measured there, and nowhere else. Drag it back up and the gap opens. That gap is not an error in the catalogue; it is the permeability doing what the datasheet says it does.

The two red marks are the only measured points. Magnetics publishes 80 % of initial permeability at 215 Oe and 50 % at 380 Oe for this core. Everything between and beyond them here is a fit, so read the shape and treat the exact percentages as an estimate.

Why more turns stop helping. Inductance goes as the square of the turns, but the field goes as the turns, and the permeability falls with the field. Write the fall as µ ∝ H⁻ᵃ and the exponents subtract: L ∝ N^(2−a). While a is under 2 the square wins; once it passes 2 the fall wins and adding turns takes inductance away. That crossing is what the marked maximum is. Its existence is firm — it depends only on the slope passing through 2. Its position is soft, because it sits well past the two points the manufacturer published.

Magnetics 0078337A7, XFlux µ26. A_L 68 nH/T² ±8 %, A_e 678 mm², l_e 324 mm, window 4710 mm². Copper figured on 7 AWG at the turn length the datasheet gives for the winding factor reached. Core loss is not shown: at this ripple it is a few watts against hundreds in the copper.

That is an identity rather than a discovery, and it is the identity that kills the idea of two checks. The inductance you lost and the permeability you lost are one number written twice. Which means the check to run is the permeability curve — H = 0.4π·N·I/l_e, read the fraction off the datasheet's plot, multiply the catalogue inductance by it. On the core worked here, 102 A through 108 turns on a 32.4 cm path is 427 Oe, and a published curve giving 80 per cent at 215 Oe and 50 at 380 puts that at 43.

And the habit of quoting both numbers at once, rather than the catalogue one, is his too. Sizing a choke for somebody else on the same forum:

a pair of the 0078912A7 wound with 32 turns would give you 116 µH at zero bias and just above 40 µH at 150 A

116 -> 40 uH
  35 % of catalogue

Two thirds of it gone, and he says so in the same breath as the first figure. That is the whole argument of this section stated by somebody who sizes these for a living: a powder-core inductance is not a value, it is a value and a current.

And B = L·I/(N·A_e) with the catalogue inductance is wrong in the direction that reassures you. It wants the chord permeability, total flux against total current, while the datasheet publishes the incremental one, the slope at the bias point, which governs ripple and nothing else. Substituting one for the other is silent and reports less flux than you have. The honest answer on a powder core is a pair of bounds; if both clear you are clear, and if they straddle you have computed nothing. B_sat for these materials is usually an inherited family figure rather than a line in the datasheet in your hand, while the curve is measured.

And turns stop helping, at a point you can compute. L goes as N² and H goes as N, so adding turns pushes you further down the curve while it raises the count. Write the fade as µ ∝ H^-a and the exponents subtract:

L ~ N^(2-a)

a < 2   turns buy inductance
a > 2   turns cost it

On this core the slope passes 2 between 181 and 220 turns, which is where the peak sits:

turns    H      mu     L
 108   427 Oe  43%   340 uH
 181   715     18%  ~395  <- here
 300  1186      6%  ~365

Past that more turns make it worse, and the approach is poor value: 108 to 181 buys about 16 per cent more inductance for 68 per cent more wire, and the figures past the first row are rounded because they come off a curve read past its published points. The location is soft — those readings are well past the two points a datasheet publishes — but the existence is not, since the criterion is only that the slope passes 2. A powder core has an inductance ceiling set by the material and not by the window, and turns cannot break it. Only a different core can.

Copper is the bottleneck and the window is not. 108 turns of 7 AWG fills about 30 per cent of a 4710 mm² window, and 50 at the 181 that would maximise the inductance. What limits you is heat and wire, not room — the opposite of the intuition a transformer gives.

Then check it stays continuous, which is what the ripple arithmetic is actually for. At 341 µH, 22 kHz and a 650 V bus:

Vout    dI      DCM at   bus I   margin
 40 V   5.0 A    2.5 A   12.5 A   5.0x
325 V  21.7 A   10.8 A  101.9 A   9.4x

Continuous everywhere, with five times the margin at its worst, and 4.9 V of ripple at the top on 25 µF — so no extra capacitance is needed to carry the doubler, which a smaller winding would have demanded.

The freewheel diode, and the trap in the usual shortcut

An asynchronous buck needs one, and the common shortcut is a half-bridge module: the top device switches and the bottom one is held off so its body diode does the freewheeling.

Shorting the bottom gate to its emitter is not enough. You need either a driver with an active Miller clamp or a very short gate loop, or dv/dt will crack the bottom device open. A TVS on the gate is secondary insurance and it has to be chosen carefully: standard parts often conduct above the gate's limit and protect nothing at all.

And the bank behind it has to keep up

gsch.labs had a weak 300 W input transformer, the bank sagged, and the current stopped following the ramp. What matters is not the average power but the ability to deliver it over the length of the ramp. There is a piece on the reservoir for that.

Where to look when it misbehaves

Watch the freewheel diode, the choke and the bank, not the resistors. The choke takes continuous heating in both copper and core, and it is the part that is quietly running warm while everything measurable looks fine.

The QCW diagram draws the buck as what is inside it, a switch, a choke, a freewheel diode and a capacitor, because the choke is the reason a QCW has a ramp at all.

buckinductorQCW