Chapter 12: Buck, Boost, and Buck–Boost Converters — Power Electronics

Where this sits

Part 3 · DC–DC Converters
Chapter 12 of 30

You should already know

Duty ratio, ripple current and volt-second balance from Chapter 11, and the switching behaviour of a MOSFET or IGBT from Chapter 5.

By the end you can

Derive any non-isolated converter's gain in four lines, size \(L\) and \(C\) from ripple specifications, find the CCM/DCM boundary, and choose between the three topologies for a real supply.

Time

≈ 60 min reading · ≈ 55 min problems

i What you'll learn

The buck converter — a chopper with an output filter, and why that filter changes the character of the circuit completely.
The boost converter, and why its input current is continuous while the buck's is not.
The buck–boost converter: \( V_o = -\dfrac{D}{1-D}V_s \), which can step up or down but inverts the polarity doing it.
One method — volt-second balance plus charge balance — that produces all three results and every one to come.
How to size \( L \) from a current-ripple spec and \( C \) from a voltage-ripple spec, and why the capacitor's ESR usually dominates the answer.
The CCM/DCM boundary, and what changes in each topology when you cross it.
Which topology to reach for, judged on switch stress, filter difficulty and the source the converter must not upset.

Section 12-1

One Method for Every Converter

Chapter 11 derived the step-up gain \( V_o = V_s/(1-D) \) using volt-second balance, and promised the same tool would handle everything else in Part 3. This section makes that promise concrete, because it is genuinely the whole of the analysis.

Every converter in this chapter is built from the same five parts — a source, a controlled switch, a diode, an inductor and a capacitor — arranged differently. And every one of them yields to the same two statements.

🔑

The two balance laws

\[ \underbrace{\int_0^T v_L\,dt = 0}_{\text{volt-second balance}} \qquad\qquad \underbrace{\int_0^T i_C\,dt = 0}_{\text{charge balance}} \]

In periodic steady state an inductor's current and a capacitor's voltage must return to where they started. Everything else in this chapter is arithmetic.

◇Intuition · the reservoir that must not drift

Picture a reservoir that is filled each morning and drained each evening. If a little more went in than came out, the level would rise a little every day — and after a year the dam overflows.

An inductor's current is a reservoir of exactly this kind, and its "filling rate" is the voltage across it:

Positive voltage across \( L \) makes the current rise.
Negative voltage makes it fall.
Steady state means the level is the same each morning — so the volt-seconds up must exactly cancel the volt-seconds down.

A capacitor's charge works the same way with current in place of voltage: whatever charge flows in during part of the cycle must flow back out during the rest, or the voltage would climb without limit.

The power of this. You never have to solve a differential equation. You only have to look at the circuit in two states — switch on, switch off — write down the inductor voltage in each, and set the weighted sum to zero. Four lines, every time.

The recipe, which we will follow three times in this chapter and repeatedly in the next two:

Redraw the circuit with the switch on. The diode will be reverse-biased; remove it. Write down \( v_L \).
Redraw it with the switch off. The diode now conducts; replace it with a short. Write down \( v_L \) again.
Set \( v_{L,on}\,DT + v_{L,off}\,(1-D)T = 0 \) and solve for \( V_o/V_s \).
Get the current relation free from \( V_sI_s = V_oI_o \) — the converter is lossless in this model.

One warning before we start: every result below assumes continuous conduction. Section 12-6 handles what happens when that fails, and the answers change.

Section recap. Two statements — the average inductor voltage is zero and the average capacitor current is zero — generate every gain expression in Part 3. Look at the circuit twice, once per switch state, and solve.

Section 12-2

The Buck Converter

Take the step-down chopper of Chapter 11 and add a capacitor across the load. That is the entire structural difference — and it changes what the circuit is for.

The chopper delivered a chopped voltage whose average was useful. The buck converter delivers a genuinely smooth DC voltage, because \( L \) and \( C \) together form a second-order low-pass filter between the switching node and the load.

The buck converter and the only two states it has. Reading the inductor voltage in each state is the whole analysis.

Switch on, for a time \(DT\) The diode is reverse-biased by \( V_s \), so it is out of the circuit. The inductor sits between the source and the output:

Working

\[ v_{L,on} = V_s - V_o \]

This is positive (since \( V_o < V_s \)), so the inductor current rises.
Switch off, for a time \((1-D)T\) The inductor current must continue, so the diode conducts and clamps the switching node to 0 V:

Working

\[ v_{L,off} = 0 - V_o = -V_o \]

Negative, so the current falls — as it must, to return to where it began.
Apply volt-second balance

Working

\[ (V_s - V_o)\,DT + (-V_o)(1-D)T = 0 \]

\[ V_sD - V_oD - V_o + V_oD = 0 \;\Longrightarrow\; V_o = D\,V_s \]
Get the current from power balance Since \( V_sI_s = V_oI_o \) in the lossless model, \( I_s = D\,I_o \). The buck draws less average current than it delivers, which is the mirror image of the boost converter in the next section.

🔑

Buck converter, continuous conduction

\[ V_o = D\,V_s, \qquad I_s = D\,I_o, \qquad \Delta I_L = \frac{V_o(1-D)}{fL} = \frac{V_s D(1-D)}{fL} \]

The same gain as the step-down chopper — the capacitor changes the output waveform, not the average. But the ripple is now a design specification rather than an accepted nuisance.

◈Why the buck's input current is the hard problem

Look at where the inductor sits: on the output side of the switch. That has a consequence people often miss until they see the EMI test results.

Output current is continuous and smooth. The inductor is in series with the load, so the load sees a gently rippling DC. Excellent.
Input current is a chopped square wave. The source is connected only while the switch is on. It supplies \( I_L \) for \( DT \) and nothing at all for the rest of the period.

So the buck converter draws pulses of current from its supply, with a peak equal to the full inductor current and edges as fast as the switch can move — which in a modern converter is nanoseconds.

The practical consequences:

An input capacitor is mandatory, placed physically adjacent to the switch, to supply those pulses locally. Without it, the pulses flow through the supply wiring, which becomes an efficient antenna.
That capacitor sees severe RMS ripple current — roughly \( I_o\sqrt{D(1-D)} \), worst at \( D = 0.5 \) — and ripple current, not capacitance, is usually what determines its size and cost.
The layout loop containing the switch, the diode and the input capacitor carries the fastest \( di/dt \) in the converter. Keeping it small is the single most important physical-design rule in a buck converter.

The boost converter has the opposite problem, and knowing which way round it is for each topology is genuinely useful when choosing one. We will tabulate it in Section 12-7.

The capacitor now needs sizing, and this is where the buck differs most sharply from the bare chopper. In steady state the load takes a constant \( I_o \) while the inductor delivers \( I_o \pm \Delta I_L/2 \). The difference flows into and out of the capacitor.

That difference is a triangular waveform of zero average. Integrating the positive half gives the charge that the capacitor must store:

The capacitor current is the inductor's ripple \( i_C = i_L - I_o \), a triangle of peak-to-peak height \( \Delta I_L \) and zero mean.
Charge accumulated over the positive half The triangle is above zero for half the period, and its area is that of a triangle with base \( T/2 \) and height \( \Delta I_L/2 \):

Working

\[ \Delta Q = \frac{1}{2}\cdot\frac{T}{2}\cdot\frac{\Delta I_L}{2} = \frac{\Delta I_L T}{8} \]
Convert charge to voltage

Result

\[ \Delta V_o = \frac{\Delta Q}{C} = \frac{\Delta I_L}{8fC} = \frac{V_s D(1-D)}{8f^2LC} \]

▲Common mistake

"Use \( \Delta V_o = \Delta I_L/(8fC) \), pick the capacitance, done."

Wrong Choosing \( C \) from the formula alone, then measuring ten times the predicted ripple on the bench.

Right Real capacitors have equivalent series resistance. The ripple current flowing through the ESR produces \( \Delta V_{ESR} = \Delta I_L \times R_{ESR} \), which usually dominates.

Put numbers on it. Take \( \Delta I_L = 1 \) A, \( f = 100 \) kHz, \( C = 100\ \mu\text{F} \):

Capacitive term: \( \dfrac{1}{8(10^5)(10^{-4})} = 12.5 \) mV.
ESR term, with a typical electrolytic \( R_{ESR} = 100\ \text{m}\Omega \): \( 1 \times 0.1 = 100 \) mV.

The ESR term is eight times larger. Sizing on capacitance alone would miss the answer by nearly an order of magnitude.

What designers actually do: choose the capacitor from a maximum permitted ESR, \( R_{ESR} \le \Delta V_{o(spec)}/\Delta I_L \), and treat the capacitance that comes with it as a bonus. This is why a 100 μF ceramic (ESR ≈ 2 mΩ) can outperform a 1000 μF electrolytic (ESR ≈ 80 mΩ) for ripple, despite a tenth of the capacitance. It is also why high-performance designs use several capacitors in parallel — the ESRs divide.

One caution the other way: ESR is not purely a nuisance. Some voltage-mode control loops rely on the ESR zero for phase margin, and replacing electrolytics with very-low-ESR ceramics has destabilised more than one working design. Chapter 15 returns to this.

Section recap. The buck is a chopper plus an output capacitor: \( V_o = DV_s \), \( \Delta I_L = V_o(1-D)/fL \), \( \Delta V_o = \Delta I_L/(8fC) \) plus an ESR term that usually dominates. Its output current is smooth and its input current is pulsed, so the input capacitor and its layout loop matter more than beginners expect.

Section 12-3

The Boost Converter

Chapter 11 derived \( V_o = V_s/(1-D) \) already. Here we complete the picture — the ripple expressions, the capacitor sizing, and the structural feature that makes the boost converter the natural choice for one very large application.

The rearrangement is: inductor first, in series with the source; switch across the line after it; diode in series to the output capacitor.

Switch on The switch shorts the node after the inductor to ground, so the inductor is directly across the source. The diode is reverse-biased by \( V_o \), and the load is fed entirely by the capacitor.

Working

\[ v_{L,on} = V_s \]
Switch off The inductor current forces the diode on, connecting the inductor to the output.

Working

\[ v_{L,off} = V_s - V_o \quad(\text{negative, since } V_o > V_s) \]
Volt-second balance

Working

\[ V_s DT + (V_s - V_o)(1-D)T = 0 \;\Longrightarrow\; V_o = \frac{V_s}{1-D} \]
Charge balance gives the capacitor ripple The capacitor supplies the entire load current \( I_o \) for the whole on-time \( DT \), because the diode is blocked.

Result

\[ \Delta Q = I_o\,DT \;\Longrightarrow\; \Delta V_o = \frac{I_o D}{fC} \]

🔑

Boost converter, continuous conduction

\[ V_o = \frac{V_s}{1-D}, \qquad I_L = \frac{I_o}{1-D}, \qquad \Delta I_L = \frac{V_s D}{fL}, \qquad \Delta V_o = \frac{I_o D}{fC} \]

Compare the output ripple with the buck's \( \Delta I_L/(8fC) \). The boost's is far larger for the same capacitance, because the capacitor must carry the whole load current alone during every on-time, not merely a triangular remainder.

◈Why solar and PFC both use a boost converter

The boost's structure is the exact mirror of the buck's, and that gives it one property worth a great deal:

Input current is continuous and smooth, because the inductor is in series with the source and never disconnects from it.
Output current is pulsed, because the diode only conducts for \( (1-D)T \).

Why that matters for a solar panel. A photovoltaic panel has a specific operating point at which it delivers maximum power, and that point is a particular voltage and current pair. Pulsed current would drag the panel back and forth across its I–V curve many thousands of times a second, and its average power would fall below the peak it is capable of. A boost converter draws a steady current, so the maximum-power-point tracker can hold the panel exactly where it wants to be. Chapter 29 develops this.

Why it matters for power factor correction. An active PFC stage must make a rectifier's input current follow a sine wave, so it must be able to control that current smoothly and continuously. Only a topology with a continuous input current can. That is why virtually every PC power supply, LED driver and appliance above about 75 W has a boost stage immediately after its bridge rectifier — its job is not to raise the voltage (though it does, usually to about 400 V) but to shape the input current.

The price is on the output side: the load must be fed from a capacitor for the whole on-time, so the output capacitor is large, and its ripple-current rating is often what sets its size.

▲Common mistake

"When the boost converter's switch is off, no current flows — so the switch never sees stress."

Wrong Rating the switch for \( V_s \), because that is the supply voltage.

Right The switch must block \( V_o \), the output voltage — which is higher than the input by definition.

Trace the loop. With the switch off, the diode conducts, so the switching node sits at \( V_o \). The switch is connected between that node and ground, so it stands off the full \( V_o \).

The design consequence is uncomfortable. A converter that boosts 12 V to 400 V needs a 400 V-rated switch (500 V in practice, for margin and spikes). Higher-voltage MOSFETs have dramatically higher on-resistance for the same die area — \( R_{DS(on)} \) scales roughly with \( V_{BR}^{2.5} \) — so the boost converter's conduction loss rises steeply with the voltage it is asked to produce.

And the current is high too. \( I_L = I_o/(1-D) \), so the switch carries the input current, not the output current. High voltage rating and high current together are exactly the combination that makes a switch expensive.

The same trap in the other topologies: the buck's switch blocks \( V_s \), which is the smallest of the three. The buck–boost's switch blocks \( V_s + |V_o| \), which is the worst of all — see Section 12-4.

Section recap. \( V_o = V_s/(1-D) \), with \( \Delta I_L = V_sD/(fL) \) and \( \Delta V_o = I_oD/(fC) \). Continuous input current makes it the topology of choice for solar MPPT and active PFC; the price is a large output capacitor and a switch that must block the full output voltage while carrying the input current.

Section 12-4

The Buck–Boost Converter

The third arrangement puts the inductor in parallel — the switch charges it from the source, and then the inductor discharges into the load through the diode, with the source disconnected.

That separation is the key to the topology. The inductor is never connected to both the source and the load at once, so it acts as an energy bucket: filled from one, emptied into the other.

◇Intuition · the bucket brigade

Water is moved from a well to a tank by one person with a bucket. They fill the bucket at the well, walk across, and empty it into the tank. The well and the tank are never connected.

Because of that separation, three things follow — and all three are properties of the buck–boost converter:

The tank's level is unrelated to the well's. Fill fast and empty slowly and the tank rises above the well; do the reverse and it stays below. Hence a converter that can step up or down.
The direction of pouring is your choice. The bucket does not remember which way up it was filled — so the output polarity is set by how the diode is oriented, and the standard arrangement inverts it.
Everything passes through the bucket. All the energy delivered is stored in \( L \) first, so the inductor must be sized for the full throughput. In the buck converter, by contrast, energy flows straight from source to load during the on-time and only the ripple is stored.

That third point is the buck–boost's fundamental cost. It needs a larger inductor than either of the other two for the same power, because it is not a filter — it is the conveyor.

Switch on The source is connected directly across the inductor; the diode is reverse-biased and the load is fed by the capacitor.

Working

\[ v_{L,on} = V_s \]
Switch off The source is disconnected. The inductor current forces the diode on, discharging into the output.

Working

\[ v_{L,off} = V_o \quad (\text{with } V_o \text{ negative}) \]
Volt-second balance

Working

\[ V_s DT + V_o(1-D)T = 0 \;\Longrightarrow\; V_o = -\frac{D}{1-D}V_s \]
Read the gain's behaviour \( D/(1-D) \) equals 1 at \( D = 0.5 \), is less than 1 below it and greater above. So the same circuit steps down for \( D < 0.5 \) and up for \( D > 0.5 \) — with the polarity inverted throughout.

🔑

Buck–boost converter, continuous conduction

\[ V_o = -\frac{D}{1-D}V_s, \qquad \Delta I_L = \frac{V_s D}{fL}, \qquad \Delta V_o = \frac{I_o D}{fC} \]

The switch and diode must each block \( V_s + |V_o| \) — the sum, not the larger. This is the highest device stress of the three basic topologies.

Buck–boost gain across the duty range, from a 12 V source
Duty ratio D	Gain \(D/(1-D)\)	Output	Switch must block	Regime
0.20	0.25	−3 V	15 V	Step down
0.33	0.50	−6 V	18 V	Step down
0.50	1.00	−12 V	24 V	Unity
0.67	2.00	−24 V	36 V	Step up
0.80	4.00	−48 V	60 V	Step up

Interactive · the three gain curves

All three topologies from a 12 V source. Move the duty ratio and watch where each curve goes — and where each stops being buildable. The buck is a straight line, the boost and buck–boost both run away as \( D \to 1 \).

Duty ratio D 0.50

Buck output6.0 V

Boost output24.0 V

Buck–boost−12.0 V

Boost switch stress24 V

—

✦Think about it

A car's 12 V battery sags to 9 V during cranking and rises to 15 V when the alternator is charging. A circuit needs a stable 12 V.

Which of the three topologies can do this, and what is the difficulty?

Work through the options

Buck alone: no. It cannot produce 12 V from 9 V. \( D \le 1 \) means \( V_o \le V_s \) always.

Boost alone: no. It cannot produce 12 V from 15 V. \( V_o \ge V_s \) always.

Buck–boost: yes, electrically — it spans the range comfortably, needing \( D = 0.57 \) at 9 V and \( D = 0.44 \) at 15 V. But it inverts the polarity, so you get −12 V, which is not what the load wants. And its switch must block \( V_s + |V_o| = 27 \) V while carrying \( I_s + I_o \).

What is actually used:

A SEPIC converter — buck–boost capability with the same output polarity. This is Chapter 13, and this exact application is why it exists.
A four-switch buck–boost, which is a buck stage and a boost stage sharing one inductor, operating as a buck when the input is high, as a boost when it is low, and blending near unity. Higher part count, but the best efficiency of the options and non-inverting.
A Ćuk converter, also Chapter 13, if the continuous input current matters more than the added capacitor.

The lesson. The three basic topologies span the arithmetic but not the requirements. Polarity, input-current continuity and device stress are constraints they cannot always satisfy — which is precisely why Chapter 13's fourth-order converters exist.

Section recap. \( V_o = -DV_s/(1-D) \), stepping down below \( D = 0.5 \) and up above it, at the price of inverted polarity, the highest device stress of the three (\( V_s + |V_o| \)), and an inductor that carries all the transferred energy rather than just the ripple.

Section 12-5

Sizing \(L\) and \(C\)

Choosing the two reactive components is most of the design work, and the procedure is the same for all three topologies once the ripple expressions are in hand.

L Choosing the inductor

Decide the ripple ratio. Industry practice is \( \Delta I_L \) between 20% and 40% of the full-load inductor current. Below 20% the inductor is needlessly large; above 40% the peak current — and so the core size and the switch rating — rise faster than the saving.
Rearrange the ripple expression for \( L \) at the worst-case duty ratio, which means the highest input voltage for a buck and the lowest for a boost.
Check the peak current, \( I_{pk} = I_L + \Delta I_L/2 \), against the core's saturation rating — with margin, because saturation is abrupt and unforgiving.
Check the RMS current for the copper loss, and the ripple for the core loss. They are different currents and they heat different parts of the component.

C Choosing the capacitor

Start from the ESR, not the capacitance: \( R_{ESR} \le \Delta V_{o(spec)} / \Delta I_{C(pk\text{-}pk)} \). This is nearly always the binding constraint.
Then check the capacitance from the charge-balance expression, and take the larger of the two requirements.
Then check the RMS ripple current rating, which is what actually determines a capacitor's lifetime. An electrolytic run at its ripple limit ages roughly twice as fast for every 10 °C of self-heating.
Finally check the transient requirement. A sudden load step is supplied by the capacitor alone until the inductor current can change — and for fast steps this demands far more capacitance than the ripple ever does.

▣In practice · the point-of-load converter

A modern processor draws 100 A at 0.9 V and can change its demand by 50 A in under a microsecond when a workload starts.

Ripple is not the difficulty. At 1 MHz, a modest inductance keeps the ripple well inside specification. The problem is the transient.

During a 50 A step the inductor current cannot follow instantly — its slew rate is limited to \( (V_s - V_o)/L \). Everything the load asks for in the meantime comes from the capacitor, and the resulting voltage dip is roughly

\[ \Delta V_{dip} \approx \frac{(\Delta I)^2 L}{2 C\,(V_s - V_o)} \]

Two things follow, and they explain the design of every server power stage:

Small \( L \) is now desirable, because it lets the inductor catch up sooner — the opposite of what ripple alone would suggest. Designers accept 50% ripple to get transient response.
Many phases in parallel. Four or eight buck converters share the load with their switching instants staggered. The ripple currents partly cancel, the effective ripple frequency is multiplied by the number of phases, and each phase's inductor can be small. This is a multiphase buck, and it is why a motherboard has a row of identical inductors beside the CPU socket.

It is a good illustration that the textbook ripple equations set a floor, not the design.

Section recap. Size \( L \) for 20–40% ripple at the worst-case duty ratio and check saturation at the peak; size \( C \) from ESR first, then capacitance, then ripple-current rating, then the load transient — which in fast digital loads dominates everything else.

Section 12-6

Discontinuous Conduction

Every gain expression so far assumed the inductor current never reaches zero. At light load it does, and the results change — differently for each topology.

The boundary is where the average inductor current equals half the ripple, so the trough just touches zero.

The continuous/discontinuous boundary and what lies beyond it
Topology	Boundary condition	Critical inductance	Gain in DCM
Buck	\( I_o = \dfrac{\Delta I_L}{2} \)	\( L_{crit} = \dfrac{(1-D)R}{2f} \)	Rises above \(DV_s\), towards \(V_s\) at no load
Boost	\( I_L = \dfrac{\Delta I_L}{2} \)	\( L_{crit} = \dfrac{D(1-D)^2R}{2f} \)	Rises above \(V_s/(1-D)\), without bound at no load
Buck–boost	\( I_L = \dfrac{\Delta I_L}{2} \)	\( L_{crit} = \dfrac{(1-D)^2R}{2f} \)	Rises in magnitude; depends on \(R\)

▲Common mistake

"Discontinuous conduction is a light-load curiosity. Real converters run in CCM."

Wrong Ignoring DCM because the datasheet operating point is at full load.

Right Every converter that can be lightly loaded will enter DCM, and the control loop must survive the transition.

The three things that change when you cross the boundary:

The gain becomes load-dependent. In CCM, \( V_o/V_s \) depends only on \( D \). In DCM it depends on \( D \), \( L \), \( f \) and \( R \). The loop's plant gain therefore changes with load.
The dynamics change order. A CCM buck is a second-order system with a resonant \( LC \) pole pair. In DCM the inductor current starts each cycle from zero, so it carries no state from one cycle to the next — and the converter becomes effectively first-order. A compensator designed for the second-order plant is now over-compensated and sluggish, or, if it was aggressive, unstable.
An unloaded boost converter has no upper bound. With no load, each cycle delivers charge into the capacitor and nothing removes it. The output rises until something breaks down. Every practical boost converter needs either a minimum load, an output voltage clamp, or a control loop that reduces \( D \) to zero — usually all three.

How real converters handle it. Most modern controllers detect the transition and change behaviour: some switch to a fixed low frequency (pulse-frequency modulation) at light load, which actually improves light-load efficiency by cutting switching losses; others force continuous conduction by using a synchronous MOSFET instead of a diode, so the current can reverse and the inductor never disconnects. The second approach keeps the dynamics constant at the cost of circulating current.

✦Think about it

Replace a buck converter's diode with a second MOSFET, switched on exactly when the diode would have conducted. This is a synchronous buck, and it is standard in low-voltage supplies.

What does it gain, and what does it change about discontinuous conduction?

Two answers, and the second is the interesting one

The obvious gain: efficiency. A Schottky diode drops perhaps 0.4 V. A MOSFET with \( R_{DS(on)} = 3\ \text{m}\Omega \) carrying 20 A drops 60 mV. At an output of 1.2 V, the diode would waste 25% of the output power in the freewheeling path; the MOSFET wastes 4%. Below about 5 V output, synchronous rectification is not optional.

The subtle change: DCM no longer happens by itself. A diode blocks reverse current, so when the inductor current reaches zero it simply stops — that is what makes conduction discontinuous. A MOSFET conducts in both directions, so at light load the current goes negative: energy flows back from the output capacitor into the input.

Is that good? It depends on what you want.

Good for control: conduction stays continuous at any load, so the plant never changes order and the loop never has to be re-tuned. Forced-CCM operation is used wherever predictable dynamics matter more than light-load efficiency.
Bad for efficiency: circulating current produces conduction loss that delivers nothing. At 1% load a forced-CCM converter can be less than 20% efficient.

What controllers actually do: offer both, selectable. A laptop supply uses diode emulation (turn the synchronous FET off when the current reaches zero, mimicking a diode) when the machine is idle, and forced CCM when it is under load and transient response matters. The mode pin on a synchronous buck controller is exactly this choice.

Section recap. Below the critical inductance — or below the critical load — the inductor current reaches zero, the gain becomes load-dependent, and the converter drops from second-order to first-order dynamics. An unloaded boost converter has no upper output bound at all, which is why every one has a minimum load or a clamp.

Section 12-7

Choosing Between Them

The three basic topologies compared
Property	Buck	Boost	Buck–boost
Gain	\(D\)	\(\dfrac{1}{1-D}\)	\(-\dfrac{D}{1-D}\)
Range	Step down only	Step up only	Either
Polarity	Same	Same	Inverted
Input current	Pulsed	Continuous	Pulsed
Output current	Continuous	Pulsed	Pulsed
Switch blocks	\(V_s\)	\(V_o\)	\(V_s + \|V_o\|\)
Switch carries	\(I_o\)	\(I_o/(1-D)\)	\(I_s + I_o\)
Inductor role	Filter	Energy transfer	Energy transfer
Output capacitor	Small	Large	Large
Typical use	Point-of-load, LED drivers, battery-to-logic	Solar MPPT, PFC front end, battery-to-bus	Small inverting rails, some LED drivers

◈Why "pulsed" appears twice for the buck–boost

Read down the input-current and output-current rows. The buck has one continuous side, the boost has the other — and the buck–boost has neither.

The reason is structural. The buck–boost's inductor is connected to the source during \( t_{on} \) and to the load during \( t_{off} \), and to only one at a time. So:

The source sees current for \( DT \) and zero for the rest — pulsed.
The load side sees current for \( (1-D)T \) and zero for the rest — also pulsed.

The consequences are real and cumulative:

Filtering is needed on both sides, so two capacitors, both carrying substantial RMS ripple.
EMI is generated at both ports rather than one.
All the transferred energy passes through the inductor, so it is larger than the buck's for the same power.
The device stress \( V_s + |V_o| \) is the highest of the three.

So why is it used at all? Because it is the only one of the three that spans both directions, and because it does so with just one inductor, one switch and one diode — the lowest possible part count for that capability. For a small auxiliary rail (a −5 V bias supply drawing 50 mA, say) the penalties are trivial and the simplicity is worth having.

At real power it is displaced by the Ćuk and SEPIC converters of Chapter 13, which fix the polarity and the input-current problems by adding one capacitor and one inductor. Understanding why the buck–boost falls short is exactly the motivation for the next chapter.

Section recap. Buck for stepping down with a clean output, boost for stepping up with a clean input, buck–boost when you must do both and can accept inverted polarity, both ports pulsed, and the highest device stress. Match the topology to the port that must stay clean.

Section 12-8

Worked Examples

1 A complete buck design

Problem. Design a buck converter: \( V_s = 24 \) V, \( V_o = 12 \) V, \( I_o = 5 \) A, \( f = 100 \) kHz. Ripple current 30% of \( I_o \); output voltage ripple below 50 mV. Find \( D \), \( L \), the peak inductor current, and the required capacitance and ESR.

Working

\[ D = \frac{V_o}{V_s} = \frac{12}{24} = 0.5 \]

\[ \Delta I_L = 0.30(5) = 1.5\ \text{A} \]

\[ L = \frac{V_o(1-D)}{f\,\Delta I_L} = \frac{12(0.5)}{10^5(1.5)} = 40\ \mu\text{H} \]

\[ I_{pk} = 5 + \frac{1.5}{2} = 5.75\ \text{A} \quad \text{(saturation rating: choose} \ge 8\ \text{A)} \]

Now the capacitor, from both constraints.

Capacitance requirement

\[ C \ge \frac{\Delta I_L}{8f\,\Delta V_o} = \frac{1.5}{8(10^5)(0.05)} = 37.5\ \mu\text{F} \]

ESR requirement

\[ R_{ESR} \le \frac{\Delta V_o}{\Delta I_L} = \frac{0.05}{1.5} = 33\ \text{m}\Omega \]

Reading the two answers together. They are not independent — the two ripple contributions add. Allocating half the budget to each gives \( C \ge 75\ \mu\text{F} \) and \( R_{ESR} \le 17\ \text{m}\Omega \).

What to actually fit: three 22 μF ceramics in parallel. That gives 66 μF with an effective ESR under 2 mΩ, so the ESR term all but vanishes and the capacitance term dominates — comfortably inside specification, in a fraction of the volume a 100 μF electrolytic would need.

2 A boost converter for a solar panel

Problem. A panel delivers 30 V at 8 A. A boost converter raises this to 60 V at 40 kHz, with inductor ripple 20% and output ripple 1%. Find \( D \), \( L \), \( C \), the output current and the switch ratings.

Working

\[ 60 = \frac{30}{1-D} \;\Longrightarrow\; D = 0.5 \]

\[ I_o = \frac{P}{V_o} = \frac{30(8)}{60} = 4\ \text{A}, \qquad I_L = I_s = 8\ \text{A} \]

\[ \Delta I_L = 0.20(8) = 1.6\ \text{A}, \qquad L = \frac{V_sD}{f\,\Delta I_L} = \frac{30(0.5)}{4\times10^4(1.6)} = 234\ \mu\text{H} \]

\[ \Delta V_o = 0.01(60) = 0.6\ \text{V}, \qquad C = \frac{I_oD}{f\,\Delta V_o} = \frac{4(0.5)}{4\times10^4(0.6)} = 83\ \mu\text{F} \]

Device ratings. The switch and the diode both block \( V_o = 60 \) V (choose 100 V devices) and carry a peak of \( I_L + \Delta I_L/2 = 8.8 \) A.

Compare the two capacitors. The boost needs 83 μF where the buck of Example 1 needed 37.5 μF — for a comparable power level and a looser ripple specification. That is the structural penalty named in Section 12-3: the boost's capacitor carries the entire load current alone during every on-time.

3 Buck–boost across a wide input

Problem. A buck–boost must produce −15 V at 2 A from an input that varies from 10 V to 30 V, at 50 kHz. Find the duty-ratio range, the worst-case device stress, and the inductance for 30% ripple.

Duty ratio at the two extremes

\[ \frac{|V_o|}{V_s} = \frac{D}{1-D} \;\Longrightarrow\; D = \frac{|V_o|}{|V_o| + V_s} \]

\[ V_s = 10\ \text{V}: \; D = \frac{15}{25} = 0.60 \qquad V_s = 30\ \text{V}: \; D = \frac{15}{45} = 0.33 \]

Device stress is \( V_s + |V_o| \), worst at the highest input: \( 30 + 15 = 45 \) V. Choose 100 V devices.

Inductance. The inductor carries \( I_L = I_o/(1-D) \), which is largest at the largest \( D \) — that is, at the lowest input voltage:

Worst case: \(V_s = 10\) V, \(D = 0.60\)

\[ I_L = \frac{2}{1-0.6} = 5\ \text{A}, \qquad \Delta I_L = 0.30(5) = 1.5\ \text{A} \]

\[ L = \frac{V_sD}{f\,\Delta I_L} = \frac{10(0.6)}{5\times10^4(1.5)} = 80\ \mu\text{H} \]

The lesson in the worst case. Only 30 W is being delivered, yet the inductor carries 5 A and peaks at 5.75 A while the devices block 45 V. Check both extremes of the input range — for this topology they place their demands on different components, and a design checked at only one end will fail at the other.

4 Finding the DCM boundary

Problem. The buck converter of Example 1 (\( L = 40\ \mu\text{H} \), \( D = 0.5 \), \( f = 100 \) kHz) is unloaded down to 200 mA. Does it enter DCM, and what is the output voltage if it does?

Working

\[ \Delta I_L = \frac{V_o(1-D)}{fL} = \frac{12(0.5)}{10^5(40\times10^{-6})} = 1.5\ \text{A} \]

\[ I_{o(crit)} = \frac{\Delta I_L}{2} = 0.75\ \text{A} \]

Yes — 200 mA is well below the 750 mA boundary, so the converter is firmly in DCM.

What happens to the output. In an open-loop converter the output would rise above 12 V, towards 24 V as the load approaches zero. In a real closed-loop converter it does not: the controller senses the rise and reduces \( D \) until the output returns to 12 V. In DCM the duty ratio needed is

\[ D_{DCM} = \sqrt{\frac{2 L f I_o}{V_s}\cdot\frac{V_o/V_s}{1 - V_o/V_s}} \]

which for these numbers gives \( D \approx 0.18 \) — very different from the 0.5 that CCM required at the same output voltage.

Why this matters for the loop. The controller must swing its duty ratio from 0.5 to 0.18 as the load falls, and the plant gain changes with it. This is the load-dependent gain of Section 12-6, and a compensator that ignores it will either be sluggish at one end of the range or unstable at the other.

5 Choosing a topology from a specification

Problem. An LED string needs 36 V at 700 mA, constant current. It is fed from a 24 V battery that falls to 20 V when depleted. EMI must be low at the battery terminals. Choose a topology and justify it.

Solution — work through the constraints in order.

Step up or down? The output is 36 V and the input is 20–24 V, so it is always a step-up. That eliminates the buck immediately.
Polarity? The LED string needs the same polarity as the battery, so the inverting buck–boost is out unless there is a good reason for it.
EMI at the battery? This is the constraint that decides it. A boost converter has a continuous input current — the inductor is permanently in series with the battery, so the battery sees smooth DC with a small triangular ripple. A buck–boost would draw pulses.
Answer: a boost converter, and every one of the three constraints agrees.

The design. \( D = 1 - V_s/V_o \), so \( D \) ranges from \( 1 - 24/36 = 0.33 \) to \( 1 - 20/36 = 0.44 \) as the battery depletes — a narrow and comfortable range, well away from the region where the gain runs away.

One change from the earlier examples. An LED driver regulates current, not voltage, so the feedback comes from a small sense resistor in series with the string rather than from a voltage divider across it. The topology and the equations are identical; only the sensed quantity differs. LEDs need this because their forward voltage varies by several volts between parts and with temperature, so a voltage-regulated supply would give an unpredictable — and possibly destructive — current.

Section 12-9

Summary & Formula Sheet

Chapter 12 in five sentences:

Volt-second balance on the inductor and charge balance on the capacitor give every gain expression in Part 3, in four lines each.
The buck steps down with a smooth output and a pulsed input; the boost steps up with a smooth input and a pulsed output.
The buck–boost does both, but inverts the polarity, pulses at both ports, and imposes the highest device stress.
Size \( L \) for 20–40% ripple at the worst-case duty ratio; size \( C \) from ESR first, then capacitance, then the load transient.
Below the critical load the converter enters DCM, its gain becomes load-dependent, and its dynamics drop from second order to first.

Formula sheet · Chapter 12

Volt-second balancethe master tool

\( v_{L,on}DT + v_{L,off}(1-D)T = 0 \)

Buck gainCCM only

\( V_o = D V_s \)

Buck inductor ripplesize \(L\) from this

\( \Delta I_L = \dfrac{V_o(1-D)}{fL} \)

Buck output ripplecapacitive part only; add ESR

\( \Delta V_o = \dfrac{\Delta I_L}{8fC} \)

Boost gainpractical limit \(\approx\) 4–5

\( V_o = \dfrac{V_s}{1-D} \)

Boost inductor ripple\(L\) carries the input current

\( \Delta I_L = \dfrac{V_s D}{fL} \)

Boost output ripple\(C\) alone feeds the load for \(DT\)

\( \Delta V_o = \dfrac{I_o D}{fC} \)

Buck–boost gainnote the sign

\( V_o = -\dfrac{D}{1-D}V_s \)

Buck–boost duty ratiosolve directly for \(D\)

\( D = \dfrac{|V_o|}{|V_o| + V_s} \)

ESR ripple contributionusually dominant

\( \Delta V_{ESR} = \Delta I_L \, R_{ESR} \)

Buck critical inductanceCCM/DCM boundary

\( L_{crit} = \dfrac{(1-D)R}{2f} \)

Boost critical inductanceCCM/DCM boundary

\( L_{crit} = \dfrac{D(1-D)^2R}{2f} \)

Transient voltage dipload step \(\Delta I\); often sizes \(C\)

\( \Delta V \approx \dfrac{(\Delta I)^2 L}{2C(V_s - V_o)} \)

Key terms

Volt-second balance: In periodic steady state the average voltage across an inductor is zero. Generates every converter gain in Part 3.
Charge balance: In periodic steady state the average current into a capacitor is zero. Gives the output ripple expressions.
Buck converter: Step-down converter, \(V_o = DV_s\). Continuous output current, pulsed input current.
Boost converter: Step-up converter, \(V_o = V_s/(1-D)\). Continuous input current, pulsed output current.
Buck–boost converter: Inverting converter, \(V_o = -DV_s/(1-D)\). Steps up or down; both ports pulsed.
ESR: Equivalent series resistance of a capacitor. Usually dominates output ripple; often the real selection criterion.
Ripple ratio: \(\Delta I_L / I_L\), conventionally 20–40%. Sets the inductance for a given frequency and voltage.
CCM / DCM: Continuous and discontinuous conduction modes. The gain expressions above hold only in CCM.
Synchronous rectification: Replacing the freewheeling diode with a MOSFET. Cuts conduction loss sharply and allows reverse current.
Diode emulation: Turning the synchronous MOSFET off at the current zero crossing, so the converter behaves as if a diode were fitted. Improves light-load efficiency.
Multiphase converter: Several converters in parallel with staggered switching. Cancels ripple and multiplies the effective ripple frequency.

Check yourself

Test Yourself

Chapter 12 · six questions answers hidden until you ask

Why does the buck's output ripple formula have an 8 in it, while the boost's does not?

Because the two capacitors are doing entirely different jobs.

In the buck, the inductor is in series with the load, so the load's current is already nearly supplied. The capacitor only has to absorb the triangular difference between the inductor's ripple and the load's steady draw. Integrating that triangle over its positive half gives

\[ \Delta Q = \frac{1}{2}\cdot\frac{T}{2}\cdot\frac{\Delta I_L}{2} = \frac{\Delta I_L T}{8} \]

and the 8 is the product of those three factors: one half for the triangle's area, one half for the half-period, one half for the half-amplitude.

In the boost, the diode is blocked for the whole on-time, so the capacitor supplies the entire load current alone:

\[ \Delta Q = I_o \, DT \]

No triangle, no factor of 8 — just a rectangle.

Put numbers on the difference. Take \( I_o = 5 \) A, \( \Delta I_L = 1.5 \) A, \( D = 0.5 \), \( f = 100 \) kHz, \( C = 100\ \mu\text{F} \):

Buck: \( \Delta V_o = 1.5/(8 \times 10^5 \times 10^{-4}) = 19 \) mV.
Boost: \( \Delta V_o = 5(0.5)/(10^5 \times 10^{-4}) = 250 \) mV.

Thirteen times larger for the same capacitor. That single structural difference is why boost converters carry conspicuously bigger output capacitors than bucks of the same power.

A buck converter is measured with 10 times more output ripple than designed. The capacitance is correct and the inductor measures right. What is the most likely cause?

ESR — almost always, and it should be the first thing checked.

The design used \( \Delta V_o = \Delta I_L/(8fC) \), which accounts only for the charge stored in the capacitance. The real ripple is

\[ \Delta V_o = \underbrace{\frac{\Delta I_L}{8fC}}_{\text{capacitive}} + \underbrace{\Delta I_L R_{ESR}}_{\text{resistive}} \]

With \( \Delta I_L = 1.5 \) A, \( f = 100 \) kHz and \( C = 100\ \mu\text{F} \), the capacitive term is 19 mV. An ordinary aluminium electrolytic with 120 mΩ of ESR contributes \( 1.5 \times 0.12 = 180 \) mV — ten times more, matching the symptom exactly.

How to confirm it on the bench. Look at the ripple waveform's shape:

Capacitance-dominated ripple is parabolic and smooth, because it is the integral of the triangular current.
ESR-dominated ripple is triangular and follows the inductor current exactly, because \( v = iR \) has no phase shift.

A triangular ripple waveform is a direct fingerprint of ESR.

Two other candidates, less likely but worth ruling out:

Equivalent series inductance. Above a few hundred kilohertz the capacitor's own lead inductance produces sharp spikes at each switching edge — visible as narrow needles rather than a raised triangle.
Measurement error. A long ground lead on the oscilloscope probe forms a loop that picks up the switching node's field. Ripple measurements must be made with a spring-tip ground or a probe socket directly across the capacitor's terminals. Many "excess ripple" reports are entirely this.

A boost converter's output is measured at 5 V with a 5 V input, and the duty ratio reads 0. Is it broken?

No — this is correct and expected behaviour. It reveals something structural about the topology that is worth understanding.

Setting \( D = 0 \) gives \( V_o = V_s/(1-0) = V_s \), so the formula agrees. But look at the circuit rather than the formula: with the switch permanently off, the source connects to the load through the inductor and the diode in series. It is simply a wire with a diode drop in it. The converter is not converting — it is passing the input straight through.

The important consequence: a boost converter cannot disconnect its load.

There is no way to get an output below \( V_s \). Even with the switch off forever, the input reaches the output.
There is no inherent short-circuit protection. Short the output and the current path from the source through \( L \) and the diode is uninterrupted — the switch is not in it. The fault current is limited only by the source and the wiring, and it will destroy the diode.
There is no inrush limiting. On power-up the output capacitor charges from empty through that same path, drawing a large uncontrolled surge.

What real designs add: a series disconnect device (a MOSFET, sometimes doubling as the inrush limiter) or a fuse chosen to clear before the diode fails. Contrast the buck converter, whose switch is in the power path, so turning it off genuinely isolates the load. This is a real advantage of the buck that its gain expression does not reveal.

Two buck converters have identical specifications, but one runs at 100 kHz and the other at 1 MHz. Compare their inductors, capacitors, efficiency and transient response.

Inductor: 10 times smaller at 1 MHz. \( L = V_o(1-D)/(f\Delta I_L) \), so \( L \propto 1/f \). A 40 μH inductor becomes 4 μH — from a wound bobbin to a small moulded part.

Capacitor: 10 times smaller for the same ripple, by the same reasoning. Often it also changes type, from electrolytic to ceramic, with an incidental ESR improvement.

Efficiency: worse at 1 MHz, but not uniformly. Split the losses:

Conduction loss (\( I^2R \) in the switch and inductor) is unchanged — it does not care about frequency.
Switching loss is roughly \( \tfrac12 V I (t_r + t_f) f \), so it is 10 times larger.
Gate drive loss, \( Q_gV_gf \), is also 10 times larger.
Core loss rises with frequency, though the smaller core partly offsets it.

At light load, where switching loss dominates, the 1 MHz converter is clearly worse. At full load, where conduction loss dominates, the difference may be a fraction of a percent.

Transient response: much better at 1 MHz, and this is usually the real reason for the choice. Two reinforcing effects: the loop bandwidth can be roughly ten times higher (it is limited to a fraction of \( f \)), and the smaller inductor lets the current slew ten times faster.

Which to choose. For a battery-powered device that spends most of its life idle, 100 kHz — efficiency dominates. For a processor core rail that must answer a 50 A step in a microsecond, 1 MHz or higher — the transient dominates, and the smaller passives are a bonus. The frequency is not a free parameter; it is where you place yourself on the efficiency–size–speed triangle.

Why does a buck–boost converter's inductor have to be larger than a buck converter's, for the same output power?

Because the two inductors are doing different jobs, and the energy each must store is genuinely different.

In a buck converter, the inductor is in series between the source and the load. During the on-time, energy flows straight through it from source to load — the inductor merely smooths the flow. It stores only the ripple energy, \( \tfrac12 L(\Delta I_L)^2 \), which is a small fraction of the energy transferred.

In a buck–boost converter, the source and load are never connected. Every joule delivered to the load is first stored in the inductor's magnetic field and then released. The energy per cycle is

\[ W = \frac{P_o}{f} \quad\text{and this must fit in}\quad \frac{1}{2}LI_{pk}^2 \]

Put numbers on it. For 60 W at 50 kHz, the buck–boost inductor stores 1.2 mJ per cycle. A buck delivering the same 60 W with 30% ripple stores only about 0.1 mJ of ripple energy — an order of magnitude less.

What that means physically. Stored energy determines core volume, because a core saturates at a fixed flux density. So the buck–boost's inductor is larger, heavier and more expensive, and it usually needs an air gap to store the energy without saturating.

The generalisation worth carrying. Converters divide into two families:

Direct converters (buck, boost, forward): the inductor filters, energy passes through, magnetics are small.
Indirect converters (buck–boost, flyback, Ćuk): the inductor is the transfer mechanism, all energy is stored en route, magnetics are large.

This is exactly why a flyback converter (Chapter 14) is limited to about 150 W while a forward converter of the same size handles several times more.

Sliding the explorer to \( D = 0.9 \) shows the boost producing 120 V from 12 V. Would a real converter do this?

No. The curve above \( D \approx 0.8 \) is drawn from the ideal equation, and reality departs from it sharply.

Three effects, all pointing the same way:

Winding and switch resistance. Including \( r_L \), the gain becomes \[ \frac{V_o}{V_s} = \frac{1}{1-D}\cdot\frac{1}{1 + r_L/[R(1-D)^2]} \] The correction factor is negligible at low \( D \) and dominant as \( D \to 1 \), because \( (1-D)^2 \) collapses. In practice the gain peaks around \( D = 0.8 \) to \( 0.9 \) and then falls — pushing the duty ratio higher makes the output lower.
Input current. \( I_L = I_o/(1-D) = 10I_o \) at \( D = 0.9 \). For a 1 A output the inductor and switch carry 10 A, and \( I^2R \) losses are a hundredfold worse than the output current alone suggests.
Off-time. At \( D = 0.9 \) and 100 kHz, \( t_{off} = 1\ \mu\text{s} \), and every joule of that cycle's energy must be delivered inside it. The diode's reverse recovery and the switch's transition occupy a growing fraction of that window.

And a control problem on top. \( dV_o/dD = V_s/(1-D)^2 = 1200 \) V per unit duty at \( D = 0.9 \). A 1% duty error moves the output by 12 V. The loop gain rises as \( 1/(1-D)^2 \), so a compensator stable at \( D = 0.3 \) may well oscillate at \( D = 0.9 \).

The practical limit is a gain of about 4 to 5 for a single boost stage. Beyond that: a transformer (Chapter 14), two cascaded stages, or a coupled-inductor topology. The explorer's note above \( D = 0.8 \) says exactly this, and it is worth trusting over the curve.

Practice

Problems

Three habits for every problem in this chapter:

Derive, do not recall. Two circuit states, volt-second balance, four lines. It is faster than remembering which formula belongs to which topology, and it never gives you the wrong one.
Check both ends of the input range. The worst case for the inductor and the worst case for the switch are often at opposite extremes.
Add the ESR term before believing any output-ripple answer.

Problems 1–5 are direct application; 6–10 need judgement; 11–12 are design questions worth discussing in a tutorial.

A buck converter has \( V_s = 48 \) V, \( V_o = 18 \) V, \( I_o = 4 \) A, \( f = 200 \) kHz. Find \( D \), the inductance for 25% ripple, the peak inductor current, and the capacitance for 30 mV of capacitive ripple.
For Problem 1, find the maximum permissible ESR if the total output ripple must not exceed 60 mV, allocating half the budget to each term.
A boost converter takes 15 V to 45 V at 3 A output, switching at 60 kHz. Find \( D \), the inductor current, the inductance for 20% ripple, and the capacitance for 2% output ripple.
A buck–boost produces −24 V at 1.5 A from 12 V at 80 kHz. Find \( D \), the inductor current, the device voltage stress, and the inductance for 40% ripple.
Show by volt-second balance that a converter with \( v_{L,on} = V_s - V_o \) and \( v_{L,off} = -V_o \) must have \( V_o = DV_s \), and state which topology it is.
A buck converter designed for 5 A is operated at 300 mA. With \( L = 22\ \mu\text{H} \), \( V_s = 12 \) V, \( V_o = 5 \) V, \( f = 250 \) kHz, determine whether it is in CCM or DCM, and describe what the control loop must do differently.
A boost converter's output capacitor is specified as 470 μF, 3 A RMS ripple. At \( V_s = 200 \) V, \( V_o = 400 \) V, \( I_o = 2 \) A, verify whether the ripple-current rating is adequate. (The capacitor's RMS current is approximately \( I_o\sqrt{D/(1-D)} \).)
Compare the total switch VA stress (blocking voltage times peak current) of a buck, a boost and a buck–boost, each delivering 100 W from 24 V to 48 V where possible. Comment on which topologies can meet the requirement at all.
A synchronous buck's controller offers forced-CCM and diode-emulation modes. For (a) a laptop CPU rail and (b) a battery-powered sensor that sleeps 99% of the time, choose a mode for each and justify it.
A designer proposes running a boost converter at \( D = 0.85 \) to obtain 200 V from 30 V. Using the resistive-loss gain expression with \( r_L = 0.2\ \Omega \) and \( R = 400\ \Omega \), find the actual gain and comment.
A 100 W converter must accept 9–36 V and produce a regulated 24 V. Compare a buck–boost, a four-switch buck–boost, and a boost-followed-by-buck cascade on part count, efficiency, device stress and control complexity. Recommend one.
A multiphase buck uses four phases at 500 kHz each, staggered by 90°. Explain why the input capacitor's RMS ripple current is far lower than four times a single phase's, identify the duty ratio at which the cancellation is perfect, and state what happens to the effective output ripple frequency.