Why the inductor current decides everything
A switching converter moves energy in small packets. Whether the inductor current ever falls back to zero changes the equations, the ripple, and how the output reacts to the load. In this module you will watch that happen, then use it to choose an inductor, a capacitor, and a switching frequency.
By the end you can
1. Trace the current path in each switch state, then derive the buck, boost, and buck-boost voltage gains from inductor volt-second balance.
2. Decide from L, f, R, and D whether a converter is in continuous (CCM) or discontinuous (DCM) conduction.
3. Show how lightening the load pushes a converter from CCM into DCM, and what that does to Vo and D.
4. Pick L and C, including the capacitor’s ESR, to meet current-ripple and voltage-ripple limits.
5. Explain why the switching frequency is a trade-off, not a “higher is better” knob.
Three rules for any steady-state converter
Every result in this module comes from three statements that hold once the circuit has settled into a repeating pattern:
Inductor volt-second balance. The average voltage across the inductor over one period is zero, so the current ends each period where it started.
Capacitor charge balance. The average capacitor current is zero, so the average inductor (or diode) current equals the load current.
Power balance. With ideal parts, the average power from the source equals the power in the load.
The three converters
| Converter | Output | CCM gain | Boundary Kcrit | DCM gain |
|---|---|---|---|---|
| Buck | Lower than Vs | Vo = D·Vs | 1 − D | Vo/Vs = 21 + √1 + 4KD2 |
| Boost | Higher than Vs | Vo = Vs1 − D | D(1 − D)2 | Vo/Vs = 1 + √1 + 4D2K2 |
| Buck-boost | Higher or lower, polarity reversed | Vo = −VsD1 − D | (1 − D)2 | |Vo|/Vs = D√K |
One number decides CCM or DCM
Define K = 2LRT = 2LfR. It compares the “stiffness” of the inductor (L·f) with the load resistance. A large K means the inductor current has a large dc level to ride on. If K is at least Kcrit from the table, the inductor current stays positive (CCM). If K is smaller, the current hits zero (DCM).
Hart’s text writes the same boundary as a minimum product (L·f)min, for example (1 − D)R/2 for the buck. Dividing by f gives Lmin.
How to use this module
Each tab has a short explanation, a live lab, and activities. Move any slider and every number and graph updates. Activities with a number to find have a Load these settings button so everyone starts from the same circuit, and a Check button that compares your answer with the model. Open-response prompts are kept for you to download on the last tab. Nothing is sent anywhere; your answers stay in this page.
Suggested path: how the circuits work (follow the current), then CCM, then DCM, then the load tab where you push a converter from one to the other, then components, frequency, and the design challenges.
How the circuits work
Before the equations, see where the current goes, why an inductor current ramps, and how the duty ratio sets the output. Everything on this page responds to your clicks and sliders.
Why switch at all?
Suppose 12 V must become 5 V for a 1 A load. A resistor or a linear regulator drops the extra 7 V while carrying the full 1 A, so the difference becomes heat. A switch behaves differently: closed, it drops almost no voltage; open, it carries almost no current. In both states very little power is lost. Try it:
Where does the power go?
The switching converter’s 90 % efficiency is a typical assumed value, not a measurement. The linear regulator’s efficiency follows from the voltages alone: η = Vout/Vin.
The inductor is the key part
An inductor’s voltage is set by how fast its current changes:
A constant voltage across an inductor makes its current change at a constant rate, which is why every inductor current in this module is made of straight ramps. An inductor also resists sudden changes in current. If a switch suddenly interrupts the current, the inductor’s voltage swings until the current has somewhere to go. That is why every converter has a diode (or a second switch) that gives the inductor current a path when the main switch opens.
Apply a voltage, watch the current
The current starts at 0 A. Choose a voltage across the inductor, the inductance, and how long the voltage is applied. A negative voltage makes the current ramp the other way.
The thin lines show all four inductances for the same voltage. The red line is your selection. A larger inductance gives a flatter ramp.
One idea powers all three converters
When a converter has settled into steady operation, the inductor current at the end of each cycle equals the current at the start. The inductor’s average voltage must therefore be zero, or its current would climb or fall a little more every cycle. With voltage Von while the switch is on for a fraction D of the period T, and Voff while it is off:
This is volt-second balance: the positive and negative areas of the inductor voltage are equal. A matching rule applies to the output capacitor: in steady state its average current is zero, so the average inductor current delivers the load.
| Part | Its job |
|---|---|
| Switch | Connects and disconnects the source. Controlled by the duty ratio D. |
| Inductor | Stores energy while current rises and releases it while current falls. Keeps current flowing when the switch opens. |
| Diode | A one-way valve. Gives the inductor current a path when the switch opens, and stops the output from draining backward. |
| Capacitor | Holds the output voltage steady and supplies the load whenever the inductor cannot. |
Follow the current in each converter
A buck converter makes a lower dc voltage from a higher one. The inductor sits in series with the output.
How it works
ONThe switch closes. The source pushes current through the inductor to the load and output capacitor. The inductor voltage is Vs − Vo, which is positive, so the inductor current rises. The diode is reverse-biased and carries nothing.
OFFThe switch opens. The inductor keeps the current flowing, and the current now returns through the diode. The inductor voltage is about −Vo, so the current falls. The source is disconnected.
Where the gain comes from
The output is always lower than the input, because D cannot exceed 1. Since the inductor is in series with the load, the current reaching the load is smooth. The input current is not smooth: it flows only while the switch is on.
A boost converter makes a higher dc voltage from a lower one. The inductor sits in series with the input.
How it works
ONThe switch closes and connects the inductor’s far end to ground. The full input voltage appears across the inductor, so its current rises and it stores energy. The diode is reverse-biased by the output voltage, so the output capacitor alone supplies the load.
OFFThe switch opens. The inductor keeps its current flowing, and the only path is through the diode into the output. Because the output is higher than the input, the inductor voltage Vs − Vo is negative, so the current falls. The input and the inductor’s stored energy together recharge the capacitor and feed the load.
Where the gain comes from
The output is always higher than the input. As D approaches 1 the ideal gain grows without bound, but real parts limit it long before that (see the dashed curve). The input current is smooth because the inductor is in series with the source. The output current is not: it flows only while the diode conducts.
No free energy
A higher output voltage does not create power. In an ideal converter VsIs = VoIo, so if the output voltage is twice the input voltage, the input current is twice the output current. The boost converter trades current for voltage.
The inverting buck-boost converter can make an output whose magnitude is lower or higher than the input. The price is that the output is negative.
How it works
ONThe switch closes and the full input voltage appears across the inductor, whose other end goes to ground. The current rises and the inductor stores energy. The diode is reverse-biased, so the output capacitor alone supplies the load.
OFFThe switch opens. The inductor keeps its current flowing, but now that current is drawn from the ground rail, up through the load, and then through the diode back to the inductor. Current flowing up through the load makes the output node negative with respect to ground. The inductor voltage equals the (negative) output voltage, so the current falls.
Where the gain comes from
With D below 0.5 the output magnitude is smaller than the input. At D = 0.5 the magnitudes are equal. With D above 0.5 the magnitude is larger. Both the input current and the output current are pulsed, which is why this converter usually needs good capacitors at both ports.
Compare the three
| Property | Buck | Boost | Buck-boost (inverting) |
|---|---|---|---|
| Output compared with input | Lower | Higher | Lower or higher |
| Output polarity | Same as input | Same as input | Inverted |
| Inductor placed | In series with the output | In series with the input | Across the switch node and ground |
| Input current | Pulsed | Smooth | Pulsed |
| Output current | Smooth | Pulsed | Pulsed |
| Voltage across the OFF switch | Vs | Vo | Vs + |Vo| |
| Same four parts | Switch, diode, inductor and capacitor. Only their positions change. | ||
Worked examples: change the answer and see what happens
Each example opens on the correct answer. Move any slider under “Your answer” to see what would change: the checklist shows what passes, and a grey dashed line on every graph marks the correct answer. Sliders under “Change the givens” change the problem itself, and the correct answer updates to match. Use “Ask a question” on any example.
Activities
Continuous conduction mode
In CCM the inductor current never reaches zero. The inductor spends part of each period charging and part discharging, and the output depends only on the duty ratio.
Two circuits in every period
Switch closed (time DT). The diode is off and the inductor sees a positive voltage, so its current ramps up.
Switch open (time (1 − D)T). The diode turns on, the inductor sees −Vo, and the current ramps down.
In steady state the two changes cancel, which gives the gain:
What you need for design
Two circuits in every period
Switch closed (DT). The source is across the inductor, the diode is off, and the capacitor alone supplies the load.
Switch open ((1 − D)T). The inductor current cannot jump, so the diode conducts and the inductor pushes energy into the output.
What you need for design
The inductor carries more than the load current: IL = Io/(1 − D). As D approaches 1 the ideal gain goes to infinity, but real inductor resistance limits it (Hart, Sec. 6.5).
Two circuits in every period
Switch closed (DT). The source charges the inductor; the capacitor supplies the load.
Switch open ((1 − D)T). The inductor current finds its path through the diode into the output, which makes the output negative.
D < 0.5 steps down, D > 0.5 steps up.
What you need for design
Lab: continuous conduction
The sliders start at a CCM design. Change one thing at a time and watch the readouts. The shaded bands show which part of the cycle you are in, using the same colors as the circuit explorer: red = switch on, blue = diode on, green = idle (only in DCM).
Worked examples: change the answer and see what happens
Each example opens on the correct answer. Move any slider under “Your answer” to see what would change: the checklist shows what passes, and a grey dashed line on every graph marks the correct answer. Sliders under “Change the givens” change the problem itself, and the correct answer updates to match. Use “Ask a question” on any example.
Activities
Discontinuous conduction mode
When the average current is too small for the ripple, the inductor current runs out before the period ends. The diode blocks it from going negative, and the converter gains a third state in which both the switch and the diode are off.
Three intervals instead of two
Interval 1 (time DT): switch on, current rises from zero to Imax = (slope)·DT. Interval 2 (time D1T): switch off, diode on, current falls to zero. Interval 3 (time (1 − D − D1)T): both off, inductor current is zero, and the capacitor alone feeds the load.
Two conditions fix the output: the average inductor voltage is still zero over all three intervals, and the average diode current must equal the load current. The second one is new, and it is what brings the load into the gain equation.
Buck in DCM
This is the same as Hart’s Eq. (6-96), written with K. Boundary with CCM: K = 1 − D (equivalently D1 = 1 − D). Notice that Vo now depends on the load through K, and Vo > D·Vs.
Boost in DCM
Same as Hart’s Eq. (6-102). Boundary with CCM: K = D(1 − D)2. As the load lightens, K falls and Vo climbs well above Vs/(1 − D). Without feedback, a boost converter at light load can reach dangerous voltages.
Buck-boost in DCM: your turn to derive it
Follow the same two conditions. (1) Volt-second balance over the three intervals gives D1 in terms of Vs, D, and |Vo|. (2) The average diode current is ½ImaxD1 and must equal |Vo|/R, with Imax = VsDT/L. (3) Eliminate D1 and solve for |Vo|.
Reveal the result and check your work
Setting this equal to the CCM result VsD/(1 − D) gives the boundary K = (1 − D)2. The lab on this page uses this result, so you can test your derivation numerically.
What changes when you cross the boundary
| Quantity | CCM | DCM |
|---|---|---|
| Gain Vo/Vs | Depends only on D | Depends on D and K, so on the load, L, and f |
| Inductor current | Never zero; Imin > 0 | Starts each period at zero; peak = VsDT/L (boost, buck-boost) |
| Switch turn-on | At a nonzero current | At zero current, so less turn-on loss |
| Control | Simple: D sets Vo | Feedback must change D as the load changes |
Lab: discontinuous conduction
The sliders start at a light-load, DCM setting of the same circuit you used in CCM. Compare the shaded “idle” band with the CCM tab, and compare the output voltage with the CCM equation (shown under the Vo readout).
Worked examples: change the answer and see what happens
Each example opens on the correct answer. Move any slider under “Your answer” to see what would change: the checklist shows what passes, and a grey dashed line on every graph marks the correct answer. Sliders under “Change the givens” change the problem itself, and the correct answer updates to match. Use “Ask a question” on any example.
Activities
How a lighter load pushes a converter into DCM
The ripple in the inductor current depends on Vs, D, L, and f. It does not depend on the load. The average current does. So when the load current falls, the average slides down while the ripple stays put, until the bottom of the ripple touches zero.
The load current at which this just fails is the critical load current. In terms of the circuit, it is Io,crit = Vo/Rcrit, with Rcrit = 2Lf/Kcrit. Loads lighter than that (R larger than Rcrit) are in DCM.
The lab sets the load as a percentage of full load. Drag the load down, or press Play to lighten it automatically. Switch between fixed duty ratio (no feedback) and regulated output (a controller adjusts D to hold Vo) and compare the two graphs on the right.
Worked examples: change the answer and see what happens
Each example opens on the correct answer. Move any slider under “Your answer” to see what would change: the checklist shows what passes, and a grey dashed line on every graph marks the correct answer. Sliders under “Change the givens” change the problem itself, and the correct answer updates to match. Use “Ask a question” on any example.
Activities
Choosing the inductor and the capacitor
The inductor sets the current ripple and keeps the converter in CCM. The capacitor and its ESR set the voltage ripple. The load decides how much margin you have.
Inductor
A larger L gives a smaller ΔiL (ΔiL ∝ 1/L), lower peak and rms current, and more room before DCM. The cost is a bigger, heavier, more resistive part. The inductor wire must carry the rms current and the core must not saturate at the peak current.
Capacitor
The ideal-capacitor ripple falls as 1/C (boost, buck-boost) or 1/(LC) (buck). A real capacitor adds ΔVo,ESR ≈ ΔiC·rC. Often the ESR term is larger, and then a bigger capacitance barely helps: you need a lower-ESR part.
The lab starts with a 100 mΩ ESR so you can see the effect. The two sweep graphs show where your current choice sits: the left graph varies L (watch for the point where Imin reaches zero), the right varies C with the ESR held where you set it.
Worked examples: change the answer and see what happens
Each example opens on the correct answer. Move any slider under “Your answer” to see what would change: the checklist shows what passes, and a grey dashed line on every graph marks the correct answer. Sliders under “Change the givens” change the problem itself, and the correct answer updates to match. Use “Ask a question” on any example.
Activities
Choosing the switching frequency
Raising f shrinks the inductor and capacitor you need, but every switching event costs energy. The right frequency is where the two pressures balance.
What f does to the circuit
With L and C fixed, K = 2Lf/R rises with f, so a higher frequency moves the converter toward CCM. In CCM, ΔiL ∝ 1/f, and the voltage ripple falls as 1/f (boost, buck-boost) or 1/f² (buck).
What f costs
The switch dissipates energy at each transition, so switching loss grows in proportion to f. Gate-drive loss also grows with f. Conduction loss does not. Typical designs stay above about 20 kHz to avoid audible noise and rarely go above a few hundred kilohertz unless size matters more than efficiency.
The lab has two parts. First, keep L and C fixed and sweep f to see the converter cross into DCM. Second, set a ripple specification and see how large the inductor and capacitor must be at each frequency, next to the loss budget. The loss model is a simple educational estimate: linear switching transitions of 60 ns, a 30 mΩ MOSFET, a 0.4 V diode, and 20 mΩ of inductor resistance.
Worked examples: change the answer and see what happens
Each example opens on the correct answer. Move any slider under “Your answer” to see what would change: the checklist shows what passes, and a grey dashed line on every graph marks the correct answer. Sliders under “Change the givens” change the problem itself, and the correct answer updates to match. Use “Ask a question” on any example.
Activities
Design challenges
Choose f, L, and a capacitor from the parts list. Each design is tested at the lightest and the heaviest load, with the duty ratio adjusted by feedback to hold the output voltage. A design passes when every check passes.
Check your understanding
Eight questions on the ideas from this module, then reflection prompts and a downloadable copy of your responses.
Reflect
Your questions for your instructor
Questions the example panels could not answer are saved here. Copy them to send to your instructor, or download them with the rest of your responses.
Save your work
This page does not store or send anything. To keep a record, or to turn it in if your instructor asks, download a text file of every answer you have entered.
Formula reference
Ideal components, steady state. K = 2L/(RT) = 2Lf/R. For the buck-boost, Vo is negative; |Vo| is shown.
Continuous conduction
| Buck | Boost | Buck-boost | |
|---|---|---|---|
| Gain |Vo|/Vs | D | 11 − D | D1 − D |
| Average inductor current IL | Io | Io1 − D | Io1 − D |
| Current ripple ΔiL | Vo(1 − D)Lf | VsDLf | VsDLf |
| Imax, Imin | IL ± ΔiL/2 | ||
| Minimum L for CCM | (1 − D)R2f | D(1 − D)2R2f | (1 − D)2R2f |
| Kcrit (CCM if K ≥ Kcrit) | 1 − D | D(1 − D)2 | (1 − D)2 |
| Ripple ΔVo/Vo (ideal C) | 1 − D8LCf2 | DRCf | DRCf |
| ESR ripple | ΔiL·rC | IL,max·rC | IL,max·rC |
Discontinuous conduction
| Buck | Boost | Buck-boost | |
|---|---|---|---|
| Gain |Vo|/Vs | 21 + √1 + 4KD2 | 1 + √1 + 4D2K2 | D√K |
| Peak current Imax | (Vs − Vo)DTL | VsDTL | VsDTL |
| Diode on-time D1 | (Vs − Vo)DVo | VsDVo − Vs | VsD|Vo| |
| Duty ratio to hold Vo = M·Vs | M√K1 − M | √K·M(M − 1) | M√K |