ISEGORIA / MATH ENCYCLOPEDIA
Switching power converters: chopping to convert
Every phone charger, laptop brick and processor voltage regulator changes one DC voltage into another without wasting the difference as heat. It does so by switching hundreds of thousands of times a second and letting an inductor and a capacitor average the chopped waveform. Volt-second balance on the inductor gives the conversion ratio in one line; the inductor’s own resistance puts a ceiling on how far a boost converter can step up; and at light load the current runs dry each cycle, which changes the rules.
Before you begin: Circuits with inductors and capacitors, and time averages
Predict, manipulate, then check your reasoning against the example and question. Graphs illustrate the mathematics; they do not replace a proof.
1. The buck converter
A switch connects the input to an inductor for a fraction D of each period and disconnects it for the rest, while a diode, or a second switch, gives the inductor’s current a path to keep flowing. The switch node therefore sees a square wave between V_in and zero, whose average is D·V_in, and the inductor and capacitor form a low-pass filter that passes that average and blocks the switching. The inductor current is the clue: it ramps up at (V_in − V)/L while the switch is on and down at V/L while it is off, and in steady state it must end each period where it began. Watch the current path change with the switch, drag the end of the on time to change D, and make the load light enough for the current to run dry.
Worked example. Step 12 V down to 5 V at 200 kHz with \(D=0.42\): the output is \(0.42\times12=5.04\) V. With \(L=10\) µH the current ripple is \((12-5.04)\times0.42/(10\,\mu\mathrm H\times200\,\mathrm{kHz})=1.46\) A, so into a 2 Ω load the current swings between 1.79 and 3.25 A about its average of 2.52 A. A 47 µF capacitor holds the output ripple to \(1.46/(8\times200\,\mathrm{kHz}\times47\,\mu\mathrm F)=19.4\) mV. The current stays continuous as long as the load draws more than half the ripple, 0.73 A, that is for any load below 6.9 Ω.
Watch out. The switches and diode are ideal here: no on-resistance, no forward drop, no switching time, so the efficiency is 100% and the steady state is exact. In a real converter the diode’s 0.4 to 0.7 V forward drop is a large loss at low output voltage, which is why modern converters use a synchronous second switch; and the duty cycle is not set by hand but by a feedback loop that measures the output and adjusts D. The ripple formula for the capacitor assumes all the current ripple flows into it and ignores its series resistance, which in practice often dominates.
Why must the inductor current end each period exactly where it started?
Because otherwise it would drift by the same amount every period and grow without limit, which is not a steady state. The change over a period is \(\frac1L\int_0^{T_s}v_L\,dt\), so the inductor’s average voltage must be zero: the volt-seconds it gains while on, \((V_{in}-V)DT_s\), must equal those it loses while off, \(V(1-D)T_s\). The same argument applied to the capacitor’s charge says its average current is zero, so the inductor’s average current equals the load current.
2. Buck, boost, buck-boost, and the limit on gain
Rearranging the same three parts, a switch, a diode and an inductor, gives converters that step down, step up, or invert. Each ratio follows from volt-second balance: the boost’s inductor sees V_in while the switch is on and V_in − V while it is off, so V = V_in/(1 − D), which seems to promise any gain as D approaches 1. It does not, because the inductor’s winding resistance R_L carries the input current, which is the load current divided by 1 − D, so its loss grows as 1/(1 − D)². Past a certain duty cycle more switching produces less output. Choose a converter, set the winding resistance as a percentage of the load, and drag the operating point along the curve.
Worked example. Boost a 3.7 V lithium cell to 12 V. The ideal formula asks for \(D=1-3.7/12=0.692\), but with \(R_L/R=2\%\) that duty cycle gives only \(M=2.68\), 9.9 V, at 82.6% efficiency. Reaching \(M=12/3.7=3.24\) takes \(D=0.784\), where the efficiency has fallen to 69.9%, and no duty cycle at all gives more than \(1/(2\sqrt{0.02})=3.54\), reached at \(D=1-\sqrt{0.02}=0.859\). A designer who needs a large step-up uses a transformer or a multi-stage converter instead.
Watch out. Only one loss is modelled, the inductor’s winding resistance, with everything else ideal; real switch on-resistance, diode drop and switching loss add similar terms and lower the curves further. All three ratios assume continuous conduction. The buck-boost’s output is negative relative to the input’s ground, and its gain curve is drawn below the axis.
Why does the buck converter’s efficiency not depend on the duty cycle, while the boost’s collapses near D = 1?
Because of where the inductor sits. In the buck it is in series with the load, so it always carries exactly the load current \(I\) and dissipates \(I^2R_L\), a fixed fraction of the output power. In the boost it is in series with the input, and power balance forces the input current up to \(I/(1-D)\); the loss \(I^2R_L/(1-D)^2\) therefore grows without bound as \(D\to1\), while the power delivered cannot.
3. When the current runs dry
With a diode as the freewheeling path, the inductor current cannot reverse. If the load is light, the current falls to zero before the period ends and stays there: the diode blocks, nothing flows, and the capacitor alone feeds the load until the switch turns on again. This is discontinuous conduction. Volt-second balance still holds, but over D and the shorter falling interval D₂ instead of 1 − D, and the output now depends on the load as well as on D. One dimensionless number decides which regime you are in, K = 2L/(R·T_s), which compares the inductor’s time constant with the switching period: the boundary is K = 1 − D. Drag the operating point across the boundary.
Worked example. Take the 12 V buck at 200 kHz with \(D=0.42\), but now \(L=2\) µH and a light 20 Ω load: \(K=2\times2\,\mu\mathrm H/(20\,\Omega\times5\,\mu\mathrm s)=0.04\), far below \(1-D=0.58\). The output is \(M=2/(1+\sqrt{1+0.16/0.1764})=0.840\), so 10.08 V instead of 5.04 V. The current peaks at \((1-M)D\,V_{in}T_s/L=2.02\) A, falls back to zero after a further \(D_2=0.08\) of the period, and then sits at zero for the remaining 50%.
Watch out. The formulas assume the output ripple is small, so the output voltage is effectively constant within a period, and ideal components. A synchronous converter, whose second switch conducts in both directions, never enters this mode: its current simply goes negative at light load and \(M=D\) holds, at the cost of circulating current. Many controllers deliberately switch to discontinuous or burst operation at light load because it reduces switching losses.
Why does the output voltage rise above D·V_in when the load gets lighter?
In discontinuous conduction the inductor delivers a fixed packet of energy each period, set by the on-time and the input-to-output voltage, rather than a steady current. A lighter load draws less, so the capacitor charges until the falling interval \(D_2\) has shrunk enough that the volt-seconds balance again: \((V_{in}-V)D=VD_2\) with a small \(D_2\) means \(V\) close to \(V_{in}\). That is why a converter left in this mode needs feedback to regulate its output.