Fundamentals of power electronicsUni Siegen · CC BY 4.0
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Current ripple
The switch is on for a fraction D of each period, so the output averages D·U₁ while the inductor current zig-zags around the load current, widest at D = ½.
Δ𝑖𝐿=𝐷(1−𝐷)𝑇𝑠𝐿𝑈1
The switchreactive
D
L
U2 = D · U₁ = drag D
Δi_L = 0.64 A Ī_L = 2.00 A · 12 V in, 100 kHz
fig 2.1 · step-down converter, ideal switch
No selectionx,y —w×h —100%8 objectsSlide 49 / 361
Current ripple
The switch is on for a fraction D of each period, so the output averages D·U₁ while the inductor current zig-zags around the load current, widest at D = ½.
Δ𝑖𝐿=𝐷(1−𝐷)𝑇𝑠𝐿𝑈1
The switchreactive
D
L
U2 = D · U₁ = drag D
Δi_L = 0.64 A Ī_L = 2.00 A · 12 V in, 100 kHz
fig 2.1 · step-down converter, ideal switch
Adapted from “Fundamentals of power electronics” by O. Wallscheid et al., University of Siegen, CC BY 4.0 (doi:10.5281/zenodo.19662909). Modified by Quodra; not affiliated with or endorsed by the university. Open this lecture in Quodra →
A Quodra slide is made of live objects. Here is one taken apart: a single parameter drives its equation, its chart and its circuit, so drag D and every layer follows.
One slide, taken apartUni Siegen · slide 49 · Current ripple
Ī_L = 2.00 A · 12 V in, 100 kHz
fig 2.1 · step-down converter, ideal switch
Ī_L = 2.00 A · 12 V in, 100 kHz
fig 2.1 · step-down converter, ideal switch
The switchreactive
D
L
U2 = D · U₁ = drag D
The switchreactive
D
L
U2 = D · U₁ = drag D
The switchreactive
D
L
U2 = D · U₁ = drag D
The switchreactive
D
L
U2 = D · U₁ = drag D
Δ𝑖𝐿=𝐷(1−𝐷)𝑇𝑠𝐿𝑈1
ΔiL = 0.64 A
Δ𝑖𝐿=𝐷(1−𝐷)𝑇𝑠𝐿𝑈1
ΔiL = 0.64 A
Current ripple
The switch is on for a fraction D of each period, so the output averages D·U₁ while the inductor current zig-zags around the load current, widest at D = ½.
Current ripple
The switch is on for a fraction D of each period, so the output averages D·U₁ while the inductor current zig-zags around the load current, widest at D = ½.
TextEquationParameterChartCircuit
The slide “Current ripple” taken apart into five layers above its paper: the text; the ripple equation, Delta i L equals D times one minus D times T s over L, times U 1, with its calculated value; a parameter panel holding the duty cycle D and the inductance L; a chart of the switch voltage and the inductor current; and the step-down converter’s circuit. Threads run from the parameter to the equation, the chart and the circuit, which read D and L.
Current rippleThe switch is on for a fraction D of each period, so the output averages D·U₁ while the inductor current zig-zags around the load current, widest at D = ½.
Δ𝑖𝐿=𝐷(1−𝐷)𝑇𝑠𝐿𝑈1ΔiL = 0.64 A
The switchdrag to try
U2 = ΔiL =
Ī_L = 2.00 A · 12 V in, 100 kHz fig 2.1 · step-down converter, ideal switch
Slide 49 of “Fundamentals of power electronics” by O. Wallscheid et al., University of Siegen, CC BY 4.0, rebuilt by this page; not affiliated with or endorsed by the university.
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Current ripple
The switch is on for a fraction D of each period, so the output averages D·U₁ while the inductor current zig-zags around the load current, widest at D = ½.
Δ𝑖𝐿=𝐷(1−𝐷)𝑇𝑠𝐿𝑈1
The switchreactive
D
L
U2 = D · U₁ = drag D
Δi_L = 0.64 A Ī_L = 2.00 A · 12 V in, 100 kHz
fig 2.1 · step-down converter, ideal switch
Presenter viewFundamentals of power electronics
12:09ElapsedSlide 49 / 361
End
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Now presenting — audience sees this
Current ripple
The switch is on for a fraction D of each period, so the output averages D·U₁ while the inductor current zig-zags around the load current, widest at D = ½.
Δ𝑖𝐿=𝐷(1−𝐷)𝑇𝑠𝐿𝑈1
The switchreactive
D
L
U2 = D · U₁ = drag D
Δi_L = 0.64 A Ī_L = 2.00 A · 12 V in, 100 kHz
fig 2.1 · step-down converter, ideal switch
Up next · 50Current ripple (cont.)
Current ripple (cont.)Δ𝐼𝐿 = Δ𝑖𝐿 ⁄ 2√3the RMS deviation from 𝑖̄𝐿
Speaker notes
Widest ripple at D = ½. Ask the room why, then drag D down and watch U₂ and the ripple follow.
?
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Current ripple
The switch is on for a fraction D of each period, so the output averages D·U₁ while the inductor current zig-zags around the load current, widest at D = ½.
Text
Δ𝑖𝐿=𝐷(1−𝐷)𝑇𝑠𝐿𝑈1Equation
The switchreactive
D
L
U2 = D · U₁ = drag D
Parameter
fig 2.1 · step-down converter, ideal switch
Circuit
New
Ask Quodra
Proposal · 4 new objectsChecks passed
CheckedCalculatesBoundFitsClear
TextThe switch is on for…Drafted
EquationCurrent rippleComputed
ParameterThe switch · D, LDrafted
CircuitStep-down converterComputed
Applies as one undo step
Slide 49 of 361 · Content · Deck design · Deck · 361 slidesYou: Explain the current ripple here: its equation, D and L as parameters, and the circuit.
An example of Ask Quodra docked beside a slide. Asked to explain the current ripple on slide 49 of a power electronics lecture, it first asks who the slide is for. For students it proposes 4 new objects: a text, the ripple equation, a parameter panel for the duty cycle D and the inductance L, and the step-down converter circuit, each tagged on the slide with its checks passed. Nothing on the slide changes until you apply, and the proposal applies as one undo step. The second example builds a six-slide tutorial: the plan comes first, then the slides land one by one, each its own undo step.
Proposal: 4 new objects, text, equation, parameter and circuit, tagged on the slide. Checks passed. Nothing on the slide changes until you apply.
An example played by this page; nothing here calls a model. Ask Quodra runs in selected beta workspaces. The slides adapt the University of Siegen’s “Fundamentals of power electronics” (CC BY 4.0), rebuilt by this page. Ask Quodra on the Product page →
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