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Power Electronics Buck & Boost Converters, Rectifiers & Online Tutoring

Power electronics is the use of switching devices to convert electrical power efficiently: AC to DC, DC to DC, and DC to AC. Most exam and design questions come down to a few calculations: duty cycle and ripple in buck and boost converters, output voltage against firing angle in controlled rectifiers, and the output of an inverter. This page works each one with numbers. FLN Tutors (Freelancer Network) also offers one-to-one help. Matched to a verified specialist who knows your course, your university's marking criteria, and your deadline.

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FLN offers one-to-one online Power electronics tutoring with verified subject specialists. Sessions focus on calculations, design problems and the theory behind them, including Power Diodes, Rectifiers, DC-DC Converters and Inverters, using your own course material, so you understand the subject and can apply it yourself.

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What power electronics covers

Power electronics switches semiconductors fully on and fully off, thousands of times a second, instead of letting them dissipate power as a linear regulator does. That is why a phone charger can be small and cool while delivering tens of watts.

ConversionCircuit familyExample
AC to DCRectifiers, uncontrolled or controlledBattery chargers, DC motor drives, HVDC links
DC to DCChoppers and switch-mode converters: buck, boost, buck-boostLaptop and phone power supplies, solar charge controllers
DC to ACInvertersSolar inverters, UPS systems, variable-speed motor drives
AC to ACAC voltage controllers, cycloconvertersLight dimmers, soft starters

The idea to state in every answer: an ideal switch has either zero voltage across it or zero current through it, so it dissipates no power. Real devices lose energy mainly while switching, which is why switching frequency and device choice matter so much.

For diode rectifier basics and ripple on a smoothing capacitor, see the analog electronics page; this page concentrates on controlled conversion.

Choosing a switching device

DeviceTurned on byTurned off byTypical strength
Power diodeForward voltageReverse voltageUncontrolled rectification, freewheeling paths
Thyristor (SCR)A gate pulse while forward biasedCurrent falling below the holding currentVery high voltage and current at mains frequency
Power MOSFETGate voltageRemoving gate voltageVery fast switching at low to medium voltage
IGBTGate voltageRemoving gate voltageMedium-to-high voltage at medium switching frequency

The thyristor's catch is that the gate can turn it on but cannot turn it off. It stays on until the current through it falls below the holding current, which happens naturally at each zero crossing of an AC supply. That is why thyristors suit AC-to-DC rectifiers and why DC choppers use MOSFETs or IGBTs instead.

Two other thyristor terms examiners ask about: the latching current is the minimum current needed to keep it on just after the gate pulse is removed, and it is larger than the holding current.

Worked example: designing a buck converter

A buck converter steps DC voltage down. In continuous conduction the ideal output is Vo = D × Vin, where D is the duty cycle, the fraction of each switching period the switch is on.

The brief. Step 24 V down to 12 V at 2 A, switching at 100 kHz, with inductor ripple current no more than 30% of the load current.

  1. Duty cycle: D = Vo ÷ Vin = 12 ÷ 24 = 0.5
  2. Allowed ripple current: ΔI = 0.3 × 2 = 0.6 A
  3. Inductance: L = (Vin − Vo) × D ÷ (ΔI × f) = (24 − 12) × 0.5 ÷ (0.6 × 100,000) = 100 µH
  4. Output voltage ripple with C = 47 µF: ΔV = ΔI ÷ (8 × f × C) = 0.6 ÷ (8 × 100,000 × 47 × 10⁻⁶) = 16 mV

What the numbers tell you. Doubling the switching frequency halves the inductance needed, which is why modern converters switch fast. The inductor must also be rated for the peak current, Io + ΔI/2 = 2 + 0.3 = 2.3 A, not just the average, or it will saturate.

The assumption to state: these formulas assume continuous conduction, meaning the inductor current never falls to zero. At very light loads it does, the converter enters discontinuous mode, and Vo = D × Vin no longer holds.

Worked example: a boost converter

A boost converter steps DC voltage up. In continuous conduction the ideal output is Vo = Vin ÷ (1 − D).

The brief. Raise 12 V to 24 V for a 1 A load, at 100 kHz with a 100 µH inductor.

  1. Duty cycle: D = 1 − Vin ÷ Vo = 1 − 12 ÷ 24 = 0.5
  2. Input current (ideal, lossless): power in equals power out, so Iin = Io ÷ (1 − D) = 1 ÷ 0.5 = 2 A
  3. Inductor ripple current: ΔI = Vin × D ÷ (L × f) = 12 × 0.5 ÷ (100 × 10⁻⁶ × 100,000) = 0.6 A

The trap. As D approaches 1, the formula says the output voltage rises without limit. In practice losses in the switch, diode and inductor resistance cap the real gain, and efficiency collapses at high duty cycles. A ratio of more than about four to one is usually done with a transformer-based design instead.

The second trap: the input current is larger than the output current. A boost converter delivering 1 A at 24 V draws 2 A from a 12 V source, before losses, so the source and the inductor must be sized for the input side.

Worked example: controlled rectifiers and firing angle

In a controlled rectifier, each thyristor is fired at a delay angle α after the point where it first becomes forward biased. Increasing α delays conduction and lowers the average DC output.

Supply: 230 V rms, so the peak is Vm = √2 × 230 = 325.27 V.

RectifierAverage output formulaAt α = 0°At α = 30°
Single-phase full-wave, fully controlled, highly inductive loadVdc = (2Vm ÷ π) × cos α207.07 V179.33 V
Single-phase semi-converter, or full-wave with resistive loadVdc = (Vm ÷ π) × (1 + cos α)207.07 V193.2 V

Working the fully controlled case: 2 × 325.27 ÷ π = 207.07 V, and cos 30° = 0.866, so Vdc = 207.07 × 0.866 = 179.33 V.

Why the two formulas differ. With a highly inductive load, current keeps flowing through the thyristors after the supply voltage goes negative, so part of each cycle contributes negative voltage. In a fully controlled bridge at α = 90°, the average output is zero, and beyond 90° it becomes negative, which is how the converter can invert and return energy to the supply. A semi-converter has a freewheeling path that prevents the negative portion, so its output never goes below zero.

The uncontrolled full-wave value of 207.07 V, which is 0.9 × 230 V, is the α = 0° limit of both formulas: a useful check on any answer.

Inverters: DC to AC

An inverter switches a DC supply to produce alternating current. In a single-phase full-bridge, four switches connect the load first one way across the DC link and then the other.

Square-wave output. Switching each pair for half a cycle produces a square wave of amplitude Vdc. Its fundamental component has an rms value of:

V₁ = 4Vdc ÷ (π√2) ≈ 0.9 × Vdc

With Vdc = 200 V, V₁ = 4 × 200 ÷ (π × 1.414) = 180.06 V rms.

Why square waves are not enough. A square wave contains large odd harmonics, the third at one third of the fundamental amplitude, the fifth at one fifth, and so on. Motors overheat and transformers hum. Pulse-width modulation (PWM) solves this by switching many times per cycle with varying pulse widths, so that the average follows a sine wave and the harmonics move to high frequencies, where a small filter removes them.

Dead time. The two switches in one leg of the bridge must never be on together, or they short the DC supply. A short delay, called dead time, is inserted between turning one off and the other on.

Solving a power electronics problem, step by step

  1. Identify the conversion: AC-DC, DC-DC, DC-AC or AC-AC, and the circuit, such as buck, boost or fully controlled bridge.
  2. State the operating assumptions: continuous or discontinuous conduction, ideal or lossy components, resistive or inductive load.
  3. Write the governing relationship before substituting, such as Vo = D × Vin or Vdc = (2Vm ÷ π) cos α.
  4. Convert units first: kHz to Hz, µH to H, µF to F. Most wrong answers are out by a factor of a thousand.
  5. Calculate, then sanity-check: a buck output must be less than its input, a boost output greater, and a rectifier output at α = 0° must equal the uncontrolled value.
  6. Size the components for the peak values, not just averages: peak inductor current, peak reverse voltage on the diode, peak switch voltage.

Where students lose marks

  • Using Vo = D × Vin for a boost converter, or Vin ÷ (1 − D) for a buck.
  • Mixing rms and peak voltage: the rectifier formulas use Vm, the peak, not the rms value.
  • Using the resistive-load rectifier formula for an inductive load, or the reverse.
  • Forgetting that a fully controlled rectifier's output reaches zero at α = 90° and goes negative beyond it.
  • Leaving kHz, µH or µF unconverted, so the answer is wrong by a power of ten.
  • Rating an inductor for the average current instead of the peak.
  • Saying a thyristor can be turned off from its gate.
  • Ignoring the continuous-conduction assumption at light loads.
  • Treating a square-wave inverter output as if it were sinusoidal.
  • Forgetting dead time between switches in the same inverter leg.

See analog electronics help for diode rectifiers, smoothing and amplifier circuits, electrical engineering help for AC power and three-phase calculations, power system help for transmission and HVDC, electrical measurements help for measuring the waveforms these circuits produce, and digital electronics help for the control logic behind PWM.

If a converter design or rectifier calculation will not come out right, send it over with your working and a tutor will show you where it goes wrong.

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Power electronics tutoring: frequently asked questions

What is power electronics?

Power electronics is the branch of electrical engineering that uses semiconductor switches such as diodes, thyristors, MOSFETs and IGBTs to convert and control electrical power efficiently. It covers rectifiers for AC to DC, choppers and switch-mode converters for DC to DC, and inverters for DC to AC conversion.

How does a buck converter work?

A buck converter steps DC voltage down by switching the input on and off rapidly and smoothing the result with an inductor and capacitor. In continuous conduction the average output equals the duty cycle multiplied by the input voltage, so a 50 per cent duty cycle halves the voltage.

How do you calculate the inductor for a buck converter?

Use L equals the input minus output voltage, multiplied by the duty cycle, divided by the allowed ripple current times the switching frequency. Stepping 24 volts to 12 volts at 100 kilohertz with 0.6 amps of ripple needs about 100 microhenries.

How does a boost converter work?

A boost converter stores energy in an inductor while the switch is on, then releases it to the output at a higher voltage when the switch turns off. In continuous conduction the ideal output equals the input divided by one minus the duty cycle, so a duty cycle of 0.5 doubles the voltage.

What is the duty cycle?

The duty cycle is the fraction of each switching period during which the switch is on, written as a decimal or a percentage. It sets the output of switch-mode converters: output voltage is duty cycle times input in a buck converter, and input divided by one minus duty cycle in a boost converter.

What is a thyristor?

A thyristor, or silicon-controlled rectifier, is a four-layer semiconductor switch that turns on when a gate pulse is applied while it is forward biased. It then stays on until the current through it falls below the holding current, so the gate can turn it on but cannot turn it off.

What is the firing angle of a thyristor?

The firing angle is the delay, measured in electrical degrees from the point where the thyristor first becomes forward biased, before its gate is pulsed. Increasing the firing angle delays conduction and reduces the average output voltage of a controlled rectifier, which is how the output is regulated.

What is the output of a fully controlled rectifier?

For a single-phase fully controlled bridge with a highly inductive load, the average output equals two times the peak voltage divided by pi, multiplied by the cosine of the firing angle. With a 230 volt rms supply and a 30 degree firing angle, that gives about 179 volts.

What is the difference between holding current and latching current?

Latching current is the minimum anode current needed to keep a thyristor on immediately after the gate pulse is removed. Holding current is the smaller minimum current below which a thyristor that is already conducting switches off. Latching current is always larger than holding current for the same device.

What is a chopper in power electronics?

A chopper is a DC-to-DC converter that switches a DC supply on and off rapidly to produce a lower or higher average voltage. A step-down chopper is the same circuit as a buck converter, and a step-up chopper corresponds to a boost converter. Choppers are widely used for DC motor speed control.

What is an inverter?

An inverter converts DC to AC by switching the DC supply across the load in alternating directions. A simple square-wave inverter gives a fundamental of about 0.9 times the DC voltage in rms terms, while pulse-width modulation produces a much cleaner, near-sinusoidal output with fewer harmonics.

What is PWM in an inverter?

Pulse-width modulation switches the inverter many times per cycle, varying the width of each pulse so that the average voltage follows a sine wave. This moves the unwanted harmonics to high frequencies, where a small filter removes them, and allows both the output voltage and frequency to be controlled.

What is the difference between a MOSFET and an IGBT?

Both are turned on and off by a gate voltage. Power MOSFETs switch very fast and suit low to medium voltages, as in phone chargers and laptop supplies. IGBTs handle higher voltages and currents at moderate switching frequencies, as in motor drives, electric vehicles and solar inverters.

Why is dead time needed in an inverter?

The two switches in one leg of an inverter bridge must never conduct at the same time, because that would short-circuit the DC supply. Dead time is a short delay inserted between turning one switch off and turning the other on, allowing the first to stop conducting completely.

Can a tutor help with power electronics?

Yes. A one-to-one tutor can work through converter design, rectifier and firing-angle calculations, device selection and simulation results with you, and check where your working goes wrong. You complete your own assessed work and lab reports in line with your college or university's academic rules.

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