Power electronics: inverters and DC-DC converters

How EV power electronics work: switching devices (IGBT, Si and SiC MOSFET), the three-phase PWM traction inverter and its bidirectional power flow, buck, boost and isolated DC-DC converters, and losses and cooling, with converter, inverter-voltage and loss numericals.

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Why it matters

Between the battery and every load in an EV sits power electronics: the traction inverter that drives the motor and recovers braking energy, the DC-DC converter that powers the 12 V system, and the on-board charger. These converters decide how efficiently battery energy becomes wheel torque, and their heat, cost and reliability are major design constraints. The move from silicon IGBTs to silicon-carbide MOSFETs and from 400 V to 800 V systems is changing EV design now.

Key ideas

Switching converters. Power semiconductors are used as fast on/off switches, not as variable resistors; an ideal switch dissipates nothing when fully on or fully off. Average voltages and currents are controlled by pulse-width modulation (PWM): the fraction of each switching period a switch is on is the duty cycle D. Inductors and capacitors filter the switched waveform.

Devices.

  • IGBT: high voltage and current capability with low conduction loss at high current; slower switching, so switching losses limit frequency (typically about 5–20 kHz in traction inverters). Long the standard for 400 V inverters.
  • Si MOSFET: very fast, used at low voltage (48 V systems, 12 V converters).
  • SiC MOSFET: wide-band-gap device with much lower switching loss, higher blocking voltage and higher temperature capability — enables 800 V systems, higher switching frequency and higher efficiency, at higher device cost.
  • Diodes (antiparallel or body diodes) carry current when the switch is off.

Traction inverter.

  • A three-phase, two-level voltage-source inverter: three half-bridge legs (six switches with antiparallel diodes) across a DC-link capacitor (usually a film capacitor) fed by the battery.
  • The controller (field-oriented control) measures phase currents and rotor angle and sets PWM so that each phase gets a sinusoidal current of the required frequency (speed) and amplitude and phase (torque). Sinusoidal PWM compares a sine reference with a triangular carrier; space-vector PWM uses the DC link more fully (about 15 % more output voltage).
  • Dead time: a short delay between turning one switch of a leg off and the other on prevents a shoot-through short across the DC link.
  • Bidirectional power flow: during regenerative braking the motor's back-emf drives current back through the switches and diodes; the inverter then acts as an active rectifier, charging the battery.
  • Also contains gate drivers with isolation, current sensors, DC-link voltage sensing, protection (over-current, over-temperature, desaturation detection), a liquid-cooled baseplate and an active discharge circuit for the DC link.

DC-DC converters.

  • Buck (step-down): output voltage = D × input voltage.
  • Boost (step-up): output = input / (1 − D). Some vehicles boost a lower battery voltage to a higher DC-link voltage for the motor.
  • Buck-boost and bidirectional converters let power flow either way (e.g. between a 48 V battery and 12 V net).
  • The auxiliary converter replaces the alternator: it converts the high-voltage battery (e.g. 300–800 V) to about 14 V for the 12 V network, typically 1–3 kW. It must be isolated (a high-frequency transformer, e.g. phase-shifted full-bridge or LLC) so the high-voltage system stays separated from the chassis-referenced 12 V system.

On-board charger. AC mains → rectifier with power-factor correction (PFC) → isolated DC-DC stage → battery. Covered further in the charging topic.

Losses and cooling. Conduction loss (current × on-state voltage or I²R_DS(on)) and switching loss (energy per switching event × switching frequency). Efficiencies of traction inverters are typically about 95–99 %, but at 100 kW even 2 % is 2 kW of heat; junction temperature must be kept below the device rating, so inverters are liquid cooled.

Formulas

D = t_on / T_s, f_s = 1 / T_s

  • D: duty cycle (dimensionless), t_on: on-time (s), T_s: switching period (s), f_s: switching frequency (Hz).

V_o = D × V_in (ideal buck, continuous conduction)

V_o = V_in / (1 − D) (ideal boost, continuous conduction)

ΔI_L = (V_in − V_o) × D / (L × f_s) (buck inductor peak-to-peak ripple)

  • ΔI_L (A), L: inductance (H).

V_ph,peak = m × V_dc / 2 (sinusoidal PWM, m ≤ 1); V_LL,rms = (√3 / (2√2)) × m × V_dc ≈ 0.612 m V_dc

  • m: modulation index, V_dc: DC-link voltage (V), V_ph,peak: peak fundamental phase voltage (V), V_LL,rms: rms line-to-line fundamental voltage (V).

V_ph,peak,max = V_dc / √3 (space-vector PWM, linear limit)

P_cond = V_on × I (IGBT, instantaneous) or I_rms² × R_DS(on) (MOSFET); P_sw = (E_on + E_off) × f_s

  • Losses (W); E_on, E_off: switching energies per event (J) at the operating current and voltage, from the datasheet.

η = P_out / (P_out + P_loss)

Worked examples

Example 1 (standard). A 48 V to 12 V buck converter in a mild-hybrid vehicle switches at 100 kHz with a 22 µH inductor. It supplies 20 A at 12 V with 95 % efficiency. Find the duty cycle, inductor ripple current, input power and input current.

  1. D = V_o / V_in = 12 / 48 = 0.25.
  2. ΔI_L = (48 − 12) × 0.25 / (22 × 10⁻⁶ × 100 × 10³) = 9 / 2.2 = 4.09 A.
  3. P_out = 12 × 20 = 240 W; P_in = 240 / 0.95 = 252.6 W.
  4. I_in = 252.6 / 48 = 5.26 A.

Answer: D = 0.25, ripple about 4.1 A, 252.6 W, 5.26 A.

Example 2 (GATE level). A traction inverter has a 350 V DC link. (a) With sinusoidal PWM at m = 0.9, find the peak phase voltage and rms line voltage. (b) Find the maximum rms line voltage with space-vector PWM. (c) Delivering 60 kW, each of the six switches has an average conduction loss of 150 W and total switching energy of 6 mJ per carrier cycle at 10 kHz. Find the total loss, efficiency and DC input current.

  1. V_ph,peak = 0.9 × 350 / 2 = 157.5 V.
  2. V_LL,rms = 0.612 × 0.9 × 350 = 192.9 V.
  3. SVPWM: V_ph,peak = 350 / √3 = 202.1 V; V_LL,rms = 202.1 × √3 / √2 = 247.5 V.
  4. Switching loss per switch = 6 × 10⁻³ × 10⁴ = 60 W; per switch total = 150 + 60 = 210 W; inverter = 6 × 210 = 1260 W.
  5. η = 60 / (60 + 1.26) = 0.979; I_dc = 61,260 / 350 = 175 A.

Answer: 157.5 V and 192.9 V; 247.5 V; 1.26 kW loss, 97.9 %, 175 A.

Common mistakes

  • Applying V_o = D × V_in to every DC-DC converter; it holds only for an ideal buck. A boost gives V_in / (1 − D).
  • Thinking the inverter only drives the motor; in regeneration it also rectifies energy back into the battery.
  • Ignoring switching losses, which grow in proportion to switching frequency.
  • Forgetting dead time, or thinking both switches in a leg can be on together.
  • Confusing peak and rms values, and phase and line voltages (a factor of √3 between line and phase, √2 between peak and rms).
  • Assuming the 12 V converter can be non-isolated; the high-voltage system must stay isolated from the chassis.

For GATE ME

Power electronics is not a core GATE ME subject, but its numericals use basic ideas that are: energy balance and efficiency, duty-cycle averaging, average power, and phase/line and peak/rms relations in three-phase systems. Practise buck and boost relations, efficiency chains and loss calculations.

Quick check

  1. What is the output of an ideal boost converter with 200 V input and D = 0.5?
  2. Why is dead time needed in an inverter leg?
  3. Why are SiC MOSFETs attractive for 800 V EVs?
  4. An inverter with 98 % efficiency delivers 80 kW. How much heat does it produce?
  5. Why must the auxiliary DC-DC converter be isolated?

Answers: 1. 400 V. 2. To prevent both switches conducting at once and shorting the DC link (shoot-through). 3. Lower switching losses, higher voltage and temperature capability, giving higher efficiency and switching frequency. 4. 80 / 0.98 − 80 = 1.63 kW. 5. To keep the high-voltage battery separated from the chassis-referenced 12 V system for safety.

Interview questions

All Automotive Electronics and Electric Vehicles interview questions

Try answering each one aloud before you open it.

  1. 1.What is an inverter in the context of electric vehicles?Concept

    An inverter in electric vehicles is a power electronic device that converts direct current (DC) from the battery into alternating current (AC) to drive the electric motor. It plays a crucial role in controlling the speed and torque of the motor by adjusting the frequency and amplitude of the AC output.

  2. 2.Explain the function of a DC-DC converter in electric vehicles.Concept

    A DC-DC converter in electric vehicles is used to convert the high voltage from the main battery pack to a lower voltage suitable for auxiliary systems, such as lighting, infotainment, and control systems. It ensures that these systems receive a stable and appropriate voltage level, improving efficiency and safety.

  3. 3.Why are IGBTs and SiC MOSFETs used in EV traction inverters?Application

    A traction inverter must switch hundreds of amperes at 400–800 V several thousand times a second with low loss. IGBTs combine high voltage and current capability with low conduction loss at high current, which made them the standard for 400 V inverters, but their switching losses limit frequency to roughly 5–20 kHz. Silicon-carbide MOSFETs switch much faster with far lower switching loss, block higher voltages and tolerate higher temperatures, so they are increasingly used, especially in 800 V vehicles, giving higher efficiency and smaller cooling despite higher device cost.

  4. 4.What happens if a DC-DC converter fails in an electric vehicle?Application

    If a DC-DC converter fails in an electric vehicle, the auxiliary systems that rely on the lower voltage supply may stop functioning. This can lead to the failure of critical systems like lighting and control units, potentially compromising vehicle safety and operation.

  5. 5.What does the inverter do during regenerative braking?Application

    During regeneration the controller commands negative torque, so the motor acts as a generator driven by the wheels. The inverter's switches, with their antiparallel diodes, now act as an active rectifier: they shape the phase currents so power flows from the motor's AC terminals into the DC link and on into the battery. The same field-oriented control sets the braking torque. The BMS limits regenerative power when the battery is full or cold, and the friction brakes provide the rest.

  6. 6.What are the main differences between a buck converter and a boost converter?Concept

    A buck converter is a type of DC-DC converter that steps down voltage from a higher level to a lower level, while a boost converter steps up voltage from a lower level to a higher level. Buck converters are used when the load requires a lower voltage than the source, whereas boost converters are used when a higher voltage is needed.

  7. 7.Calculate the output voltage of a buck converter with an input voltage of 48 V, a duty cycle of 0.5, and assuming ideal conditions.Numerical

    Under ideal conditions, the output voltage (Vout) of a buck converter is given by Vout = Vin × Duty Cycle. Therefore, Vout = 48 V × 0.5 = 24 V.

  8. 8.What is the role of a snubber circuit in power electronics?Concept

    A snubber circuit in power electronics is used to protect semiconductor devices from voltage spikes and to reduce electromagnetic interference (EMI). It typically consists of resistors and capacitors and helps in dissipating energy from inductive loads, thereby enhancing the reliability and longevity of the devices.

  9. 9.Explain why thermal management is crucial for inverters in electric vehicles.Application

    Thermal management is crucial for inverters in electric vehicles because excessive heat can degrade the performance and reliability of power electronic components. Effective thermal management ensures that the inverter operates within safe temperature limits, preventing overheating and potential failure, and thus maintaining efficiency and longevity.

  10. 10.Determine the power loss in an IGBT with a forward voltage drop of 2 V and a current of 50 A.Numerical

    The instantaneous conduction loss is P = V_CE(on) × I = 2 V × 50 A = 100 W while the IGBT is conducting. The average conduction loss is lower if the device conducts only part of the time, for example about 50 W at a 50 % duty cycle. Switching losses come on top of this, (E_on + E_off) × f_s, and can be comparable at high switching frequency, so both are needed to size the heat sink.

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