Electric vehicle architecture and drivetrain sizing

The building blocks and layouts of a battery-electric vehicle and how to size its motor, gear ratio and battery from the road-load equation (rolling, aero, grade and inertia), with steady-speed consumption, range, gear-ratio and gradeability numericals.

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Why it matters

India's electric two-, three- and four-wheeler market is growing fast, and every new EV programme starts with the same questions: how big must the motor be, what gear ratio, and how many kWh of battery for the target range and price? These answers come from a simple road-load model that every automotive engineer should be able to set up and solve on paper before any simulation.

Key ideas

Main blocks of a battery-electric vehicle (BEV).

  • Traction battery: typically about 300–400 V in cars (800 V in some), 48–72 V in two- and three-wheelers, with a battery management system (BMS), main contactors, a pre-charge circuit, fuses and a service disconnect.
  • Traction inverter: converts battery DC into variable-frequency, variable-voltage AC for the motor and works in reverse during regenerative braking.
  • Traction motor: usually a permanent-magnet synchronous motor (PMSM), sometimes an induction motor; hub BLDC motors in many two-wheelers.
  • Transmission: a single-speed reduction gear (overall ratio typically about 7–11:1 in cars) plus differential. Motor, inverter and gearbox are often integrated into one "e-axle".
  • On-board charger (OBC) for AC charging, DC fast-charge inlet direct to the battery, and a DC-DC converter that replaces the alternator by feeding the 12 V system.
  • Thermal management: coolant loops for battery, motor and inverter, often a heat pump for the cabin.
  • Vehicle control unit (VCU): turns pedal position into a torque request, blends regenerative and friction braking, and manages energy.

Layouts. Single front or rear motor driving one axle; dual motor (one per axle) for all-wheel drive; two motors on one axle for torque vectoring; in-wheel (hub) motors, which free space but add unsprung mass.

Why one gear is enough. A traction motor gives its maximum torque from zero speed up to its base speed, then constant power up to maximum speed (often 3–4 times base speed). This curve already resembles the ideal tractive-effort hyperbola, so a single ratio can meet both launch/gradient torque and top speed. Two-speed gearboxes appear only in some high-performance vehicles.

Road-load model. At the wheels the drive must overcome:

  • Rolling resistance, roughly constant with speed (C_rr about 0.008–0.015 for car tyres on asphalt).
  • Aerodynamic drag, rising with the square of speed (C_d about 0.25–0.35 for cars).
  • Grade resistance, the weight component along the slope.
  • Inertia, mass times acceleration, increased by a rotational-inertia factor δ (about 1.03–1.1) for the motor, gears and wheels.

Sizing steps.

  1. Top speed → continuous power and maximum motor speed, which with wheel radius sets the gear ratio.
  2. Gradeability (e.g. a given % slope from rest) → peak wheel torque, which with the ratio sets peak motor torque.
  3. Acceleration (e.g. 0–100 km/h time) → peak power.
  4. Range on a drive cycle → energy per km, which sets usable battery energy; total energy then follows from the allowed depth of discharge. Then iterate: a bigger battery adds mass, which raises the road load. Efficiencies of gearbox, motor and inverter (each about 90–97 %) and auxiliary loads (HVAC can be several kW) must be included.

Formulas

F_tr = m g C_rr cos θ + ½ ρ C_d A v² + m g sin θ + δ m a

  • F_tr: tractive force at the wheels (N), m: vehicle mass (kg), g = 9.81 m/s², C_rr: rolling-resistance coefficient, θ: grade angle (tan θ = grade/100), ρ: air density (kg/m³, about 1.2), C_d: drag coefficient, A: frontal area (m²), v: speed (m/s), δ: rotational-inertia factor, a: acceleration (m/s²).

P_wheel = F_tr × v, P_batt = P_wheel / η_drive

  • P (W); η_drive: battery-to-wheel efficiency (gearbox × motor × inverter). For regeneration multiply instead of divide.

v = ω_m × r / G, T_wheel = T_m × G × η_g

  • ω_m: motor speed (rad/s), r: wheel rolling radius (m), G: overall gear ratio, T_m: motor torque (N·m), η_g: gear efficiency.

P = T × ω

e = P_batt / v → in Wh/km: e = P_batt (W) / v (km/h)

  • Energy per distance at steady speed.

E_batt = Range × e / DoD

  • E_batt: installed battery energy (Wh), Range (km), e (Wh/km), DoD: usable fraction.

Worked examples

Example 1 (standard). A 1500 kg EV has C_rr = 0.012, C_d = 0.30, A = 2.2 m², ρ = 1.2 kg/m³ and battery-to-wheel efficiency 85 %. Find the battery power and energy per km at a steady 90 km/h on level road, and the range from 40 kWh usable energy (ignore auxiliaries).

  1. v = 90 / 3.6 = 25 m/s.
  2. F_roll = 1500 × 9.81 × 0.012 = 176.6 N.
  3. F_aero = 0.5 × 1.2 × 0.30 × 2.2 × 25² = 247.5 N. Total F = 424.1 N.
  4. P_wheel = 424.1 × 25 = 10,602 W; P_batt = 10,602 / 0.85 = 12,473 W.
  5. e = 12,473 W / 90 km/h = 138.6 Wh/km.
  6. Range = 40,000 / 138.6 = 289 km.

Answer: about 12.5 kW, 139 Wh/km, about 289 km.

Example 2 (GATE level). The same EV has a wheel radius of 0.30 m and a motor with a maximum speed of 12,000 rpm; gear efficiency 95 %. (a) Choose the gear ratio for a top speed of 150 km/h. (b) With G = 9, find the motor torque needed to hold the car on a 20 % grade (just moving, ignore aero and acceleration). (c) What is the actual top speed with G = 9?

  1. ω_m = 12,000 × 2π / 60 = 1256.6 rad/s; v = 150 / 3.6 = 41.67 m/s.
  2. G = ω_m r / v = 1256.6 × 0.30 / 41.67 = 9.05 → choose G = 9.
  3. Grade 20 %: θ = tan⁻¹ 0.2 = 11.31°, sin θ = 0.1961, cos θ = 0.9806.
  4. F = m g (sin θ + C_rr cos θ) = 1500 × 9.81 × (0.1961 + 0.012 × 0.9806) = 3059 N.
  5. T_wheel = F r = 3059 × 0.30 = 917.7 N·m; T_m = 917.7 / (9 × 0.95) = 107.3 N·m.
  6. v_max = ω_m r / G = 1256.6 × 0.30 / 9 = 41.89 m/s = 150.8 km/h.

Answer: G ≈ 9; about 107 N·m at the motor; about 151 km/h.

Common mistakes

  • Using the grade percentage as the angle in degrees; a 20 % grade means tan θ = 0.2 (θ ≈ 11.3°).
  • Forgetting that aerodynamic force goes with v² and power with v³, so range drops sharply at high speed.
  • Dividing by efficiency when motoring and also when regenerating; in regeneration the losses reduce what reaches the battery, so multiply.
  • Mixing rpm and rad/s, or km/h and m/s.
  • Sizing the battery on nominal capacity without allowing for depth of discharge, auxiliary loads and ageing.
  • Ignoring the added mass of a larger battery when increasing range.

For GATE ME

This topic maps directly onto engineering mechanics and machine-design style numericals: forces on an inclined plane, power = force × velocity, torque and speed through a gear pair with efficiency, and energy = power × time. Practise the full road-load equation and unit conversions; expect questions on required motor torque, gear ratio for a top speed, and range from a given consumption.

Quick check

  1. Why do most EVs use a single-speed reduction gear?
  2. Which road-load term dominates at motorway speed?
  3. A motor gives 200 N·m at 400 rad/s. What is its power?
  4. A 60 kWh pack (100 % usable) and a consumption of 150 Wh/km give what range?
  5. What component replaces the alternator in an EV?

Answers: 1. The motor delivers full torque from zero speed and a wide constant-power range, so one ratio covers launch and top speed. 2. Aerodynamic drag. 3. 80 kW. 4. 400 km. 5. The DC-DC converter from the traction battery.

Interview questions

All Automotive Electronics and Electric Vehicles interview questions

Try answering each one aloud before you open it.

  1. 1.What is the basic architecture of an electric vehicle?Concept

    The basic architecture of an electric vehicle includes the electric motor, battery pack, power electronics controller, transmission system, and auxiliary systems. The electric motor converts electrical energy into mechanical energy to drive the wheels. The battery pack stores electrical energy, while the power electronics controller manages the flow of electricity between the battery and the motor. The transmission system may be simpler than in internal combustion vehicles, often involving a single-speed gearbox. Auxiliary systems include HVAC, lighting, and infotainment.

  2. 2.Explain the role of the power electronics controller in an electric vehicle.Concept

    The power electronics controller in an electric vehicle manages the flow of electrical energy between the battery and the electric motor. It converts the DC power from the battery into AC power for the motor, and vice versa during regenerative braking. It also controls the motor's speed and torque by adjusting the frequency and amplitude of the AC power. Additionally, it ensures efficient energy use and protects the system from electrical faults.

  3. 3.Why is a single-speed transmission often used in electric vehicles?Application

    A traction motor delivers its maximum torque from zero speed up to its base speed and then roughly constant power up to a maximum speed that is often three to four times base speed. That curve is already close to the ideal tractive-effort curve, so one reduction ratio, typically about 7–11:1 in cars, can give enough wheel torque for launch and gradients and still reach top speed within the motor's speed limit. There is also no idle speed or clutch requirement. A single-speed unit is lighter, cheaper, quieter and more efficient, and shifts give no torque interruption; only some high-performance EVs use two speeds.

  4. 4.What happens if the battery management system (BMS) fails in an electric vehicle?Application

    If the battery management system (BMS) fails in an electric vehicle, it can lead to several issues. The BMS is responsible for monitoring and managing the battery's state of charge, temperature, and health. A failure could result in overcharging or deep discharging, which can damage the battery cells. It may also lead to inaccurate range estimation and, in severe cases, pose safety risks such as thermal runaway.

  5. 5.What are the advantages of using a lithium-ion battery in electric vehicles?Application

    Lithium-ion batteries are preferred in electric vehicles due to their high energy density, which allows for longer driving ranges. They also have a relatively low self-discharge rate, maintaining charge over time. Additionally, lithium-ion batteries are lightweight, which contributes to better vehicle efficiency. They also have a longer lifespan compared to other battery types, making them cost-effective over the vehicle's life.

  6. 6.Calculate the energy consumption of an electric vehicle that travels 100 km with an average energy efficiency of 15 kWh/100 km.Numerical

    To calculate the energy consumption, use the formula: Energy Consumption = Distance × Energy Efficiency. Here, Distance = 100 km and Energy Efficiency = 15 kWh/100 km. Therefore, Energy Consumption = 100 km × (15 kWh/100 km) = 15 kWh.

  7. 7.What is the impact of temperature on the performance of an electric vehicle's battery?Application

    Temperature significantly impacts the performance of an electric vehicle's battery. High temperatures can increase the battery's capacity temporarily but may lead to faster degradation over time. Low temperatures can reduce the battery's capacity and efficiency, leading to decreased range and performance. Battery management systems are designed to maintain optimal temperature ranges to ensure longevity and efficiency.

  8. 8.Explain the concept of torque vectoring in electric vehicles.Concept

    Torque vectoring in electric vehicles refers to the ability to independently control the torque delivered to each wheel. This enhances vehicle handling, stability, and performance, especially during cornering. By adjusting the torque distribution, the vehicle can maintain better traction and reduce understeer or oversteer. Electric vehicles can implement torque vectoring more easily due to the independent control of electric motors.

  9. 9.If an electric vehicle's motor has a power output of 100 kW and operates at an efficiency of 90%, what is the input power required from the battery?Numerical

    To find the input power required from the battery, use the formula: Input Power = Output Power / Efficiency. Here, Output Power = 100 kW and Efficiency = 90% = 0.9. Therefore, Input Power = 100 kW / 0.9 = 111.11 kW.

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