Battery Fundamentals and Performance Metrics

Introduction

A battery can be described as high-voltage, high-capacity or high-power, but those claims refer to different physical quantities. Capacity counts stored charge, energy accounts for the voltage at which charge is delivered, and power measures how quickly energy is delivered. Efficiency metrics add still another dimension.

Core explanation

Cell voltage is the potential difference between terminals. Open-circuit voltage is related to reversible electrode potentials and composition, while loaded voltage also includes overpotentials and iR loss. It changes during discharge, so one nominal voltage does not always represent the full curve. A battery pack voltage depends on how cells are connected in series and on their individual states.

Capacity Q is the integrated discharge current, Q = ∫I dt. Common units are ampere-hours, with 1 Ah = 3600 C. Capacity is not energy; a 1 Ah cell at an average 3 V delivers approximately 3 Wh, whereas a 1 Ah cell at 1 V delivers about 1 Wh. More exactly, delivered energy is ∫V(t)I(t)dt. If current is constant, integrate voltage over discharged capacity rather than multiplying by an unsuitable open-circuit voltage.

Specific capacity divides charge by mass, often reported in mAh g−1 for an active material. Specific energy divides delivered energy by mass, often Wh kg−1, and can refer to an active material, electrode or complete cell. Those bases differ substantially because separators, current collectors, packaging and electrolyte contribute mass but may not store the same charge. Always state the denominator before comparing values.

Power is energy delivery rate, P = VI, in watts. Specific power is power per mass. A cell can have high theoretical energy but low practical power if ion transport or electrode kinetics limits current. Increasing discharge current may reduce terminal voltage and accessible capacity, so power and energy trade off. Temperature and aging further alter the relationship.

Coulombic efficiency of a charge–discharge cycle is discharge charge divided by charge input, multiplied by 100%, with a clear cycle convention. Side reactions can consume charge, making it less than one. Energy efficiency is discharge energy divided by charge energy and is generally lower because charging voltage exceeds discharging voltage under losses. A cell can have high coulombic efficiency yet appreciable voltage hysteresis and lower energy efficiency.

Step-by-step reasoning

Identify whether a question asks for voltage, charge, energy or power. Convert hours to seconds when using coulombs or Faraday's constant. Integrate current for capacity and voltage times current for energy, or use a justified average voltage. State whether mass refers to active material or complete cell. Calculate coulombic and energy efficiencies from corresponding charge and energy quantities rather than substituting one for the other.

Visual explanation

Draw a discharge curve of voltage against capacity. The area beneath it is energy; the horizontal extent is capacity; a point's V times I is instantaneous power. Beside it draw charge and discharge voltage curves at the same state of charge, with the area difference representing energy loss.

Real-world analogy

A water reservoir's volume is like capacity, its height gives pressure analogous to voltage, and the rate of water delivery resembles power. The energy available depends on both volume and height. The analogy helps distinguish metrics, though electrochemical voltage varies with composition and load.

Real-world example

Suppose a cell supplies a steady 2.0 A for 3.0 hours at an average terminal voltage of 3.5 V. Its discharge capacity is 6.0 Ah, and approximate delivered energy is 21 Wh. If it weighs 0.20 kg, its cell-level specific energy is about 105 Wh kg−1. The calculation uses average loaded voltage, not a theoretical electrode potential.

Why?

Current is charge flow per time and voltage is energy per unit charge. Multiplying voltage by current gives energy per time; integrating across time gives total energy. Mass normalization changes the meaning of a performance figure, while side reactions and polarization explain why charged input and discharged output differ.

Common misconception

Capacity in Ah is not energy in Wh. A high open-circuit voltage also does not guarantee high delivered energy under heavy load, because voltage and accessible capacity can fall. Coulombic efficiency is not the same as round-trip energy efficiency.

Worked example

Question: A battery receives 5.0 Ah during charge and returns 4.8 Ah during discharge. The average charging and discharging voltages are 4.0 and 3.6 V. Estimate coulombic and energy efficiencies.

Reasoning: Coulombic efficiency is 4.8/5.0 = 0.96, or 96%. Input energy is approximately 5.0 Ah × 4.0 V = 20 Wh. Output energy is 4.8 Ah × 3.6 V = 17.28 Wh. Energy efficiency is 17.28/20 = 0.864, or 86.4%. The difference reflects both charge loss and voltage hysteresis.

Answer: About 96% coulombic efficiency and 86.4% energy efficiency.

Quick check

1. How many coulombs are in 2.0 Ah? Answer: 2.0 × 3600 = 7200 C.

Exam focus

Keep Ah, Wh and W distinct. Use average loaded voltage only when justified, and label active-material versus full-cell mass. Calculate coulombic and energy efficiencies from their own numerators and denominators.

Advanced insight

At pack level, thermal management, electronics and enclosure add mass and consume energy. A cell-level Wh kg−1 figure therefore cannot be used directly as a whole-device specific energy without accounting for those components and operating conditions.

Summary

Voltage is energy per charge, capacity is delivered charge, energy is the integral of voltage over charge, and power is VI. Specific metrics require a stated mass basis. Coulombic efficiency tracks charge recovery, while energy efficiency also includes voltage losses.

Practice questions

1. What is the energy of 2 Ah delivered at constant 3 V? Answer: 6 Wh. 2. Why can energy efficiency be below coulombic efficiency? Answer: Charging occurs at higher voltage than discharge because of overpotentials and resistance. 3. Does mAh g−1 always refer to a whole battery? Answer: No. It often refers only to an active material; the mass basis must be stated. 4. What is instantaneous electrical power at 4 V and 3 A? Answer: 12 W.