Understanding Cell Efficiency: Ampere-hour vs. Watt-hour

When we talk about the efficiency of a cell or battery, we are usually interested in how much energy or charge we get out compared to how much we put in. There are two main ways to measure this: Ampere-hour efficiency and Watt-hour efficiency.

What is Ampere-hour (Ah) Efficiency?

Ampere-hour efficiency, often called Coulombic efficiency, measures the ratio of the total charge removed from the cell during discharge to the total charge supplied to the cell during charge. It's essentially a measure of charge throughput efficiency.

The formula for Ampere-hour efficiency ($\eta_{Ah}$) is:

\begin{equation*} \eta_{Ah} = \frac{\text{Charge extracted during discharge}}{\text{Charge supplied during charge}} \times 100\% \end{equation*}

Charge is typically measured in Ampere-hours (Ah), which is current multiplied by time.

What is Watt-hour (Wh) Efficiency?

Watt-hour efficiency, also known as energy efficiency, measures the ratio of the total energy removed from the cell during discharge to the total energy supplied to the cell during charge. It's a measure of energy throughput efficiency.

The formula for Watt-hour efficiency ($\eta_{Wh}$) is:

\begin{equation*} \eta_{Wh} = \frac{\text{Energy extracted during discharge}}{\text{Energy supplied during charge}} \times 100\% \end{equation*}

Energy is typically measured in Watt-hours (Wh), which is power (Watt) multiplied by time. Power is voltage multiplied by current. So, energy is voltage multiplied by current multiplied by time (Volts $\times$ Amps $\times$ hours).

\begin{equation*} \text{Energy} = \text{Average Voltage} \times \text{Ampere-hours} \end{equation*}

Comparing Ampere-hour and Watt-hour Efficiency

Let's think about what happens during charging and discharging of a cell:

• During charging, you apply a voltage across the cell terminals to force current into it. This voltage needs to be slightly higher than the cell's open-circuit voltage to overcome internal resistance and drive the charging reaction.
• During discharging, the cell itself provides a voltage to drive current through an external circuit. This voltage is slightly lower than the cell's open-circuit voltage due to internal resistance and polarization losses within the cell.

Consider the energy transfer:

• Energy supplied during charge = Average charging voltage $\times$ Charge supplied (Ah)
• Energy extracted during discharge = Average discharging voltage $\times$ Charge extracted (Ah)

Therefore, Watt-hour efficiency can also be expressed as:

\begin{equation*} \eta_{Wh} = \frac{\text{Average discharging voltage} \times \text{Charge extracted}}{\text{Average charging voltage} \times \text{Charge supplied}} \times 100\% \end{equation*}

We can rearrange this formula:

\begin{equation*} \eta_{Wh} = \left( \frac{\text{Charge extracted}}{\text{Charge supplied}} \right) \times \left( \frac{\text{Average discharging voltage}}{\text{Average charging voltage}} \right) \times 100\% \end{equation*}

We know that $\frac{\text{Charge extracted}}{\text{Charge supplied}} = \eta_{Ah}$. So,

\begin{equation*} \eta_{Wh} = \eta_{Ah} \times \left( \frac{\text{Average discharging voltage}}{\text{Average charging voltage}} \right) \end{equation*}

As discussed, the average charging voltage is always higher than the average discharging voltage because of voltage drops during discharge and voltage rises required during charge due to internal resistance and other factors. Thus, the ratio \( \frac{\text{Average discharging voltage}}{\text{Average charging voltage}} \) is always less than 1.

Since $\eta_{Wh}$ is equal to $\eta_{Ah}$ multiplied by a factor that is less than 1, it logically follows that Watt-hour efficiency is always less than Ampere-hour efficiency.

For example, if a cell has an Ampere-hour efficiency of 95% and the average discharging voltage is 1.2V while the average charging voltage is 1.3V, the Watt-hour efficiency would be:

\begin{equation*} \eta_{Wh} = 0.95 \times \left( \frac{1.2V}{1.3V} \right) \approx 0.95 \times 0.923 \approx 0.877 \end{equation*}

So, the Watt-hour efficiency would be about 87.7%, which is less than the 95% Ampere-hour efficiency.

This voltage difference is due to internal losses within the cell during energy conversion, which impacts the voltage available during discharge and the voltage required during charge.

Summary Comparison

Therefore, Watt-hour efficiency, which accounts for energy (Volts x Amps x Time), is always impacted more by the internal losses that cause voltage differences between charge and discharge than Ampere-hour efficiency, which only considers charge (Amps x Time).

The Watt-hour efficiency of a cell is always less than its Ampere-hour efficiency.

Revision Table: Cell Efficiency Analysis

Additional Information on Battery Efficiency Factors

Several factors influence both Ampere-hour and Watt-hour efficiency:

• Internal Resistance: Resistance within the cell causes voltage drops during discharge ($V_{discharge} = V_{oc} - I \times R_{int}$) and voltage increases during charge ($V_{charge} = V_{oc} + I \times R_{int}$), where $V_{oc}$ is open-circuit voltage and $R_{int}$ is internal resistance. This directly impacts Watt-hour efficiency.
• Charge/Discharge Rate (C-rate): Higher currents (higher C-rates) lead to larger voltage drops/rises due to internal resistance, further reducing Watt-hour efficiency. Ah efficiency can also be affected by rate if side reactions become more prominent at higher rates.
• Temperature: Temperature affects internal resistance and the rate of side reactions, influencing both efficiencies.
• Age and Cycle Life: As a cell ages and is cycled, internal resistance often increases, and side reactions may become more significant, leading to decreased efficiency.
• State of Charge (SOC): Voltage characteristics can vary with SOC, which affects the average voltages during charge and discharge cycles.
• Side Reactions: Undesirable chemical reactions within the cell consume charge or energy without contributing to the main charge/discharge process, reducing both efficiencies, particularly Ah efficiency.

Understanding these efficiencies and the factors affecting them is crucial for optimizing battery performance and lifespan in various applications.