How to Calculate Battery Capacity for Design Engineers (Step-by-Step Guide)

By | 23/10/2025

If you’re a design engineer, you’ve likely received the infamous spec sheet from marketing that says:

“Maximize run time, minimize battery size and cost.”

No mention of how much runtime, how small the battery should be, or what cost is acceptable. It’s almost as if they’re hoping for a miracle—and you’re the one expected to deliver it.

So how do you turn that vague directive into an actual battery specification? This guide walks you through the engineering process of calculating required battery capacity, helping you build a battery runtime matrix that marketing can’t argue with.

Step 0: Understanding Battery Capacity and Energy Units

Before you grab your calculator, let’s review a few fundamentals.

Charge and Current Basics

  • Charge (Q) is measured in coulombs (C).
  • 1 amp (A) flowing for 1 second equals 1 coulomb of charge: Q = I \times t

Since one hour is 3600 seconds, 1 amp-hour (Ah) equals 3600 coulombs.
That’s why battery capacity is measured in amp-hours (Ah) or milliamp-hours (mAh) for smaller devices.

From Amp-Hours to Watt-Hours

Because voltage changes during discharge, amp-hours alone don’t tell you total energy. Multiply by average voltage to estimate energy capacity: E = C \times V_{avg}

Where:

  • E = Energy in watt-hours (Wh)
  • C = Capacity in amp-hours (Ah)
  • V_{avg}​ = Average discharge voltage

Example:
A 2Ah, 3.7V Li-ion battery = 7.4 Wh of stored energy.

Step 1: Back-of-the-Envelope Battery Capacity Calculation

The simplest estimation assumes a constant current draw: C = I \times T

Where:

  • C = capacity (Ah)
  • I = current (A)
  • T = time (hours)

Example:
If your circuit draws 120mA and must run for 24 hours:

  • C = 0.12 \times 24 = 2.88 \text{ Ah}

That’s your minimum capacity—but we’re just getting started.

Step 2: Adjust for Cycle Life (Don’t Discharge Fully)

Running a battery to zero each cycle drastically reduces its lifespan. For example:

  • Lead-acid batteries should not be discharged beyond 80% depth.
  • Lithium-ion and NiMH cells also last longer with shallower discharges.

So, adjust for usable capacity:

  • C' = \frac{C}{0.8}

Continuing our example:

  • C' = \frac{2.88}{0.8} = 3.6 \text{ Ah}

Now you have a battery size that preserves cycle life.

Step 3: Consider the Peukert Effect (Discharge Rate)

Not all amp-hours are created equal. When you discharge a battery quickly, its effective capacity decreases—a phenomenon known as the Peukert effect.

  • Lead-acid and alkaline batteries are most affected.
  • Lithium-ion and NiMH cells handle high discharge rates better.

For example, a lead-acid battery rated for 10Ah at a 20-hour rate might deliver only 5Ah at a 1-hour discharge rate.

Rule of thumb:

  • C'' = \frac{C'}{\text{efficiency at discharge rate}}

If your system draws 20A for one hour, you’ll likely need:

  • C'' = \frac{25Ah}{0.5} = 50Ah

So, a 50Ah sealed lead-acid battery will sustain 20A for about one hour.

Step 4: When the Load Isn’t Constant

Most devices don’t draw a steady current—think pumps, motors, or transmitters.

To estimate average current, use:

  • I_{avg} = \frac{\sum (I_n \times t_n)}{T_{total}}

Example:
Your device pulls 20A for 1 second and 0.1A for the next 3599 seconds.

  • I_{avg} = \frac{(20 \times 1) + (0.1 \times 3599)}{3600} = 0.1044A

Then, use I_{avg} in your Step 1–3 calculations.

The good news: brief current spikes are less punishing than sustained high-current discharge, so capacity estimates tend to be conservative.

When your design specifications list power in watts rather than current in amps, use the following method to determine the battery capacity.

Battery Sizing from Power Ratings: Converting Watts to Amps

If your design specifies power (W) instead of current, convert it using the battery voltage.

  • \text{Amp-hours} = \frac{\text{Watts} \times \text{Hours}}{\text{Battery Voltage} \times \text{Efficiency}}

Example:
You need to run a 250W, 110V AC light for 5 hours through an inverter (85% efficient) using a 12V battery.

  • \text{Watt-hours} = \frac{250 \times 5}{0.85} = 1470 Wh
  • \text{Amp-hours} = \frac{1470}{12} = 122.5Ah

So, you’d need a 12V 125Ah battery for that runtime.

Key Takeaways for Design Engineers

  1. Start simple: C = I \times T gives a baseline.
  2. Account for usable depth: Divide by 0.8 (or appropriate derating factor).
  3. Adjust for discharge rate: Apply the Peukert correction.
  4. For variable loads: Calculate the average current.
  5. For watt-based specs: Convert using battery voltage and efficiency.

By combining these steps, you’ll have a reliable battery sizing estimate ready for presentation—no miracles required.

Conclusion

Engineering battery capacity isn’t about guessing—it’s about balancing performance, cost, and practicality. Whether you’re powering an IoT sensor or a portable amplifier, these formulas will help you justify your design decisions and communicate effectively with marketing, procurement, and production.

So next time they ask for “maximum runtime, minimum size,” you’ll be ready—with the math to prove it.

Author: Zakaria El Intissar

Automation and industrial computing engineer passionate about innovation at the heart of the energy sector, I am a recognized specialist in control systems for electrical substations and SCADA solutions. With expertise forged by years of practical experience, I excel in the design, implementation, and optimization of these critical technologies, ensuring optimal reliability of energy infrastructures, including those integrating renewable energy sources, where I have contributed to innovative projects for a sustainable energy transition.

Leave a Reply

Your email address will not be published. Required fields are marked *