categories: physics, automation & robotics, electrical engineering
New sub-series: back-of-the-envelope engineering calculations you actually need before building something, not just analyzing it. First up — the question every embedded/robotics project runs into: what battery capacity do I actually need so the thing doesn't die mid-run?
Task: choosing a LiPo battery capacity for an embedded system
Setup: an embedded system draws power through a DC-DC converter from a single-cell (1S) LiPo battery. Onboard: an ESP32 microcontroller ($$3.3\text{ V}, 160\text{ mA}$$), a servo ($$5.0\text{ V}, 250\text{ mA}$$), and a LiDAR sensor ($$3.3\text{ V}, 20\text{ mA}$$). Find the minimum battery capacity for 8 hours of continuous operation.
Step 1 — total load power. Each component draws power at its own voltage rail, so before anything else, convert every current draw to watts and add them — power is what actually has to come out of the battery, current alone can't be summed across different voltage rails: $$P_{ESP32}=3.3\text{ V}\cdot0.160\text{ A}=0.528\text{ W}$$ $$P_{Servo}=5.0\text{ V}\cdot0.250\text{ A}=1.250\text{ W}$$ $$P_{Sensor}=3.3\text{ V}\cdot0.020\text{ A}=0.066\text{ W}$$ $$P_{load}=0.528+1.250+0.066=1.844\text{ W}$$
Step 2 — power actually drawn from the battery. The DC-DC converter isn't free — it wastes some power as heat, at efficiency $$\eta=85\%=0.85$$. The battery has to supply more than $$P_{load}$$ to make up for that loss: $$P_{bat}=\frac{P_{load}}{\eta}=\frac{1.844\text{ W}}{0.85}=2.169\text{ W}$$ Skipping this step is the single most common mistake in a first-pass battery estimate — it silently underestimates real draw by the converter's loss percentage, which compounds badly over a long runtime.
Step 3 — battery current and required capacity. For a 1S LiPo cell at nominal $$V_{bat}=3.7\text{ V}$$, convert battery power to battery current, then to a capacity, over the target runtime $$t=8\text{ h}$$. One more correction is needed: LiPo cells shouldn't be run down to empty — a maximum depth of discharge $$DoD=80\%$$ leaves a 20% margin that protects the cell's lifespan and voltage stability: $$I_{bat}=\frac{P_{bat}}{V_{bat}}=\frac{2.169\text{ W}}{3.7\text{ V}}=0.586\text{ A}=586\text{ mA}$$ $$C_{min}=\frac{I_{bat}\cdot t}{DoD}=\frac{586\text{ mA}\cdot8\text{ h}}{0.80}=5860\text{ mAh}$$ Practical choice: a 1S LiPo battery with a minimum capacity of 6000 mAh — rounding up to the nearest common off-the-shelf size, never down.
Three corrections stacked on top of the naive "just add up the currents" approach — mixed voltage rails, converter losses, and depth-of-discharge margin — and each one alone can throw a real design off by 15-20%. Skip all three and you get a robot that mysteriously dies two hours early. Thank you for reading! Next post: sizing a mechanical shaft instead of a battery, using a combined-stress strength criterion.
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