Lead-acid batteries: construction, rating and testing

Why it matters

Almost every car, two-wheeler and truck on Indian roads still carries a 12 V (or 24 V) lead-acid battery to crank the engine and to power lights, ECUs and alarms when the engine is off. Even battery-electric vehicles keep a 12 V lead-acid or similar auxiliary battery for low-voltage loads. Knowing how the battery is built, how it is rated and how to test it lets you size a replacement correctly and tell a flat battery from a dead one.

Key ideas

Cell chemistry. Each cell has a positive plate of lead dioxide (PbO₂), a negative plate of spongy lead (Pb) and an electrolyte of dilute sulfuric acid (H₂SO₄ in water). On discharge both plates turn into lead sulfate (PbSO₄) and the acid is consumed, producing water, so the electrolyte becomes weaker and its specific gravity falls. Charging reverses the reaction. The open-circuit voltage of a cell is about 2.1 V, so a "12 V" battery is six cells in series (about 12.6–12.7 V when fully charged and rested).

Construction.

• Grids: lead alloy (lead–antimony in older designs, lead–calcium in maintenance-free batteries) that hold the active material (paste) and carry current.
• Plates and plate groups: several positive and negative plates are interleaved and connected in parallel inside each cell; more plate area gives more current, more cells give more voltage.
• Separators: porous polyethylene envelopes (or glass mat in AGM batteries) that stop the plates touching but let ions pass.
• Container and lid: polypropylene case divided into six compartments, cell connectors, terminal posts (positive post is slightly larger), vent caps or a flame-arresting vent.
• Types: flooded (topped up with distilled water), sealed maintenance-free (flooded, lead–calcium, low water loss), VRLA — AGM and gel — where oxygen recombination keeps the battery sealed. AGM types are used in start-stop vehicles.

Ratings.

• Nominal voltage: 12 V (six cells).
• Capacity (Ah): usually the 20-hour rate (C20) — the charge delivered at a constant current for 20 h at about 25 °C until the voltage falls to 10.5 V (1.75 V per cell). A 60 Ah battery delivers 3 A for 20 h. At higher current the delivered capacity is lower (Peukert effect).
• Cold cranking amps (CCA): the current a new, fully charged battery can deliver at −18 °C for 30 s while holding at least 7.2 V (SAE definition; EN and JIS use different test conditions, so compare like with like).
• Reserve capacity (RC): minutes a battery can supply 25 A at about 27 °C before falling to 10.5 V — how long the car runs if the alternator fails.

State of charge and testing.

• Open-circuit voltage (OCV): measured after the battery has rested several hours. Roughly 12.6–12.7 V full, about 12.4 V at 75 %, 12.2 V at 50 %, 12.0 V at 25 %, 11.9 V or less discharged (typical figures; check the maker's chart).
• Hydrometer: specific gravity about 1.260–1.280 fully charged and about 1.100–1.120 discharged in temperate climates (tropical batteries use weaker acid). Readings are referred to 27 °C: add about 0.0007 per °C above, subtract below (take the exact correction from your data book). A spread of more than about 0.050 between cells points to a bad cell.
• Load (high-rate discharge) test: on a charged battery apply a load of about half the CCA for 15 s; at about 21 °C the voltage should stay at or above 9.6 V.
• Conductance / internal-resistance testers: electronic testers estimate CCA capability without a heavy load.

Failure modes. Sulfation (hard PbSO₄ crystals after long periods discharged), grid corrosion, active-material shedding, water loss from overcharging, and internal short circuits. Charging voltage is held near 14.2–14.4 V at room temperature; above about 2.4 V per cell the cell gasses hydrogen and oxygen, so keep sparks away and ventilate.

Links. The starter motor (next topic) draws hundreds of amperes from this battery, and the alternator and regulator recharge it.

Formulas

PbO₂ + Pb + 2H₂SO₄ ⇌ 2PbSO₄ + 2H₂O (left to right = discharge)

Q = I × t

• Q: charge (A·h or C; 1 A·h = 3600 C), I: current (A), t: time (h or s). Valid for constant current.

E = V × Q

• E: energy (W·h or J), V: nominal voltage (V), Q: capacity (A·h or C). Gives nominal stored energy; usable energy depends on depth of discharge and rate.

V_t = E_oc − I × R_int

• V_t: terminal voltage under load (V), E_oc: open-circuit voltage (V), I: discharge current (A), R_int: internal resistance (Ω). Use for cranking voltage drop.

t = H × (C / (I × H))^k (Peukert)

• t: run time at current I (h), C: rated capacity (A·h) at the H-hour rate, H: rated discharge time (h), I: actual current (A), k: Peukert exponent (dimensionless, about 1.1–1.3 for lead-acid; take from the maker's data).

SG_27 = SG_read + 0.0007 × (T − 27)

• SG_27: specific gravity corrected to 27 °C, SG_read: hydrometer reading, T: electrolyte temperature (°C). The coefficient is approximate.

Worked examples

Example 1 (standard). A 12 V, 60 Ah (C20) battery. Find (a) the 20-hour discharge current, (b) the nominal energy stored, (c) how long it runs a 6 A parking load from full charge if only 50 % of capacity may be used (ignore Peukert).

1. I_20 = C / 20 = 60 A·h / 20 h = 3 A.
2. E = V × Q = 12 V × 60 A·h = 720 W·h = 720 × 3600 J = 2.592 MJ.
3. Usable charge = 0.5 × 60 = 30 A·h; t = Q / I = 30 A·h / 6 A = 5 h.

Answer: 3 A, 720 W·h (2.592 MJ), 5 h.

Example 2 (GATE level). A 12 V battery has an open-circuit voltage of 12.6 V and internal resistance 8 mΩ. While cranking, the starter draws 250 A. (a) Find the terminal voltage and the power lost inside the battery. (b) The same battery is rated 100 A·h at the 20-hour rate with Peukert exponent k = 1.2. How long will it supply a steady 25 A, and what capacity does it actually deliver?

1. V_t = E_oc − I × R_int = 12.6 − 250 × 0.008 = 12.6 − 2.0 = 10.6 V.
2. P_loss = I² × R_int = 250² × 0.008 = 500 W.
3. t = H × (C / (I × H))^k = 20 × (100 / (25 × 20))^1.2 = 20 × 0.2^1.2 = 20 × 0.1450 = 2.90 h.
4. Delivered capacity = I × t = 25 × 2.90 = 72.5 A·h, compared with 100 A·h at the 20-hour rate.

Answer: 10.6 V and 500 W lost; about 2.90 h, delivering about 72.5 A·h.

Example 3 (hydrometer). A cell reads SG 1.230 at 37 °C. Corrected value SG_27 = 1.230 + 0.0007 × (37 − 27) = 1.237. With linear interpolation between 1.120 (discharged) and 1.280 (full), SoC = (1.237 − 1.120)/(1.280 − 1.120) = 0.731, i.e. about 73 %.

Common mistakes

• Treating "Ah" as current, or multiplying Ah by 12 V and calling the result Joules (it is W·h; multiply by 3600 for J).
• Assuming capacity is the same at every discharge rate — a 60 Ah battery will not give 60 A for 1 h.
• Reading OCV straight after charging or driving; surface charge makes it read high. Let the battery rest.
• Forgetting the temperature correction on hydrometer readings, or reading the float with the eye off the liquid level.
• Comparing CCA figures measured to different standards (SAE, EN, JIS).
• Topping up with acid instead of distilled water; only water is lost in gassing.
• Disconnecting the positive terminal first when working on the car — remove the negative (earth) lead first to avoid shorting the spanner to the body.

For GATE ME

This topic sits on the edge of the GATE syllabus, so expect it through basic electrical numericals: charge and energy from A·h ratings, run time at a given current, terminal voltage with internal resistance, I²R losses, and series-parallel combinations of cells for a required voltage and capacity. Practise unit conversion between A·h, C, W·h and J, and work Peukert-type problems with the formula given in the question.

Quick check

1. How many cells does a 12 V lead-acid battery have, and what is the approximate full-charge OCV?
2. What happens to electrolyte specific gravity during discharge, and why?
3. What does a CCA rating of 500 A mean?
4. A 12 V, 45 A·h battery stores how much nominal energy in kJ?
5. Why does a load test use about half the CCA for 15 s?

Answers: 1. Six cells, about 12.6–12.7 V. 2. It falls, because sulfuric acid is consumed and water is produced. 3. The battery can deliver 500 A at −18 °C for 30 s while staying at or above 7.2 V. 4. 12 × 45 × 3600 = 1944 kJ (540 W·h). 5. It imitates the cranking load; a healthy charged battery holds at least 9.6 V.