In this article, Dr Mike McDonagh examines the practical considerations when seeking to minimise the negative impacts of the sources of battery formation inefficiency.
In the winter 2026 edition of BEST, I described the chemical and structural changes that occur in the positive and negative plates when converting the active material from the dry-cured to the as-formed condition. That article showed that the positive plate consumed part of the available active material to create a structure that combines physical integrity and electron conduction.
This use of a significant portion of the positive active material (PAM), reduces its ability to provide coulombic capacity. This relative inefficiency means that there is a need to provide an excess of PAM compared to the theoretical amount.
In practical terms this also affects the amount of energy needed to convert the dry-cured active materials into the formed PbO2 structure of the positive plate. To this end, additions of red lead or minium to the positive plate, help reduce the conversion energy and formation time for the positive active material (PAM).
In this article we will look at the practical considerations when seeking to minimise the negative impacts of the sources of formation inefficiency. Fig 1 shows two methods of reducing the inefficiency side effects. Fig 1a is a very slow formation program taking 72 hours. This low current reduces the heat and keeps the voltage low to minimise the parasitic reactions. Fig 1b is a water bath with gas extraction. This additional capital equipment reduces the operating temperature of the batteries and removes the hazardous gases and acid fumes associated with faster, high voltage charging.
The starting point is to look at the theoretical coulombic input required to transform the dry cured PAM in the positive plate to the formed active material PbO2. Table 1 is a simplified list of the dry cured/pickled PAM constituents. Its purpose is to illustrate how to calculate the theoretical ampere hours needed for the formation conversion process. The reader is left to ascertain the composition of their company’s PAM and follow the calculation method given here to predict the effect of process changes and additions to their products’ active materials.
For this exercise we will be looking at two PAM ingredients – PbO and Pb3O4 (red lead). PbO is a standard ingredient arising from the ball mill and Barton pot lead oxide production processes. Red lead is a known addition used by many lead-acid tubular plate battery manufacturers. The benefits of red lead are described in many articles and Penox GMBH have spent many years researching this topic. However, for this article we are concerned simply with the reduction of coulombic input that is possible by the addition of red lead for forming positive plates.
Theoretical Ah required for conversion of dry cured PAM components to formed PAM
When considering the reactions of formation, we need to identify the numbers of electrons needed to convert one molecule or atom of a substance into another compound. As an example, we will use the formation equations for converting the dry cured components, PbO and Pb3O4 (red lead) to the formed PAM – PbO2
PbO
Step 1 – Number of electrons used for the conversion of PbO to PbO2
PbO + H2O → PbO2 + 2H+ + 2e– eq. 1
In this case lead has moved from a valency of two in PbO to a valency of four in PbO2. This requires the transference of two electrons as shown in eq. 1. This means that two electrons are transferred for every molecule of PbO transformed into PbO2. To bring this from the molecular level to the real world of weights of substances, we need to work with gram moles (g-mol). That is the chemical term for the molecular or atomic weight of materials expressed in grams. For example:
2H2 + O2 → 2H2O
In this gaseous reaction that produces a liquid, we have two moles of molecular hydrogen (H2 gas) (4g) + one mole of molecular oxygen (O2) (32g) gas . The atomic weights of hydrogen and oxygen are one and 16 respectively. Because both hydrogen and oxygen exist as gases which are diatomic molecules, the above reaction represents the true situation. The molecular weights of the reacting gases are twice that of the elements, i.e. two and 32. Therefore, for this real-world reaction we have 4g of hydrogen reacting with 32g of oxygen to give 2 x 36 = 72g or 2g-mol of water.
Step 2 – Calculate the electrical charge in coulombs per mole of lead monoxide.
For this we use the Faraday constant (F) for the number of electrons transferred in a reaction to produce one mole of any substance. It is defined as:
F = e– x A
Where
e– = charge on 1 electron
(1.6 x 10-19C).
A = Avogadros number, i.e. the number of atoms or molecules in 1g-mol of a substance (6.022 x 1023).
Faraday constant = 96,485 coulombs, where one coulomb could be defined as one amp second.
To convert this to amp hours (Ah), we simply divide by 3,600 (number of seconds in one hour).
In the case of eq. 1 we have a valency of two, therefore two electrons are transferred. The total number of coulombs is:
2 x 96,485 = 192,970 coulombs (C) of charge.
Since 1Ah is 3,600 coulombs (one coulomb/sec x 3,600), then the Ah required for one mole of PbO to be converted to one mole of PbO2 is:
192,970/3,600 = 53.6Ah
Step 3 – Calculate the Ah per kilogram
For this we need to know the number of gram-moles of a substance in a kilogram. Firstly, a gram mole is the molecular weight of a substance expressed in grams. To find the molecular weight of a substance, we simply add up the atomic weights of all the constituent parts. For convenience Table 1 also gives the atomic and molecular weights of some lead acid PAM constituents. For lead monoxide we have the following rounded up weights:
Pb = 207
O = 16
One gram mole (g-mol)of PbO is therefore 207 + 16 = 223g
The number of g-mol in one kilo of PbO is therefore 1000/223 = 4.48g-mol
The number of theoretical Ah to convert 1Kg of lead monoxide to lead dioxide is:
4.48 x 53.6 = 240Ah/Kg of PbO
A note of caution here: the weight of the resultant lead dioxide will be more than that of the monoxide due to the addition of an O atom into the molecular structure i.e.
223 + 16 = 239
The ratio is 239/223 = 1.07. Therefore 1Kg PbO will yield 1.07Kg of PbO2. Also bear in mind these are theoretical amounts; the more accurate picture with allowances for PAM material utilisation and formation energy inefficiency will be described later.
Ah/kg required for lead monoxide compared to red lead – Pb3O4
Looking specifically at the effect of adding red lead to the paste mix, or dry blend of a tubular plate, we can apply the preceding method to calculate the Ah required for its conversion to lead dioxide. The first point to note is that red lead is a combination of PbO and PbO2 usually written as 2PbO.PbO2. Following this, it is hardly surprising that there would be less energy required to convert the red lead molecule to a lead dioxide molecule. The overall formation reaction is:
Pb3O4 + 2H20 = 3PbO2 + 4H+ + 4e–
Simply put there are two more oxygen atoms added to the Pb3O4 molecule. This requires four electrons compared to the two electrons in the PbO reaction. That would need more Ah per molecule for the formation process. Following the procedure above we see that we need 4 x 96,485 coulombs or 107Ah/g-mol when divided by 3,600.
Total Coulombs requires = 4 x 96485 = 385,940 coulombs
This means to convert 1g-mol of red lead to PbO2, we need 385,940/3600 = 107.21Ah
However, 1g-mol of Pb3O4 weighs (207×3) + (16 x 4) = 685g
One kilo of red lead will contain 1000/685 = 1.46g-mol of Pb3O4
This will require 107.21 x 1.46 = 156.5Ah per kg of Pb3O4
This exercise shows clearly that the theoretical Ah input to convert dry cured PAM can be reduced by adding lead oxides with a greater degree of oxidation. This is also true of lead sulphate PbSO4, which, like lead monoxide, has a lead valency of two. That would give the same Ah input as PbO. Unfortunately lead acid battery unformed PAM is rather more complex, as shown in Table 1.
The percentage weight of each of the constituents is important when calculating its effect of increasing or decreasing one or more of the PAM. The total theoretical energy requirement is simply: the sum of the percentage of each constituent x the calculated formation energy:
FE = Sum (%X1xQ1 + %X2xQ2 + %XnxQn) x Y
Where FE is the theoretical formation energy for one positive plate, %X is the percentage of a dry charged component in the PAM, Q the theoretical Ah calculated for that component and Y the total weight of the dry cured PAM.
Unfortunately, the amounts of each are variable according to paste mixing and curing processes. For this reason, it is necessary to ascertain the composition of a particular manufacturer’s PAM in order to assess the effectiveness of changing the constituents.
Inefficiency of material conversion and the battery formation process
It is a known problem that the formation of PAM, both materially, as measured by the g/Ah available from the formed active mass, and energetically, as measured by the formation energy input, is far from efficient.
As explained in BEST magazine Winter 2026, almost 50% of the PAM structure is employed as a conductor and a framework to support the electron producing portion of the material Fig 2. This factor alone means that for every Ah of formed capacity, around 50% of the formation energy input is used for passive components in the PAM.
Likewise, the energy of formation is not entirely consumed in converting the dry cured positive paste into formed active material. This is a double whammy of increasing internal resistance, plus the energy drain from the parasitic reactions of water electrolysis, gas evolution and heat generation.
As formation progresses the voltage rises due to increasing internal resistance. It is this voltage rise that triggers several reactions Fig 3:
Water electrolysis 2H2O → 4H+ + 2O– + 4e–
Hydrogen gas evolution 2H+ + 2e– → H2
Oxygen gas evolution O2- → ½O2 + 2e–
The coulombs required to produce these reactions plus the increased voltages needed to drive the beneficial conversion reactions, push up the watts (V x A) and therefore the energy required to complete the formation process, or at least 90% of it.
Control of formation battery temperatures
Another practical consideration is the heat generated from the sources identified above.
Heat is generated according to the following relation:
Heat generated (Q) in Joules
Q = I2Rt where I = Amps,
R = ohms and t = time
For convenience the equation can be written as watts rather than joules:
Power = watts = V x I, since V = I x R, then watts = I x R x I = I2R. Therefore Q = watts x t
Since J = W.s (Watt seconds) the total heat generated (energy) can be expressed as watts x t, or watt- hours where 1Wh = 3,600J
However, we need only consider the magnitude and rate of heat generation when considering the water flow needed to remove the heat during battery formation.
First step is to calculate the maximum heat generated at the end of the formation period. This is because the internal resistance of the cells will vary for the reasons given earlier. A simplified method to calculate the heat generated per cell is to limit the heat factors to the current input and the charging voltage:
Heat Q = (Voc – Vch) x I (watts)
where Voc = open circuit cell voltage and Vch = cell charging voltage
Second step is to determine the water flow rate necessary to remove the heat generated (Q)
Q = m x Cp x dT
where m= the mass flow rate of water (Kg/s), dT = temperature difference between added water and heated water in degrees Kelvin (°K), Cp = specific heat capacity of water in J/(kg.°K) (approximately 4.200), Q is the heat generated from one cell in watts (J/s)
The required mass water flow rate to remove the generated heat from one cell is m = Q/(Cp x dT)
The required volume of flow in litres/min = m x 60/Dw, where Dw is the density of water = 1Kg/L.
It is important to note here that the internal cell resistance will change during the formation schedule. It is possible to have a variable flow rate in order to conserve water and/or energy, based on knowledge of the change in resistance during the various formation phases. This is useful provided that the water flow is controllable.
There are more considerations to take into account in order to calculate very accurately the heat generated. Such considerations include the endothermic electrochemical reactions of water electrolysis and gas evolution. However, these are quite small compared to that of the joule heating from the applied current and voltage from the charging reactions. In fact, towards the end of the process, most of the current is consumed in those parasitic reactions, which is another major source of inefficiency. These, along with the low PAM utilisation described earlier push up the formation energy requirement to around 4–6 times the cell’s C5 capacity.
As regular readers of BEST magazine are aware this has been a major focus of R&D by myself and partners over the last four years. This very practical research is continuing to make substantial advancement in reducing the energy, time and cost of this highly inefficient process. Our work and progress in formation efficiency, as always, will be reported in this magazine.
