25-05-2012, 01:54 PM
Battery Sizing Example
Sizing batteries is a 4-step process.
Step 1 – compute the characteristics of the load.
Before you start selecting batteries, you have to size the load. You need a guess at the things you need to power and how long they will be on. Specific parameters of interest include what its operating voltage will be, how much current does it need, and what is the power level needed. Other secondary features of interest would include the “quality” of the power – the input voltage must be maintained within +/- 0.2 Volts, etc.
The minimum you need is any two of Voltage, Current, Power in addition to time on.
Step 2 – select an operating voltage.
From your load characterization, you need to select a voltage. This is important especially for small systems because most of the time, the battery size is driven by the system (volts and amps) more than the energy storage. If your battery is composed of a number of say D-cells where each D-Cell has 3 volts, then you need 10 D-Cells in series to provide the standard 28Volts for an electrical distribution bus. The example of the DS-2 probes, indicated they had either 6 volts or 14 volts so one battery would have needed 2 cells (at 3-volts each) while the other battery would have needed 5 cells.
Step 3 – select a current
The selected load has a power, and when coupled with the voltage gives you a current. You then need to sum up the total current you expect to need at any given time. Then you have to look at the battery cells you have selected. Since you have cells in series to provide a voltage, you now have to add stacks of these series cells in parallel to provide the current. Each string or stack of series connected cells is limited in its current providing capability by the single cell output capability. Each cell has a maximum output current (continuous as well as pulse). From the total current you need for your load, you must divide by the continuous current capability of your cells. This tells you the number of parallel strings you need to provide the current. The assemblage of series and parallel strings then constitutes your battery.
Step 4 – determine the number of cells to provide the time phased load
The final step in the battery sizing is to make sure there are enough cells to meet the demand. Use the load determined in step 1 and multiply by the hours of operation. This gives you the total integrated energy the battery pack must provide (watt-hours, or amp-hours – if you divide by the voltage). From the catalog information, determine the energy content of a single cell. If you are using a primary battery, you can only get between 60% and 80% of this energy out before the battery overheats and exhibits thermal runaway (gets hot and explodes). If you are using a secondary battery, then you typically use 40% Depth of Discharge. Compute the effective energy content by cell by multiplying this depth of discharge by the energy content of a single cell. Divide the effective (or usable) energy content into the total integrated energy level you need. This tells you the minimum number of cells you need. However, you must remember the cells must come in integer multiples of what you need for your voltage. So if you have 3 cells in series to provide your specified voltage, and your energy content calculation says you need only 4 cells, then you actually need 6 cells to maintain the proper voltage and provide the energy content.
Example:
You have a set of devices to power using secondary cells. The power you need is 2 watts. The devices want to operate at 6 volts. You need to provide 100 hours of operation.
Power – 2 Watts
Voltage – 6 Volts
Current – I = P/E = 2/6 = 0.33 Amps
Energy – Energy = 2 Watts * 100 hrs = 200 W-hrs
You have selected a battery cell from SAFT – Lithium Sulfur Dioxide, model LO-35-SX. This battery has a 2.9 volt recommended voltage, with a maximum continuous 2 amp drain per cell. Each cell is 30 grams and contains 2 amp-hours (at 120 ma) of energy.
Cell voltage = 2.9 volts
Max. Current = 2 amps
Energy content = 5.8 watt-hours (this assumes you can extract the same energy at 2 amps as you can 120 milli-amps – not really a valid assumption) – Primary cell so only use 60% Depth of Discharge = 0.6*5.8 = 3.48 Watt-hrs
How many cells for voltage? N = 6/2.9 = 2.1 cells – round up to 3 cells in series
How many strings for current? N = 0.33 amps/2 amps = 0.16 cells – round up to a single string of cells.
How many cells for energy content? N = 200 W-hrs/3.48 W-hrs = 57.5 cells. Now divide by 3 since must come in even multiples of 3 = 57.5/3 = 19.2 strings – round to 20 strings.
Final Battery:
20 strings of 3 cells each. Each string of 3 connected in parallel. Total of 60 cells at 30 grams each results in 1800 grams for just the cells. Need to add about 15% to the mass to account for interconnecting wiring, packaging, and thermal dissipation. Total battery mass = 1800*1.15 = 2070 grams.