Power characteristic test method of lithium ion battery
A technology of lithium-ion battery and power characteristics, which is applied in the direction of electric power measurement through current/voltage, measurement of electricity, measurement device, etc., can solve problems such as large deviation from the actual situation, and achieve the effect of facilitating the comparison of product power characteristics
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Embodiment 1
[0055] Taking a 18650-1100mAh-3.2V battery as an example, the test steps for continuous charge and discharge power characteristics are as follows:
[0056] 1. Continuous charging power characteristic test:
[0057] (1) The battery is placed at room temperature for more than 12 hours, and the battery is discharged at 1C to the cut-off voltage, and the battery is left standing for 10 minutes.
[0058] (2) Charge according to the 2C current respectively, and switch to constant voltage charging when the constant current charging reaches the charging termination voltage specified by the product, until the charging current drops to 0.05C 1 (mA) when charging is stopped, such as figure 1 .
[0059] (3) Calculate the ratio of the constant current charged electricity C1=1.017Ah to the total charged electricity C0=1.075Ah, that is, α=C1 / C0=1.017 / 1.075=0.946.
[0060] (4) Calculate the average voltage U1 during the constant current process, U1=W1 / C1=3.478Wh / 1.017Ah=3.420V.
[0061] 2...
Embodiment 2
[0068] Taking a 26650-2300mAh-3.2V battery as an example, the pulse power characteristic test steps are as follows:
[0069] (1) Take two batteries and charge them with a constant current to 50% SOC.
[0070] (2) Place the battery at -20°C for 12 hours.
[0071] (3) Discharge the battery according to the current of 20C, the discharge time is 10 seconds, and the discharge stop interval is 30 seconds. Discharge was then performed for 10 seconds, and the discharge was stopped for 30 seconds.
[0072] (4) Select different time points a during the test as 0.1 seconds, 2 seconds, 5 seconds, 10 seconds, 40.1 seconds, 42 seconds, 45 seconds, and 50 seconds. The corresponding current is Ia, voltage is Ua, and power Pa= Ua×Ia, as shown in Table 1.
[0073] (5) Repeat the above steps to test and set the temperature in step (2) to 25°C. The test results are shown in Table 1.
[0074] From the above tests and the data in the table, it can be seen that under low temperature conditions,...
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