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# Two test methods of lithium battery power

2022/04/08

Author ：Iflowpower – Portable Power Station Supplier

It is often two ways to detect whether the power of ordinary zinc manganese dry batteries is sufficient. The first method is to estimate the internal resistance of the battery by measuring the instantaneous short circuit current of the battery, and then determine whether the battery is sufficient; the second method is a suitable resistance appropriate resistance in series with a current table, and calculate the battery by measuring the discharge current of the battery. Internal resistance, to determine if the battery is sufficient.

The greatest advantage of the first method is simple. It can directly determine the power of dry batteries. The disadvantage is that the test current is very large, far exceeds the limit value of the discharge current to a certain extent affects the dry battery.

life. The advantage of the second method is that the test current is small and safe. It generally does not affect the service life of the dry battery, and the disadvantage is more troublesome.

The author used the MF47 million meter to test a new 2 dry battery and the old No. 2 dry battery, respectively. It is assumed that RO is a dry battery internal resistance, and RO is a current table internal resistance.

When using the second test method, RF is an additional series resistance, 3Ω, power 2W. The measured results are as follows. New No.

2 battery E = 1.58V (measured with 2.5V DC voltage file), the internal resistance is 50kΩ, which is much larger than RO, so it can be approximately considered 1.

58V is the electric potential of the battery, or the open circuit voltage. When using the first method, the multimeter is set 5A DC current gear, the mesh internal resistance RO = 0.06 Ω, the current is 3.

3A. So RO + RO = 1.58V ÷ 3.

3A≈0.48Ω, RO = 0.48-0.

06 = 0.42Ω. When using the second method, the current is 0.

395a, RF + RO + RO = 1.58V ÷ 0.395a = 4Ω, the current 500mA file internal resistance is 0.

6Ω, so RO = 4-3-0.6 = 0.4Ω.

When the first method is measured with the first method, the open circuit voltage E = 1.2V is measured. The internal resistance RO = 6Ω, the reading is 6.

5mA, the mean table is 50mA DC current gear, RO + RO = 1.2V ÷ 0.0065A ≈184.

6Ω, RO = 184.6-6 = 178.6Ω.

With the second method, the measured current is 6.3mA, RO + RO + RF = 1.2V ÷ 0.

0063A = 190.5Ω, RO = 190.5-6-3 = 181.

5Ω. Obviously the results of the two test methods are basically. The minor difference of the final calculation results is caused by multiple factors such as reading errors, errors of resistance RF, and contact resistance.

This tiny error does not affect the determination of battery power. If the battery is measured, the voltage is high (for example, 15V, 9V stacking battery), the resistance of RF should be increased.