li-ion 18650 battery

Why is it inappropriate to use li-ion 18650 battery in parallel? Do you understand the characteristics of parallel circuits?

For a long time, whether it is domestic or foreign, whether it is a communication system or a UPS system, people have been accustomed to using two sets of batteries in parallel for use with a UPS or a communication device. I don’t know whether it is because of habit or other reasons, but this method of parallel use has become a principle that designers and users must follow. However, the author believes that it is not necessary, as long as the user can follow the battery The manufacturer's instructions for battery maintenance are good. It is enough to use only one set of batteries. Not only is it enough, but also the use effect of this set of batteries (such as: battery stability, reliability, balance, especially the battery's service life, etc.) will be much better than when two sets of batteries are used in parallel. This is especially true for valve-regulated sealed lead-acid batteries. So, why does the author actively advocate (or even disapprove) that li-ion 18650 battery should not be used in parallel? What are the pros and cons of using them in parallel?

First, let's review the characteristics of parallel circuits. In a parallel circuit, the total voltage is equal to the voltage of each branch. That is to say, the charging voltage applied to each group of batteries in the two groups of batteries connected in parallel is equal to the total charging voltage, that is, Utotal=U1=U2. According to the formula of I=U/R, it can be known after calculation that I1≠I2 (because the internal resistance of the two sets of batteries will definitely not be the same, that is, R1≠R2. In the case of U1=U2, I1 ≠The result of I2). That is to say, under the condition of the same charging voltage, two groups of li-ion 18650 battery used in parallel will receive different charging currents. The one with a large internal resistance will have a small charging current, and the one with a small internal resistance will have a small charging current. The current is large. In this way, it is possible that the battery set with a small charging current is often undercharged. Over time, this set of batteries may become sulfated due to long-term power loss and increase its internal resistance. The greater the internal resistance, the higher the charging current. Because it creates such a vicious cycle, the service life of this battery pack is greatly shortened. This does not exist with just one set of batteries. This point is enough to show that the effect of using a single battery pack is much better than using it in parallel. Therefore, the author recommends that users never use two sets of batteries in parallel when one set of batteries can meet the needs of the device. Otherwise, it will shorten the service life of the battery, increase the cost of use, and reduce the overall performance of the battery. You shouldn't do such wasteful things. If due to the high power of the equipment, two groups of batteries connected in parallel still cannot meet the power requirements of the equipment, and more than 2 groups, such as 3 groups, 4 groups, or even more groups of batteries are used in parallel, it is even more unnecessary. , the parallel use of two sets of batteries has brought many disadvantages, and the parallel use of more sets of batteries will be more complicated and even more disadvantageous. In this case, you must choose a large-capacity battery that can meet the power needs of the equipment. If there is no large-capacity battery in the 12V series battery, you can choose a 2V series battery. Among the 2V series batteries, various large-capacity batteries are available. They are all available. You can make them as large as you need. As far as the author knows, the largest 2V series batteries currently available in China can reach 6000Ah.

Of course, it is understandable that designers and users consider improving the reliability of backup power supply. If one of the two sets of batteries fails to supply power in the event of an AC power outage, another set of batteries can be used to ensure that. Even dry? ? ? It's worth the trouble for people and money. If you consider using li-ion 18650 battery in parallel from this perspective, the author only agrees to use up to two groups of batteries in parallel. If more than two groups are connected in parallel, it is definitely harmful and unhelpful. If it is necessary to use two sets of batteries in parallel, please also follow the following principles: First, the batteries used in parallel must be produced by the same manufacturer, and be of the same model and specification; The batteries must be in the same new and old condition; third, they must be from the factory with the same batch number at the same time; fourth, they must be installed and used at the same time.

The lead-acid battery is a water electrochemical system with a positive electrode and limited liquid mass transfer. During the operation of this system, gas will be produced (hydrogen evolution, oxygen evolution), resulting in water loss. Therefore, maintenance by adding water and replenishing fluid is required.

Maintenance-free (meaning no need to add water to replenish fluids) is the simplest instinctive requirement of people. In the process of realizing maintenance-free lead-acid batteries, we have gone through a long and tortuous road, many of which use catalytic hydrogen elimination, auxiliary electrodes and other methods. .

4. Take advantage of power supply peak charging

For users whose UPS power supply is powered by low voltage from the mains for a long time or has frequent power outages, in order to prevent the battery from premature damage due to long-term insufficient charging, the battery should be fully charged during the power supply peak period (such as late at night) to ensure that the battery is discharged every time. There will be plenty of charging time afterwards. Generally, after a battery is deeply discharged, it takes at least 10 to 12 hours to recharge it to 90% of its rated capacity.

5. Pay attention to the selection of charger

Maintenance-free sealed batteries used in UPS power supplies cannot be charged with thyristor-type fast chargers. This is because this kind of charger will cause the battery to be in a bad charging state of both instantaneous overcurrent charging and instantaneous overvoltage charging at the same time. This state will greatly reduce the usable capacity of the battery, and in severe cases, the battery will be scrapped. When using a UPS power supply with a constant voltage cut-off charging circuit, be careful not to adjust the low battery voltage protection operating point too low, otherwise, overcurrent charging will easily occur in the early stages of charging. Of course, it is best to use a charger with both constant current and constant voltage to charge it.

6. Ensure the ambient temperature of the power supply

The usable capacity of the battery is closely related to the ambient temperature. Under normal circumstances, the performance parameters of the battery are calibrated at room temperature of 20°C. When the temperature is lower than 20°C, the available storage capacity will be reduced, and when the temperature is higher than 20°C, its available storage capacity will be reduced. The capacity used will increase slightly. Batteries of different manufacturers and models are affected by temperature to varying degrees. According to statistics, at -20°C, the usable capacity of the battery can only reach about 60% of the nominal capacity. It can be seen that the influence of temperature cannot be ignored.

Of course, to extend the service life of the battery pack, not only must attention be paid to maintenance and use, but also the load characteristics (resistance, inductance, capacitance) and size should be fully considered when selecting. Do not run the battery under excessive light load for a long time to avoid the battery being scrapped due to low battery discharge current.

There are usually two methods.

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 power is sufficient; the second method is to use an ammeter in series with a resistor of appropriate resistance, and calculate the battery by measuring the discharge current of the battery. Internal resistance to determine whether the battery power is sufficient.

The biggest advantage of the first method is that it is simple. You can directly determine the power of the dry battery by using the high current setting of the multimeter. The disadvantage is that the test current is very large, which far exceeds the limit of the allowable discharge current of the dry battery, which affects the use of the dry battery to a certain extent. life. The advantage of the second method is that the test current is small, the safety is good, and generally it will not have any adverse effect on the service life of the dry battery. The disadvantage is that it is more troublesome.

The author used the MF47 multimeter to test and compare a new AA dry battery and an old AA dry battery using the above two methods. Assume that ro is the internal resistance of the dry cell and RO is the internal resistance of the ammeter. When using the second test method, RF is an additional series resistor with a resistance of 3 ohms and a power of 2W.

The actual measurement results are as follows. The new No. 2 battery E=1.58V (measured with 2.5V DC voltage range), the internal resistance of the voltmeter is 50k ohms, which is much larger than ro, so 1.58V can be approximately considered to be the electromotive force of the battery, or open circuit voltage. When using the first method, the multimeter is set to the 5A DC current setting, the internal resistance RO of the meter is 0.06 ohms, and the measured current is 3.3A. So ro+RO=1.58V÷3.3A≈0.48 ohm, ro=0.48-0.06=0.42 ohm. When using the second method, the measured current is 0.395A, RF+ro+RO=1.58V÷0.395A=4 ohms, and the internal resistance at the 500mA current range is 0.6 ohms, so ro=4-3-0.6=0.4 ohms.

When measuring the old AA battery using the first method, first measure the open circuit voltage E=1.2V, the internal resistance RO=6 ohms, and the reading is 6.5mA. Set the multimeter to the 50mA DC current setting, ro+RO=1.2V÷0.0065 A≈184.6 ohms, ro=184.6-6=178.6 ohms. Using the second method, the measured current is 6.3mA, ro+RO+RF=1.2V÷0.0063A=190.5 ohms, ro=190.5-6-3=181.5 ohms.

Obviously, the results of the two test methods are basically consistent. The slight difference in the final calculation result is caused by various factors such as reading error, resistance RF error and contact resistance. This small error will not affect the judgment of battery power. If the capacity of the battery under test is small and the voltage is high, the resistance of RF should be increased appropriately.

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