![]() ![]() ![]() Shutting off completely when charge current drops to ~10% of the set current.Ĭharging at a lower rate when voltage is below 3.0V, to prevent damaging a cell which has been deeply discharged. This circuit is about the minimum required to safely charge a Lipo cell, but dedicated charger ICs provide more features to enhance safety and cell life, including:. C2 maintains high frequency stability if the battery is disconnected. Q3 and Q4 compare the Lipo voltage to the reference voltage, turning on Q5 and reducing the charge current when battery voltage reaches 4.2V. The following circuit provides the required CVCC (Constant Voltage Constant Current) charge profile, using all discrete components except for the TL431 'Precision Programmable Reference' (I could have used a plain Zener diode here, but low voltage Zeners have poor regulation, and we need a very precise reference voltage):. So you will need a circuit which reduces the current when battery voltage reaches 4.2V. This circuit alone is obviously not sufficient for charging a Lipo battery, as the voltage must not be allowed to go above 4.20V. 5V - 0.4V (across R3) - 0.2V (across Q2) = 4.4V maximum output voltage at 100mA. Q2 has a saturation voltage of <0.2V at 100mA. At 100mA Collector current Q2's Base-Emitter bias is ~0.8V, leaving ~0.4V across R3. Q1, R1, and R2 act as a "Vbe multiplier" which drops about 1.2V (the exact voltage can be adjusted by varying the ratio of R1/R5). Simulate this circuit – Schematic created using CircuitLab Here's one that regulates current to 100mA at up to 4.4V on a 5V supply:. ![]() It could be if you used an appropriate circuit. ![]() so if you charge from a power bank or some solar power installation or some source where it's critical to be as efficient as possible, it may make more sense to use the 5v only charger.I have tried using discrete BJT to do it. but converting 12v to 3.7v.4.2v may be only 92-95% efficient. for example using 5v to charge a 3.7v.4.2v battery may be done with 97% efficiency. If you charge them slower, at only 10 watts, it will take longer to charge them but the batteries will be cooler and they may charge up to 98-100% of their rated capacity.Ĭharging with just 5v can also be more efficient. Yes, you may push 18w into the battery and charge it fast, but the batteries may heat up a lot and at the end of the charging process, the batteries may hold let's say 90% of the maximum capacity. If your phone does not support Quick Charge or an alternative method, then the fancy charger will only provide 5v at 1.5A, so less than 5v at 2A.Īlso, it's debatable if it's a good idea to shove so much power into a battery very fast. However, the device (phone or whatever) must either be capable of working directly with 12v, or must support Quick Charge or USB Power delivery or some other method of signaling the charger to raise the voltage from the default of 5v to 9v or 12v.īy default, the charger will output 5v - the phone or whatever you have normally "talks" to the charger and tells the charger it's safe to raise the voltage from 5v up to 12v. The 5v 2A charger can only give 10 watts. In theory the 12v 1.5A will charge a battery faster, because it can provide up to 18 watts to your device. ![]()
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