The mini power supply uses a built-in battery (18650 and polymer) with a nominal voltage of 3.7V, while digital products such as mobile phones have a standard input voltage of 5V. Therefore, the mini power supply needs to output 5V through the boost management board to charge the mobile phone and other devices.
According to the law of conservation of energy, the energy of the battery is 9000mAh×3.7V=33.3 (Watt hours). After the boost is converted to 5V, the overall energy is unchanged. The power of the mini power supply is 33.3Wh÷5V=6660mAh, which is like a cup of 3.7 liters. The water was poured into a 5 liter cup. After a one-step conversion, the original 9000mAh power is only 74%, which is the actual power that can be supplied to the phone. Someone may ask 6660mAh to get all the batteries in the phone? Of course it is no. Because, during the charging process, there are still many losses in these batteries:
1, circuit conversion loss. I believe everyone knows that the mini power supply will heat up when charging. According to the law of conservation of energy, the heat dissipated is actually the energy of the built-in battery.
2. Resistance loss. When charging the mobile phone, the built-in battery of the mini power supply and the mobile phone battery are connected by a circuit board and a wire. We know that mini-power supplies have electrical resistance for conductors and conductors without zero resistance (superconductors are beyond the scope of this discussion). The main physical characteristic of the resistor is to convert the electric energy into heat energy. Therefore, during the charging process, the mini-power source will generate heat, and the heat will have the dissipation and loss of energy.
