Greetings. A vendor selling TP4056-based 5 V charging modules for Li-ion cells specified the following resistance values and resulting current for selecting the current-programming resistor:

On account of Ohm's law the range, median, and mean of the calculated voltage are, respectively, 1145.4 V - 1500 V; 1200 V; 1247.5 V. I have interpolated the values and fitted a cubic polynomial regression thus:

The stock resistor resistance is 1.2 kΩ and, on the basis of its length and width of 1.6 mm × 0.8 mm, its power rating should be 1/16 W or thereabouts. With this resistor the resulting current is 1 A according to the TP4056 specifications.

Now, why does the calculated voltage have such high values and such a wide range? How should one calculate resistance values in order to get lower amperages, e.g. 4 mA? If the actual voltage across the current-programming resistor is 5 V or 4.2 V—which one is it?—and the current is 1 A, should not the resistor power rating be greater than or equal to 5 W or 4.2 W?

Thanks.

R (kΩ) | I (mA) |

30 | 50 |

20 | 70 |

10 | 130 |

5 | 250 |

4 | 300 |

3 | 400 |

2 | 580 |

1.66 | 690 |

1.5 | 780 |

1.33 | 900 |

1.2 | 1000 |

The stock resistor resistance is 1.2 kΩ and, on the basis of its length and width of 1.6 mm × 0.8 mm, its power rating should be 1/16 W or thereabouts. With this resistor the resulting current is 1 A according to the TP4056 specifications.

Now, why does the calculated voltage have such high values and such a wide range? How should one calculate resistance values in order to get lower amperages, e.g. 4 mA? If the actual voltage across the current-programming resistor is 5 V or 4.2 V—which one is it?—and the current is 1 A, should not the resistor power rating be greater than or equal to 5 W or 4.2 W?

Thanks.

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