Series resistor & diode - powering TP4056 from USB? - Page 1
Forum section: electronics / projects

author: Peabody
author_rank: Super Contributor
timestamp: April 29, 2020, 04:48:11 pm
content: The datasheet for the TP4056 LIPO charger I/C shows a 0.4Ω inline resistor between the USB 5V source and the TP4056 Vcc pin. It shows the resistor with a range of 0.2 - 0.5Ω, and it's in a dotted-line box, which may mean it's optional.I've also found a schematic for the 18650 charger/regulator shield V3, and the relevant part shows both the 0.4Ω resistor and an inline Schottky diode.I don't understand the purpose of these parts. It's hard to see what effect a 0.4Ω resistor would have on anything. And it's hard to see that the diode has any purpose other than to drop the voltage and generate some heat. Even without the diode, I don't think it's possible for current to flow back through the TP4056 from the battery or load to the VBUS pin. So is it just there to move some heat from the TP4056 (which is a linear part) to the diode, and keep the TP4056 from overheating? I didn't think overheating was a typical problem with the TP4056.Anyway, if someone could enlighten me, I would appreciate it. I'm going to modify the shield, and could make use of the diode for another purpose if it isn't really needed where it is. If it matters, the charging current is set at 600mA.

author: mikerj
author_rank: Super Contributor
timestamp: April 29, 2020, 07:59:38 pm
content: Probably to reduce power dissipation in TP4056. When battery voltage is low, the TP4056 will be delivering its maximum current and dropping the greatest voltage so power dissipation will be highhighest power. By adding an external resistor to drop the input voltage under the high current conditions power dissipation of the IC is reduced.Of course if you are using a cheap and nasty USB cable you'll likely already have half an ohm or more in the circuit...

author: THDplusN_bad
author_rank: Regular Contributor
timestamp: April 29, 2020, 08:18:10 pm
content: After a brief look into the (original) LTC4056 data sheet specs from Linear Technologies, I am confident that the diode shall protect from any reverse polarity.My bet is that the 0.4R resistor is intended to act as a fuse...Cheers,THDplusN_bad

author: Peabody
author_rank: Super Contributor
timestamp: April 29, 2020, 10:23:28 pm
content: Quote from: mikerj on April 29, 2020, 07:59:38 pmThe datasheet also says:"Thermal feedback regulates the charge current to limit the dietemperature during high power operation or high ambient temperature."But I guess if the resistor dissipated some of that heat, charging could continue at maximum current rather than at a reduced level. So this would explain the resistor, but not the diode. You could adjust the value of the resistor to achieve the same voltage drop at maximum current, and avoid the need for the diode at all.

author: Peabody
author_rank: Super Contributor
timestamp: April 29, 2020, 10:57:20 pm
content: Quote from: THDplusN_bad on April 29, 2020, 08:18:10 pmYes, I saw that. But this device is powered from USB. You will not have reversed polarity at the USB connector. The only thing one could do is insert the battery backwards, but it's not clear whether the diode would have any benefit where it is placed.