Could you believe it? Affordable charger circuits on the market still don’t come with load sharing components built in these day and age. So I had to build my own. I’m somewhat new to this so bear with me.
- Charging IC - TP4056
- P-MOSFET - DMP1045UQ-7
- Schottky diode - PMEG6010ELRX
Here’s a schematic of how the load-sharing circuit would look, albeit wired based on the TP4056 rather than the MCP73872:
When researching which pull-down resistor I should purchase, I ran into an issue due to the PMEG6010ELRX Schottky diode that I have.
This diode is rated with a reverse leakage current(Ir) of 5nA@Tj 25°C; 5V, which is pretty low.
According to a design guide posted here, it is recommended to keep the Rpull resistor at no more than 100kΩ and when selecting the Rpull resistor we should use the formula;
Rpull = Vtarget * RD / (VBATmax – Vtarget) where, RD= VBATmax / IR.
In this case Vtarget is a range from 1-3V, IR=5nA and VBATmax=4.2V. In most cases including in the example they had in the guide, IR is in the 100-200uA range which yields something in 6-100kΩ range.
However, PMEG6010ELRX has an incredibly low leakage current in the single digit nano Ampere range.
How would that affect my choice of selecting a resistor to pull down the voltage so the P-MOSFET will function normally?
Otherwise I’d get something in the MΩ or even GΩ region which is much more than the recommended 100kΩ. I’m under the impression that this resistor’s rating doesn’t have to be near the actual value obtained from the formula. So long as it doesn’t exceed (<100kΩ) regardless of how small the leakage current of D1 is, will the PMOSFET still work correctly?