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I am using the BQ24075 battery charger IC (datasheet here). If I am reading the datasheet correctly, on page 6 it says that the absolute maximum sink current for the CHG output pin is 15mA. I require this pin to sink at least 40mA so I assume I will need to use a P-channel MOSFET whose gate is controlled by the CHG pin.

The circuit I am considering resembles the following:

schematic

simulate this circuit – Schematic created using CircuitLab

When the device is not charging, I expect CHG to be high impedance. When the device is charging, CHG will be pulled to GND which will switch the MOSFET on and allow current to flow into the load. Please ignore the load resistor, I have placed it there only for visual reference.

From reading other posts that concern the gate resistor, it appears as though having a large (>100Ω) resistor may cause a significant delay switching the MOSFET on. I am not too concerned with the delay, but are there other concerns with using this resistor? Could I increase the value so that less current is used when the CHG pin is pulled to GND? Finally, would this circuit work correctly as I have described it?

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  • \$\begingroup\$ This should work fine. I would suggest you change R1 to around 100k. There is a wide range of parts that would work for M1. The one you show, 2SJ305, should work fine. Typically, the CHG output would be used to sink current from an LED to indicate charge. It can be desirable for the charger IC to do this autonomously without any intervention from the processor, because when VBATT is very low, the processor may not be able to boot up. \$\endgroup\$ – mkeith Sep 10 '16 at 23:50
  • \$\begingroup\$ @mkeith Thanks for the input. I would like to use the CHG pin to control a status LED and the LED of an optoisolator. Both of them together will take ~30mA (I put 40mA in the question for headroom). My issue is the effect of R1 on the MOSFET. \$\endgroup\$ – John P. Sep 11 '16 at 0:04
  • \$\begingroup\$ Don't worry about it. It will work fine. When CHG is high impedance, the PMOS will be off. When CHG is asserted (driven low), the PMOS gate will be pulled down low enough to turn on the channel. Like I said, I see no reason to use 1k, but it will work. I would use 100k. Give some thought to what voltage you want to use for your pull-up. It may make more sense to pull it up to the USB voltage rail rather than the battery (because battery voltage is more variable). CHG will never be asserted when USB is absent, so it is OK to do this. \$\endgroup\$ – mkeith Sep 11 '16 at 0:16
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    \$\begingroup\$ Why are you trying to drive the LEDs so hard? You can get away with a few mA of draw for each of the optocoupler LED and the status LED -- in fact, a modern GaN blue or green LED can be driven adequately at the submilliamp level. \$\endgroup\$ – ThreePhaseEel Sep 11 '16 at 0:41
  • \$\begingroup\$ @ThreePhaseEel I need the status LED to be as bright as possible. I will probably use a different circuit at this point but I'm still really curious if this would have worked. I don't see why it wouldn't but someone with a good amount of rep said it wouldn't work and then deleted his/her answer. \$\endgroup\$ – John P. Sep 11 '16 at 5:59

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