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I'm designing a charger circuit. When the input (+5v) is available it should disable the booster regulator by setting Booster_en to low. I used a single P-ch MOSFET like below circuit. I added pull-down to the gate of FET for when there is no (+5v) it turns on the FET and Booster_en gets high. I found many examples that use a pull-up circuit on FET's gate and some of them use a Schottky diode before VCC. Do I need to use two resistors or there are simpler ways? How about the diode? enter image description here EDIT: Here is the first version of circuit that the problem was on Booster_en that becomes floating when VUSB is off. On the other side, I want to fully isolate VBat when +5v (VUSB) exists. I also removed the Q4 transistor and connected R9 to ground. I think it is useless because when there is no VDD for charge it would turn off. enter image description here

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    \$\begingroup\$ Need the rest of the circuit. You want the gate referenced to the source and not to ground. \$\endgroup\$ Oct 22, 2018 at 14:49
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    \$\begingroup\$ What is your battery voltage? If you want Q1 to operate as an inverter then you have the drain and source reversed (as a hint, look at the intrinsic diode and ask yourself what it is going to do to Booster_en all the time). \$\endgroup\$
    – TimWescott
    Oct 22, 2018 at 14:51
  • \$\begingroup\$ @GeorgeHerold Added the circuit. \$\endgroup\$ Oct 22, 2018 at 15:04

2 Answers 2

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Personally, I find the most reliable method of switching a PMOSFET is by using an NMOSFET. The circuit I generally tend to use is like this:

schematic

simulate this circuit – Schematic created using CircuitLab

With this circuit, R1 is pulling up the gate of M1, keeping it off. R2 is pulling down the gate of M2, also keeping it off. Once the 5V is applied to the gate of M2, it will pull the gate of M1 low, which will allow M1 to enable the Boost_EN pin.

This is a circuit I have used many times and it has always been reliable.

Edit due to addition of circuit in original question

The original wording of the question sounded like the EN pin was needing to go HIGH when 5V was applied. Now it seems that the intention is for the pin to go LOW when 5V is applied. This simplifies things. Replace M1 with a pull-up resistor:

schematic

simulate this circuit

With this circuit, your EN pin is always pulled HIGH by the resistor R1. When 5V is applied, M2 will pull the pin LOW.

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  • \$\begingroup\$ Do you still think this fit with the new circuit? Is there any way that I can replace M2 with Q4? I want reverse logic. When 5v exist the Booster pin must be set low. \$\endgroup\$ Oct 22, 2018 at 15:05
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    \$\begingroup\$ Ahh. The addition of your circuit has changed things. Your original description looked like you wanted the EN pin to go high when 5V was there. I'll edit my answer \$\endgroup\$
    – MCG
    Oct 22, 2018 at 15:15
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    \$\begingroup\$ @MasoudR. the answer is now updated \$\endgroup\$
    – MCG
    Oct 22, 2018 at 15:24
  • \$\begingroup\$ Thanks, but is there any way that I can isolate VBat when the +5v is on. In the above circuit when +5V applies the R1 would drain VBat. But in my first circuit, VBat would be totally off the circuit. Am I right? \$\endgroup\$ Oct 22, 2018 at 15:34
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    \$\begingroup\$ Tie it elsewhere that's always high then \$\endgroup\$
    – MCG
    Oct 22, 2018 at 16:02
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Use a 1-gate inverter with 5V tolerant input and a VCC range that accommodates a single LiPo, powered off of VBatt, like an NC7SZ04. No resistors, no transistor weirdness, just a working part.

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