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I have a package-integrated charging module connected to my 2-cell battery. When the charger is inactive (disconnected from input voltage), the charger eats power from my battery. I thus want to prevent reverse current from going to the charger when it is inactive. When the charger is active, it puts out ~ 8.8 V and 500 mA max.

I'm playing around with modeling of an n-channel MOSFET and it seems like tying the gate to the source (which in this case would be connected to the charger) would work. It is capable of putting through enough power to the battery when the battery is at 8 V, and according to the model, it drops the amperage to <1 μA when the source voltage drops to 0. Is there any reason I should not implement it this way? I'm just surprised I haven't seen this done in any examples of reverse current protection online.

Here are some screenshots of the model:

When charger is active, and assuming the battery is at 8 V, while the charger is at 8.8 V, the mosfet will let through 1.1 A, which is enough for this application:

charger on mode

When charger becomes inactive, and voltage drops to 0 (or less than Vd), the MOSFET switches off and current goes to <1 μA

enter image description here

This solution almost seems too simple. Any reason I shouldn't do this? I could also make this model work with a p-channel MOSFET. Any reason I should pick one over the other for this?

EDIT: @PStechPaul has pointed out that this is only working because the power is flowing through the diode of the MOSFET, so it is essentially just a diode. I don't want to use a diode because of the power losses involved. The charger is being powered from a second battery (3 cell) that is charged by a solar panel, so wasting power must be minimized.

The usual solution to reverse voltage protection is to use a PMOS MOSFET with the source connected to the load, and a gate to ground like this:

enter image description here

However, I don't believe I can do this because I have a battery instead of a load here. See edited system below:

enter image description here

So while this will work to conduct power while the charger is on, the issue arrises when the power to the charger is off, like so:

enter image description here

In this situation, according to the simulator, I will have a reverse current of 422 mA as follows:

enter image description here

If you can offer any suggestions to make this system work while minimizing power losses that would be very helpful. I know a simple diode would work here, but I would prefer to avoid those power losses.

EDIT 2:

@mkeith I've simulated the design you show. At a base level it works with a few caveats:

When charger is off and voltage is absolute 0, the current is indeed stopped: Charger off with 0V

However if there is any floating voltage above 1.5 V, there will be a reverse current. (I'm not sure if anything would induce a floating voltage, and I don't know if I should be concerned). Here is an example with 2 V floating, and a reverse current. Perhaps a pull-down resistor would solve this.

floating voltage reverse current

When the charger is on, it does indeed work. However I again wonder what the difference is between this working, and the first figure I posted with just the n-channel MOSFET. Is this running in with a low Rds, vs the circuit I showed is running like a diode? Or is this also running like a diode (and thus has power losses like a diode)?

proper operation

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  • \$\begingroup\$ I am not reading this. It is too long. Here is how you do it: components101.com/articles/… \$\endgroup\$
    – user57037
    Aug 12, 2022 at 6:00
  • \$\begingroup\$ @mkeith If you read it, you might understand why I can't do this... My load is a battery... so the voltage stays positive on the source even after the drain goes to 0, and current will continue to flow (according to models). Please don't answer questions you don't read. \$\endgroup\$
    – Troy Cados
    Aug 12, 2022 at 6:31
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    \$\begingroup\$ You have the gate shorted to source, so the MOSFET will always be OFF. But with the source higher than the drain, the body diode will conduct. You could just use a diode for this. \$\endgroup\$
    – PStechPaul
    Aug 12, 2022 at 6:39
  • \$\begingroup\$ @PStechPaul Ah, so it is working because it is just flowing through the mosfet diode then? The reason I didn't want to use a diode is because I am powering the charger from a solar powered battery and I don't want to waste power by running it through a diode. But it appears I am effectively just doing that right now. Any suggestions? \$\endgroup\$
    – Troy Cados
    Aug 12, 2022 at 6:42
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    \$\begingroup\$ ti.com/power-management/power-switches/… \$\endgroup\$
    – user57037
    Aug 13, 2022 at 2:49

2 Answers 2

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If you don't mind using a $3 comparator (LT1716), here is a perfect solution:

Solar Charger Protection

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  • \$\begingroup\$ I wouldn't, mind, however I think that this is essentially an equivalent circuit to my figure1/ figure 2 circuit, just with a 1k resistor between the gate and source. I think the op amp you're using is just going full on when the charger is on, and full off when the charger is off, with the output controlling the gate. Isn't this just the same as controlling the gate using the power state of the charger as I showed? Another difference is that you appear to be using a PMOS. Does that make a difference? \$\endgroup\$
    – Troy Cados
    Aug 12, 2022 at 8:02
  • \$\begingroup\$ The main difference is that you are using a PMOS. Does that make a difference in terms of it functioning as a mosfet with low Rds vs a diode? In your model, when V1 goes to 0, does the voltage on the gate go to 8V? If so, wouldn't this be an equivalent circuit to just tying the gate to the source V2? Otherwise, if the voltage on the gate goes to 0 volts, then I think it is equivalent to the last figure I showed where I get a negative current. \$\endgroup\$
    – Troy Cados
    Aug 12, 2022 at 8:18
  • \$\begingroup\$ I think this will work but VDD for the comparator is connected directly to the battery. The datasheet said 35 uA Iq. That is pretty low so it may be OK depending on the size of the battery and the expectation for how long it can sit on the shelf or whatever. \$\endgroup\$
    – user57037
    Aug 12, 2022 at 8:22
  • \$\begingroup\$ In the end phase of the charge process (<100 mA) this might turn off because the voltage across the MOSFET (12 mohm) is below the possible offset error of the OpAmp (1.6 mV). \$\endgroup\$
    – Jens
    Aug 12, 2022 at 17:53
  • \$\begingroup\$ You could use a TLV3701 ti.com/lit/ds/slcs137d/slcs137d.pdf (about $1.50) with just 1 uA maximum supply current. And it's easy enough to add hysteresis or offset to avoid shutdown at low charging current. The inputs should have resistors anyway, and a bypass capacitor from Vcc to GND. This device also can work with common mode voltage 5V above Vcc, \$\endgroup\$
    – PStechPaul
    Aug 12, 2022 at 21:59
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THIS WON'T WORK BECAUSE IT WON'T TURN OFF UNLESS VCHARGE SOMEHOW DROPS TO A VERY LOW VOLTAGE. LEAVING IT HERE BECAUSE OP IS REFERRING TO IT.

When VCHARGE is 8.8 V, it will turn on NMOS M2 which will turn on PMOS M1, providing a low resistance path from charger to battery.

When VCHARGE is 0 V, M2 will be off, so R1 will pull up the gate of M1, turning it off. The body diode of M1 is oriented in such a way that it will be reverse biased when VCHARGE is 0 and BAT1 is above 0 V.

UNFORTUNATELY, ONCE M1 TURNS ON, IT WILL NEVER TURN OFF.

You have to choose a suitable PMOS for M1 based on how much current you need etc. There may be a few PMOS out there that can't handle Vgs of -8.8 V, so keep an eye on that. Make sure Vgs max is +/- 12 V or so.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ I simulated this model and added it into my main question edits. As I asked in the edit, Is this working like a diode? what will the power loss be? How is it different than my figure 1 example? Thank you for your response. \$\endgroup\$
    – Troy Cados
    Aug 12, 2022 at 8:52
  • \$\begingroup\$ Once turned on it will never turn off. A conducting M1 will keep the gate of M2 at BAT+ level, with or without any charger activity. \$\endgroup\$
    – Jens
    Aug 12, 2022 at 14:50
  • \$\begingroup\$ @Jens I guess you are right. OP needs an ideal diode. \$\endgroup\$
    – user57037
    Aug 12, 2022 at 16:38
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    \$\begingroup\$ @TroyCados let me know if you want me to delete this answer. Since it won't actually work I need to either delete it or add a big note to it saying it won't work. \$\endgroup\$
    – user57037
    Aug 12, 2022 at 16:43
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    \$\begingroup\$ @mkeith. I think your note is sufficient. It is still an insightful answer. \$\endgroup\$
    – Troy Cados
    Aug 16, 2022 at 6:06

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