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I have this working on/off latch circuit below that works on a 8.4V battery.

enter image description here

I can charge and bypass the battery when I connect the circuit to a 5V charger, while still providing supply to the load when the USB 5V is connected.

I need to be sure that every time I connect the 5V supply (USB charger), the latch on/off circuit will be bypassed and the circuit will be turned on even if the push-button is not pressed.

So is this solution below a correct one? In a protoboard it's working fine, but I'd like to be sure if there's any problems that may arise.

I added a 1uF capacitor, two resistors and a diode, to give a small "pulse" to "activate" the P Mosfet P2, using the 9V DC from the 5V to 9V Boost converter.

enter image description here

Both circuits can be tested in realtime here:

Circuit 1: Falstad Circuit 1

Circuit 2: Falstad Circuit 2

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  • \$\begingroup\$ A safe turn on of the latch at the boost converter start depends on a fast rising boost output voltage. If this is always the case, the solution looks fine to me. \$\endgroup\$
    – Jens
    Mar 29, 2023 at 22:16
  • \$\begingroup\$ In my protoboard, it's working all the time perfectly. I tried to connect it to the 5V directly from the USB port, but then it doesn't worked reliably. From the 9V after the buck converter it's working all the time! \$\endgroup\$
    – Rodrigo
    Mar 31, 2023 at 12:12

1 Answer 1

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Like Jens mentions in the comment, your solution depends on a fast rising boost output voltage. Since you want to override the latch when the boost output starts, you could also use the following option which does not depend on the boost voltage rise time. Note that I added an extra NMOS to the gate of the PMOS in the latch.

enter image description here

Update: Since you mentioned that you can't keep the latch on forever when the doubler is ON, you need a pulse generation for sure. The scheme I proposed does not work for you.

You need to find the slowest rise time of the doubler and check that your scheme passes in that case also. If it fails, instead of connecting the doubler output directly, to the pulse generator, you can make the rising edge sharper using say a schmitt trigger and feeding that to the pulse generator.

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  • \$\begingroup\$ In fact, I need to always "turn on" the circuit, but can't keep it always on forever. I need to be able to turn it off by a MCU (on the load part) anytime. That's why I wanted just a "pulse" to turn in on, bypassing the latch circuit, but then the latch circuit could be used normally then. \$\endgroup\$
    – Rodrigo
    Mar 31, 2023 at 12:10
  • \$\begingroup\$ @Rodrigo, ok, if you can't keep it on forever when the doubler is ON, you need a pulse generation for sure. You need to find the slowest rise time of the doubler and check that your scheme passes in that case also. If it fails, instead of connecting the doubler output directly, to the pulse generator, you can make the rising edge sharper using say a schmitt trigger and feeding that to the pulse generator. \$\endgroup\$
    – sai
    Mar 31, 2023 at 16:36
  • \$\begingroup\$ I had previously used a XOR schmitt trigger just between the capacitor and the diode, with the other input to ground. It worked also, but I wanted to use fewer components. (tinyurl.com/2exnkrc2). It seems that's working well with these resistor and capacitor values in my protobard so far... \$\endgroup\$
    – Rodrigo
    Apr 4, 2023 at 12:08

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