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I have a unique situation where I will lose Vin to my device/voltage regulator for about 10 seconds.

I tried the following circuit thinking it would solve the issue by using a 9V battery and enabling the MOSFET for those 10 seconds or so. VIN is tied into my VIN to my voltage regulator circuit. 12V through another Schottky diode is applied to VIN as well. However, this 12V can drop for about 10 seconds which kills all power to my switching voltage regulator. I am controlling the MOSFET with IO7 from an esp32. I only want to use the 9v battery as an uninterrupted battery backup for those 10 seconds.

The problem I run into is that anytime the battery is connected the board powers up. The voltage regulator I am using is TPS54202DDCR. D1 is SS2210. I do not understand how this is possible.

enter image description here

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    \$\begingroup\$ Try and add a schematic with the embedded CircuitLab option on the textbox. I am I to understand you turn the mosfet on before you lose power so that the battery takes over? Something like that?. Please provide a more complete schematic and description. \$\endgroup\$
    – garvamel
    May 17 at 3:57
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    \$\begingroup\$ Your schematic doesn't make sense. Your description indicates you have the battery in parallel with VIN, not series as shown. Please try to correct your schematic. \$\endgroup\$
    – DoxyLover
    May 17 at 5:27
  • \$\begingroup\$ What's IO7? Is R1 a series resistor to some signal or a pull resistor? The N MOSFET needs a pull down but also a positive voltage on the gate in order to conduct - where does it get a positive voltage from if all voltage sources are gone? And how does the zener make any sense? \$\endgroup\$
    – Lundin
    May 17 at 6:32
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    \$\begingroup\$ Anyway, how much current consumption are we talking of here? Supercaps might be an option in order to replace the battery entirely. \$\endgroup\$
    – Lundin
    May 17 at 6:33

2 Answers 2

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You can do this, but you need two MOSFETs.
The battery is 9V (it doesn't show), the 40Hz signal is standing in for and IO from your uController. The P-Channel controls the flow of current from the battery. The diode is a schotky.

If the uController drives the pin high, the n-Channel opens, pulling down the gate of the P-channel and hence creating the negative VGS required for the P-channel to open. If the uController is off, or the pin is low, the n-channel opens, causing VGS of the P-channel to drop to zero, hence closing the P-channel and disconnecting the battery. enter image description here

Simulate this circuit
You need the N-channel mosfet, to convert the (presumably) 0 - 3.3 V output from the uController, to a 9 - 0 V (or at least down to 9-VGS) needed for the P-channel gate.
Hence at power on the battery will normally be disconnected. You can enable the battery using the uController, and disable it. If you disable it while it's being driven by the battery, it'll stay off until the 10V is connected again.

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  • \$\begingroup\$ Any reason for the 5K ohm instead of using another 10k for the pull-up resistor? \$\endgroup\$
    – outgoingerror
    May 18 at 4:24
  • \$\begingroup\$ None at all, I just slapped them in, I'd probably go even higher than 10k as it's just wasting power \$\endgroup\$
    – LordTeddy
    May 18 at 16:23
  • \$\begingroup\$ I really appreciate the response. I didn't figure the 5k had a real purpose. \$\endgroup\$
    – outgoingerror
    May 19 at 3:36
  • \$\begingroup\$ Yes, post it as a new question \$\endgroup\$
    – LordTeddy
    May 19 at 14:51
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For that layout to work you need a P channel MOSFET and need to use a negative voltage on the gate to enable the battery.

probably you should look for a different solution.

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