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I want to use LTC4418 fuctionality of detecting which one of power supplies is on at this moment. Here is my approach to it:

enter image description here

PS_VALID1 is connected directly to VALID1 of LTC4418 (without any additional components), same for PS_VALID2. BLE_GPIO is a GPIO of 3.3V powered microcontroller.

What I got confused by is the documentation stating:

Output Currents
VALID1, VALID2, CAS..............................–2mA/+5mA

VALID1 and VALID2 are open-drain outputs, fine, they pull-down to GND or are Hi-Z, but why there is a +5mA output current stated in the docs? Is this circuit correct then?

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    \$\begingroup\$ Please explain why use transistors? Is it not possible to directly send a signal to the GPIO? \$\endgroup\$
    – AltAir
    Apr 5, 2019 at 10:00
  • \$\begingroup\$ Good point, actually I could use internal pull up resistors and connect it directly to GPIO without any external components. \$\endgroup\$
    – pbn
    Apr 5, 2019 at 12:08
  • \$\begingroup\$ Provided that you know the level of the GPIO is tri-state/low at power up. \$\endgroup\$
    – Huisman
    Apr 5, 2019 at 17:38
  • \$\begingroup\$ @Huisman I found a comment from manufacturer when starting up the GPIO pin will be in input mode, but input buffer is disconnected. So basically it's a floating pin with high impedance. origin \$\endgroup\$
    – pbn
    Apr 5, 2019 at 21:03

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You're looking at the Absolute Maximum Ratings. Those are the values you should not exceed when forcefully sinking or sourcing that pin.
So, don't draw more than 5 mA from the pin, don't source it with more than 2mA.

Without this "Q?A" or "Q?B" transistor, you would be sourcing 3.3V/10k = 0.33 mA, so, the circuit is fine to me.

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