PNP BJT switch not opening

[okw]

Hi! I have two 3.3V switched rails in a circuit. One slowly drops to 0V when turned off, but the other drops quickly to 2.4V and settles.
They are both driven individually by nRF9160.
Could it be best fixed with recalculating the base and pull-up resistor? Or by adjusting the MCU drive strength?
This is for an ultra-ultra-low power project, so which ever method using lowest possible power when MCU is sleeping is preferred.

I also have a 12V MOSFET switch which is driven by nRF52840. Is this switch designed to work properly?

[moffy]

Perhaps the 2.4V is a backfeed from the circuit being switched. Is it connected to any other power sources even through resistors?

[DavidAlfa]

Let's put a capacitor at the gate for no reason?
Remove C2 from Q3!
Also C1 from Q1 makes little sense.

Are you testing the circuits with some load?
The voltage you're measuring might be caused by leakage currents.
Try adding a 1-10K resistor to the outputs!

[Doctorandus_P]

Let's put a capacitor at the gate for no reason?
Remove C2 from Q3!
Also C1 from Q1 makes little sense.

Are you testing the circuits with some load?
The voltage you're measuring might be caused by leakage currents.
Try adding a 1-10K resistor to the outputs!

In the circuit on the left there is text that supposes C1 on the left circuit is for reducing inrush current (which it does not do), but C2 in the right circuit makes use of the Miller Effect to slow down the switching, and thus reduce inrush currents.
To make use of the Miller Effect C1 should be between the base and the emitter of the BJT.

And inrush currents such sub circuits can be severe enough to starve the main circuit of their power supply voltage.

----------------
As for the original question. It is quite possible that the "turned off" circuit is being powered by a logic line. Normally IC's have ESD diodes between (nearly) all pins and the GND and Power pins. This means that if an input voltage is higher then the power supply, then the ESD diode starts conducting and this can be enough to power the whole (low power) circuit. The 2V4 you mention is also a diode drop lower then your 3V (Not 3V3?) and this also points in the direction of a conducting diode.

[Zero999]

Let's put a capacitor at the gate for no reason?
Remove C2 from Q3!
Also C1 from Q1 makes little sense.

Are you testing the circuits with some load?
The voltage you're measuring might be caused by leakage currents.
Try adding a 1-10K resistor to the outputs!

In the circuit on the left there is text that supposes C1 on the left circuit is for reducing inrush current (which it does not do), but C2 in the right circuit makes use of the Miller Effect to slow down the switching, and thus reduce inrush currents.
To make use of the Miller Effect C1 should be between the base and the emitter of the BJT.

And inrush currents such sub circuits can be severe enough to starve the main circuit of their power supply voltage.

----------------
As for the original question. It is quite possible that the "turned off" circuit is being powered by a logic line. Normally IC's have ESD diodes between (nearly) all pins and the GND and Power pins. This means that if an input voltage is higher then the power supply, then the ESD diode starts conducting and this can be enough to power the whole (low power) circuit. The 2V4 you mention is also a diode drop lower then your 3V (Not 3V3?) and this also points in the direction of a conducting diode.

My guess is C1 is a decoupling capacitor, for the load.

No, to mak use of the Miller effect, the capacitor needs to be between the base and collector, in order to provide AC negative feedback.

[okw]

Perhaps the 2.4V is a backfeed from the circuit being switched. Is it connected to any other power sources even through resistors?

Silly overlook, that particular rail was powering the SD card circuit (amongst others) which requires quite a lot of decoupling (almost 50uF in the current design). Without the card, the charged capacitors would just "hang". I'll insert an SD card and see if it drops when I turn off the rail.
I'm not sure if the rail "hanging on" would pose problems for the other circuits on the same rail (V_in for si4455 sub-ghz-radio and V_io for tcn455x canbus controller). In some cases they might be sleeping/not turned on, and not drawing anything.
What is the preferred way of discharging the rail (if SD card isn't present), while maintaining ultra-low power usage (both while on and off)?

[moffy]

Perhaps the 2.4V is a backfeed from the circuit being switched. Is it connected to any other power sources even through resistors?

What is the preferred way of discharging the rail (if SD card isn't present), while maintaining ultra-low power usage (both while on and off)?

A switched resistor, off when power is on and on when power is off. Not sure what sort of power loss/leakage you can tolerate.