Calculating charge current of a bypass capacitor, choosing series resistance

Question

I'd like to switch power supply of two sensors, each of them with a bypass capacitor at their supply pin, with an ATmega328P MCU output pin with an absolute maximum rating of 40mA.

The sensors themselves together draw only about 230µA.

The capacitors are 100nF, the supply voltage is 4.8V max:

Image source: Analog Devices TMP36 datasheet

To reduce charge current I'm adding a series resistor of 68Ω to the capacitor.

So I calculate for one sensor:

Charge energy/time: 1.15µJ/68μs
Power: 0.17W
Current: 35.2mA

For two sensors this would be about 70mA - well above the 40mA the MCU pin can switch.

Provided I calculated correctly, what would be the preferred solution?

1. Increase series resistance to 150Ω to get a charge current of about 32mA. What is the consequence - just a slight delay until the bypass capacitors have their full effect?
2. Use a MOSFET to switch power supply of the sensors and no series resistors at all?

Answer 1

You have many options.

One is to have single current limiting resistor of 4.8V/40mA = 120 ohms for both sensors, each with their own 100nF bypass cap. The only downside is the small voltage drop but it's insignificant.

Even simpler, no reasistor at all.

The IO pin has approximately 35 ohms output impedance at 5V, even higher output impedanve at lower supply voltage.

If you only do the cap charging occasionally and not thousands of times per second, it might just be fine and won't heat up too much locally.

You can always pre-charge the capacitor by using the internal pull-up which is about 20k to 50k, sourcing about max 130uA of current.

Answer 2

Maybe you can precharge the bypass cap with pin1 and then switch to pin2 if you have enough I/O-s.

^{simulate this circuit – Schematic created using CircuitLab}