Electrical – Advice on battery powered ESP32 with NFC Design

Using that series resistor would be a bad way to regulate voltage, but I assume you do this to keep the 7805 from dissipating too much. But 400 mA \$\times\$ 60 Ω is 24 V drop across that power resistor, so at 24 V input you won't have nothing left. Decrease the resistor value to maximum 35 Ω, then 400 mA will cause a 14 V drop, and you'll have enough left to feed the 7805, which needs 8 V in. If you expect more current to be drawn by your circuit you can calculate the resistor value as

\$ R_{MAX} = \dfrac{24 V - 8 V}{I_{MAX}} \$

But for the 4050 you use a voltage divider to get 3.3 V. Don't. The voltage will vary widely with the load, and you want you power supply to behave predictably, i.e. be rock steady. Use an LDO regulator to get your 3.3 V from the 5 V, like the MCP1700, which is cheap, and has a low ground current.

edit after you uploaded the schematic
A number of things are not quite clear to me. I couldn't find the schematic of the Adafruit breakout board, but it seems to be 5 V, then why do you need the HEF4050 buffer, and why do you use a 3.3 V supply for that, if both breakout board and AVR are 5 V? If the breakout board would run at 3.3 V, like when it has an LDO on board, then I would suggest to run the AVR on 3.3 V as well, so you won't have connection problems. If you need the 7805 for the breakout board you can use the MCP1700 I mentioned earlier to power the AVR from.

R1 and R2 don't seem to serve a function. You're not using the pin as reset pin (make sure you enable the internal reset), then program it as output, and you won't need a pull-up. (I don't see why you use two 22 kΩs here instead of one 47 kΩ either.)

You are trying to make high side current measurements. In order to achieve this, you have selected ZXCT1009 high-side current monitor, which is available in a very small footprint; a 3-pin SOT23.

A high-side current monitor consists a sense resistor and a differential amplifier. These two and some other helping circuity creates a current sense amplifier.

A sense resistor is used to create a voltage drop between its terminals, then, a differential amplifier will amplify the voltage that is dropped on the sense resistor and create an output that is proportional to the current passing through the sense resistor.

For example, if you have a sense resistor of 0.1 \$\Omega\$ and you have a current of 1A passing through, then you have a voltage drop on the sense resistor of:

\$V_{drop}= R_{sense}*I_{sense}=0.1*1=0.1 V\$

So, there is a 0.1V voltage drop that is going to be fed into the differential amplifier. Say that this differential amplifier is configure with a gain of 10. If we input 0.1V, it will give us 1V as output voltage. If you take a look at the whole thing now, we have 1V output for a 1A of current that is passing through our sense resistor.

Coming back to you requirements, you need to sense the current from 150uA to 30mA. These are small currents! Let's use our 0.1 \$\Omega\$ sense resistor and calculate how much current will drop on it for 30mA:

\$V_{drop}= R_{sense}*I_{sense}=0.1*0.03=0.003 V=3 mV\$

With the differential amplifier that is configured as x10, the output is 30mV. This is a very low output voltage, and it is going to be lower when you lower the current.

Let's look at the output characteristics of the IC you have selected:

From page 3 of the datasheet:

As you can see, for a 10mV of voltage drop on sense resistor, output current is about 100uA. If you connect a 100 \$\Omega\$ resistor on the output, it will give you;

\$V_{out}=I_{out}*R_{out}=100*10^{-6}*100=10mV\$

So, for a 10mV of voltage drop on the sense resistor, we get a 10mV of output voltage with these values. What should our sense resistor be, to have a voltage drop of 10mV with 150uA:

\$V_{sense}=I_{sense}*R_{sense}=10*10^{-3}=150*10^{-6}*R_{sense}\$

\$R_{sense}=66 \Omega\$

But why do we care so much about the voltage drop? Because the voltage drop will affect the voltage that embedded PCB sees on the power rail. The voltage drop with the 66 \$\Omega\$ sense resistor and 30mA current will be about 2V. That means your 3V rail will see only 1V as supply voltage.

A solution would be to have different ranges where you switch sense resistors in order to switch ranges. You can use simple mechanical switches or you can use MOSFETs if you use a microcontroller, like so, component selections may vary: