Problem switching boost converter using 2N2222, which turns on an RF transmitter
You're using that NPN as an &quot;emitter follower&quot; which means that the voltage at the emitter will be 0.7 V lower than the base voltage which is 3.3 V in your circuit. Therefore the boost converter only gets 3.3 V - 0.7 V = 2.6 V.
This also means that a large part of the battery's energy is wasted in that NPN.
Also, the 2N3904 isn't a very &quot;strong&quot; transistor, I would use it only for currents below 50 mA. Your boost converter might need more current, especially just after it is switched on.
Consider adding a PMOS or PNP transistor with a higher current capability to properly switch the 3.7 V without dropping so much voltage, I would use this circuit:



simulate this circuit &ndash; Schematic created using CircuitLab
I am suggesting the AO3401 P-channel MOSFET as it is cheap, can handle up to 4 A and has a low threshold voltage. A small disadvantage of the A03401 is that it comes inside a very small SMD case. There are many other P-MOSFETS that are equally suitable.
Note how I also added a resistor in series with the LED at the output of the 7805 regulator. Connecting an LED directly to 5 V is asking for trouble.