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I am using a 2N2222 transistor to switch on and off the input supply of boost converter. The output of boost converter is powering my RF transmitter and voltage regulator which is used to provide 5V to four addressable LEDs.

schematic

simulate this circuit – Schematic created using CircuitLab

My controller works on 3.3V logic level, so the transistor's base is operated at 3.3V. The output of the boost converter is set to 12 volts.

My problem is I could not get a stable voltage at the boost converter output.

My transmitter does not work properly. It is missing data. (It works just fine when I don't use a transistor to switch the supply.)

Another problem is around 0.7 volt is dropping across the transistor, which is not suitable while using 3.7 volt Li-Ion cell as it further reduces the input voltage to the boost converter which is used to boost the voltage to 12 volts.

The LEDs I am using are fluctuating when data is transmitted.

My first guess was that it is due to the current capacity of transistor, so I switched to a TIP122 which is a Darlington pair transistor. Thr problem is still there.

My aim is to reduce power consumption when the RF transmitter is not used so I want to cut off power from rest circuitry.

What circuitry or component or method I should be using to switch the power supply without affecting the transmitted data?

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  • \$\begingroup\$ What kind of boost converter are you using? What is it's minimum rated input voltage? What is the current requirement of your load? \$\endgroup\$ – Math Keeps Me Busy Mar 3 at 14:11
  • \$\begingroup\$ How often do you switch things on and off? And how much time do you have for the output of the boost converter to settle before you need to transmit RF. \$\endgroup\$ – SteveSh Mar 3 at 14:19
  • \$\begingroup\$ A better (maybe "another" would be a better term) way to do this would be to put the switch between the output of the boost regulator and the RF transmitter. We do it this way all the time for controlling RF amplifiers/transmitters. \$\endgroup\$ – SteveSh Mar 3 at 14:20
  • \$\begingroup\$ My problem is I could not get a stable voltage at the boost converter output. You could have stopped right there. You can skip explaining that the rest of the circuit isn't working properly as all circuitry needs a proper supply voltage. You measured that the supply is unstable so focus on that. I already explained the reason in my answer. \$\endgroup\$ – Bimpelrekkie Mar 3 at 14:22
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You're using that NPN as an "emitter follower" 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 "strong" 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:

schematic

simulate this circuit – 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.

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    \$\begingroup\$ Another way is to replace the boost converter with one which has a /EN input and drive that directly with your signal. Note that the action will be reversed: the converter powered on when the signal is low. \$\endgroup\$ – DoxyLover Mar 3 at 15:40
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Unregulated boost converter holds output voltage for "some" load range only. I expect you do not have a sophisticated control loop involved (just some 555 open loop switcher), there is a chance so solve it connecting dummy load on boost output. This of course will decrease efficiency.

Regarding the 2N2222, common mistakes are not enough base current to let the transistor fully open or too high switching frequency for slow transistor.

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    \$\begingroup\$ Regarding the 2N2222, common mistakes are not enough base current That is not the issue in this circuit. In this circuit, the base current will be large enough as the base resistor has a very low value of 100 ohms. \$\endgroup\$ – Bimpelrekkie Mar 3 at 16:08
  • \$\begingroup\$ How do you know? \$\endgroup\$ – Michal Podmanický Mar 3 at 17:56
  • \$\begingroup\$ If you understand how bipolar transistors work, it makes perfect sense. Hint: \$I_c = \beta * I_b\$ \$\endgroup\$ – Bimpelrekkie Mar 3 at 18:31

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