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I plan on making a version of this circuit to control a 12V PC fan using an ESP32. The ESP32 GPIO pins are 3.3v logic level.

https://imgur.com/a/iqOfcME enter image description here

You can ignore the lower part of the circuit in that diagram. That part is stepping down 5.5V to 3.3V because he wants to use only one 5.5V power supply to power both his board and his LED strip. I don't have that problem, my board has its own power supply, and I plan to use a separate 12V wall wart to power the fan via the TIP120

Here are my questions:

  • Will the 3.3V logic level at point (1) from my ESP32 board be sufficient to switch the TIP120?
  • What is the purpose of the resistor at point (2)?
  • Is combining the grounds of the 3.3V and 5.5V sides of the circuit at point (3) intentional, and is that a best practice? If so, why is that done? Should I also do that with my 3.3V/12V circuit?
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  1. Bipolar transistors switch on from a current, not a voltage. As this is a NPN darlington, any positive voltage more than a few hundred millivolt will lead to a positive current into the base and a voltage higher than it's VCE will make the transistor turn on fully.
  2. It's there so the transistor base doesn't burn out from overcurrent.
  3. Yes. Yes. You have to connect the grounds so the current can flow back to its source.
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  • \$\begingroup\$ thank you! one point of clarification regarding "You have to connect the grounds so the current can flow back to its source." note that the OP specifies that there will be two power supplies, one for the board and one for the 12v fan, each with their own ground, (and each side of the circuit terminating to its respective ground.). The question #3 is more about if connecting the grounds of two different power sources of differing voltages is a good idea? I'm not an expert, but I've never seen that before. \$\endgroup\$
    – 3z33etm
    Oct 11 '18 at 3:03
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    \$\begingroup\$ You have to connect the grounds. It will not work if you don't do it. \$\endgroup\$
    – Janka
    Oct 11 '18 at 3:07
  • \$\begingroup\$ also, regarding question #2, i notice that is a 100Ω resistor. How is the correct value determined to use on the base leg? \$\endgroup\$
    – 3z33etm
    Oct 11 '18 at 3:08
  • \$\begingroup\$ This depends on the collector current, but in general, darlingtons hardly need more than 10mA base current to be steered fully open. 100Ω@3.3V gives at maximum 33mA. The ESP32 had to supply this current on its GPIO. Check if it can do it. \$\endgroup\$
    – Janka
    Oct 11 '18 at 3:17
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    \$\begingroup\$ A npn darlington would need more than a few hundred millivolt to turn on fully. More like, 1.4V to 2V or higher depending on the transistor. \$\endgroup\$
    – Passerby
    Oct 11 '18 at 4:01
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Why have a tie your grounds together? You normally connect your common grounds together on circuits that have 2 separate voltages. This is so that each part of your circuit will not float with respect to the other one. If the common connection between the 2 parts is only the gate of your transistor then the whole side of the circuit that is delivering current into your transistor may creep up (due to temp change, residual capacitance etc) and now you find out that your base voltage is changing over time and delivers different amount of current into your transistor and that drives changes to the rest of your circuit.

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