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I am very new to electronics as a whole and this is my first somewhat advanced circuit using transistors. (Also I am new to this community, so if I messed up somewhere in this post feel free to ask me to clarify). Here is my sketch:

The sketch

This is supposed to control a motor for an airsoft gun. The gun always needs to stay on a whole cycle. The end of the cycle is indicated by a short 5V signal that runs into a NOT gate. This is supposed to cut off the "feedback loop" from the 12V rail to the main transistor in order to stop the firing (if the signal below is still high, another cycle will be initiated.) The 5V is provided by a power supply connected to the 12V batteries.

Here are the issues I experience whentrying to build this circuit on a breadboard (with buttons as inputs for the signals):

  • The motor gets basically no current and the voltage at the motor is not the expected 12V but 1.2V. I thought that could be because the motor might split the voltage with the voltage going in the back of Q1, but that does not make sense at all since Q2 is not even on.
  • The NOT gate works, but the output cannot turn on the feedback line (into the base of Q1).
  • Q2 is off all the time.

I really don't know if I messed up with the components I selected or the basic laws of physics (or both.)

Help would really be appreciated.

Of course if you could explain your solution it would be great, but right now I just want it to work.

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    \$\begingroup\$ You are making the extremely common mistake of using an NPN to switch on the "high" side (+12 V). For that NPN to go fully on, you would need more than 12 V at its base. I suggest that you study other circuits to learn how others do this successfully. Making your own "design" as a beginner will be a huge challenge and give you lots of headaches. And btw, the circuit is working but it just doesn't do what you want. \$\endgroup\$
    – Bimpelrekkie
    Nov 4, 2020 at 11:00
  • \$\begingroup\$ Thanks! So a transistor cannot switch a voltage greater than the base voltage? Now that you say it - it makes sense since the voltage from the emitter to the base would be positive...right? The thing is that I don't quite know how to find a fitting circuit for this problem, but I will probably have to continue looking. I don't really have a problem with my circuits failing though, since I always learn something new! \$\endgroup\$
    – Jay
    Nov 4, 2020 at 11:05
  • \$\begingroup\$ So a transistor cannot switch a voltage greater than the base voltage? It depends on the situation / circuit. Instead of looking for general statements like that, go find some circuits to see how it is done. Almost all circuits use only very few circuit configurations. Maybe this helps: learn.sparkfun.com/tutorials/transistors/… and this: electronics-tutorials.ws/transistor/tran_4.html \$\endgroup\$
    – Bimpelrekkie
    Nov 4, 2020 at 11:18
  • \$\begingroup\$ Thanks, I'll try that! \$\endgroup\$
    – Jay
    Nov 4, 2020 at 11:20
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    \$\begingroup\$ @Bimpelrekkie So now that you have answered the question, which answer do you think OP should accept? \$\endgroup\$
    – pipe
    Nov 4, 2020 at 11:29

1 Answer 1

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The motor gets basically no current and the voltage at the motor is not the expected 12V but 1.2V?

Your NPN transistor is acting as an emitter follower and, if the 100 kΩ resistor were more like 100 Ω then you might see 4 volts on the motor when it is activated. This is what happens when you use an emitter follower - the emitter follows the base voltage (with a slight voltage reduction). However, when using a 100 kΩ resistor, the current into the base multiplied by the transistors β (circa 100) doesn't produce much emitter current and it will appear that the motor is not powered.

If I were you I'd look into using a P channel common source MOSFET to switch the motor on and off. Or, use an N channel common source switching the negative supply to the motor. Get this bit working before adding any sophistication.

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  • \$\begingroup\$ Thank you, I will try that! \$\endgroup\$
    – Jay
    Nov 5, 2020 at 12:02
  • \$\begingroup\$ @Jay is there a reason for removing the answer accept? \$\endgroup\$
    – Andy aka
    Nov 6, 2020 at 14:09

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