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I have following schematic:

schematic

simulate this circuit – Schematic created using CircuitLab

But the motor is not working, the NPN-Transistor does not seem to let any current through. What am I doing wrong? the transistor does work, I checked, but there may be something very basic wrong, electronics is not really my strong suit.

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    \$\begingroup\$ As shown, BAT2 and the diode are reversed. \$\endgroup\$
    – Nedd
    Dec 5, 2015 at 23:56
  • \$\begingroup\$ there are 2 systems, one working with 5v and the other with 9v. The part with 5v is switching the 9v circuit which is being shown. I though I have to link the grounds. \$\endgroup\$
    – Daniel
    Dec 6, 2015 at 0:06
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    \$\begingroup\$ If you have to link the grounds, you have to use a PNP in the high side or move the NPN to the low side. So if you use an NPN, run 9V to the + motor lead, the - motor lead to the collector, the emitter to ground and the 5V through the resistor to the base. Put the diode across the motor with the anode to the - lead and the cathode to the + lead. \$\endgroup\$
    – John D
    Dec 6, 2015 at 0:09

4 Answers 4

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You're very close. instead of having the NPN in its present location, it need to be between the motor and GND. So disconnect the NPN's emitter from the motor, its collector from the 9 V supply, and connect: Emitter -> GND Collector -> Motor Base -> 1k & 5 V as you have it.

The motor will go between the 9 V supply and the NPN's collector.

Whether or not 1k will work depends on the motor current. 5 V and 1k will give about 4.3 mA of base current -- probably about 100 mA of collector current. This sounds low for even a small motor.

schematic

simulate this circuit – Schematic created using CircuitLab

Also, it is possibly you have (partly) damaged your NPN -- the way you have the circuit now, the base-emitter junction is reverse biased. These usually break down around 6 V, and when they do, the beta (gain) of the transistor gets degraded.

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  • \$\begingroup\$ The motor is a RS-360SH-2885 (datasheet) \$\endgroup\$
    – Daniel
    Dec 6, 2015 at 8:43
  • \$\begingroup\$ I connected it like that, but it's still not working. I also tried to reduce the resistance :( \$\endgroup\$
    – Daniel
    Dec 6, 2015 at 9:43
  • \$\begingroup\$ I think I burnt out my 5v circuit :( but this part works correctly now. Thank you very much \$\endgroup\$
    – Daniel
    Dec 6, 2015 at 15:11
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    \$\begingroup\$ That motor will require 1.7 A at peak efficiency, and 8 A when stalled (can your 9 V give that ?). At even 2-3 A current, the NPN will only have a current gain of about 50, and so will need > 60 mA of base current. With 5 V, you'll need a base R < 70 ohm. Even then, if you stall the motor, the current will rise; the NPN will come out of saturation and will overheat. Best to use a MOSFET instead. \$\endgroup\$
    – jp314
    Dec 6, 2015 at 17:19
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As in my comment, BAT2 and diode are connected in reverse. (as shown in your schematic).

However, you might have the schematic drawn incorrectly as it is more common for the second voltage source to be on the transistor's collector line (at the far right, with + facing up in this case). The bottom lead of the motor would then be connected to the common - terminals, the diode on the motor still needs to be reversed.

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  • \$\begingroup\$ It is not drawn incorrectly since it's my own circuit (I just did it with very limited knowledge). I was trying to switch the motor on and off with the 5v battery \$\endgroup\$
    – Daniel
    Dec 6, 2015 at 0:17
  • \$\begingroup\$ Then reversing BAT2 and the diode should work, (provided that nothing has been burnt out). \$\endgroup\$
    – Nedd
    Dec 6, 2015 at 0:19
  • \$\begingroup\$ Wouldn't inverting BAT2 make the voltage at the resistance be higher? since the batteries would be like in series? \$\endgroup\$
    – Daniel
    Dec 6, 2015 at 0:22
  • \$\begingroup\$ Yes, and the initial base current into the transistor will be near 14ma. \$\endgroup\$
    – Nedd
    Dec 6, 2015 at 0:30
  • \$\begingroup\$ but I need the voltage there to stay about 5v, there is a whole circuit there that probably would burn out if the current was higher \$\endgroup\$
    – Daniel
    Dec 6, 2015 at 0:32
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What am I doing wrong?

Pretty much everything. You are reverse biasing the collect-emitter polarity - an npn wants to see positive on the collector relative to emitter but you have this reversed.

Try putting the emitter to 0V and putting the motor in series with the collector up to 9V. keep the diode connected the same way round (about the only bit you got correct).

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  • \$\begingroup\$ c'mon dude.. at least all he has to do is flip around a couple of components.. just the battery and diode. at least there's a schematic, which is about 100x more effort than most people put in here \$\endgroup\$
    – Daniel
    Dec 6, 2015 at 1:47
  • \$\begingroup\$ @Daniel if you don't like my answer you have the option of reporting it dude! \$\endgroup\$
    – Andy aka
    Dec 6, 2015 at 15:12
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    \$\begingroup\$ I could but why when I could just directly appeal to you. I'm all for hammering on the low effort / no effort Qs, but this seems legit. \$\endgroup\$
    – Daniel
    Dec 6, 2015 at 15:35
  • \$\begingroup\$ I'm not changing my answer!! \$\endgroup\$
    – Andy aka
    Dec 6, 2015 at 15:43
  • \$\begingroup\$ Ok! I didn't ask you to \$\endgroup\$
    – Daniel
    Dec 6, 2015 at 15:50
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As another alternative (and close to what Andy has said), keep your original schematic but just swap the Emitter and Collector pins. That gives you the correct polarity for the batteries and the diode.

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