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I'm struggling with a NPN transistor switching problem, and in the process I've burned out a board elsewhere in my machine. Electronics is not my strong suit. I want to use a 5V signal to switch 12V. Problem is that the grounds between the 5V and 12V are not close in potential. They are about 7V apart, so I cannot connect them. I have the circuit configured like this:

enter image description here

Which, based off of my research on this site - should work right? And it did work for a period of time! A board that I'm drawing my variable voltage from got fried somehow down the line though.

Are there any dangers to setting the circuit up as such? I feel like some excess current/voltage got drained into the variable return - messing something up down the line. Or maybe it was that I'm only using the high line from the power into the circuit (return line not attached to anything?).

When I get my board back from repair, I want to retest this configuration but with some sort of safeguard installed to protect the other components in the machine. What suggestions are there for this?

I'm very concerned about the potential difference (about 7V) between the negative line on the power, and the negative line on the variable. On most examples I've seen recommended (necessary maybe?) that the grounds be connected somewhere in the circuit.

Could be that this is indeed correct! And that I burned my board while going through the experimentation process that led me to this haha.

NOTE: diagram is wrong. I'm using R1 = 500 Ohm, R2 = 1k Ohm

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    \$\begingroup\$ The circuit on the diagram is open. There is no path from "Power+" to GND. \$\endgroup\$ – Eugene Sh. Mar 4 at 18:47
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    \$\begingroup\$ Currents (like the current needed to make the LED light up) flow in loops. You do not have a loop which involves the LED. No current can flow through the LED until you connect the emitter of the NPN to ground. \$\endgroup\$ – Bimpelrekkie Mar 4 at 18:49
  • \$\begingroup\$ @Bimpelrekkie if the two ground potentials are different how can I connect them? \$\endgroup\$ – denbjornen505 Mar 4 at 18:52
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    \$\begingroup\$ So you connect them and they will be the same potential. Well, I a supposing that at least one of them is floating. \$\endgroup\$ – Eugene Sh. Mar 4 at 18:54
  • \$\begingroup\$ @EugeneSh. can't this create a ground loop? \$\endgroup\$ – denbjornen505 Mar 4 at 19:03
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You must connect the transistor's emitter to the ground (negative end) of the 12V supply. If you cannot connect the switching signal source directly to the emitter then you probably want to use an optoisolator to drive the transistor's base.

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  • \$\begingroup\$ I will do this! Will the base see +5V though if it's negative is 7V above the 12V "ground"? I imagine that the base will see 0V in reference to the 12V ground. I apologize if I seem extra confused. Just trying to make sense of these potential differences. \$\endgroup\$ – denbjornen505 Mar 4 at 18:59
  • \$\begingroup\$ The base will see whatever voltage, with respect to the emitter, that the optoisolator provides to it. There are a variety of optoisolators that can be used in a case like this. \$\endgroup\$ – Elliot Alderson Mar 4 at 19:29
  • \$\begingroup\$ I happen to have a 4N35 laying around so I'll give it a try! \$\endgroup\$ – denbjornen505 Mar 4 at 20:23

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