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How to close a circuit using a 5V power signal?

Essentially the goal is turn on an AV receiver, using the internal power ON/OFF switch, to activate using a 5V control signal. The receiver now uses a mechanical switch to bridge a signal wire to ground to somehow activate the system power relay. So internal ON/OFF signal wire to gnd starts system. I tested a 220 ohm resistor across switch, which did not work. Only shorting switch pins turns on the system, where one pin is gnd.

When I tested a 2N3904 transistor, 5V & 20k ohm on B, CE resistance still was ~300 ohm measured on a multimeter. Which is not grounded.

If you can help I will truly be eternally grateful. Thanks, Mike

Michael Reno
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  • Would you consider a solution where the mechanical power switch always ON and power is applied and removed by a circuit between the AV box and the mains? – AlmostDone May 27 '18 at 04:07
  • Actually measuring the resistance of a BJT doesn't give an accurate picture of what it can do. This isn't how BJT's work. A 2N3904 should do the job perfectly. Only you may need to use a bit smaller resistor in the base.

    See my answer below and select it as best if it works (I am sure it will).

     – paki eng May 27 '18 at 10:57

1 Answers1

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You were going in the correct direction. Try this circuit, it should work like what you need. It will accept 1-12V at input.

This is the most basic configuration. It should work in most cases. If you don't want to load the control signal, you can increase the base resistor to 2.2k, 4.7k, 10k or 22k. But to use larger resistor than this, you may also need to consider how much current you need at the output.

schematic

simulate this circuit – Schematic created using CircuitLab

paki eng
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  • Yes! It worked like a charm with 22k at R1. Now a smart switch will trigger my music on voice command. Thank you! – Michael Reno May 27 '18 at 15:55
  • @MichaelReno If this answer solved your problem, please "accept" it. This gives the rep credit to the person who answered and tells the rest of us that the question is taken care of. – Phil C May 27 '18 at 18:38