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Lets say I have a switch with the usual 2 connections + and - I want when the switch if off (power doesnt pass) the circut to close with output wire 1 and when the switch is on(power passes) the circut to open with wire 1 and to close with wire 2 thus changing the output wire.

I tried a few ideas I had with transistors but didnt work out.What is the best way to make one of those?

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  • \$\begingroup\$ Any reason you can't use a SPDT switch? Can you explain the circuit you have in mind (load connection) and whether there is a power supply available when the switch is "off"? \$\endgroup\$ Commented Feb 16, 2014 at 18:08
  • \$\begingroup\$ @SpehroPefhany Its not a actual switch to use a SPDT it is a magnetic switch and I haven't found SPDT version of those. and yes when its off there is power available. \$\endgroup\$
    – user36976
    Commented Feb 16, 2014 at 18:17
  • \$\begingroup\$ Sounds like a relay. And those do come in SPDT. \$\endgroup\$ Commented Feb 16, 2014 at 18:28
  • \$\begingroup\$ @IgnacioVazquez-Abrams I went to 3 different shops 3 days ago they told me they didn't know if that exists... where can I get SPDT version of those... \$\endgroup\$
    – user36976
    Commented Feb 16, 2014 at 18:32
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    \$\begingroup\$ SPDT relays are almost as common as water. Find better shops. \$\endgroup\$ Commented Feb 16, 2014 at 18:33

2 Answers 2

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If it's a "reed switch" you're talking about, they do indeed exist in SPDT form, though they're not as common as SPST N.O.. Also in SPST N.C. (normally closed) form. For example:

enter image description here

If you want to reverse the action of a SPST N.O. switch, you may be able to do something like this:

In the case of the BJT, make Rx about 10-20 times RL. In the case of the MOSFET, make Rx a few K ohms.

schematic

simulate this circuit – Schematic created using CircuitLab

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If u want to conditionally connect an output circuit to either of two circuits depending on the position of your switch, you should consider using a 2:1 MUX for your design or you can do it using 2 transistors, if you don't want to use an SPDT that is. I am doing only the transistor design for you, since there is almost nothing that is needed to be done in case of the MUX.

schematic

simulate this circuit – Schematic created using CircuitLab

If the switch SW1 is on, then the output load is connected to Circuit2 since Q1 is closed and Q3 is open. Whereas if the switch SW1 is off, then the output load is connected to Circuit1 since Q3 is closed and Q1 is open.

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  • \$\begingroup\$ If the switch is open, both transistor conduct. The base current for Q2 has to flow through the base of Q1, right? \$\endgroup\$ Commented Feb 17, 2014 at 22:48
  • \$\begingroup\$ Oops sorry, yeah you are right. I'll edit and use an npn inverter for Q2. Thanks for the insight, in the process I learn too...:) \$\endgroup\$
    – Ghosal_C
    Commented Feb 17, 2014 at 23:05
  • \$\begingroup\$ There you go, I think this should work, although I should R1=R3=1k or 10k depending on the individual impedance of the external circuits. It's not really possible to figure out without that. I am assuming my base current when Q1 is on even when divided will not be high enough to drive on circuit C1. \$\endgroup\$
    – Ghosal_C
    Commented Feb 17, 2014 at 23:42
  • \$\begingroup\$ Q3 will still be mostly 'on' when the switch is closed, assuming "output circuit" is just a short. Maybe if you make R1 a short and D1 a Schottky diode. R3 can go to ground rather than P1. \$\endgroup\$ Commented Feb 17, 2014 at 23:51
  • \$\begingroup\$ I'm scared to short R1, because I might fry up my Q1, had some really bad experiences in the past, although I don't have my emitter grounded here, but this circuit should work, provided my 3 circuits in question have decent impedance ratings. It's really difficult to design without knowing my impedance caps on the circuits I'm working with. Plus designs generally have their restrictions too \$\endgroup\$
    – Ghosal_C
    Commented Feb 17, 2014 at 23:58

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