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Please excuse what may be a simple question, I am very rusty when it comes to electronics having not looked at them at all for about 25 years! I hope someone can resolve my query.

I have a circuit design that takes a switch connected to ground. When closed the LED is on, when open the LED is off.

current circuit

In this diagram the shaded switch and LED are part of a larger existing circuit which cannot be changed. The LED is just an example of my load.

What I want to achieve is to swap the action of the switch around. I want the LED on when the switch is open.

I could do this easily with a relay but I haven't got the physical space for a relay, I'm sure it can be achieved with transistors but I'm not sure how!

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    \$\begingroup\$ If you don't care about energy waste, just wire it in a way that you shunt the base to gnd \$\endgroup\$
    – PlasmaHH
    Feb 8, 2016 at 13:34

3 Answers 3

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The easiest method is to remove the transistor and both resistors connected to the base of the transistor (10k, 47k). Then connect the free end of the switch to where the collector used to be.

If this is an existing circuit board, remove the above components. Then install a jumper where the 47k resistor used to be and another jumper between where transistor leads B & C used to be.

The circuit function remains similar with similar power consumption.

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  • \$\begingroup\$ Fantastic, simple resolution. I was obviously overthinking it. Thank you. \$\endgroup\$ Feb 8, 2016 at 14:13
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No transistors need to be abused in the making of your device. All you need is a resistor:

This causes the full LED current to be drawn all the time, whether it is on or not, but it's really simple and robust. In fact, the current drawn is slightly higher when the switch is closed and the LED off.

Here is a way to use transistors to minimize the current when the switch is closed:

You can also use a single transistor to get a sortof in between version. The right answer depends on how important current draw is versus complexity.

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You want to connect the LED to the emitter instead of the collector, and the switch decides the base voltage i.e. if higher than 0.7V, the transistor canal is opened, else it is closed.

enter image description here

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  • \$\begingroup\$ Also, the LED has a forward voltage, so the voltage at the base needs to be higher than 2V in worst case. \$\endgroup\$
    – lucas92
    Feb 8, 2016 at 13:43

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