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I have 3904 and 3906 transistors, LEDs, and an SPDT switch.

I tried to use the 3904 so that when voltage is applied to the base, it flows from the power source into the LED and then to ground.

I have difficulty to use another transistor so that when 0 V or ground is applied to the base of the transistor, the LED will turn on. How would I do this?

In other words, the SPDT switch will toggle 2 LEDs; one LED will show the on status and the other LED the off status.

The schematic I drew below is working, but I need to add another LED that will switch on if the voltage is 0 V or ground.

enter image description here

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    \$\begingroup\$ Do you need to use transistors? It doesn't seem necessary to me, with what little you wrote. \$\endgroup\$
    – jonk
    Commented Jul 2, 2020 at 7:23
  • \$\begingroup\$ i was using transistor to avoid noise from the led going in the rest of the circuit which is for audio purposes \$\endgroup\$
    – noisanceritual
    Commented Jul 3, 2020 at 3:56
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    \$\begingroup\$ Well, it just seems all you need are the two LEDs, two resistors, and the SPDT switch. \$\endgroup\$
    – jonk
    Commented Jul 3, 2020 at 6:36
  • \$\begingroup\$ Could you use a DPDT switch - one pole to switch the audio, and the other to switch the LEDs? \$\endgroup\$
    – Peter Bennett
    Commented Jul 4, 2020 at 23:18

1 Answer 1

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A basic solution would be to just make a copy of this circuit and invert the driving signal which is being applied to the base of the second transistor. Here is a small schematic:

circuit

You could also spare one transistor and use just a standard diode instead. Here is an example:

circuit_2

Edit #1

Since the driving signal is an audio signal, the following circuit can be used to detect the presence of an audio. It basically consists of amplifying the audio signal and using a peak detector to store keep the output DET high if the audio signal is present. You could use this detector to drive one of the circuits above. Maybe a buffer withh be required between both circuits in order to not overload the detected signal.

circuit_3

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  • \$\begingroup\$ Hi thanks for the help, but sorry that will not work with my intended purpose. The driving signal is what determines if one led is on or the other led. Can't have both leds on at the same time, i cant invert the driving signal it will make no sense in my needed purpose. I really need to just use the 3906 or similar transistor to turn on and off the led when signal at the base is near 0v or ground. \$\endgroup\$
    – noisanceritual
    Commented Jul 2, 2020 at 19:17
  • \$\begingroup\$ I am not so sure whether you understood the proposed circuits. You do not have to change the driving signal, since the additional transistors will do it for you. As you can see from the waveforms, at all times only one of the LEDs is ON. These circuits ensure that the operate in a complementary way. BTW, by driving signal I meant the common pin of your SPDT switch. \$\endgroup\$
    – vtolentino
    Commented Jul 3, 2020 at 7:02
  • \$\begingroup\$ Thanks Vtolentino, i tried your circuit on breadboard, it works like intended. Now i realise that problem is my driving signal. The driving signal for the moment is an audio signal so using your circuit makes it that both leds lights up following the audio signal and when the audio signal is off, just one led is on. I'm using a spdt swithc to activate or cut the audio signal, I think i would need to detect if the audio is active or note and have a steady DC signal for the driving signal, right? Any ideas? \$\endgroup\$
    – noisanceritual
    Commented Jul 4, 2020 at 4:20
  • \$\begingroup\$ I thought i could use the spdt pin that is connected to gnd on my schematic as the driving signal but since it's audio, it is driving both leds of your circuit following the audio + and - voltages so both leds turn on but i would need just 1 led to turn on when audio is active. Using your circuit when its muted there is just 1 led on that's fine, but both leds turn on when audio is active because it's ac and its gets below 0v as the audio waveform is generated. \$\endgroup\$
    – noisanceritual
    Commented Jul 4, 2020 at 4:22
  • \$\begingroup\$ The audio part was not included in your question. But, I updated my answer with an audio detector circuit which can be used to drive the other proposed circuits. \$\endgroup\$
    – vtolentino
    Commented Jul 4, 2020 at 22:32

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