Electronic – Solid State Relay two switch based on headphone audio output to control dc signal

audioledrelay

Need to turn on an LED if there is any audio present out of the headphone jack of a cell phone (0.6v peak voltage). If there's no audio, need to turn off led. My plan is to hook up audio signal to the input of the relay and have the relay control a dc current of 5V that lights an led if relay is ON.

Any ideas what kind of relay or specs I should look for?

Best Answer

LEDs are driven with a constant current source. Adding a relay with a constant current source is one way to power an LED, and probably not the cheapest or simplest way. However, you mentioned using a relay, so I'll assume that you need to use a relay for isolation or something.

A relay needs fairly significant current levels to energize the coil. It is usually powered with a constant voltage source, unlike an LED. Furthermore, this voltage must be in one direction, that is, it must be a DC signal. On small relays of the type which might be used to drive a typical LED, this creates a current which is typically about 50-100 mA. An audio signal doesn't have enough power to drive this kind of current or voltage.

Additionally, audio isn't a DC signal, it's an AC signal. The cell phone audio port is likely to have an output capacitor to be incapable of outputting a DC signal. To detect the presence of audio, then, you'll need to rectify the audio signal, that is, to convert it from AC to DC.

At 0.6V output levels, a simple diode rectifier won't work, those need significantly more headroom. You may be able to get away with a Schottky diode rectifier if costs are tight, but this will need to be amplified. Again, you may be able to get away with DC-offsetting the output of this regulator to slightly less than the threshold voltage of a FET or the base-emitter voltage of a BJT, and slightly more than the required voltage when increased by 0.2V, but that would require some precision where there's usually not much precision.

If cost isn't a huge factor, it will be far easier to get yourself an op-amp and build a precision rectifier like the following example:

Wikipedia

As described on the Wikipedia page, the output of this system has a gain of \$ R_2 / R_1 \$ when the input is negative. This output can then be used to switch a transistor, or, with a sufficiently strong opamp, to drive the relay directly.