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Here is a simple circuit that will cause a relay to be closed when a sufficiently strong audio signal is fed in.

A small 12 V relay should only need 10-15 mA to turn on, and then be able to switch a few amps of current. You probably need to turn up the volume on the computer output as high as it goes.

Added:

Here is a circuit that is slightly more complex but will require a lower audio level to trigger it in most cases:

The difference is that instead of just using the positive peak voltage of the input signal, this uses the peak-to-peak voltage (minus two Schottky diode drops). It also relies on the input signal being AC. The two diodes D1 and D3 with C1 form a charge pump that will produce the p-p voltage on C1 minus the two diode drops. For a normal tone of at least a few hundred Hz, that will provide a bit more base drive to the transistor than the first circuit.

I don't think this second circuit is really necessary, and the first one will most likely work fine. I am showing this second option because I see that gain has become a issue. Try this if you have trouble with the first one needing just a bit louder audio than your PC normally produces.

Here is a simple circuit that will cause a relay to be closed when a sufficiently strong audio signal is fed in.

A small 12 V relay should only need 10-15 mA to turn on, and then be able to switch a few amps of current. You probably need to turn up the volume on the computer output as high as it goes.

Here is a simple circuit that will cause a relay to be closed when a sufficiently strong audio signal is fed in.

A small 12 V relay should only need 10-15 mA to turn on, and then be able to switch a few amps of current. You probably need to turn up the volume on the computer output as high as it goes.

Added:

Here is a circuit that is slightly more complex but will require a lower audio level to trigger it in most cases:

The difference is that instead of just using the positive peak voltage of the input signal, this uses the peak-to-peak voltage (minus two Schottky diode drops). It also relies on the input signal being AC. The two diodes D1 and D3 with C1 form a charge pump that will produce the p-p voltage on C1 minus the two diode drops. For a normal tone of at least a few hundred Hz, that will provide a bit more base drive to the transistor than the first circuit.

I don't think this second circuit is really necessary, and the first one will most likely work fine. I am showing this second option because I see that gain has become a issue. Try this if you have trouble with the first one needing just a bit louder audio than your PC normally produces.

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Here is a simple circuit that will cause a relay to be closed when a sufficiently strong audio signal is fed in.

A small 12 V relay should only need 10-15 mA to turn on, and then be able to switch a few amps of current. You probably need to turn up the volume on the computer output as high as it goes.