Given that the diode can be made to conduct, or not to conduct, simply by the choice of applied voltage polarity, then it must be possible to use any diode as a switch. So if you want to follow me, then just wire together the following circuit on a bit of breadboard, or something.
Set the signal generator to 1v RMS (3v peak-to-peak) at 1kHz. Set the oscilloscope to 1v/cm and set the timebase to 200uS/cm (0.2uS/cm). You will see an undistorted sinewave on the scope, and having an amplitude of 3 divisions.
Replace the 9v battery with a 1.5v AA cell. With the battery polarity as shown in the diagram, the waveform should still be an undistorted sinewave. But when the battery is reversed you will see just small pulses – nothing like a sinewave.
This demonstrates that the signal waveform must be lower than the switching voltage. For the switch to work perfectly the voltage must be a minimum of 2.2v – 1/2 of the 3v pk-to-pk plus the 0.7v needed to make the diode conduct.
When the circuit was connected as per the diagram, the voltage reading was 0.7vDC. This is because the diode was conducting and the 0.7v is the diode forward bias voltage-drop. When the battery was reversed, the voltage reading was 10v. The diode is open-circuit and the full battery voltage is across the diode.
