Phase Shift effect on diode conduction time

[Lunat1c]

Ok, so let's say I have a simple half-wave rectifier like the one shown in the image below:

http://img585.imageshack.us/img585/1904/unledmwc.jpg

When it comes to the analysis of the circuit, during the positive cycle of the input voltage, the diode is forward biased and the output voltage (across RL) is equal to the input voltage. So far so good. Now when the input voltage goes negative, the diode should be reverse biased, however since the circuit is predominantly inductive , the voltage is leading the current by some angle (found to be 57.52 degrees). Will the diode stop conducting when the voltage reaches '0' and starts going -ve or when the current reaches 0? i.e. If I were to see the waveform of the output on an oscilloscope, would I see it at 0 when Vin starts the negative cycle? or some time later due to the phase shift?

 

[uart]

Will the diode stop conducting when the voltage reaches '0' and starts going -ve or when the current reaches 0?

The diode does not stop conducting until the current reaches zero. This however does not happen in the manner that you are suggesting, as the current will not reach AC steady state.

To find when the conduction stops you must solve the transient response for the zero.

That is,

$$i_L(\theta) = \frac{V}{|Z|} \left( \sin(\theta - \phi) + \sin(\phi) e^{-\theta/(w \tau)} \right) = 0$$

BTW. In the above $\theta = wt$ is the variable you're trying to solve for, (the other symbols are constants defined in the obvious way).