Phase Shift Question

[Lunat1c]

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



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?


In my opinion it makes much more sense that it becomes reverse biased when the voltage reaches 0 but I'd like to be sure.

Note:The diode is assumed to be ideal that's why I said V>0 will make it FB and V<0 will make it RB

[ejeffrey]

The diode stops conducting when the voltage across the diode drops below zero.  However, that won't be the same as the time the source voltage drops below zero.  When the source voltage becomes negative, the current through the inductor starts decreasing.  The voltage drop across the inductor is proportional to the rate of change of current.  Since the current is decreasing, there is a negative voltage across the inductor, which keeps the diode forward biased and conducting.  At a certain point, the current through the inductor will drop to zero, the diode will switch off, and the current will remain zero for the rest of the cycle.

[Zero999]

The diode will blow up as the voltage across it will theoretically peak at infinity when the diode interrupts the current or tries to.

[Mechatrommer]

The diode will blow up as the voltage across it will theoretically peak at infinity when the diode interrupts the current or tries to.

yes theoritically, since you say so my master (i've no idea), which in turn... the following.

In my opinion it makes much more sense that it becomes reverse biased when the voltage reaches 0 but I'd like to be sure.
Note:The diode is assumed to be ideal that's why I said V>0 will make it FB and V<0 will make it RB

you make no sense by assuming the diode is ideal. the best thing is build it and see it on oscilloscope. as i've learnt recently, that "real" diode has internal reverse capacitance which will avoid inf. di/dt which belong to the "fantasy".

err.... assuming the diode and components (and dso probe too!) are rated accordingly.

[sacherjj]

In my opinion it makes much more sense that it becomes reverse biased when the voltage reaches 0 but I'd like to be sure.
Note:The diode is assumed to be ideal that's why I said V>0 will make it FB and V<0 will make it RB

you make no sense by assuming the diode is ideal. the best thing is build it and see it on oscilloscope. as i've learnt recently, that "real" diode has internal reverse capacitance which will avoid inf. di/dt which belong to the "fantasy".

err.... assuming the diode and components (and dso probe too!) are rated accordingly.

Or build it in LTSpice and do a quick simulation. 

[Mechatrommer]

In my opinion it makes much more sense that it becomes reverse biased when the voltage reaches 0 but I'd like to be sure.
Note:The diode is assumed to be ideal that's why I said V>0 will make it FB and V<0 will make it RB

you make no sense by assuming the diode is ideal. the best thing is build it and see it on oscilloscope. as i've learnt recently, that "real" diode has internal reverse capacitance which will avoid inf. di/dt which belong to the "fantasy".
err.... assuming the diode and components (and dso probe too!) are rated accordingly.

Or build it in LTSpice and do a quick simulation.

The diode stops conducting when the voltage across the diode drops below zero.  However, that won't be the same as the time the source voltage drops below zero.  When the source voltage becomes negative, the current through the inductor starts decreasing.  The voltage drop across the inductor is proportional to the rate of change of current.  Since the current is decreasing, there is a negative voltage across the inductor, which keeps the diode forward biased and conducting.  At a certain point, the current through the inductor will drop to zero, the diode will switch off, and the current will remain zero for the rest of the cycle.

[ejeffrey]

I am not sure what you are intending to show there, but it is exactly what I said -- the diode conducts and is forward biased for the entire positive half-cycle, and part of the negative half-cycle of the power supply.  At a certain point the inductor current drops to zero.  At this point, the diode shuts off and prevents reverse conduction.  dI/dt is now zero, so the voltage across the inductor instantly drops to zero, and the diode goes instantly from zero bias to carrying the entire reverse bias of the supply.  You can clearly demonstrate this by adding taps to monitor the voltage across only the diode and the current and voltage from the source.

For absolute clarity: even when implemented with ideal components, the circuit described never generates an infinite current or current. In fact, it never generates a voltage greater than the peak of the supply voltage.  Nor is there an infinite dI/dt -- the only infinite value is the dVdt when the reverse voltage is instantly transfered from the inductor to the diode at cutoff.

Everyone is of course right that there is no such thing as an idea diode.  They always have some junction capacitance, turn-on and turn-off time, non-zero forward drop and non-zero reverse leakage.  However, this circuit drawn is mostly well behaved without any of that.

[Mechatrommer]

i included your quote, indicating its tally with the sim i've just made. i was wrong
if we reduce the inductance, there will be less -ve voltage undershoot in the picture. and if its zero then we got ideal rectifier diode.