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After the forward conduction, when the diode polarity is reversed, the diode goes to reverse recovery condition.
What will happen if we give forward voltage again during the reverse recovery time of the diode?
I mean before the end of the reverse recovery time of the diode.
Will the diode again start to conduct in forwarding recovery and if yes, then will the forward recovery time be the same as if it goes to forward recovery from completely off state?
A general answer is, it depends.
The physics is diffusion of charge within the junction: when turning on, free charges (minority carriers) are emitted, and propagate through the depletion region, un-depleting it. The momentary lack of carriers gives a high voltage drop, or an inductive characteristic, at turn-on: forward recovery.
Conversely, the excess of carriers holds the device on (conductive) until they are cleared from the depletion region by force (recovery current) or by recombination. (The junction acts like a very small, very leaky battery, indeed this is no accident as the charges in batteries have similar statistics.)
What's interesting is the transient behavior, for on and off pulses shorter than the transit time (that is, the time for charges to diffuse into the junction). This is modeled poorly by SPICE for example (which uses a single element bulk equivalent -- basically modeling that battery (mentioned above) as a nonlinear capacitor), and is given by diffusive transport.
For example, with a short enough on-time, charges are emitted, but they haven't saturated the junction; if reversed during this time, the charges cross the junction briefly, then suddenly clear -- snap recovery. That is, the softness factor goes down considerably for short t_on.
This is a very hand-waved explanation, and I don't know the dynamics of the physics myself. Heck, it's been a long time since I did any homework problems with semiconductors in the first place...
In any case, suffice it to say, it is a well understood problem, to the extent that very accurate simulations can be performed -- given you have an exact model of the material under test (i.e., which semiconductor and bulk properties, dopants and concentration, etc.).
Notice you will not have these data for commercial parts, so somewhere between a basic SPICE model (manufacturer-provided, fitted to the real component; but doesn't model diffusion at all), and research-grade material physics model (arbitrarily accurate to a real component, given the correct dimensions and profiles; but those data are strictly unavailable for commercial components), lies what happens in reality.
Anyway, I will demonstrate recovery for an example part below.
Background
Some time ago, I constructed a diode recovery test fixture: Diode Recovery Tester
A direct link to the schematic on my website: https://www.seventransistorlabs.com/Images/Diode_Recovery_Tester.pdf
This generates the two-pulse waveform used for measuring recovery and switching loss in diodes, MOSFETs, etc. The first pulse charges an inductor up to the desired test current; the switch turns off, dumping that current into the diode, which clamps the flyback. The switch turns on again, causing hard switching (reverse recovery). The switch turns off some time after recovery has been observed, and the inductor slowly discharges into the diode. Eventually, current goes to zero, and a new cycle begins.
Experimental
There are minor differences between published circuit and what was tested. Specifically, the gate drive has an additional complementary emitter follower after Q7/Q9 (slightly increases dI/dt), and uses a STP19NM50N for Q4.
Here is a standard recovery waveform. I am testing a standard-recovery high voltage rectifier, H1B120 (probably similar to 1N5408; measured breakdown: 1460V).
Ch.3: gate drive waveform (uncal; timing is subject to delays through the drive circuitry). (This is the voltage on R28.)
Ch.1: DUT current (10x probe, -10A/div): current is zero initially, forward-biased (for about 1µs), then reversed, then forward again.
Ch.4: DUT voltage (10x probe, 50V/div): voltage is zero initially, then reversed (113V, just left of view), then the forward-bias pulse sequence.
Forward bias is about 4A. Reverse begins with a ramp of about 500A/µs, shooting up to a 33A peak before dropping (coincident with voltage reversing as the DUT finally turns off).
Same scale as above, forward reduced to 500ns. You can see the recovery tail is shorter, shallower. Peak current is also lower; it seems to be switching better.
Same scales, horizontal zoom. Forward reduced to 100ns. trr is only 40ns, Irr, 16A, but the softness -- it's completely disappeared and ringing an extreme amount instead.
It's hard to see in B&W, but the voltage and current waveforms do seem to be ringing against each other, which makes sense, this should be the junction capacitance ringing with lead inductance. (Ballpark figures of 110MHz, 20nH and 100pF are plausible.)
This short-forward regime seems to be the more interesting case, as you can see. What about reverse?
Brief Reverse
Ch.3: gate drive waveform.
Ch.1: -5A/div; current is zero initially, forward-biased briefly (about 370ns), then reversed very briefly (a hump of about 200ns, clearly reversing current flow), then forward again.
Ch.4: DUT voltage.
Due to limitations in my circuit, I can't get a very crisp re-forward-bias pulse, but thanks to the slowness of this DUT, it seems to be showing the desired waveform. There is a small twiddle in Ch.4, as it's briefly reversed; but the voltage change is small, and likely that's all attributable to stray inductance.
So, nothing interesting seems to be happening. Which is reasonable. We expect from the physics that, as long as a fair number of carriers remain in the junction, it's still "on" (voltage near zero for most any current +/-), and likewise, forward recovery is largely avoided by already having charges in the junction. Any excess voltage will be pretty modest.
Here's a zoom on diode voltage (Ch.4):
Initial forward recovery seems to be around 20V, dropping to 4V after 134ns, and 2V (very near DC Vf) after 400ns. Whereas, after the brief reversal, the peak voltage is only 3.4V; meanwhile dI/dt is around 60A/µs, or 1.2V across 20nH. So probably the intrinsic Vfr during this moment is more like 2.2V -- not much above Vf(DC).
At least for pulse widths and peak currents within the range of most applications / datasheet range values, brief reversal should be ignorable -- it doesn't turn far enough off, or do anything really interesting as far as transient effects, to notice anything.
As pointed out by @Jonk, certainly a "physics question" ... (a big number of parameters should be known).
For reference, here are three cases of simulation.
The normal behavior of a switched diode (square wave), recovery diode time ~ 6 us.
The other mode could be a switching current with a defined "slew-rate" (to do).
Switching at the "limit" ...
Switching too fast (internal capacitors involved), diode always ON.
