
The basic situation is shown in
the figure in the backbone
module which will be repeated here in a somewhat more detailed
fashion. 




We switch from a forward condition to a reverse
condition at some time. The external voltage (blue lines in the diagram) is
supposed to change suddenly (we have an ideal switch) 


What we would measure in terms of the junction voltage and the
junction current is shown in magenta or red, respectively. 


The outstanding feature is the "reverse recovery",
the reverse current flowing for some time after we switched the voltage. Right
after the switching it will be limited to U_{R}/R
for a time t_{s}, because we can not drive more current
than that through the circuit. But after t_{s} seconds,
the current decays with some time constant t_{r} until it
reaches the small (zero in the picture) static reverse current of the
junction. 


If we look at t_{r} quantitatively, we
take it to be the time it takes the current to decay to 10% of the
plateau value. 


Can we calculate this behavior, which of course is the crucial
behavior for the large signal switching of a pnjunction? 

Well  not without some problems. But we can
understand what others have calculated. Let's see. 


During static forward behavior, we have a surplus of minority
carriers a the edge of the space charge region, and this surplus concentration
has to disappear after we switch to reverse conditions. We looked at that in
some details
before, and we already have some equations for this case 


We have to solve the relevant diffusion equation as given in
the link above, but now for different conditions. Before, we looked at the
static case (i.e. ¶n^{min}(x,t) /
¶t = 0, now we want to calculate how
the minority carrier concentration changes in time. 


So, once more, we have to solve the relevant continuity
equation. We do it for one side of the junction only; the other side then is
trivial. 


¶n^{min}(x,t)
¶t 
= 
D · 
¶^{2}n^{min}(x,t)
¶x^{2} 
– 
n^{min}(x,t)
– n_{0}
t_{eff} 




The last term simply governs the disappearance of carriers by
recombination; otherwise we just have Ficks second law. For t_{eff} we have to take the minority carrier
lifetime t or the transit time t_{trans} as the geometry demands (inbetween
situations are messy!). 

If we have the solution for
n^{min}(x,t), we can calculate everything
else easily, the voltage across the junction. e.g.
is always 


U_{junct} 
= 
kT
q 
· ln 
Dn^{min}(x, t)
n_{0} 



Now we have to look at the boundary conditions for
the problem 


If you look at the picture above long enough, you realize that
as long as U_{junct} is positive, the boundary conditions
are 


I_{R} 
= 
U_{R}
R_{R} 
= q · D · 
¶n^{min}(x,t)
¶x 




As soon as U_{junct} = 0 V, the boundary
conditions need to be changed to 





You don't see it? That's OK, at least for the second case. The
boundary conditions are actually only approximations, and would take a lengthy
discussion to justify them (in particular the second one and the switch over
point) in detail. So just believe it. 

Now it is math  solving differential equations
with certain boundary conditions. Not so easy, but doable. According to
Kingston (1953), the solutions for the two time constants
t_{s} and t_{r} are (in implicit
form) 


1
1 + I_{R}/I_{F} 
= 
erf 
æ
ç
è 
t_{s}
t_{eff} 
ö
÷
ø 
1/2 
erf 
æ
ç
è 
t_{r}
t_{eff} 
ö
÷
ø 
1/2 
+ 
exp –(t_{r}/t_{eff} )
p · t_{r}/t_{eff} 
= 
1 + 
0,1 · I_{R}
I_{F} 




with erf = error function as
we know it from
diffusion problems. 

OK. May the force be with you when you try to
prove these solutions or just to extract data. Only one thing is clear: We
better look at the ratios t_{s}/t_{eff} and t_{r}/t_{eff} than at the t's
directly. 


Well, there are always the approximations, which we are going
to use here: 


t_{s} + t_{r}
» 
t_{eff}
2_{ } 
· 
I_{F}
I_{R} 



Even better, there are complete solutions in
graphical form: 











The solid lines are for the "small" diode, where we
have to take the transit time for t_{eff}, the dashed line indicate the
"large" diode case. 

It is clear that you really can achieve much
larger switching speeds for a given t_{eff} by being smart about
I_{R}/I_{F}, i.e. if you increase
I_{R} (or decrease I_{F}, but that
is rarely an option) 


However, don`t forget the prize you have to pay: Large reverse
currents while "idling" = large losses = heating your device. 

This is the first inkling we get that there is
some trade off between speed and
power. 


