Re: Can the PIV of a diode be "safely" exceeded?

From: Terry Given (my_name_at_ieee.org)
Date: 11/20/04 

Date: Sun, 21 Nov 2004 03:12:51 +1300

ChrisGibboGibson wrote:
> Terry Given wrote:
>
> [snip]
>
>
>>Chris is making an underlying assumption - that the diode is being
>>supplied by a "stiff" voltage source (ie one which can supply more than
>>enough current to, say, melt the semiconductor). In such a case, he is
>>pretty much bang on (pun intentional). Because this was not explicitly
>>stated by the OP, Chris is incorrect.
>>
>
>
> But if the voltage source isn't stiff enough to supply enough current to
> destroy the semiconductor, then it will not be stiff enough to actually
> *exceed* the breakdown voltage. In which case the diode will be *at* the
> breakdown voltage, not above it.
>
> Gibbo

Hi Gibbo,

One can easily argue that (at least until the diode is destroyed) it
NEVER exceeds its breakdown voltage - although the measured voltage is
of course a function of current (model it as an ideal zener in series
with a resistor). The breakdown voltage remains constant, but the series
resistance has a current-dependant voltage drop, and you measure the sum
of the two.

It is not the "exceeding" the breakdown voltage that causes the problem
- its V*I*t = the energy dumped into the silicon. (If the current
density is sufficient, interconnects may fuse, but by that stage you've
already got a serious problem).

Consider this: a 400V DC supply with 10 Ohm impedance across a 5V1 zener
will pump about 40A thru the zener, which will dissipate something like
200W. If the junction is 1mm x 1mm x 1mm, V = 1e-9m^3. density =
2330kg/m^3 so m = 2.33e-6kg. specific heat capacity cp = 700J/(kg*K) so
m*cp = 1.63e-3J/K. If we assume the diode started out at 25C, and we'll
pick 200C as the "stuffed" point (Si is pretty much a conductor by then,
ie kaboom) then dT = 175C. m*cp*dT = 0.285J.

So if we shove 285mJ into the junction (and dont allow the heat to flow
out) it will go kaboom. 1J = 1W*1s so 285mJ/200W = 1.43ms. If the diode
has bond wires (ie isnt one of these monstrous high-current
thingummies), little heat will flow from the junction in this short time
  so the adiabatic approximation will be valid (IOW all the heat dumps
into the silicon). So this zener will *** itself in about 1.5ms or so
(which is kinda what you'd expect for shoving 40A up the arse of a 5V
zener).

Of course the actual zener voltage will be much higher because of its
series impedance). But its not too hard to (iteratively if need be)
calculate the actual zener drop and hence dissipation, before doing the
dT calcs.

OTOH if we did the 200W splat for say 100us the total energy dumped into
the zener dE = 20mJ so dT = 20mJ/1.63mJ/K = 12.3C. The real question is
thus "can we limit the avalanche breakdown energy in this diode"

Normally when a zener diode is used there is usually some impedance
between the supply and the zener. This sets the zener current, which
then determines the actual zener voltage. Thats why if you want good
line regulation (zener voltage doesnt change when supply voltage
changes) you use a constant current source. And sometimes if you pick
the current just right, you can get a zero tempco (or some other value
that happens to be convenient).

Another way is to control the energy - for example a zener clamp across
a single-ended flyback smps primary winding with peak current Ipk will
clamp at (Vz + Ipk*Rz), and will dissipate all of the energy stored in
the primary leakage inductance Pz = 0.5*Lleak*Ipk^2*Fsmps. This is the
average power loss. The peak power loss will be much higher - Pzpk =
0.5*Vz*Ipk - but for a short duration. In this case the zener
requirements are:

1) peak current capacity > Ipk

2) thermal resistance when mounted can cope with Pz
[funny story: first flyback I ever designed had three SOD87 smt zeners
and a series diode for the primary clamp. The pcb stood on its edge. The
layout guy used the standard SOD87 footprint and 10mil tracks to connect
the diodes in series. When I first ran the smps, they got so hot they
melted the solder, fell off the pcb and kaboom! When I went to diagnose
the fault, I noticed the diodes sitting on the bench. I checked the
thermal resistance, it was like 300C/W or so! We re-did the PCB with
some meaty copper strips attached to the diodes, and they ran nice and
cool]

3) diode transient thermal impedance can cope with Pzpk. If you dont
have a transient thermal impedance curve, use the junction dimensions to
calculate the adiabatic temperature rise, as above.

This also works for a FET driving an unclamped inductive load.

Cheers
Terry

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