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

[KILOWATT]

Hi everyones thanks to read. Since a few weeks, i'm using the LightKeeper
Pro from Ulta-Lit Tree Company ( http://www.lightkeeper.biz/default.asp ) to
test and repair miniature lights sets (series-wired). I've found the Quick
Fix Trigger utility very useful. as you can see on their website, it's based
on a piezo igniter that sends a high voltage pulse through the lights set to
activate (short) the shunt inside the bulb, that didn't do so when the
bulb's filament burned out. I wanted to see how this system is built so i
opened the unit. Here's two photos plus a schematic i've drawed for the
igniter section. (Sorry, the symbol for the piezo igniter is probably
wrong...i don't know the correct one).
http://www3.sympatico.ca/kilo.watt/images/lightkeeper1.JPG
http://www3.sympatico.ca/kilo.watt/images/lightkeeper2.JPG
http://www3.sympatico.ca/kilo.watt/images/hv_pulse_gen.bmp
As i can see, the four diodes allow a peak reverse voltage of approximately
4Kv, wich is sufficient to break down the shunt's insulation inside the
burned bulb(s) and complete the series circuit. Once completed, half of an
AC cycle can flow through the four diodes, allowing the set to glow (dimly)
and show which bulbs are burned and needs replacement.
What leads me to my question (please look at the schematic) is: why the
reverse voltage doesn't seem to damages those diodes? According to the piezo
igniter's website ( http://www.yiqiang-piezo.com/english/production2.htm ),
the output voltage is at least 15Kv. With the four 1n4007 wired in series, i
should get a peak reverse voltage of about 4Kv isn't? For what i know, a
diode is destroyed once it's PIV is exceeded. I think i missed something
when studying the basic operation theory of a diode. ;-) TIA for any
useful reply.
--
Alain(alias:Kilowatt)
Montréal Québec
PS: 1000 excuses for errors or omissions,
i'm a "pure" french canadian!
Come to visit me at: http://kilowatt.camarades.com
(If replying also by e-mail, remove
"no spam" from the adress.)

 

[ChrisGibboGibson]

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

 

[Terry Given]

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 shit 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

 

[ChrisGibboGibson]

Terry Given wrote:

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 shit 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.

Yes. To all of it.

And you *still* will *never* exceed the breakdown voltage of the diode without
zapping it.

Gibbo

 

[Winfield Hill]

ChrisGibboGibson wrote...

Terry Given wrote:

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.

Still trolling?

KILOWATT asked "For what i know, a diode is destroyed once it's PIV
is exceeded." That's the question we're addressing here.

The PIV or "Peak Inverse Voltage" is merely a manufacturer's
specification. It's an "Absolute Maximum Rating" and is called
the "DC Blocking Voltage," "DC Reverse Voltage," "Working Inverse
Voltage" "Continuous Reverse Voltage," "Working Peak Reverse
Voltage," or "Peak Repetitive Reverse Voltage" rating.

This is the maximum voltage at which a worst-case part (one with
a minimum value of breakdown voltage) will have a leakage of some
specified value, like 5uA. So clearly one can go above the PIV,
even for a worst-case part, and accept the resulting higher leakage
current, without damage. For example, the dissipation for 100uA
leakage at 1000V is only 100mW, and will not heat up the die much.

If we have managed to find one of the (very rare) worst-cast parts,
conducting 5uA at exactly 1000V, then to force 100uA through it
we'll have to raise the voltage above 1000 volts. Perhaps not much
higher, but higher nonetheless. (Keep in mind this is for the very
rare worst-cast part, real parts will require you to go to voltages
well above PIV to experience 5uA or 1000uA of leakage.)

The same diode type will often have a second higher-voltage spec on
the datasheet, e.g. called "Non-Repetitive Peak Reverse Voltage."
This is for a special case, e.g. halfwave, single phase, 60 Hz.

For example, for a 1n4007 the first spec is 1000V, and the second is
1200V. The manufacturer is telling us the PIV is 1kV for continuous
DC, but we're allowed to go up to 1.2kV for a single 8.33ms half-sine.
That's an extra 200V, or 20% higher. It's a lot higher than the PIV,
and it's done safely according to the specification. Lower-voltage
diodes, such as a 100V diode, will often allow an extra 100% over PIV
for a single 8.33ms half-sine. Again, safely according to the spec.

Some diodes also have a "Repetitive Peak Reverse Voltage" spec,
which will generally have a value equal to the PIV.


--
Thanks,
- Win

 

[Terry Given]

ChrisGibboGibson wrote:

Terry Given wrote:


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 shit 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.



Yes. To all of it.

And you *still* will *never* exceed the breakdown voltage of the diode without
zapping it.

Gibbo

Semantics, but yes. (or a very terse way of summarising the last couple
of paragraphs). Not helped by the OP question - he's really asking "is
it OK to slap a voltage > Vbreakdown_rated across a diode" (and
Vbreakdown_rated is usually at some weeny current). The answer to that
question is yes, provided you dont let it melt (or conversely no, if you
cant stop it from melting).

But yes, technically if it doesnt melt, you will indeed *never* exceed
the breakdown voltage, merely sit there. Terminological Inexactitude
strikes again.

Cheers
Terry

ps have you ever noticed "casino" is an anagram for "is a con" - I just
thought of that yesterday.

 

[Winfield Hill]

ChrisGibboGibson wrote...

And you *still* will *never* exceed the breakdown voltage
of the diode without zapping it.

You appear to be using a silly definition of breakdown voltage:
whatever voltage the part happens to have when it's conducting a
measureable amount of leakage current. That's a nearly useless
definition, and not what most of understand by the term. It is
certainly not what's meant by PIV.


--
Thanks,
- Win

 

[ChrisGibboGibson]

Winfield Hill wrote:

ChrisGibboGibson wrote...

And you *still* will *never* exceed the breakdown voltage
of the diode without zapping it.

You appear to be using a silly definition of breakdown voltage:
whatever voltage the part happens to have when it's conducting a
measureable amount of leakage current. That's a nearly useless
definition, and not what most of understand by the term. It is
certainly not what's meant by PIV.

Correct. PIV is a rating given by the manufacturer, so the user knows what is a
safe reverse voltage across a diode without running the risk of entering
breakdown and thus destroying the diode.

Gibbo

 

[Winfield Hill]

Terry Given wrote...

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.

Actually the breakdown voltage has a rather high positive temperature
coefficient. This can amount to dozens of volts increase for a diode
with its junction temp close to the rated maximum. The diode-breakdown
tempco is the same as a zener diode with that rated voltage. A table
of these tempcos is in the Motorola 1n5221B family datasheet. There
you'll see that above 100V the tempco is stable at 0.11% per degree C.
This means a 1n4007 1kV diode, with an actual breakdown voltage of say
1100 volts, will have a breakdown tempco of 1.21V/C and the voltage
will increase by 211V for a junction heated to 200C from 25C. That's
a pretty substantial increase, and should not be called "constant."

It is not the "exceeding" the breakdown voltage that causes the problem
- its V*I*t = the energy dumped into the silicon.

Yes, exactly.

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.

Actually silicon junctions continue to work at well far above 200C.
The 150, 175 or 200C temperature given as an operating limit is more
a package limit for reliable repeated and longterm use. A 5V1 zener
will "merely" have a different breakdown voltage at 200C -- about 6V
plus a resistive drop, which would be considerable in your scenario
(see the curves in AoE page 332).

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.

Not Kaboom, exactly. This can be rather misleading. I like your
little story, and the issues it raises, but I disagree with your
calculations and assumptions. It'd be more instructive to present
such a disaster scenario and show how the diode can survive nicely.

[I have done these calculations plus transient thermal modeling,
combined with actual high-energy-breakdown junction-temperature
measurements on a microsecond timescale. My in-situ dynamic
junction-temperature measurements were made several ways. One
was by interrupting the high current and placing the diode into
low-current breakdown and using the voltage tempco to self-measure
the junction temperature. These measurements matched my modeling.

I also did a 10,000-cycle test with rapid (100us) jumps to 250C,
and saw no damage, nor any observable spec change to the part!

Of course this is far above the rated temp, and one should not use
such capabilities in production designs, but it _was_ interesting.

Someday I'd like to return to the issue and find out just where one
does get into trouble. My understanding from the work of others is
that this varies from component type to type, and from manufacturer
to manufacturer - some are VERY good, and some are not.]


--
Thanks,
- Win

 

[Terry Given]

Hi Win,

Winfield Hill wrote:

Terry Given wrote...

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.


Actually the breakdown voltage has a rather high positive temperature
coefficient. This can amount to dozens of volts increase for a diode
with its junction temp close to the rated maximum. The diode-breakdown
tempco is the same as a zener diode with that rated voltage. A table
of these tempcos is in the Motorola 1n5221B family datasheet. There
you'll see that above 100V the tempco is stable at 0.11% per degree C.
This means a 1n4007 1kV diode, with an actual breakdown voltage of say
1100 volts, will have a breakdown tempco of 1.21V/C and the voltage
will increase by 211V for a junction heated to 200C from 25C. That's
a pretty substantial increase, and should not be called "constant."

I was waiting for that. Thanks for the info - I'll go dig up the data
sheet (and archive it). I didnt want to further complicate the waffly
story, and had no idea what the tempco might be so I ignored it. Shoulda
written (ignoring tempco) in there though. Bad Terry

Nor did I think it would be the same as a "zener" diode. Why is that?

It is not the "exceeding" the breakdown voltage that causes the problem
- its V*I*t = the energy dumped into the silicon.


Yes, exactly.


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.


Actually silicon junctions continue to work at well far above 200C.
The 150, 175 or 200C temperature given as an operating limit is more
a package limit for reliable repeated and longterm use. A 5V1 zener
will "merely" have a different breakdown voltage at 200C -- about 6V
plus a resistive drop, which would be considerable in your scenario
(see the curves in AoE page 332).

"merely" - LOL

200C is a number I have used in the past, plucked from waffly stories
that make the blatant assertion "silicon goes intrinsic at around 200C"
without providing supporting proof. I once tried to find some hard data
on this, but to no avail. Many years ago I read Dye and Granberg's book
on RF transistors, and discovered that Tjmax = 125C is a kinda arbitrary
thing (apparently RF Si-abusers beat their silicon much harder, 175C not
being uncommon) but as you say, reliability/lifetime goes down - high dT
exacerbates TCE mismatch related thermal fatigue, (IIRC metal migration
speeds up too).

So: What number should I use? And at what temperature does silicon go
"intrinsic". Obviously carrier mobility increases with temperature, but
I guess this boils down to:

a) where you arbitrarily define "too much" conductivity (one persons
excess leakage is another persons bias current)

b) whats in the actual silicon

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.


Not Kaboom, exactly. This can be rather misleading. I like your
little story, and the issues it raises, but I disagree with your
calculations and assumptions. It'd be more instructive to present
such a disaster scenario and show how the diode can survive nicely.

Si melts at 1410C, so it will melt in 11ms or so. Ignoring the incorrect
zener dissipation calc (Vz hasn't been corrected for Iz) of course

[I have done these calculations plus transient thermal modeling,
combined with actual high-energy-breakdown junction-temperature
measurements on a microsecond timescale. My in-situ dynamic
junction-temperature measurements were made several ways. One
was by interrupting the high current and placing the diode into
low-current breakdown and using the voltage tempco to self-measure
the junction temperature. These measurements matched my modeling.

I also did a 10,000-cycle test with rapid (100us) jumps to 250C,
and saw no damage, nor any observable spec change to the part!

Wow! Now thats what I call characterising your semiconductors.

Of course this is far above the rated temp, and one should not use
such capabilities in production designs, but it _was_ interesting.

Someday I'd like to return to the issue and find out just where one
does get into trouble. My understanding from the work of others is
that this varies from component type to type, and from manufacturer
to manufacturer - some are VERY good, and some are not.]

That is a good question. I have a job coming up where I will be making a
3-phase ZCS converter using an auxiliary commutation network. It gets a
0.5us pulse every 50us, so knowing the peak "acceptable" (whatever that
means) Tj will allow me to minimise the Si cost. Luckily the
manufacturers provide no data whatsoever in this regard.

My background is big motor drives, and thermal cycling is perhaps the
major determining factor in IGBT lifetime (assuming the design actually
works, not always a valid assumption) - the mechanical forces caused by
mismatched TCE cause voids to form in the die-to-DCB-substrate solder
joint, increasing Rtheta and hence dTj. But these are BIG junctions -
1cm x 1cm is not uncommon (rough dimensions only), so the forces
involved are quite significant. There have been a couple of lovely
papers in IEEE trans. industry apps, showing (IIRC) xrays of the voids
forming. hockey-puck thyristors et al are immune to this type of
failure, as they aint soldered!

Yee-ha - I've been up for 30 minutes, and have already learned something
useful. Thanks Win. Todays shaping up to be a great day. Oh wait, thats
right, I have to mow the lawns - a 3 hour job. doh. Still, I'm now a
slightly smarter labourer


Cheers
Terry

 

[Winfield Hill]

ChrisGibboGibson wrote...

Correct. PIV is a rating given by the manufacturer, so the user
knows what is a safe reverse voltage across a diode without
running the risk of entering breakdown and thus destroying the
diode.

<sigh> I thought we might have by now disabused you of such
bold yet incorrect language, with its overly alarmist message.


--
Thanks,
- Win

 

[Pooh Bear]

Terry Given wrote:

200C is a number I have used in the past, plucked from waffly stories
that make the blatant assertion "silicon goes intrinsic at around 200C"
without providing supporting proof. I once tried to find some hard data
on this, but to no avail. Many years ago I read Dye and Granberg's book
on RF transistors, and discovered that Tjmax = 125C is a kinda arbitrary
thing (apparently RF Si-abusers beat their silicon much harder, 175C not
being uncommon) but as you say, reliability/lifetime goes down - high dT
exacerbates TCE mismatch related thermal fatigue, (IIRC metal migration
speeds up too).

Interesting to note that packaging influences Tj max.

On Semi make a couple of devices ( typicaaly used in high power audio output
stages ) that are available in TO-3 or TO-3P. MJ(L)21193/4

The TO-3 version is rated for Tj = 200C but the plastic packaged part is just
150C.

Clearly if the device is specced to operate OK at 200C - intrinsic conduction
must occur at a somewhat higher temp.


Many decades ago RCA wrote an app note relating to reliability vs thermal
cycling too. Lower temp rise results in less stress on the die attach.


Graham

 

[Winfield Hill]

Pooh Bear wrote...

Many decades ago RCA wrote an app note relating to reliability vs thermal
cycling too. Lower temp rise results in less stress on the die attach.

Do you know which one that was?


--
Thanks,
- Win

 

[Pig Bladder]

On Sat, 20 Nov 2004 15:58:00 -0800, Winfield Hill wrote:

ChrisGibboGibson wrote...

Correct. PIV is a rating given by the manufacturer, so the user
knows what is a safe reverse voltage across a diode without
running the risk of entering breakdown and thus destroying the
diode.

sigh> I thought we might have by now disabused you of such
bold yet incorrect language, with its overly alarmist message.

Kids these days!
--
The Pig Bladder From Uranus, still waiting for
some hot babe to ask what my favorite planet is.

 

[Fred Bloggs]

According to the piezo
igniter's website ( http://www.yiqiang-piezo.com/english/production2.htm ),
the output voltage is at least 15Kv. With the four 1n4007 wired in series, i
should get a peak reverse voltage of about 4Kv isn't? For what i know, a
diode is destroyed once it's PIV is exceeded. I think i missed something
when studying the basic operation theory of a diode.

Yeah you did- the PIV is the reverse voltage a diode can withstand
indefinitely without breakdown- it is not the voltage at which the diode
will breakdown. Clearly the diode design has to take into account
manufacturing process variations so that despite these an acceptably
large percentage of the diodes meet this PIV specification- and under
worst case conditions such as Tj=70oC or something- who knows. Therefore
it would not be unlikely that a typical 1N4007 breakdown at a much
larger voltage than 1KV. The product specification was written by some
sap with less than perfect command of the English language, so that the
real meaning may have been the output voltage is no greater than 15KV-
meaning it is an insignificantly rare occurrence that a stack of four
randomly selected 1N4007's withstand that voltage without breakdown. One
thing that is certain is that the output will be significantly greater
than 4KV.

 

[Rich Grise]

On Sun, 21 Nov 2004 10:51:07 +1300, Terry Given wrote:

Winfield Hill wrote:

Actually the breakdown voltage has a rather high positive temperature
coefficient. This can amount to dozens of volts increase for a diode
with its junction temp close to the rated maximum. The diode-breakdown
tempco is the same as a zener diode with that rated voltage. A table
of these tempcos is in the Motorola 1n5221B family datasheet. There
you'll see that above 100V the tempco is stable at 0.11% per degree C.
This means a 1n4007 1kV diode, with an actual breakdown voltage of say
1100 volts, will have a breakdown tempco of 1.21V/C and the voltage
will increase by 211V for a junction heated to 200C from 25C. That's
a pretty substantial increase, and should not be called "constant."

I was waiting for that. Thanks for the info - I'll go dig up the data
sheet (and archive it). I didnt want to further complicate the waffly
story, and had no idea what the tempco might be so I ignored it. Shoulda
written (ignoring tempco) in there though. Bad Terry

Nor did I think it would be the same as a "zener" diode. Why is that?

My take on this is that they're the same thing, it's just a naming thing.
Probably for marketing purposes. ;-)

I also had at least one tech school teacher who said that in some cases an
ordinary rectifier is put in series with a zener and so their tempcos, if
not exactly cancel out, at least mostly do. Which makes sense, and I kind
of wonder if there are a lot of them already being made like that - the
die could be just an ordinary transistor, calibrated. ;-)

Cheers!
Rich

 

[Winfield Hill]

Rich Grise wrote...

I also had at least one tech school teacher who said that in some cases an
ordinary rectifier is put in series with a zener and so their tempcos, if
not exactly cancel out, at least mostly do. Which makes sense, and I kind
of wonder if there are a lot of them already being made like that - the
die could be just an ordinary transistor, calibrated. ;-)

Yep, but only at a specific voltage. Look at the zener tempco curve
in AoE page 333, and note the +2.2mV/C tempco for a 5.8V zener running
at 10mA. Put a -2.2mV/C tempco in series with that (a diode-connected
transistor) and now you've got a zero-tempco 6.4V reference. Curves on
page 332 show it'll have a comparatively-low dynamic impedance as well.


--
Thanks,
- Win

 

[legg]

On 20 Nov 2004 15:58:00 -0800, Winfield Hill
<hill_a@t_rowland-dotties-harvard-dot.s-edu> wrote:

ChrisGibboGibson wrote...

Correct. PIV is a rating given by the manufacturer, so the user
knows what is a safe reverse voltage across a diode without
running the risk of entering breakdown and thus destroying the
diode.

sigh> I thought we might have by now disabused you of such
bold yet incorrect language, with its overly alarmist message.

There is more than one way that a diode can 'break down'.

Zener behaviour is presented by only one of these three mechanisms.

A very short summary, with diags:

http://www.voltagemultipliers.com/html/bvr.html

RL

 

[Winfield Hill]

ChrisGibboGibson wrote...

PIV is a rating given by the manufacturer, so the user knows what
is a safe reverse voltage across a diode without running the risk
of entering breakdown and thus destroying the diode.

Voltage Multipliers is a company that makes HV diodes. They perform
PIV tests on each and every HV diode they ship, described this way:

"Rectifiers are generally subjected to a peak inverse voltage
(PIV) test to identify their breakdown characteristics. This
test is performed by applying 60-hertz half-wave reverse voltage
of sufficient amplitude to initiate breakdown. During the test,
the reverse current is usually limited to 50ľA. The resulting
waveform is observed on an oscilloscope to determine the
sharpness of the 'knee' at the point of breakdown."

So either the statement "entering breakdown and thus destroying
the diode" is wrong, or VMI is selling 100% destroyed HV diodes.


--
Thanks,
- Win

 

[legg]

On 21 Nov 2004 07:54:18 -0800, Winfield Hill
<hill_a@t_rowland-dotties-harvard-dot.s-edu> wrote:

ChrisGibboGibson wrote...

PIV is a rating given by the manufacturer, so the user knows what
is a safe reverse voltage across a diode without running the risk
of entering breakdown and thus destroying the diode.

Voltage Multipliers is a company that makes HV diodes. They perform
PIV tests on each and every HV diode they ship, described this way:

"Rectifiers are generally subjected to a peak inverse voltage
(PIV) test to identify their breakdown characteristics. This
test is performed by applying 60-hertz half-wave reverse voltage
of sufficient amplitude to initiate breakdown. During the test,
the reverse current is usually limited to 50ľA. The resulting
waveform is observed on an oscilloscope to determine the
sharpness of the 'knee' at the point of breakdown."

So either the statement "entering breakdown and thus destroying
the diode" is wrong, or VMI is selling 100% destroyed HV diodes.

They also list the three mechanisms of reverse leakage that would be
identified in such a test. They also state that there are two
requirements to avoid destruction - only one of which is the control
of reverse current.

Unless the diode is manufactured with this in mind, there is no
likelihood that controlled avalanche or resistive zenering will occur.

They also state that " high voltage zener diodes are less practical
because the higher-resistivity silicon and deeper-diffusion depths
required to achieve the higher voltage ratings make it difficult to
predict the voltage at which breakdown occurs." .....ie avalanche or
punch through are the most likely limitations in testing or employing
HV parts.

As a manufacturer of HV devices, VMI will be conscious of the need to
control all three, so that series applications have some chance of
success. The planar or deep-diffused wafer construction, the mesa
wafer-etch technology and the glass or other surface passivation
methods employed, will produce a more expensive part that other
manufacturers need not duplicate to produce a similar PIV book rating,
for less stringent commercial applications.

Assuming that 1N4000 series parts are so designed, in reproducing or
modifying a piezo-electric light bulb tester outputing some kilovolts,
will not neccessarily produce predictable results. Being able to
identify the required characteristics from a diode's spec sheet or
test results, will offer that possibility.

To assume that zenering will occur, with it's characteristic PTC above
6V, is unwarranted., though controlled avalanche also seems to have a
PTC, where it occurs.

RL