Maker Pro
Maker Pro

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

K

KILOWATT

Jan 1, 1970
0
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.)
 
R

Rich Grise

Jan 1, 1970
0
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.

Just "exceeding the PIV" doesn't automatically destroy a diode.
In fact, that's how Zener diodes are _intended_ to be operated -
in inverse breakdown.

What lets the smoke out is exceeding the power dissipation
ability of the chip. A 1N4007 with 1KV across it and 50
microamps flowing through it would have to dissipate 50
milliwatts, which it would hardly notice, especially
considering that they'll dissipate right about a watt in
the forward direction, at one amp, just in normal operation.
With a proper heat sink, or if the pulse doesn't last very
long, they can pass a surprising amount of current and live
to tell about it. ;-)

Hope This Helps!
Rich
 
T

Terry Given

Jan 1, 1970
0
Rich said:
Just "exceeding the PIV" doesn't automatically destroy a diode.
In fact, that's how Zener diodes are _intended_ to be operated -
in inverse breakdown.

What lets the smoke out is exceeding the power dissipation
ability of the chip. A 1N4007 with 1KV across it and 50
microamps flowing through it would have to dissipate 50
milliwatts, which it would hardly notice, especially
considering that they'll dissipate right about a watt in
the forward direction, at one amp, just in normal operation.
With a proper heat sink, or if the pulse doesn't last very
long, they can pass a surprising amount of current and live
to tell about it. ;-)

Hope This Helps!
Rich

A better way to look at it is the total energy Elost dumped into the
device, in Joules. Then if you know the die area, thickness and material
you can calculate volume and mass, then look up specific heat capacity
cp (J/kg/K). Adiabatic temperature rise = Elost/(m*cp). Voila. You can
also make a fair estimate at the thermal resistance, and calculate the
thermal time constant:

Tau = (m*cp)*Rtheta = [J/K]*[K/W] = [W*s/K]*[K/W] =

if the pulse is less than Tau then the adiabatic approximation is good -
the die absorbs pretty much all the heat. If the pulse is longer than
Tau, then some (or perhaps almost all) of the heat flows out via the
surface and the end-caps.

I read a fascinating paper a few years back on electronic one-shots -
literally! They were electronically fired single-shot guns. I forget
exactly how the circuit worked, but basically they turned a switch on
and dumped all of the energy from a cap into a short section of pcb
track sitting below a projectile. All the energy dumps into the track,
and gets converted into heat and projectile motion. They used a 1N4007
in series with a FET as the switch. IIRC the fet leakage current was
controlled/chosen to hold the 1N4007 on the cusp of avalanche breakdown,
but keep dissipation low. To fire the device, the FET was turned hard
on, and the 1N4007 broke down. The large amount of energy involved
created a ball of plasma which swept thru the diode, turning it into a
very effective short. I must dig that back out.....

Cheers
Terry
 
R

Rich Grise

Jan 1, 1970
0
Rich Grise wrote: ....

A better way to look at it is the total energy Elost dumped into the
device, in Joules.

Terry, thanks for picking up the ball here! :)
Rich
[etc.]
Then if you know the die area, thickness and material
you can calculate volume and mass, then look up specific heat capacity
cp (J/kg/K). Adiabatic temperature rise = Elost/(m*cp). Voila. You can
also make a fair estimate at the thermal resistance, and calculate the
thermal time constant:

Tau = (m*cp)*Rtheta = [J/K]*[K/W] = [W*s/K]*[K/W] =

if the pulse is less than Tau then the adiabatic approximation is good -
the die absorbs pretty much all the heat. If the pulse is longer than
Tau, then some (or perhaps almost all) of the heat flows out via the
surface and the end-caps.

I read a fascinating paper a few years back on electronic one-shots -
literally! They were electronically fired single-shot guns. I forget
exactly how the circuit worked, but basically they turned a switch on
and dumped all of the energy from a cap into a short section of pcb
track sitting below a projectile. All the energy dumps into the track,
and gets converted into heat and projectile motion. They used a 1N4007
in series with a FET as the switch. IIRC the fet leakage current was
controlled/chosen to hold the 1N4007 on the cusp of avalanche breakdown,
but keep dissipation low. To fire the device, the FET was turned hard
on, and the 1N4007 broke down. The large amount of energy involved
created a ball of plasma which swept thru the diode, turning it into a
very effective short. I must dig that back out.....

Cheers
Terry
 
W

Winfield Hill

Jan 1, 1970
0
KILOWATT wrote...
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.

The other replies in this thread have shown you how the four diodes can
happily survive breakdown if the breakdown energy they experience is
below a critical value. Of course they will breakdown if the voltage is
high enough, and thereby limit the piezo output, which cannot be 15kV in
this instance. It's also likely that with a bulb in place the intended
breakdown processes in the bulb (the clearing of the shunt) will limit
the piezo's output voltage before the diode breakdown voltage is reached.

You can measure the actual voltage with an oscilloscope, even if you don't
have a high-voltage probe, by using a capacitive divider, if you can get
hold of a small high-voltage capacitor. Here's a 1000:1 divider circuit.

.. 1000:1 high-voltage probe
..
.. 10pF 20kV ______________ 330
.. <------||----)_____________)--+-/\/\--+----o ) scope input
.. coax cable | _|_ _|_ | Rin = 1M
.. capacitance | --- |___| | Cin = 20pF (etc)
.. 8 ft = 220pF | | | |
.. <--------------------------+--+-------+-----+------ ground
.. ground clip ~9750pF series back-to
.. select back 20V zeners

To avoid a shock hazard, be sure to connect the ground clip and plug
the probe into the scope before probing anything! And don't touch
that 10pF cap!

The probe has a high-frequency response of 10MHz, due to the 330-ohm
input-protection resistor and the ~25pF of capacitance of two 1n5250B
zener diodes in series. It also has a -3dB low-frequency rolloff of
16Hz, due to the 1M scope Zin and the 10,000pF divider capacitance.
You're measuring pulses, so the low-freq rolloff won't be an issue.
Calibrate the probe with a 10V sine wave using the 10mV scope range.
 
W

Winfield Hill

Jan 1, 1970
0
Winfield Hill wrote...
You can measure the actual voltage with an oscilloscope, even if you don't
have a high-voltage probe, by using a capacitive divider, if you can get
hold of a small high-voltage capacitor. Here's a 1000:1 divider circuit.

I should add that most of the HV meter probes one finds around a typical
workbench are *not* suitable for pulse measurements. I'm referring to
the DC probes with the 6" proboscis tip and a double banana plug on the
other end of the cable, for use with a multimeter. These employ a long
high-voltage resistor and do not have proper internal shielding, so they
will give you an erroneous (e.g., too high) result for short pulses.
 
F

Fred Bloggs

Jan 1, 1970
0
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.
 
F

Fred Bloggs

Jan 1, 1970
0
Terry said:
I read a fascinating paper a few years back on electronic one-shots -
literally! They were electronically fired single-shot guns. I forget
exactly how the circuit worked, but basically they turned a switch on
and dumped all of the energy from a cap into a short section of pcb
track sitting below a projectile. All the energy dumps into the track,
and gets converted into heat and projectile motion. They used a 1N4007
in series with a FET as the switch. IIRC the fet leakage current was
controlled/chosen to hold the 1N4007 on the cusp of avalanche breakdown,
but keep dissipation low. To fire the device, the FET was turned hard
on, and the 1N4007 broke down. The large amount of energy involved
created a ball of plasma which swept thru the diode, turning it into a
very effective short. I must dig that back out.....

Nothing new- the electronically fired catridge-less projectile has been
around forever- but usually to achieve extraordinarily high rates of
fire like millions of rounds per minute.
http://www.cnn.com/2003/BUSINESS/06/26/australia.metalstorm/
 
J

Joel Kolstad

Jan 1, 1970
0
Winfield Hill said:
I should add that most of the HV meter probes one finds around a typical
workbench are *not* suitable for pulse measurements. I'm referring to
the DC probes with the 6" proboscis tip and a double banana plug on the
other end of the cable, for use with a multimeter. These employ a long
high-voltage resistor and do not have proper internal shielding, so they
will give you an erroneous (e.g., too high) result for short pulses.

....due to capacitive coupling 'around' the resistor? Or is there some other
mechanism at work?
 
R

Rich Grise

Jan 1, 1970
0
...due to capacitive coupling 'around' the resistor? Or is there some other
mechanism at work?

As well as 'through' the resistor, yes. :) Or maybe 'in' it. Win will
know. ;-)

Cheers!
Rich
 
L

legg

Jan 1, 1970
0
Can the PIV of a diode be "safely" exceeded?...........snip
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.

Yes it can,........ but it is only guaranteed within known limits for
devices characterized to do so. For a rectifier, this characteristic
is known as 'soft avalanche' and will be part of it's specification.

A standard rectifier may exhibit a degradation in breakdown voltage
level, following the first avalanche event, even in 'energy-limited'
situations. If parallel capacitance is signifigant, (ie in common EMI
component locations) or nonlinear series loads are present in series,
this break-back of voltage represents a potentially serious, unplanned
energy dump.

In a series string, the break-back of one series element can set the
others off in a chain.

RL
 
N

Nicholas O. Lindan

Jan 1, 1970
0
legg said:
In a series string, the break-back of one series [diode] can set the
others off in a chain.

Not only can, but does. Always. Found out the hard way ...

The reverse voltage across a string of diodes will not be equally divided.
One diode will accumulate the majority of the drop, go into avalanche -
creating a short and increasing the voltage across the remaining diodes,
then the next diode will do the same, and the next ... till you have a dead
short across the string and something goes poof.

High value resistors, so that Ires >> Ileakage, in series with each
diode will equalize the voltage along the diode string. Be sure
not to exceed the voltage rating on the resistors; resistors can
be safely strung for high voltage.
 
R

Rich Grise

Jan 1, 1970
0
legg said:
In a series string, the break-back of one series [diode] can set the
others off in a chain.

Not only can, but does. Always. Found out the hard way ...

The reverse voltage across a string of diodes will not be equally divided.
One diode will accumulate the majority of the drop, go into avalanche -
creating a short and increasing the voltage across the remaining diodes,
then the next diode will do the same, and the next ... till you have a dead
short across the string and something goes poof.

High value resistors, so that Ires >> Ileakage, in series with each
diode will equalize the voltage along the diode string. Be sure
not to exceed the voltage rating on the resistors; resistors can
be safely strung for high voltage.

I've even seen it done with capacitors. This can be useful for a
HV, low-current supply, like a focus electrode.

Cheers!
Rich
 
L

legg

Jan 1, 1970
0
legg said:
In a series string, the break-back of one series [diode] can set the
others off in a chain.

Not only can, but does. Always. Found out the hard way ...

The reverse voltage across a string of diodes will not be equally divided.
One diode will accumulate the majority of the drop, go into avalanche -
creating a short and increasing the voltage across the remaining diodes,
then the next diode will do the same, and the next ... till you have a dead
short across the string and something goes poof.

High value resistors, so that Ires >> Ileakage, in series with each
diode will equalize the voltage along the diode string. Be sure
not to exceed the voltage rating on the resistors; resistors can
be safely strung for high voltage.

I've even seen it done with capacitors. This can be useful for a
HV, low-current supply, like a focus electrode.
This is also another inadvertent source of capacitive energy for parts
with unstable breakdown characteristics.

It appears that the correct term is 'controlled avalanche', not soft
avalanche, as originally indicated. This is searchable.

RL
 
W

Winfield Hill

Jan 1, 1970
0
Joel Kolstad wrote...
Winfield Hill wrote ...

...due to capacitive coupling 'around' the resistor? Or is there some
other mechanism at work?

Capacitive coupling to the 1/1000 side from all over. The portion from
the resistor's body could be compensated (to first order only) with
capacitance on the output side, but additional exposure from E fields
from wiring near the measured voltage create large indeterminate errors
that cannot be compensated. It's a mess and cannot work as delivered.
 
Top