Reply by Lasse Langwadt Christensen●November 25, 20152015-11-25
Den onsdag den 25. november 2015 kl. 21.09.06 UTC+1 skrev John Larkin:
> On Wed, 25 Nov 2015 00:57:46 -0800 (PST), habib.bouaziz@gmail.com
> wrote:
>
> >Le mardi 24 novembre 2015 18:00:43 UTC+1, John Larkin a �crit�:
> >> On Tue, 24 Nov 2015 00:54:38 -0800 (PST), habib.bouaziz@gmail.com
> >> wrote:
> >>
> >> >Le lundi 23 novembre 2015 21:09:11 UTC+1, John Larkin a �crit�:
> >> >> On Mon, 23 Nov 2015 20:15:49 +0100, Habib Bouaziz-Viallet
> >> >> <habib@nowhere.com> wrote:
> >> >>
> >> >> >On 23/11/2015 18:49, John Larkin wrote:
> >> >> >> On Mon, 23 Nov 2015 09:19:02 -0800 (PST), habib.bouaziz@gmail.com
> >> >> >> wrote:
> >> >> >>
> >> >> >>> Le lundi 23 novembre 2015 17:57:39 UTC+1, Winfield Hill a �crit :
> >> >> >>>> habib.bouaziz@gmail.com wrote...
> >> >> >>>>>
> >> >> >>>>> --> Vgs becomes smaller when current is increasing through
> >> >> >>>>> 100 Ohm resistor
> >> >> >>>>> --> then vgs becomes smaller ... There it may have needs
> >> >> >>>>> some math or Spice simulation isn't it ?
> >> >> >>>>
> >> >> >>>> This is likely intentional, because it effectively creates
> >> >> >>>> a current limit, which is desirable. Of course the FETs
> >> >> >>>> have to be able to handle this. Ultimately the 1.5kV
> >> >> >>>> supply will sag and shut down. The current limit would
> >> >> >>>> act until the HV byass capacitors are drained.
> >> >> >>>>
> >> >> >>>>
> >> >> >>>> --
> >> >> >>>> Thanks,
> >> >> >>>> - Win
> >> >> >>>
> >> >> >>> Of course as you say "1.5kV supply will sag and shut down".
> >> >> >>> E(capacitors) = 1/2 * C * V *V = 0.5J
> >> >> >>> P max admissible for th MOSFET = 80W
> >> >> >>>
> >> >> >>> Let's say the discharge time is about Td = 10ms
> >> >> >>>
> >> >> >>> E = P.t then P = 50W --> Ok for the MOSFET
> >> >> >>>
> >> >> >>> But if Td = 1ms the P = 500W --> The MOSFET should blow
> >> >> >>>
> >> >> >>> May be i missed something ...
> >> >> >>>
> >> >> >>> Habib.
> >> >> >>
> >> >> >> The 100 ohm resistors and the opto CTR limit the mosfet currents.
> >> >> >> That's not so much to protect the fets as to limit power supply dip.
> >> >> >>
> >> >> >> One 2SK4177 can dissipate about a kilowatt for one millisecond, and
> >> >> >> around 3KW for 100 us. It could easily discharge all the energy that I
> >> >> >> have available.
> >> >> >Ok i missed the fig. ASO on page p.4/7
> >> >> >http://www.onsemi.com/pub_link/Collateral/ENA0869-D.PDF
> >> >> >
> >> >> >Since it is a one shot overload, it's ok. Apologies.
> >> >> >
> >> >> >Habib
> >> >> >>
> >> >> >>
> >> >>
> >> >> Oh, don't apologize, we all keep learning stuff.
> >> >>
> >> >> I just did this, to better understand the mosfet.
> >> >>
> >> >> https://dl.dropboxusercontent.com/u/53724080/Parts/Fets/2SK4177_Measurements.JPG
> >> >>
> >> >> They don't guardband that 1500 volt rating much!
> >> >>
> >> >>
> >> >>
> >> >>
> >> >>
> >> >>
> >> >> --
> >> >>
> >> >> John Larkin Highland Technology, Inc
> >> >> picosecond timing precision measurement
> >> >>
> >> >> jlarkin att highlandtechnology dott com
> >> >> http://www.highlandtechnology.com
> >> >
> >> >John,
> >> >
> >> >You measure VGss = +-42V ! OnSemi gives absolute VGss = +-20V
> >> >
> >> >Strange ! Habib.
> >>
> >> They built a gate protection zener into the chip, and it zeners at 42
> >> volts. They don't want you to apply a low-impedance voltage source to
> >> that zener, or you'll fry it. The 20 volt limit is very conservative.
> >>
> >> Most mosfets have 20 volt abs max gate voltages, even unprotected
> >> ones.
> >>
> >> I've tested a few non-protected mosfets, and the gates typically blow
> >> out around 70 volts. The test destroys the fets.
> >
> >As you say we all learn something new every day ...
> >
> >Best regards, Habib.
>
> It's fun to blow up parts and measure things.
>
On Wed, 25 Nov 2015 00:57:46 -0800 (PST), habib.bouaziz@gmail.com
wrote:
>Le mardi 24 novembre 2015 18:00:43 UTC+1, John Larkin a �crit�:
>> On Tue, 24 Nov 2015 00:54:38 -0800 (PST), habib.bouaziz@gmail.com
>> wrote:
>>
>> >Le lundi 23 novembre 2015 21:09:11 UTC+1, John Larkin a �crit�:
>> >> On Mon, 23 Nov 2015 20:15:49 +0100, Habib Bouaziz-Viallet
>> >> <habib@nowhere.com> wrote:
>> >>
>> >> >On 23/11/2015 18:49, John Larkin wrote:
>> >> >> On Mon, 23 Nov 2015 09:19:02 -0800 (PST), habib.bouaziz@gmail.com
>> >> >> wrote:
>> >> >>
>> >> >>> Le lundi 23 novembre 2015 17:57:39 UTC+1, Winfield Hill a �crit :
>> >> >>>> habib.bouaziz@gmail.com wrote...
>> >> >>>>>
>> >> >>>>> --> Vgs becomes smaller when current is increasing through
>> >> >>>>> 100 Ohm resistor
>> >> >>>>> --> then vgs becomes smaller ... There it may have needs
>> >> >>>>> some math or Spice simulation isn't it ?
>> >> >>>>
>> >> >>>> This is likely intentional, because it effectively creates
>> >> >>>> a current limit, which is desirable. Of course the FETs
>> >> >>>> have to be able to handle this. Ultimately the 1.5kV
>> >> >>>> supply will sag and shut down. The current limit would
>> >> >>>> act until the HV byass capacitors are drained.
>> >> >>>>
>> >> >>>>
>> >> >>>> --
>> >> >>>> Thanks,
>> >> >>>> - Win
>> >> >>>
>> >> >>> Of course as you say "1.5kV supply will sag and shut down".
>> >> >>> E(capacitors) = 1/2 * C * V *V = 0.5J
>> >> >>> P max admissible for th MOSFET = 80W
>> >> >>>
>> >> >>> Let's say the discharge time is about Td = 10ms
>> >> >>>
>> >> >>> E = P.t then P = 50W --> Ok for the MOSFET
>> >> >>>
>> >> >>> But if Td = 1ms the P = 500W --> The MOSFET should blow
>> >> >>>
>> >> >>> May be i missed something ...
>> >> >>>
>> >> >>> Habib.
>> >> >>
>> >> >> The 100 ohm resistors and the opto CTR limit the mosfet currents.
>> >> >> That's not so much to protect the fets as to limit power supply dip.
>> >> >>
>> >> >> One 2SK4177 can dissipate about a kilowatt for one millisecond, and
>> >> >> around 3KW for 100 us. It could easily discharge all the energy that I
>> >> >> have available.
>> >> >Ok i missed the fig. ASO on page p.4/7
>> >> >http://www.onsemi.com/pub_link/Collateral/ENA0869-D.PDF
>> >> >
>> >> >Since it is a one shot overload, it's ok. Apologies.
>> >> >
>> >> >Habib
>> >> >>
>> >> >>
>> >>
>> >> Oh, don't apologize, we all keep learning stuff.
>> >>
>> >> I just did this, to better understand the mosfet.
>> >>
>> >> https://dl.dropboxusercontent.com/u/53724080/Parts/Fets/2SK4177_Measurements.JPG
>> >>
>> >> They don't guardband that 1500 volt rating much!
>> >>
>> >>
>> >>
>> >>
>> >>
>> >>
>> >> --
>> >>
>> >> John Larkin Highland Technology, Inc
>> >> picosecond timing precision measurement
>> >>
>> >> jlarkin att highlandtechnology dott com
>> >> http://www.highlandtechnology.com
>> >
>> >John,
>> >
>> >You measure VGss = +-42V ! OnSemi gives absolute VGss = +-20V
>> >
>> >Strange ! Habib.
>>
>> They built a gate protection zener into the chip, and it zeners at 42
>> volts. They don't want you to apply a low-impedance voltage source to
>> that zener, or you'll fry it. The 20 volt limit is very conservative.
>>
>> Most mosfets have 20 volt abs max gate voltages, even unprotected
>> ones.
>>
>> I've tested a few non-protected mosfets, and the gates typically blow
>> out around 70 volts. The test destroys the fets.
>
>As you say we all learn something new every day ...
>
>Best regards, Habib.
It's fun to blow up parts and measure things.
--
John Larkin Highland Technology, Inc
picosecond timing precision measurement
jlarkin att highlandtechnology dott com
http://www.highlandtechnology.com
Reply by John Larkin●November 25, 20152015-11-25
On 25 Nov 2015 03:32:13 -0800, Winfield Hill
<hill@rowland.harvard.edu> wrote:
I haven't measured that much, but we did recently blow up some 100
volt fets with inductive kicks of 110 or so. These parts have no
avalanche energy specs. We were making 100 volt pulses, through a
transmission-line transformer, into 50 ohms in under 1 ns.
On the other hand, mesfets and phemts seem to be OK at over-voltages
like 2x to 3x specified max. I think the RF boys automatically cut the
real abs max in half, assuming that X volts of power supply makes 2x
volts of swing into a resonant tank. Or something. The specs on RF
part data sheets are usually miserable.
Reply by Winfield Hill●November 25, 20152015-11-25
That's not so bad, 5% is what we usually get for MOSFETs.
It's more than we get for electrolytic capacitors. With
a MOSFET we're not in too dangerous territory using it all
the way up to its rating, but with electrolytic caps we'd
better keep a 10% safety margin below the official rating.
BTW, in any design where the MOSFET's junction temp rises,
we get as much as an extra 10% in the avalanche voltage.
--
Thanks,
- Win
Reply by ●November 25, 20152015-11-25
Le mardi 24 novembre 2015 18:00:43 UTC+1, John Larkin a �crit�:
> On Tue, 24 Nov 2015 00:54:38 -0800 (PST), habib.bouaziz@gmail.com
> wrote:
>
> >Le lundi 23 novembre 2015 21:09:11 UTC+1, John Larkin a �crit�:
> >> On Mon, 23 Nov 2015 20:15:49 +0100, Habib Bouaziz-Viallet
> >> <habib@nowhere.com> wrote:
> >>
> >> >On 23/11/2015 18:49, John Larkin wrote:
> >> >> On Mon, 23 Nov 2015 09:19:02 -0800 (PST), habib.bouaziz@gmail.com
> >> >> wrote:
> >> >>
> >> >>> Le lundi 23 novembre 2015 17:57:39 UTC+1, Winfield Hill a �crit :
> >> >>>> habib.bouaziz@gmail.com wrote...
> >> >>>>>
> >> >>>>> --> Vgs becomes smaller when current is increasing through
> >> >>>>> 100 Ohm resistor
> >> >>>>> --> then vgs becomes smaller ... There it may have needs
> >> >>>>> some math or Spice simulation isn't it ?
> >> >>>>
> >> >>>> This is likely intentional, because it effectively creates
> >> >>>> a current limit, which is desirable. Of course the FETs
> >> >>>> have to be able to handle this. Ultimately the 1.5kV
> >> >>>> supply will sag and shut down. The current limit would
> >> >>>> act until the HV byass capacitors are drained.
> >> >>>>
> >> >>>>
> >> >>>> --
> >> >>>> Thanks,
> >> >>>> - Win
> >> >>>
> >> >>> Of course as you say "1.5kV supply will sag and shut down".
> >> >>> E(capacitors) = 1/2 * C * V *V = 0.5J
> >> >>> P max admissible for th MOSFET = 80W
> >> >>>
> >> >>> Let's say the discharge time is about Td = 10ms
> >> >>>
> >> >>> E = P.t then P = 50W --> Ok for the MOSFET
> >> >>>
> >> >>> But if Td = 1ms the P = 500W --> The MOSFET should blow
> >> >>>
> >> >>> May be i missed something ...
> >> >>>
> >> >>> Habib.
> >> >>
> >> >> The 100 ohm resistors and the opto CTR limit the mosfet currents.
> >> >> That's not so much to protect the fets as to limit power supply dip.
> >> >>
> >> >> One 2SK4177 can dissipate about a kilowatt for one millisecond, and
> >> >> around 3KW for 100 us. It could easily discharge all the energy that I
> >> >> have available.
> >> >Ok i missed the fig. ASO on page p.4/7
> >> >http://www.onsemi.com/pub_link/Collateral/ENA0869-D.PDF
> >> >
> >> >Since it is a one shot overload, it's ok. Apologies.
> >> >
> >> >Habib
> >> >>
> >> >>
> >>
> >> Oh, don't apologize, we all keep learning stuff.
> >>
> >> I just did this, to better understand the mosfet.
> >>
> >> https://dl.dropboxusercontent.com/u/53724080/Parts/Fets/2SK4177_Measurements.JPG
> >>
> >> They don't guardband that 1500 volt rating much!
> >>
> >>
> >>
> >>
> >>
> >>
> >> --
> >>
> >> John Larkin Highland Technology, Inc
> >> picosecond timing precision measurement
> >>
> >> jlarkin att highlandtechnology dott com
> >> http://www.highlandtechnology.com
> >
> >John,
> >
> >You measure VGss = +-42V ! OnSemi gives absolute VGss = +-20V
> >
> >Strange ! Habib.
>
> They built a gate protection zener into the chip, and it zeners at 42
> volts. They don't want you to apply a low-impedance voltage source to
> that zener, or you'll fry it. The 20 volt limit is very conservative.
>
> Most mosfets have 20 volt abs max gate voltages, even unprotected
> ones.
>
> I've tested a few non-protected mosfets, and the gates typically blow
> out around 70 volts. The test destroys the fets.
As you say we all learn something new every day ...
Best regards, Habib.
Reply by Jim Thompson●November 24, 20152015-11-24
On 24 Nov 2015 17:17:52 -0800, Winfield Hill
<hill@rowland.harvard.edu> wrote:
[snip]
>
> This is a pulsed power-dissipation situation, and thermal
> mass is your friend.
[snip]
Yep. Solved a thermal dissipation problem on the TOW missile (~1972
when I ran the Dickson Electronics hybrid facility) by switching a
hybrid (three-legged regulator) from a Aluminum TO-3 package to a
steel TO-3... thermal mass keeps the temperature rise under control
until target is reached >:-}
...Jim Thompson
--
| James E.Thompson | mens |
| Analog Innovations | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| San Tan Valley, AZ 85142 Skype: skypeanalog | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |
I love to cook with wine. Sometimes I even put it in the food.
Reply by Winfield Hill●November 24, 20152015-11-24
Winfield Hill wrote...
>
> habib.bouaziz@gmail.com wrote...
>>
>> Le lundi 23 novembre 2015 17:57:39 UTC+1, Winfield Hill
>>> habib.bouaziz@gmail.com wrote...
>>>>
>>>>--> Vgs becomes smaller when current is increasing through
>>>> 100 Ohm resistor
>>>> --> then vgs becomes smaller ... There it may have needs
>>>> some math or Spice simulation isn't it ?
>>>
>>> This is likely intentional, because it effectively creates
>>> a current limit, which is desirable. Of course the FETs
>>> have to be able to handle this. Ultimately the 1.5kV
>>> supply will sag and shut down. The current limit would
>>> act until the HV byass capacitors are drained.
>>
>> Of course as you say "1.5kV supply will sag and shut down".
>> E(capacitors) = 1/2 * C * V *V =3D 0.5J
>> P max admissible for th MOSFET =3D 80W
>>
>> Let's say the discharge time is about Td =3D 10ms
>>
>> E = P.t then P = 50W --> Ok for the MOSFET
>> But if Td = 1ms the P = 500W --> The MOSFET should blow
>>
>> May be i missed something ...
>
> This is a pulsed power-dissipation situation, and thermal
> mass is your friend. You want to look at the Effective
> Transient Thermal Impedance plots, which show the extra
> power you can safely dissipate for a short time. But oops,
> Sanyo's datasheet doesn't have one for this part. :-(
>
> We do have an SOE (safe operating-area) plot, called ASO.
> Looking at the 1ms line, this says we can do 100V and 8A
> for 1ms. That's 800W, so the MOSFET shouldn't blow.
>
> Actually, the capacitor voltage is dropping as we go, so
> we'd be safe for 2x that rating, or 3x your 500W example.
Oops, I was looking at the 2sk4117LS datasheet. That's a
400V part, with a different plot, weird. It's likely the
same die, with the same transient thermal properties. The
2sk4117 ASO plot says 800V and 1A for 1ms, or also 800W.
--
Thanks,
- Win
Reply by Winfield Hill●November 24, 20152015-11-24
John Larkin wrote...
>
> I checked out AOE3 as you suggested, and then put the
> book down on my desk, opened to page 210. Now the book
> is making very strange crackling noises.
Avalanche, no doubt.
--
Thanks,
- Win
Reply by Winfield Hill●November 24, 20152015-11-24
habib.bouaziz@gmail.com wrote...
>
> Le lundi 23 novembre 2015 17:57:39 UTC+1, Winfield Hill
>> habib.bouaziz@gmail.com wrote...
>>>
>>>--> Vgs becomes smaller when current is increasing through
>>> 100 Ohm resistor
>>> --> then vgs becomes smaller ... There it may have needs
>>> some math or Spice simulation isn't it ?
>>
>> This is likely intentional, because it effectively creates
>> a current limit, which is desirable. Of course the FETs
>> have to be able to handle this. Ultimately the 1.5kV
>> supply will sag and shut down. The current limit would
>> act until the HV byass capacitors are drained.
>
> Of course as you say "1.5kV supply will sag and shut down".
> E(capacitors) = 1/2 * C * V *V =3D 0.5J
> P max admissible for th MOSFET =3D 80W
>
> Let's say the discharge time is about Td =3D 10ms
>
> E = P.t then P = 50W --> Ok for the MOSFET
> But if Td = 1ms the P = 500W --> The MOSFET should blow
>
> May be i missed something ...
This is a pulsed power-dissipation situation, and thermal
mass is your friend. You want to look at the Effective
Transient Thermal Impedance plots, which show the extra
power you can safely dissipate for a short time. But oops,
Sanyo's datasheet doesn't have one for this part. :-(
We do have an SOE (safe operating-area) plot, called ASO.
Looking at the 1ms line, this says we can do 100V and 8A
for 1ms. That's 800W, so the MOSFET shouldn't blow.
Actually, the capacitor voltage is dropping as we go, so
we'd be safe for 2x that rating, or 3x your 500W example.
--
Thanks,
- Win
Reply by Winfield Hill●November 24, 20152015-11-24
habib.bouaziz@gmail.com wrote...
>
> Le lundi 23 novembre 2015 17:57:39 UTC+1, Winfield Hill
>> habib.bouaziz@gmail.com wrote...
>>>
>>>--> Vgs becomes smaller when current is increasing through
>>> 100 Ohm resistor
>>> --> then vgs becomes smaller ... There it may have needs
>>> some math or Spice simulation isn't it ?
>>
>> This is likely intentional, because it effectively creates
>> a current limit, which is desirable. Of course the FETs
>> have to be able to handle this. Ultimately the 1.5kV
>> supply will sag and shut down. The current limit would
>> act until the HV byass capacitors are drained.
>
> Of course as you say "1.5kV supply will sag and shut down".
> E(capacitors) = 1/2 * C * V *V =3D 0.5J
> P max admissible for th MOSFET =3D 80W
>
> Let's say the discharge time is about Td =3D 10ms
>
> E = P.t then P = 50W --> Ok for the MOSFET
> But if Td = 1ms the P = 500W --> The MOSFET should blow
>
> May be i missed something ...