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Driver for IGBT? (-ve off and +ve on from CMOS)

W

Winfield Hill

Jan 1, 1970
0
Ignoramus14838 wrote...
Terry, I am thinking of buying a Semikron 23.
http://www.semikron.com/internet/gecont/pdf/SKHI_23_12.pdf
What would you say about it? Seems like it can drive up to 200A IGBT.

A key capability in their module for protecting the IGBTs
can be seen in the "Vce monitoring" item, pages 3 and 7.
If you were to roll your own, you'd want to include such a
feature in your circuit. Their isolation circuitry is no
slouch either, see their 75kV/us input-output dv/dt spec.

Note the maximum frequency vs gate-charge plot, figure 15.
 
F

Fred Bloggs

Jan 1, 1970
0
Winfield said:
A key capability in their module for protecting the IGBTs
can be seen in the "Vce monitoring" item, pages 3 and 7.
If you were to roll your own, you'd want to include such a
feature in your circuit.

That's largely a sales gimmick, it only detects a dead short.
Their isolation circuitry is no
slouch either, see their 75kV/us input-output dv/dt spec.

Yeah- but look at that delay time.
Note the maximum frequency vs gate-charge plot, figure 15.

Oh who cares about that cryptic little thing...
 
S

Spehro Pefhany

Jan 1, 1970
0
That's largely a sales gimmick, it only detects a dead short.

Well, sure, but it protects an expensive drive from complete failure
when that short occurs. That's why the semiconductors are often SC
rated, because it's an important consideration for the customers.

Turn off conditions from a dead short are particularly demanding, and
part of the reason for the negative gate bias.



Best regards,
Spehro Pefhany
 
F

Fred Bloggs

Jan 1, 1970
0
Spehro said:
Well, sure, but it protects an expensive drive from complete failure
when that short occurs. That's why the semiconductors are often SC
rated, because it's an important consideration for the customers.

Turn off conditions from a dead short are particularly demanding, and
part of the reason for the negative gate bias.

How they keep that bipolar from going into secondary breakdown is the
question. There must be a diode from G-E to pull it out with the reverse
current. Now things are starting to fall in place- I thought that Miller
effect story was hokus-pokus.
 
F

Fred Bloggs

Jan 1, 1970
0
Fred said:
How they keep that bipolar from going into secondary breakdown is the
question. There must be a diode from G-E to pull it out with the reverse
current. Now things are starting to fall in place- I thought that Miller
effect story was hokus-pokus.

I mean a diode from G-B....
 
I

Ignoramus1797

Jan 1, 1970
0
Ignoramus14838 wrote...

A key capability in their module for protecting the IGBTs
can be seen in the "Vce monitoring" item, pages 3 and 7.
If you were to roll your own, you'd want to include such a
feature in your circuit. Their isolation circuitry is no
slouch either, see their 75kV/us input-output dv/dt spec.

Note the maximum frequency vs gate-charge plot, figure 15.

Yes, I really like their feedback features and safety features and
output power. My time is not free. If I could spend $50 instead of
spending a month endlessly ordering parts and such to add another
output stage to an IC, I would rather do that. Obviously, there is a
limit to how much I want to spend, as well.

i
 
W

Winfield Hill

Jan 1, 1970
0
Spehro Pefhany wrote...
Well, sure, but it protects an expensive drive from complete failure
when that short occurs. That's why the semiconductors are often SC
rated, because it's an important consideration for the customers.

Turn off conditions from a dead short are particularly demanding,
and part of the reason for the negative gate bias.

I believe the Vce-monitoring catches much more than dead shorts.
Any time the IGBT's Vce is significantly in excess of the normal
saturation voltage, for an adjustable time, the IGBT error shutoff
will trigger. This occurs for Vce > 5.2V for the SKHI 23/12, with
Rce = 18k. The time is also user-set, with the Rce and Cce values.
Semikron's point is that if the time is short enough, the full IGBT
self-current-limit can be sustained at short circuit without damage.
Any current that forces the IGBT out of saturation beyond 5.2V for
more than the set time will trigger a fault shutoff for that cycle.
 
I

Ignoramus1797

Jan 1, 1970
0
Spehro Pefhany wrote...

I believe the Vce-monitoring catches much more than dead shorts.
Any time the IGBT's Vce is significantly in excess of the normal
saturation voltage, for an adjustable time, the IGBT error shutoff
will trigger. This occurs for Vce > 5.2V for the SKHI 23/12, with

High Vce could be a result of low gate/emitter voltage. Many chips
have protection against that, but it is nice to see a gate driver that
protects against actual problem (high Vce) rather than one of its
causes, low Vge.

High Vce leaves a little bit of time to shut down the IGBT.

Rce = 18k. The time is also user-set, with the Rce and Cce values.
Semikron's point is that if the time is short enough, the full IGBT
self-current-limit can be sustained at short circuit without damage.
Any current that forces the IGBT out of saturation beyond 5.2V for
more than the set time will trigger a fault shutoff for that cycle.

Yep... that's very nice in fact...

It is, though, a half bridge driver, I think that I need two to drive
a full bridge.

i
 
W

Winfield Hill

Jan 1, 1970
0
Winfield Hill wrote...
Spehro Pefhany wrote...

I believe the Vce-monitoring catches much more than dead shorts.
Any time the IGBT's Vce is significantly in excess of the normal
saturation voltage, for an adjustable time, the IGBT error shutoff
will trigger. This occurs for Vce > 5.2V for the SKHI 23/12, with
Rce = 18k. The time is also user-set, with the Rce and Cce values.
Semikron's point is that if the time is short enough, the full IGBT
self-current-limit can be sustained at short circuit without damage.
Any current that forces the IGBT out of saturation beyond 5.2V for
more than the set time will trigger a fault shutoff for that cycle.

BTW, Semikron has a nice 269-page power IGBT manual on their website
(although it's broken into a million pieces). They discuss the Vce-
monitoring scene on page 210, and rules for protecting IGBTs against
shorts on page 207. "The following important boundary conditions
have to be fulfilled to guarantee safe operation:
- the short circuit has to be detected and turned off within 10 µs,
- the time between two short circuits has to be at least 1 second,
- the IGBT must not be subjected to more than 1000 short circuits
during its total operation time."

The technique is not able to function as a universal power-limit
detector. E.g., if the IGBT is desaturated, but under 5.2V, it can
still quickly exceed its thermal power limit. For example, consider
Ignoramus14838's Toshiba MG200Q2YS40 IGBT module, datasheet page 3.
With 15V gate drive, the collector current goes offscale at 400A
for Vce above 4 volts. The current at Vce = 5V (just below the 5.2V
trigger) could be as high as 550A, which would be a power dissipation
level of 2750 watts. The transient thermal resistance curves show
this power level can be sustained for only about 15-milli-seconds,
with a 100C junction-temperature rise. A cautious user could reduce
the Vce-fault value, e.g. 4V (400A, 1600W, 0.1s) with Rce = 14k, etc.,
but the IGBT could still be damaged with exactly the right conditions.
But this exact condition (a sustained current of more than say 250A,
but less than 400 to 500A) may be rather unusual and hard to create.
 
I

Ignoramus1797

Jan 1, 1970
0
Winfield Hill wrote...

BTW, Semikron has a nice 269-page power IGBT manual on their website
(although it's broken into a million pieces). They discuss the Vce-
monitoring scene on page 210, and rules for protecting IGBTs against
shorts on page 207. "The following important boundary conditions
have to be fulfilled to guarantee safe operation:
- the short circuit has to be detected and turned off within 10 µs,
- the time between two short circuits has to be at least 1 second,
- the IGBT must not be subjected to more than 1000 short circuits
during its total operation time."

The technique is not able to function as a universal power-limit
detector. E.g., if the IGBT is desaturated, but under 5.2V, it can
still quickly exceed its thermal power limit. For example, consider
Ignoramus14838's Toshiba MG200Q2YS40 IGBT module, datasheet page 3.
With 15V gate drive, the collector current goes offscale at 400A
for Vce above 4 volts. The current at Vce = 5V (just below the 5.2V
trigger) could be as high as 550A, which would be a power dissipation
level of 2750 watts. The transient thermal resistance curves show
this power level can be sustained for only about 15-milli-seconds,
with a 100C junction-temperature rise. A cautious user could reduce
the Vce-fault value, e.g. 4V (400A, 1600W, 0.1s) with Rce = 14k, etc.,
but the IGBT could still be damaged with exactly the right conditions.
But this exact condition (a sustained current of more than say 250A,
but less than 400 to 500A) may be rather unusual and hard to create.

It is impossible to create in my welder, which is limited to 200A. A
welding machine shorts all the time (like when an electrode sticks to
the workpiece), and my machine handles it just fine.

My situation is 100% different from, say, a power supply using utility
power or a big battery. My situation is a "constant current" power
supply.

So, while I follow the shorting discussion with interest, it does not
apply to my situation. If I cannot find these semikrons at a good
price, perhaps I would need to make an output stage for a small chip,
but this would not be my preference.

I am actually negotiating with someone regarding buying them. He wants
too much money.

I want to clarify something. To drive a full bridge, it is my
understanding that I need TWO of these semikron 23 boards.

Is that correct?

i
 
R

Rich Grise

Jan 1, 1970
0
Of course. Chips always fail where they do the most damage. Some kind of
law of nature.

"Law of Murphy." ;-)

Cheers!
Rich
 
R

Rich Grise

Jan 1, 1970
0
Terry, I will answer the rest of your post separately, but I would like to
know which semikron gate driver do you recommend. I am fully in agreement
with the idea of buying more expensive, but perfectly well designed gate
drivers. I mean, $3 or $15, who cares. I paid $9.99 for my TIG welder and
$60 for four IGBTs and can splurge a bit on a gate driver.

I've been following this thread, and I'm getting a bad feeling. Unless
you really dramatically hack the welder, no matter how fast you chop
the current, it isn't going to switch from DCEN to DCEP - it's just
going to be chopped. To go from DCEN to DCEP, you have to reverse the
whole 200 (or whatever) amps. And you have to do this FAST. I think
you might be barking up a tree, so to speak, but hey, it's your
welder, and your IGBTs, so you can do whatever you want, and you can't
help but learn a lot one way or the other. :)

Good Luck!
Rich
 
I

Ignoramus1797

Jan 1, 1970
0
I've been following this thread, and I'm getting a bad feeling. Unless
you really dramatically hack the welder, no matter how fast you chop
the current, it isn't going to switch from DCEN to DCEP - it's just
going to be chopped. To go from DCEN to DCEP, you have to reverse the
whole 200 (or whatever) amps. And you have to do this FAST.

Yes, that's what I am planning, to reverse the whole 200 amps, using a
full bridge made up of two separate IGBT bricks (each is a half
bridge).

Ultimately, this inverter will become part of the welder and will be
mounted inside.
I think you might be barking up a tree, so to speak, but hey, it's
your welder, and your IGBTs, so you can do whatever you want, and
you can't help but learn a lot one way or the other. :)

I agree, worst case, I will "learn a lot" and lose some money. That
welder cost me $9.99 (no kidding). That said, that's why I am looking
at gate drivers, to invert the current the right way, fast etc.

i
 
W

Winfield Hill

Jan 1, 1970
0
Ignoramus1797 wrote...
It is impossible to create in my welder, which is limited to 200A.
A welding machine shorts all the time (like when an electrode sticks
to the workpiece), and my machine handles it just fine.

My situation is 100% different from, say, a power supply using utility
power or a big battery. My situation is a "constant current" power
supply.

So, while I follow the shorting discussion with interest, it does not
apply to my situation. If I cannot find these semikrons at a good
price, perhaps I would need to make an output stage for a small chip,
but this would not be my preference.

What kind of output stage does it have? How fast does its
voltage change to accommodate an instantaneous change in load?
Generally there're capacitive and inductive aspects to any
large power source's output. A capacitive aspect means the
current would be much higher than the "current limit" for a
short time after an open circuit (during which the output
voltage was at the maximum). If this happens, and lasts too
long, an IGBT operating into a short without fault shutoff
could be damaged.

An inductive aspect means the voltage will rapidly soar after
the load is interrupted, as the inductance tries to enforce
no change in the current.
I am actually negotiating with someone regarding buying them.
He wants too much money.

$50 was too much?
I want to clarify something. To drive a full bridge, it is
my understanding that I need TWO of these semikron 23 boards.
Is that correct?

Yes, although the switching delay times could mean that the
output will be disconnected for a time on each cycle change.
This is more tightly controlled in full H-bridge drivers.
 
G

Glen Walpert

Jan 1, 1970
0
Ignoramus1797 wrote...

What kind of output stage does it have? How fast does its
voltage change to accommodate an instantaneous change in load?
Generally there're capacitive and inductive aspects to any
large power source's output. A capacitive aspect means the
current would be much higher than the "current limit" for a
short time after an open circuit (during which the output
voltage was at the maximum). If this happens, and lasts too
long, an IGBT operating into a short without fault shutoff
could be damaged.

An inductive aspect means the voltage will rapidly soar after
the load is interrupted, as the inductance tries to enforce
no change in the current.

I haven't seen the schematic for the welder in question, but Miller DC
TIG welders in the pre-inverter era typically used a magnetic
amplifier for AC current regulation prior to rectification, with a
smoothing inductor following rectification. (The mag amp
"transformer" accounts for about 300 lbs of the 400 lb welder weight
in these now antiquated workhorses, which will probably still be
working 50 years from now when someone finally carts it off to the
scrapyard.) In the Millers I have worked on there is no output
current feedback to the mag amp DC control winding, (which is varied
to set operating current); with a current setting of 200 A the machine
might put out about 200 A with an 18 volt arc, 250 A into a dead short
and 60 V on open circuit (after the inductor current dies). Might be
a set of I/V curves in the welder manual but probably nothing on
response times.
 
I

Ignoramus1797

Jan 1, 1970
0
I haven't seen the schematic for the welder in question, but Miller DC
TIG welders in the pre-inverter era typically used a magnetic
amplifier for AC current regulation prior to rectification, with a
smoothing inductor following rectification. (The mag amp
"transformer" accounts for about 300 lbs of the 400 lb welder weight
in these now antiquated workhorses, which will probably still be
working 50 years from now when someone finally carts it off to the
scrapyard.) In the Millers I have worked on there is no output
current feedback to the mag amp DC control winding, (which is varied
to set operating current); with a current setting of 200 A the machine
might put out about 200 A with an 18 volt arc, 250 A into a dead short
and 60 V on open circuit (after the inductor current dies). Might be
a set of I/V curves in the welder manual but probably nothing on
response times.

my welder puts out exactly what is specified, regardless of arc
resistance (if the arc does not extinguish, that is). It is a hobart
cybertig.

i
--
 
W

Winfield Hill

Jan 1, 1970
0
Ignoramus1797 wrote...
my welder puts out exactly what is specified, regardless
of arc resistance (if the arc does not extinguish, that is).
It is a hobart cybertig.

No inductive flyback? I fear that statement, my dear
Ignoramus1797, in the context of our discussion, is the
embodiment of ignorance.

<Sorry, I couldn't resist, pardon the pun.>
 
W

Winfield Hill

Jan 1, 1970
0
Ignoramus1797 wrote...
Yes, I really like their feedback features and safety features and
output power. My time is not free. If I could spend $50 instead of
spending a month endlessly ordering parts and such to add another
output stage to an IC, I would rather do that. Obviously, there is
a limit to how much I want to spend, as well.

$50 for a Semikron 23, Bwaahahaha! Oh, wait, I got one for $39
on eBay. Nice.
 
I

Ignoramus1797

Jan 1, 1970
0
Ignoramus1797 wrote...

$50 for a Semikron 23, Bwaahahaha! Oh, wait, I got one for $39
on eBay. Nice.

So, was that you who won the last auction?

In any case... I will soon receive some very nice IR2114SS chips, that
do, basically, something very similar to Semikrons. They are designed
for driving large IGBTs. And the nice part is, I will get them for
free as samples. If all works out well, that is. Check out their
datasheet, maybe you will like what it says.

i
 
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