Maker Pro
Maker Pro

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

R

Richard

Jan 1, 1970
0
I would like to drive an IGBT from a microcontroller (CMOS).
Is there a driver circuit that can outputs around +15v (on) and a -5v (off)
for switching frequencies around 40Khz.
I think -ve voltage supply is avaliable
What is this type of circuit called?

Thanks
Richard
 
I

Ignoramus26315

Jan 1, 1970
0
I would like to drive an IGBT from a microcontroller (CMOS).
Is there a driver circuit that can outputs around +15v (on) and a -5v (off)
for switching frequencies around 40Khz.
I think -ve voltage supply is avaliable
What is this type of circuit called?

I am also interested for something that makes negative off
signal... If you find anything, let me know.

i
 
M

Mook Johnson

Jan 1, 1970
0
I'm not sure why you'd need to drive the gate negative. below the threshold
voltage should be fine to turn it off (just like a Mosfet). if you insist
on turning swingint it negative, you might look at a gate drive transformer
that naturally swings negative (ac coupling the gate signal)

A discrete design wouldn't be to difficult to come up with.
 
T

Terry Given

Jan 1, 1970
0
Mook said:
I'm not sure why you'd need to drive the gate negative. below the threshold
voltage should be fine to turn it off (just like a Mosfet). if you insist

because negative gate bias can prevent dV/dt thru Cmiller from turning
the darned thing back on again.

IR wrote a nice paper about 6-7 years ago, where they go through and
show at what point -ve bias becomes necessary. its mandatory for
man-sized IGBTs, but baby IGBTs can often do without.
 
M

Martin Riddle

Jan 1, 1970
0
I found this some where, I forget where, but its a novel approach.
A MAX232, +/-10 volt swing.

Cheers
 
R

Robert Baer

Jan 1, 1970
0
Martin said:
I found this some where, I forget where, but its a novel approach.
A MAX232, +/-10 volt swing.

Cheers
I would never even *think* of Maxim or Dallas VaporWare.
 
B

Bob Monsen

Jan 1, 1970
0
I would never even *think* of Maxim or Dallas VaporWare.

Those MAX232 chips have been around for a while, so they probably aren't
too dangerous to depend on (although I've had my run-ins with Maxim and
Dallas). However, getting them to drive an IGBT at 40kHz may not be
possible, due to the large input capacitance, which may require lots of
current to overcome quickly. They use a charge pump, which may not be able
to supply the needed current.

There are dedicated IGBT driver chips available. There are also integrated
H-Bridge chips that would work nicely. Using one of those is probably the
simplest and safest course.
 
F

Fred Bloggs

Jan 1, 1970
0
I would like to drive an IGBT from a microcontroller (CMOS).
Is there a driver circuit that can outputs around +15v (on) and a -5v (off)
for switching frequencies around 40Khz.
I think -ve voltage supply is avaliable
What is this type of circuit called?

Surprisingly it is called a gate driver circuit. IGBTs imply large power
switching so that even though the switching frequency is a modest 40KHz,
the driver is still required to switch the device through ON/OFF
transitions quickly to reduce losses and component heating. Micrel at
http://www.micrel.com manufactures a line of powerful gate driver
products capable of handling nearly any situation. The most basic part
in their line-up is the MIC4120, a non-inverting driver capable of
supporting 6A peak gate drive currents. A high speed interface to a
microcontroller might look like this:
View in a fixed-width font such as Courier.
 
P

Paul Burke

Jan 1, 1970
0
Bob said:
Those MAX232 chips have been around for a while, so they probably aren't
too dangerous to depend on (although I've had my run-ins with Maxim and
Dallas).

Remember the early ones, that typically lasted about 3 months? They
later changed the circuit, so that the high side output capacitor went
to 5V instead of ground. Before that, it would latch up unpredicatbly,
usually when you'd just delivered a lot to an important customer.

Paul Burke
 
I

Ignoramus14838

Jan 1, 1970
0
I would like to drive an IGBT from a microcontroller (CMOS).
Is there a driver circuit that can outputs around +15v (on) and a -5v (off)
for switching frequencies around 40Khz.
I think -ve voltage supply is avaliable
What is this type of circuit called?

Check out Fuji's gate driver:

http://www.fujielectric.co.jp/eng/fdt/scd/pdf/Manual/REH924.pdf

Apparently, it makes negative reverse bias of -5V and has a built in
opto coupler.

i
 
K

Ken Smith

Jan 1, 1970
0
Martin said:
I found this some where, I forget where, but its a novel approach.
A MAX232, +/-10 volt swing.
[...]
I would never even *think* of Maxim or Dallas VaporWare.

Use an LT1081. They are easy to get. I think they are about as slow and
weak as the MAX232 so it really isn't such a good idea.
 
S

Spehro Pefhany

Jan 1, 1970
0
On Mon, 10 Oct 2005 14:05:58 +0000 (UTC), the renowned
Martin said:
I found this some where, I forget where, but its a novel approach.
A MAX232, +/-10 volt swing.
[...]
I would never even *think* of Maxim or Dallas VaporWare.

Use an LT1081. They are easy to get. I think they are about as slow and
weak as the MAX232 so it really isn't such a good idea.

MAX232s are multiple-sourced these days, so you're not stuck with
Maxim, any more than you need fear the uA709A because you don't trust
Fairchild. ;-)


Best regards,
Spehro Pefhany
 
S

Spehro Pefhany

Jan 1, 1970
0
i really don't understand why you need that?, normally pulling the
gate low will do how ever if you insist.

For big IGBTs, some negative gate drive is an advantage.
simply provide a +15 volt source via a resistor sufficient enough
to drive the gate in the on mode, use something like a simply NPN
transistor to pull it low at the gate to shut the IGBT off, put the
emitter to a - supply.
since these Gates are H Z, the - source does not need to be much.
using either a high freq OSC from something simple like a 555 or
an output from the MC, you can generate - voltage via a Cap and diode et..
the base of the transistor will get switched from the MC signal line.

There's a lot of capacitance.


Best regards,
Spehro Pefhany
 
I

Ignoramus14838

Jan 1, 1970
0
i really don't understand why you need that?, normally pulling the
gate low will do how ever if you insist.

I am actually playing with IGBTs.

I have Toshiba-MG200Q2YS40 IGBT.

It does NOT turn off itself if I simply remove Gate/Emitter voltage.

I have to apply reverse voltage to turn it off. I actually tried it.

It would, possibly, switch itself off with a small resistor between
gate and emitter, but it is just a hypothesis of mine.


i
 
J

Jamie

Jan 1, 1970
0
Richard said:
I would like to drive an IGBT from a microcontroller (CMOS).
Is there a driver circuit that can outputs around +15v (on) and a -5v (off)
for switching frequencies around 40Khz.
I think -ve voltage supply is avaliable
What is this type of circuit called?

Thanks
Richard
i really don't understand why you need that?, normally pulling the
gate low will do how ever if you insist.
simply provide a +15 volt source via a resistor sufficient enough
to drive the gate in the on mode, use something like a simply NPN
transistor to pull it low at the gate to shut the IGBT off, put the
emitter to a - supply.
since these Gates are H Z, the - source does not need to be much.
using either a high freq OSC from something simple like a 555 or
an output from the MC, you can generate - voltage via a Cap and diode et..
the base of the transistor will get switched from the MC signal line.
 
F

Fred Bloggs

Jan 1, 1970
0
Spehro said:
There's a lot of capacitance.

Gate charge is what you really mean...capacitance can be misleading
because it ignores threshold voltage. After looking at that Fugi
Electric line, it seems they require hundreds of ohms of gate series
damping resistance- because of the bipolar current gain multiplying the
lead inductance reflected into the gate circuit most likely- and this
explains the negative bias requirement, as a lower impedance pulldown
would shunt the Miller transient- at turn-off.
 
T

Terry Given

Jan 1, 1970
0
Ignoramus14838 said:
I am actually playing with IGBTs.

I have Toshiba-MG200Q2YS40 IGBT.

It does NOT turn off itself if I simply remove Gate/Emitter voltage.

I have to apply reverse voltage to turn it off. I actually tried it.

a decent short from gate to emitter would do it.
It would, possibly, switch itself off with a small resistor between
gate and emitter, but it is just a hypothesis of mine.


i

what you are loking at is a combination of things - miller effect
mostly. There is a non-linear capacitance between gate and collector,
call it Ccg. Whenever the collector voltage changes value (say from
Vcesat to Vdc when turning off) the voltage slews at some rate, dV/dt.
This causes current to flow thru Ccg, Icg = Ccg*dVce/dt. This current
flows into the gate circuit, charging up Cge. If the gate driver is not
very low impedance, Icg will happily charge Cge up to Vth, and beyond.
when turning the IGBT on, Vce falls so Icg flows out of Cge, discharging
it - IOW trying to turn it back off.

This can greatly increase switching time (and hence losses) and, if the
gatedrive is bad enough, can make the device "latch" in the on-state,
with Vge sitting around Vth - but you need a seriously piss-weak gate
driver to do that. It can even over-voltage the gate, poke a hole thru
the thin gate oxide layer, and KABOOM, one dead IGBT.

a negative gate bias means the switch-off miller current has to supply
more charge to reach Vth - instead of supplying Qg = Vth*Cge, it must
supply (Vth + |Vnegative|)*Cge. This allows the use of higher gate
impedance circuitry. Its not just the resistance, its the inductance
too. bad gate drives have high resistance and high inductance; good gate
drives have low resistance and low inductance.

to give you an idea, I've worked on gatedrives for 0.5kW - 2.2kW drives
that have about -3V reverse-bias, and -15V for 400kW drives (6 300A fuji
IGBTs directly paralleled).

modern IGBT designs have much better packaging - far lower inductance,
both Lce and Lge. with some of these IGBTs it is theoretically possible
to keep the gate impedance sufficiently low that negative bias isnt
necessary. Like I said, IR wrote a pretty good paper on the topic - why
you dont, in theory, need -ve bias for little IGBTs.

In practice, if you screw up the gatedriver, your circuit is going to go
BANG. If you really know what you are doing, and are very good at taking
measurements, its fairly easy to dewsign gatedrivers and make them work.
OTOH if you are not too experienced, its very easy to destroy many, many
IGBTs.

Generally the gatedrive is a lot cheaper than the IGBTs, so its sensible
to throw -ve bias at it. For little IGBTs, it can be as simple as a
zener in series with the gate resistor, with a 100nF cap across it.

Cheers
Terry
 
I

Ignoramus14838

Jan 1, 1970
0
a decent short from gate to emitter would do it.


what you are loking at is a combination of things - miller effect
mostly. There is a non-linear capacitance between gate and collector,
call it Ccg. Whenever the collector voltage changes value (say from
Vcesat to Vdc when turning off) the voltage slews at some rate, dV/dt.
This causes current to flow thru Ccg, Icg = Ccg*dVce/dt. This current
flows into the gate circuit, charging up Cge. If the gate driver is not
very low impedance, Icg will happily charge Cge up to Vth, and beyond.
when turning the IGBT on, Vce falls so Icg flows out of Cge, discharging
it - IOW trying to turn it back off.

This can greatly increase switching time (and hence losses) and, if the
gatedrive is bad enough, can make the device "latch" in the on-state,
with Vge sitting around Vth - but you need a seriously piss-weak gate
driver to do that. It can even over-voltage the gate, poke a hole thru
the thin gate oxide layer, and KABOOM, one dead IGBT.

a negative gate bias means the switch-off miller current has to supply
more charge to reach Vth - instead of supplying Qg = Vth*Cge, it must
supply (Vth + |Vnegative|)*Cge. This allows the use of higher gate
impedance circuitry. Its not just the resistance, its the inductance
too. bad gate drives have high resistance and high inductance; good gate
drives have low resistance and low inductance.

to give you an idea, I've worked on gatedrives for 0.5kW - 2.2kW drives
that have about -3V reverse-bias, and -15V for 400kW drives (6 300A fuji
IGBTs directly paralleled).

modern IGBT designs have much better packaging - far lower inductance,
both Lce and Lge. with some of these IGBTs it is theoretically possible
to keep the gate impedance sufficiently low that negative bias isnt
necessary. Like I said, IR wrote a pretty good paper on the topic - why
you dont, in theory, need -ve bias for little IGBTs.

In practice, if you screw up the gatedriver, your circuit is going to go
BANG. If you really know what you are doing, and are very good at taking
measurements, its fairly easy to dewsign gatedrivers and make them work.
OTOH if you are not too experienced, its very easy to destroy many, many
IGBTs.

Generally the gatedrive is a lot cheaper than the IGBTs, so its sensible
to throw -ve bias at it. For little IGBTs, it can be as simple as a
zener in series with the gate resistor, with a 100nF cap across it.

Terry, it is great that you participate in this thread and actually
have experience with power switching. I am building a DC -> AC
inverter for a TIG welder and seems like I can learn a lot from you.

This is a 200A constant current, 28 V welding current, 80 v open
circuit voltage tig welder.I have aforementioned IGBTs.

Can you give a little schematic of how go give the gates negative bias
during turn off. I definitely want to turn IGBTs off in the best
possible way. I want the simplest solution.

Another question, would a resistor between gate and emitter solve ths
issue? Or is that not the case?

thanks a lot for your time!

i
 
T

Terry Given

Jan 1, 1970
0
Ignoramus14838 said:
Terry, it is great that you participate in this thread and actually
have experience with power switching. I am building a DC -> AC
inverter for a TIG welder and seems like I can learn a lot from you.

This is a 200A constant current, 28 V welding current, 80 v open
circuit voltage tig welder.I have aforementioned IGBTs.

Can you give a little schematic of how go give the gates negative bias
during turn off. I definitely want to turn IGBTs off in the best
possible way. I want the simplest solution.

here's one simple solution: use an optocoupler to drive a FET driver
chip, running from a +10V/-5V supply. connect the supply 0V to the IGBT
emitter, +10V to the driver chip Vcc and -5V to the driver chip gnd.
that way the driver chip "sees" a 15V supply, whereas the IGBT "sees"
+10V and -5V.

we used to use tens of thousands of UC3842 smps chips as gate drivers.
An opto drove the 3842, which ran from a 15V isolated supply. the 3842
Vref output was connected to the IGBT emitter, giving +10V/-5V gate
drive. We had hefty caps from +10V to 0V to -5V. for little drives. for
big drives, we used +/-15V supplies.

for really little drives (< 2kW) we used the series zener trick. get a
15V FET driver, with a 15V supply. bung a 5V zener in series with the
output, with a 100nF cap across it. When the gatedrive output is high,
5V is dropped across the zener, charging the cap - the cathode end of
the cap is 5V below the anode end, and the gate gets 10V.

when the gatedrive output switches to 0V, the anode end of the cap is at
0V. The cathode end of the cap is 5V more negative than the anode end,
so the gate sees -5V. The 100nF cap needs to be at least 10x the gate
capacitance. For larger IGBTs, 1uF would be better.

this is probably the easiest option for you, and works with *any* gate
driver. Make sure you use a stompy zener though - you dont want its
series impedance to be large compared to your gate resistor. Think BZT03
or suchlike.

gatedrives are tricky, and until they work you kill a lot of IGBTs. why
not just buy one from Semikron.
Another question, would a resistor between gate and emitter solve ths
issue? Or is that not the case?

yes but no. You would need a seriously low valued resistor (< 10R) to
even attempt to hold the IGBT off wrt Cmiller, which would quite
successfully bugger up the gatedrive. not to mention the 15V^2/10R = 22W
or more power dissipation when the gatedrive is ON.
thanks a lot for your time!

i

Cheers
Terry
 
I

Ignoramus14838

Jan 1, 1970
0
here's one simple solution: use an optocoupler to drive a FET driver
chip, running from a +10V/-5V supply. connect the supply 0V to the IGBT
emitter, +10V to the driver chip Vcc and -5V to the driver chip gnd.
that way the driver chip "sees" a 15V supply, whereas the IGBT "sees"
+10V and -5V.

That's good to know. I am looking at using an IR21094 chip. Is it true
that it supports that configuration (it seems to be the case to me,
but I want to confirm).
we used to use tens of thousands of UC3842 smps chips as gate drivers.
An opto drove the 3842, which ran from a 15V isolated supply. the 3842
Vref output was connected to the IGBT emitter, giving +10V/-5V gate
drive. We had hefty caps from +10V to 0V to -5V. for little drives. for
big drives, we used +/-15V supplies.

I will double check, but I think that I have a 12V with +-5V power
supply lying around. If not, I can make one from something, even a
dual power supply for starters.

Now, I want to make sure that I understand what you mean.

Question: So, the above should work for a "big drive" (which, I
assume, applies to my 200 A system). What you write below is simply
another option that applies to "smaller drives".

Is that correct?
for really little drives (< 2kW) we used the series zener trick. get a
15V FET driver, with a 15V supply. bung a 5V zener in series with the
output, with a 100nF cap across it. When the gatedrive output is high,
5V is dropped across the zener, charging the cap - the cathode end of
the cap is 5V below the anode end, and the gate gets 10V.
OK.

when the gatedrive output switches to 0V, the anode end of the cap is at
0V. The cathode end of the cap is 5V more negative than the anode end,
so the gate sees -5V. The 100nF cap needs to be at least 10x the gate
capacitance. For larger IGBTs, 1uF would be better.

I think that I understand.
this is probably the easiest option for you, and works with *any* gate
driver. Make sure you use a stompy zener though - you dont want its
series impedance to be large compared to your gate resistor. Think BZT03
or suchlike.

gatedrives are tricky, and until they work you kill a lot of IGBTs. why
not just buy one from Semikron.

Which one would you suggest?
yes but no. You would need a seriously low valued resistor (< 10R)
to even attempt to hold the IGBT off wrt Cmiller, which would quite
successfully bugger up the gatedrive. not to mention the 15V^2/10R =
22W or more power dissipation when the gatedrive is ON.

Not very feasible... I was hoping to be able to get away with a much
smaller resistor...

Thanks a lot...

i
Cheers
Terry


--
 
Top