Application note, MOSFET Power Losses
- Thread starter Lagoethe
- Start date
Lagoethe
10 W
- Joined
- Nov 10, 2010
- Messages
- 89
From the previous Application note, I made an exemple.
But I'll need your help to fill (and fix) things I don't know.
The I'll be able to make an nice excel wich will save time for controller designs.
And for easier reading, you'll have to read the application note.
And this isn't a funny thread but it'll help to make good and precise work (as long as there are not mistakes). So I hope you'll help me.
CONDUCTION LOSSES
Assume that we use
a 6 Mosfet IRFB4468 design: http://www.irf.com/product-info/datasheets/data/irfp4468pbf.pdf
a 48V 20A Battery.
UDD = 48V.
Is it right to take 48V ? no more, no less?
We also use a Brushless e-bike Motor. Here I fail because I have absolutely no Idea of electric values for a motor.
Let say it's a 1500 W motor. How do we find (or choose) Uo and Iorms?
Does any motor dealer give the following informations: displacement factor, equivalent inductance ?
So missing information are:
Uo= ? (linked to Uan1)
Iorms= ?
cosφ1= ?
L = ?
What is ma ?
Output Ripple Current = (Udd - sqrt(2)xUo )xUo / (2 x L x Udd x fsw)
Io= sqrt(2) x Iorms
Mosfet conduction losses:
Pcm = RDSon x IDrms² = RDSon x Io² x (1/8 + ma x Cos φ1/ (3 x pi))
And RDSon depends on Temperature.
RDSon calculation:
RDSon (Tj) = RDson (25°C) x (1+ alpha/100)^(Tj-25)
From another source (I can't remember which one), RDSon 175° was computed this way:
RDSon(175) = (175-25)/ (Rth(JC) x ID²)
Although I didn't find it anywhere else, this formula works very well
RDSon 4468 (175) = 6,4
Then
alpha = 100 x ( (RDSon(175)/RDSon(25))^(1/175-25) -1) x 100
Diode conduction losses
Pcd= Ud0 x I fav + Rd x I²frms = Ud0 x Io x (1/(2Pi) - ma x cos φ1/8) + Rd x Io² x (1/8 - ma x Cos φ1/ (3 x pi))
From the Fig 7 of the IRFP4468 we read
Ud0 depends on temperature. Worst case is 25°
Ud0= 0.5V
Rd can be estimated as Vsd/Isd = 4.5mOhms
THEN
Pcd= ....................
SWITCHING LOSSES
Does that mean that switching losses will have to be doubled (1/2 fo)?
Mosfet Switching Losses
PswM=(EonM+EoffM) x fsw
EonM = UDD x Idon x (tri+tfu)/2 + Qrr x UDD = Udd x 1/Pi x Io x (tri+tfu)/2 + Qrr x Udd
EoffM = UDD x Idon x (tru+tfi)/2 = UDD x 1/Pi x Io x (tru+tfi)/2
Assume we drive the mosfet at 12V and Rdrive = 10ohms
Udr = 12V
Rdrive = 10ohms
From the IRFP4468 we read Qrr at 152°C worst case
Qrr =420nc
tfu =Rdrivetot x Qgd /(Udr - Uplateau)
tru = Rdrivetot x Qgd / Uplateau
Rdrivetot = Rgate + Rdrive
I'm not 100% sure of those formulas so i'd be happy if they were confirmed.
From The IRF4468:
Qgd = 89nC
Rgate = 0.8Ohms
Uplateau = Vgs = 2 to 4V
Then
tfu=
tru=
tfu= tfi?
tru= tri?
then
EonM=
EoffM=
And
PswM=
Diode Switching Losses
EonD= 1/4 Qrr x Udrr
worst case:
EonD = 1/4 Qrr x UDD
EoffD= UDD x 1/Pi x Io x (tru+tfi)/2
Then
EonD =
EoffD=
then
PswD=
then
Psw =
TOTAL LOSSES
P=
But I'll need your help to fill (and fix) things I don't know.
The I'll be able to make an nice excel wich will save time for controller designs.
And for easier reading, you'll have to read the application note.
And this isn't a funny thread but it'll help to make good and precise work (as long as there are not mistakes). So I hope you'll help me.
CONDUCTION LOSSES
Assume that we use
a 6 Mosfet IRFB4468 design: http://www.irf.com/product-info/datasheets/data/irfp4468pbf.pdf
a 48V 20A Battery.
UDD = 48V.
Is it right to take 48V ? no more, no less?
We also use a Brushless e-bike Motor. Here I fail because I have absolutely no Idea of electric values for a motor.
Let say it's a 1500 W motor. How do we find (or choose) Uo and Iorms?
Does any motor dealer give the following informations: displacement factor, equivalent inductance ?
So missing information are:
Uo= ? (linked to Uan1)
Iorms= ?
cosφ1= ?
L = ?
What is ma ?
Output Ripple Current = (Udd - sqrt(2)xUo )xUo / (2 x L x Udd x fsw)
Io= sqrt(2) x Iorms
Mosfet conduction losses:
Pcm = RDSon x IDrms² = RDSon x Io² x (1/8 + ma x Cos φ1/ (3 x pi))
And RDSon depends on Temperature.
RDSon calculation:
RDSon (Tj) = RDson (25°C) x (1+ alpha/100)^(Tj-25)
From another source (I can't remember which one), RDSon 175° was computed this way:
RDSon(175) = (175-25)/ (Rth(JC) x ID²)
Although I didn't find it anywhere else, this formula works very well
RDSon 4468 (175) = 6,4
Then
alpha = 100 x ( (RDSon(175)/RDSon(25))^(1/175-25) -1) x 100
Diode conduction losses
Pcd= Ud0 x I fav + Rd x I²frms = Ud0 x Io x (1/(2Pi) - ma x cos φ1/8) + Rd x Io² x (1/8 - ma x Cos φ1/ (3 x pi))
From the Fig 7 of the IRFP4468 we read
Ud0 depends on temperature. Worst case is 25°
Ud0= 0.5V
Rd can be estimated as Vsd/Isd = 4.5mOhms
THEN
Pcd= ....................
SWITCHING LOSSES
Does that mean that switching losses will have to be doubled (1/2 fo)?
Mosfet Switching Losses
PswM=(EonM+EoffM) x fsw
EonM = UDD x Idon x (tri+tfu)/2 + Qrr x UDD = Udd x 1/Pi x Io x (tri+tfu)/2 + Qrr x Udd
EoffM = UDD x Idon x (tru+tfi)/2 = UDD x 1/Pi x Io x (tru+tfi)/2
Assume we drive the mosfet at 12V and Rdrive = 10ohms
Udr = 12V
Rdrive = 10ohms
From the IRFP4468 we read Qrr at 152°C worst case
Qrr =420nc
tfu =Rdrivetot x Qgd /(Udr - Uplateau)
tru = Rdrivetot x Qgd / Uplateau
Rdrivetot = Rgate + Rdrive
I'm not 100% sure of those formulas so i'd be happy if they were confirmed.
From The IRF4468:
Qgd = 89nC
Rgate = 0.8Ohms
Uplateau = Vgs = 2 to 4V
Then
tfu=
tru=
tfu= tfi?
tru= tri?
then
EonM=
EoffM=
And
PswM=
Diode Switching Losses
EonD= 1/4 Qrr x Udrr
worst case:
EonD = 1/4 Qrr x UDD
EoffD= UDD x 1/Pi x Io x (tru+tfi)/2
Then
EonD =
EoffD=
then
PswD=
then
Psw =
TOTAL LOSSES
P=
rhitee05
10 kW
Good piece-by-piece explanation of how to calculate the various loss components:
http://www.fairchildsemi.com/an/AN/AN-6005.pdf
Note: they assume a synchronous design, so you'd have to modify the loss calculation for using the freewheel diode instead.
http://www.fairchildsemi.com/an/AN/AN-6005.pdf
Note: they assume a synchronous design, so you'd have to modify the loss calculation for using the freewheel diode instead.
Lagoethe
10 W
- Joined
- Nov 10, 2010
- Messages
- 89
I made some modifications (in blue). Questions are still in red.
From the previous Application note, I made an exemple.
But I'll need your help to fill (and fix) things I don't know.
The I'll be able to make an nice excel wich will save time for controller designs.
And for easier reading, you'll have to read the application note.
And this isn't a funny thread but it'll help to make good and precise work (as long as there are not mistakes). So I hope you'll help me.
CONDUCTION LOSSES
"Input parameters for the calculation: Input voltage (UDD), output line-to-line voltage (Uo) or output phase voltage (Uan1), rms value of the output current (Iorms) or output apparent power (So=3Uan1Iorms), motor displacement factor (cosφ1), equivalent stator inductance (L), switching frequency (fsw), output (motor electrical) frequency fo and an inverter amplitude modulation index ma"
Assume that we use
a 6 Mosfet IRFB4468 design: http://www.irf.com/product-info/datashe ... 468pbf.pdf
a 48V 20A Battery.
UDD = 48V.
Is it right to take 48V ? no more, no less?
We also use a Brushless e-bike Motor. Here I fail because I have absolutely no Idea of electric values for a motor.
Let say it's a 1500 W motor. How do we find (or choose) Uo and Iorms?
Does any motor dealer give the following informations: displacement factor, equivalent inductance ?
So missing information are:
Uo= ? (linked to Uan1)
Iorms= ?
cosφ1= ?
L = ?
What is ma ?
Other question:
I've got a motor at home
How can I find L between phases?
(for R, I have my idea
)
Output Ripple Current = (Udd - sqrt(2)xUo )xUo / (2 x L x Udd x fsw)
Io= sqrt(2) x Iorms
Mosfet conduction losses:
Pcm = RDSon x IDrms² = RDSon x Io² x (1/8 + ma x Cos φ1/ (3 x pi))
And RDSon depends on Temperature.
RDSon calculation:
RDSon (Tj) = RDson (25°C) x (1+ alpha/100)^(Tj-25)
From another source (I can't remember which one), RDSon 175° was computed this way:
RDSon(175) = (175-25)/ (Rth(JC) x ID²)
Although I didn't find it anywhere else, this formula works very well
RDSon 4468 (175) = 6,4
Then
alpha = 100 x ( (RDSon(175)/RDSon(25))^(1/175-25) -1) x 100
=0,602
RDSon (Tj) = RDson (25°C) x (1,00602)^(Tj-25)
Pcm =2,6 x (1,00602)^(Tj-25) x 2 x Iorms² x (1/8 + ma x Cos φ1/ (3 x pi))
But still miss Iorms, ma, Cos φ1
Diode conduction losses
Pcd= Ud0 x I fav + Rd x I²frms = Ud0 x Io x (1/(2Pi) - ma x cos φ1/8) + Rd x Io² x (1/8 - ma x Cos φ1/ (3 x pi))
From the Fig 7 of the IRFP4468 we read
Ud0 depends on temperature. Worst case is 25°
Ud0= 0.5V
Rd can be estimated as Vsd/Isd = 4.5mOhms
THEN
Pcd= ....................
miss Iorms, ma, Cos φ1.
SWITCHING LOSSES
"In order to find a simple solution for the switching loss calculation, it is supposed that the losses generated in the inverter in one half-wave of the output frequency (1/(2 fo) ) correspond to the losses generated if instead of AC output current a DC equivalent output current is applied."
Does that mean that switching losses will have to be doubled (1/2 fo)?
The equivalent DC output current value is:
Idc = 1/Pi x Io
Mosfet Switching Losses
PswM=EswM x fsw
EswM = UDD x Idon x (tr+tf) + Qrr x UDD = Udd x 1/Pi x Io x (tr+tf) + Qrr x Udd
Assume we drive the mosfet at 12V and Rdrive = 10ohms
Udr = 12V
Rdrive = 10ohms
From the IRFP4468 we read Qrr at 152°C worst case
Qrr =420nc
Rdrivetot = Rgate + Rdrive
We use a mosfet driver: IRS2186
http://www.irf.com/product-info/datasheets/data/irs2186pbf.pdf
tf= Qgsw /Idriver
tr= Qgsw /Idriver
Qgsw = Qgd + Qgs/2 (approximation)
from IRF4468
Qgd = 89nC
Qgs = 81nC
Rgate = 0.8Ohms
Uplateau = 4V
Qgsw=130nC
tf =Rdrivetot(up) x Qgsw /(Udr - Uplateau)
=10,8 x 130 /(12-4)= 175ns
tr = Rdrivetot(down) x Qgsw / Uplateau
= 350ns
Note: Something suprise me. Using this method, changing mosfet driver don't impact rise and fall time. Maybe we could use AN 799 Microchip:
with I driver = 4A (IRS2186)
dT = Qtotal /Idriver = 90ns
What should I do with these 90ns???
From IRF4468 we read
Qrr=370-420 nC (depending on temperature)
Worst case Qrr = 420nC
then
EswM=Udd x 1/Pi x Io x (tr+tf) + Qrr x Udd
Io missing
And
PswM=
Diode Switching Losses
EonD= 1/4 Qrr x Udrr
worst case:
EonD = 1/4 Qrr x UDD
EoffD= UDD x 1/Pi x Io x (tr+tf)/2
Then
EonD =
EoffD=
then
PswD=
then
Psw =
TOTAL LOSSES
P=
From the previous Application note, I made an exemple.
But I'll need your help to fill (and fix) things I don't know.
The I'll be able to make an nice excel wich will save time for controller designs.
And for easier reading, you'll have to read the application note.
And this isn't a funny thread but it'll help to make good and precise work (as long as there are not mistakes). So I hope you'll help me.
CONDUCTION LOSSES
"Input parameters for the calculation: Input voltage (UDD), output line-to-line voltage (Uo) or output phase voltage (Uan1), rms value of the output current (Iorms) or output apparent power (So=3Uan1Iorms), motor displacement factor (cosφ1), equivalent stator inductance (L), switching frequency (fsw), output (motor electrical) frequency fo and an inverter amplitude modulation index ma"
Assume that we use
a 6 Mosfet IRFB4468 design: http://www.irf.com/product-info/datashe ... 468pbf.pdf
a 48V 20A Battery.
UDD = 48V.
Is it right to take 48V ? no more, no less?
We also use a Brushless e-bike Motor. Here I fail because I have absolutely no Idea of electric values for a motor.
Let say it's a 1500 W motor. How do we find (or choose) Uo and Iorms?
Does any motor dealer give the following informations: displacement factor, equivalent inductance ?
So missing information are:
Uo= ? (linked to Uan1)
Iorms= ?
cosφ1= ?
L = ?
What is ma ?
Other question:
I've got a motor at home
How can I find L between phases?
(for R, I have my idea
)Output Ripple Current = (Udd - sqrt(2)xUo )xUo / (2 x L x Udd x fsw)
Io= sqrt(2) x Iorms
Mosfet conduction losses:
Pcm = RDSon x IDrms² = RDSon x Io² x (1/8 + ma x Cos φ1/ (3 x pi))
And RDSon depends on Temperature.
RDSon calculation:
RDSon (Tj) = RDson (25°C) x (1+ alpha/100)^(Tj-25)
From another source (I can't remember which one), RDSon 175° was computed this way:
RDSon(175) = (175-25)/ (Rth(JC) x ID²)
Although I didn't find it anywhere else, this formula works very well
RDSon 4468 (175) = 6,4
Then
alpha = 100 x ( (RDSon(175)/RDSon(25))^(1/175-25) -1) x 100
=0,602
RDSon (Tj) = RDson (25°C) x (1,00602)^(Tj-25)
Pcm =2,6 x (1,00602)^(Tj-25) x 2 x Iorms² x (1/8 + ma x Cos φ1/ (3 x pi))
But still miss Iorms, ma, Cos φ1
Diode conduction losses
Pcd= Ud0 x I fav + Rd x I²frms = Ud0 x Io x (1/(2Pi) - ma x cos φ1/8) + Rd x Io² x (1/8 - ma x Cos φ1/ (3 x pi))
From the Fig 7 of the IRFP4468 we read
Ud0 depends on temperature. Worst case is 25°
Ud0= 0.5V
Rd can be estimated as Vsd/Isd = 4.5mOhms
THEN
Pcd= ....................
miss Iorms, ma, Cos φ1.
SWITCHING LOSSES
"In order to find a simple solution for the switching loss calculation, it is supposed that the losses generated in the inverter in one half-wave of the output frequency (1/(2 fo) ) correspond to the losses generated if instead of AC output current a DC equivalent output current is applied."
Does that mean that switching losses will have to be doubled (1/2 fo)?
The equivalent DC output current value is:
Idc = 1/Pi x Io
Mosfet Switching Losses
PswM=EswM x fsw
EswM = UDD x Idon x (tr+tf) + Qrr x UDD = Udd x 1/Pi x Io x (tr+tf) + Qrr x Udd
Assume we drive the mosfet at 12V and Rdrive = 10ohms
Udr = 12V
Rdrive = 10ohms
From the IRFP4468 we read Qrr at 152°C worst case
Qrr =420nc
Rdrivetot = Rgate + Rdrive
We use a mosfet driver: IRS2186
http://www.irf.com/product-info/datasheets/data/irs2186pbf.pdf
tf= Qgsw /Idriver
tr= Qgsw /Idriver
Qgsw = Qgd + Qgs/2 (approximation)
from IRF4468
Qgd = 89nC
Qgs = 81nC
Rgate = 0.8Ohms
Uplateau = 4V
Qgsw=130nC
tf =Rdrivetot(up) x Qgsw /(Udr - Uplateau)
=10,8 x 130 /(12-4)= 175ns
tr = Rdrivetot(down) x Qgsw / Uplateau
= 350ns
Note: Something suprise me. Using this method, changing mosfet driver don't impact rise and fall time. Maybe we could use AN 799 Microchip:
with I driver = 4A (IRS2186)
dT = Qtotal /Idriver = 90ns
What should I do with these 90ns???
From IRF4468 we read
Qrr=370-420 nC (depending on temperature)
Worst case Qrr = 420nC
then
EswM=Udd x 1/Pi x Io x (tr+tf) + Qrr x Udd
Io missing
And
PswM=
Diode Switching Losses
EonD= 1/4 Qrr x Udrr
worst case:
EonD = 1/4 Qrr x UDD
EoffD= UDD x 1/Pi x Io x (tr+tf)/2
Then
EonD =
EoffD=
then
PswD=
then
Psw =
TOTAL LOSSES
P=
Lagoethe
10 W
- Joined
- Nov 10, 2010
- Messages
- 89
Here is the last review of the mosfet losses calculation
Sources:
http://www.btipnow.com/library/white_papers/MOSFET%20Power%20Losses%20Calculation%20Using%20the%20Data-Sheet%20Parameters.pdf
http://www.nxp.com/documents/application_note/APPCHP3.pdf
http://skory.gylcomp.hu/alkatresz/IRFP4468pbf.pdf
http://www.fairchildsemi.com/an/AN/AN-6005.pdf
There is a attached document which is a EXCEL which would help to choose IRF mosfet from INput parameters
GENERAL SPEAKING
"Input parameters for the calculation: Input voltage (Vdc), output line-to-line voltage (Vline) or output phase voltage (rms) (Vph), rms value of the output current (Il), motor displacement factor(or power factor) (cosφ1), equivalent stator inductance (L), switching frequency (fsw), output (motor electrical) frequency fo and an inverter amplitude modulation index ma"
Assume that we use
a 6 Mosfet IRFB4468 design
a 48V 20A Battery.
VDC = 48V.
??? Is it a good idea or can the Battery Voltage be more than that ???
Let say it's a 1500 W motor.
Vph = Vdc/(2 x sqrt(2))
= 17V
Vline = sqrt(3) x Vph
= 29V
??? This is the Vline RMS i'd like to know what is the max voltage which can be seen by a MOSFET. I deeply hope that it hasn't any link with input bulk capacitors
VmaxMosfet = Vline
and VBRDSS > VmaxMosfet ???
For IRFP4468 VBRSS= 100V > 29V
So it's OK
Power(W) = n x cosφ1 x VA
Generally for small induction motor n x cosφ1 = 0,55 to 0,65
??? I don't know how this term can be calculated or found. I don't think the motor supplier provide this information ???
We assume a 0,55 for worst case
VA = 2727 W
VA = Vline x Il x sqrt(3)
Il = VA / (Vline x sqrt(3))
=2727/(29 x sqrt(3))
=54A
Due to current ripple
Imax=sqrt(2) x Il
=sqrt(2) x 54
= 76 A
ID(T°)>76A
For IRFP4468
ID(100°C)= 200A > Imax
So it's OK
Output Ripple Current = (Vdc - sqrt(2)xVl )xVl / (2 x L x Vdc x fsw) = K x Vline/ (L x Fsw)
??? K = 0,14434 my math are no more what they were. This value was found through excel. To be confirmed, shame on me ???
fsw=12kHz (chosen)
??? L=10µH. I took this value, choosing with my nose, because I don't have any idea of it. I'd like to measure it on my own motor, but i don't know how to do. I don't even know if the phase to phase inductance is the same as the equivalent one. ???
SO here
Output current ripple = 35A
Mosfet conduction losses:
From this source: http://www.btipnow.com/library/white_papers/MOSFET%20Power%20Losses%20Calculation%20Using%20the%20Data-Sheet%20Parameters.pdf
Pcm = RDSon x IDrms² = RDSon x Imax² x (1/8 + ma x Cos φ1/ (3 x pi))
I don't know a lot of information here and I don't even understand where this formula come from. So I'll forget this one and use another one coming from this source:
http://www.nxp.com/documents/application_note/APPCHP3.pdf
Pcm = It² x RDSon
where IT is the rms value of the half sinusoid MOSFET current envelope.
It=Imax/2 for a sinusoidal envelope.
??? What would it be for a trapezoidal envelope???
Assume we do a sinusoidal command.
It = Imax/2 = 76/2
It = 38A
And RDSon depends on Temperature.
RDSon calculation:
RDSon (Tj) = RDson (25°C) x (1+ alpha/100)^(Tj-25)
From another source (I can't remember which one), RDSon 175° was computed this way:
RDSon(175) = (175-25)/ (Rth(JC) x ID²)
Although I didn't find it anywhere else, this formula works very well
RDSon IRFP4468 (175) = 6,4
Then
alpha = 100 x ( (RDSon(175)/RDSon(25))^(1/175-25) -1) x 100
=0,602
RDSon (Tj) = RDson (25°C) x (1,00602)^(Tj-25)
Assume we Tj = 150°C
RDSon(150) = 5,3mOhms
Pcm = It² x RDSon = 7,7W
Diode conduction losses
From what I can read in the application note used, diode losses can be neglected.
I'll assume that Conduction Diode losses = 10% of conduction losses
Then Total conduction losses = Pcm x 1,1 = 8,4W
SWITCHING LOSSES
"In order to find a simple solution for the switching loss calculation, it is supposed that the losses generated in the inverter in one half-wave of the output frequency (1/(2 fo) ) correspond to the losses generated if instead of AC output current a DC equivalent output current is applied."
??? Does that mean that switching losses will have to be doubled (1/2 fo)???
The equivalent DC output current value is:
Idc = 1/Pi x Imax = 24A
Mosfet Switching Losses
PswM=EswM x fsw
EswM = Vdc x 1/Pi x Io x (tr+tf) + Qrr x Vdc
Assume we drive the mosfet at 12V and Rdrive = 10ohms
Udr = 12V
Rdrive = 10ohms
From the IRFP4468 we read Qrr at 152°C worst case
Qrr =420nc
Rdrivetot = Rgate + Rdrive
We use a mosfet driver: IRS2186
http://www.irf.com/product-info/datashe ... 186pbf.pdf
tf= Qgsw /Idriver
tr= Qgsw /Idriver
Qgsw = Qgd + Qgs/2 (approximation)
from IRF4468
Qgd = 89nC
Qgs = 81nC
Rgate = 0.8Ohms
Uplateau = 4V
Qgsw=130nC
tf =Rdrivetot(up) x Qgsw /(Udr - Uplateau)
=10,8 x 130 /(12-4)= 175ns
tr = Rdrivetot(down) x Qgsw / Uplateau
= 350ns
??? Note: Something suprise me. Using this method, changing mosfet driver don't impact rise and fall time. Maybe we could use AN 799 Microchip:
with I driver = 4A (IRS2186)
dT = Qtotal /Idriver = 90ns
What should I do with these 90ns???
From IRF4468 we read
Qrr=370-420 nC (depending on temperature)
Worst case Qrr = 420nC
then
EswM=Vdc x Idc x (tr+tf) + Qrr x Vdc
=48 x 24 x (525) x 10^(-9) + 420 x 48 x (10^-9)
=0,62mJ
Note: Qrr energy is negletible compared to transition energy.
??? Is it J unit ???
And
PswM= EswM x Fsw =0,62 x 12000 /1000
PswM =7,5W
Diode Switching Losses
EonD= 1/4 Qrr x Udrr
worst case:
EonD = 1/4 Qrr x VDC = 1/4 x 420 x 48 x 10^-9
EonD = 5 x 10^-6
EoffD= Vdc x Idc x (tr+tf)/2
EoffD= 48 x 24 x 525 /2 x 10^-9
EoffD=25 x 10^-6
EswD = 30 x 10^-6
EswD<<EswM (to keep in mind for the excel calculation)
PswD=0,36W
Psw = PswM + PswD = 7,9W
TOTAL LOSSES
P= 7,9W + 8,4W =16,3W
HEAT DISSIPATION:
P x Rth (J-A)= Tj-Ta
Assume Ta = 40°C
P x (Rth (J-C) + Rth (C-S) + Rth (S-A) = 150 - 40 =110
16,3 x (0,29+0,5+Rth(S-A))=110
Rth(S-A)=5,96 °C/W Max
??? I wonder If all mosfet are on the same dissipator:
Rthglobal = Rth(S-A) /(number of fets) = 0,993°C/W
Or because fet aren't always the warmer at the same time
Rthglobal = Rth(S-A) /(number of fets) x 3/2 = 1,49°C/W ???
ABOUT EXCEL SPREADSHEET
It's a first shot. I'll need your help to confirm all this.
Thanks and have a nice day
Sources:
http://www.btipnow.com/library/white_papers/MOSFET%20Power%20Losses%20Calculation%20Using%20the%20Data-Sheet%20Parameters.pdf
http://www.nxp.com/documents/application_note/APPCHP3.pdf
http://skory.gylcomp.hu/alkatresz/IRFP4468pbf.pdf
http://www.fairchildsemi.com/an/AN/AN-6005.pdf
There is a attached document which is a EXCEL which would help to choose IRF mosfet from INput parameters
GENERAL SPEAKING
"Input parameters for the calculation: Input voltage (Vdc), output line-to-line voltage (Vline) or output phase voltage (rms) (Vph), rms value of the output current (Il), motor displacement factor(or power factor) (cosφ1), equivalent stator inductance (L), switching frequency (fsw), output (motor electrical) frequency fo and an inverter amplitude modulation index ma"
Assume that we use
a 6 Mosfet IRFB4468 design
a 48V 20A Battery.
VDC = 48V.
??? Is it a good idea or can the Battery Voltage be more than that ???
Let say it's a 1500 W motor.
Vph = Vdc/(2 x sqrt(2))
= 17V
Vline = sqrt(3) x Vph
= 29V
??? This is the Vline RMS i'd like to know what is the max voltage which can be seen by a MOSFET. I deeply hope that it hasn't any link with input bulk capacitors
VmaxMosfet = Vline
and VBRDSS > VmaxMosfet ???
For IRFP4468 VBRSS= 100V > 29V
So it's OK
Power(W) = n x cosφ1 x VA
Generally for small induction motor n x cosφ1 = 0,55 to 0,65
??? I don't know how this term can be calculated or found. I don't think the motor supplier provide this information ???
We assume a 0,55 for worst case
VA = 2727 W
VA = Vline x Il x sqrt(3)
Il = VA / (Vline x sqrt(3))
=2727/(29 x sqrt(3))
=54A
Due to current ripple
Imax=sqrt(2) x Il
=sqrt(2) x 54
= 76 A
ID(T°)>76A
For IRFP4468
ID(100°C)= 200A > Imax
So it's OK
Output Ripple Current = (Vdc - sqrt(2)xVl )xVl / (2 x L x Vdc x fsw) = K x Vline/ (L x Fsw)
??? K = 0,14434 my math are no more what they were. This value was found through excel. To be confirmed, shame on me ???
fsw=12kHz (chosen)
??? L=10µH. I took this value, choosing with my nose, because I don't have any idea of it. I'd like to measure it on my own motor, but i don't know how to do. I don't even know if the phase to phase inductance is the same as the equivalent one. ???
SO here
Output current ripple = 35A
Mosfet conduction losses:
From this source: http://www.btipnow.com/library/white_papers/MOSFET%20Power%20Losses%20Calculation%20Using%20the%20Data-Sheet%20Parameters.pdf
Pcm = RDSon x IDrms² = RDSon x Imax² x (1/8 + ma x Cos φ1/ (3 x pi))
I don't know a lot of information here and I don't even understand where this formula come from. So I'll forget this one and use another one coming from this source:
http://www.nxp.com/documents/application_note/APPCHP3.pdf
Pcm = It² x RDSon
where IT is the rms value of the half sinusoid MOSFET current envelope.
It=Imax/2 for a sinusoidal envelope.
??? What would it be for a trapezoidal envelope???
Assume we do a sinusoidal command.
It = Imax/2 = 76/2
It = 38A
And RDSon depends on Temperature.
RDSon calculation:
RDSon (Tj) = RDson (25°C) x (1+ alpha/100)^(Tj-25)
From another source (I can't remember which one), RDSon 175° was computed this way:
RDSon(175) = (175-25)/ (Rth(JC) x ID²)
Although I didn't find it anywhere else, this formula works very well
RDSon IRFP4468 (175) = 6,4
Then
alpha = 100 x ( (RDSon(175)/RDSon(25))^(1/175-25) -1) x 100
=0,602
RDSon (Tj) = RDson (25°C) x (1,00602)^(Tj-25)
Assume we Tj = 150°C
RDSon(150) = 5,3mOhms
Pcm = It² x RDSon = 7,7W
Diode conduction losses
From what I can read in the application note used, diode losses can be neglected.
I'll assume that Conduction Diode losses = 10% of conduction losses
Then Total conduction losses = Pcm x 1,1 = 8,4W
SWITCHING LOSSES
"In order to find a simple solution for the switching loss calculation, it is supposed that the losses generated in the inverter in one half-wave of the output frequency (1/(2 fo) ) correspond to the losses generated if instead of AC output current a DC equivalent output current is applied."
??? Does that mean that switching losses will have to be doubled (1/2 fo)???
The equivalent DC output current value is:
Idc = 1/Pi x Imax = 24A
Mosfet Switching Losses
PswM=EswM x fsw
EswM = Vdc x 1/Pi x Io x (tr+tf) + Qrr x Vdc
Assume we drive the mosfet at 12V and Rdrive = 10ohms
Udr = 12V
Rdrive = 10ohms
From the IRFP4468 we read Qrr at 152°C worst case
Qrr =420nc
Rdrivetot = Rgate + Rdrive
We use a mosfet driver: IRS2186
http://www.irf.com/product-info/datashe ... 186pbf.pdf
tf= Qgsw /Idriver
tr= Qgsw /Idriver
Qgsw = Qgd + Qgs/2 (approximation)
from IRF4468
Qgd = 89nC
Qgs = 81nC
Rgate = 0.8Ohms
Uplateau = 4V
Qgsw=130nC
tf =Rdrivetot(up) x Qgsw /(Udr - Uplateau)
=10,8 x 130 /(12-4)= 175ns
tr = Rdrivetot(down) x Qgsw / Uplateau
= 350ns
??? Note: Something suprise me. Using this method, changing mosfet driver don't impact rise and fall time. Maybe we could use AN 799 Microchip:
with I driver = 4A (IRS2186)
dT = Qtotal /Idriver = 90ns
What should I do with these 90ns???
From IRF4468 we read
Qrr=370-420 nC (depending on temperature)
Worst case Qrr = 420nC
then
EswM=Vdc x Idc x (tr+tf) + Qrr x Vdc
=48 x 24 x (525) x 10^(-9) + 420 x 48 x (10^-9)
=0,62mJ
Note: Qrr energy is negletible compared to transition energy.
??? Is it J unit ???
And
PswM= EswM x Fsw =0,62 x 12000 /1000
PswM =7,5W
Diode Switching Losses
EonD= 1/4 Qrr x Udrr
worst case:
EonD = 1/4 Qrr x VDC = 1/4 x 420 x 48 x 10^-9
EonD = 5 x 10^-6
EoffD= Vdc x Idc x (tr+tf)/2
EoffD= 48 x 24 x 525 /2 x 10^-9
EoffD=25 x 10^-6
EswD = 30 x 10^-6
EswD<<EswM (to keep in mind for the excel calculation)
PswD=0,36W
Psw = PswM + PswD = 7,9W
TOTAL LOSSES
P= 7,9W + 8,4W =16,3W
HEAT DISSIPATION:
P x Rth (J-A)= Tj-Ta
Assume Ta = 40°C
P x (Rth (J-C) + Rth (C-S) + Rth (S-A) = 150 - 40 =110
16,3 x (0,29+0,5+Rth(S-A))=110
Rth(S-A)=5,96 °C/W Max
??? I wonder If all mosfet are on the same dissipator:
Rthglobal = Rth(S-A) /(number of fets) = 0,993°C/W
Or because fet aren't always the warmer at the same time
Rthglobal = Rth(S-A) /(number of fets) x 3/2 = 1,49°C/W ???
ABOUT EXCEL SPREADSHEET
It's a first shot. I'll need your help to confirm all this.
Thanks and have a nice day
