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Hi,
Page 15 Datasheet: Power dissipation, per input part is rated with 100mW and 12mW/°C derating above 65°C.
So the maximal ambient temperature could not be higher than 65+100/12 = 73.3°C ? but the device is rated for 40…125°.
Do I understand this rating in a wrong manner?
To calculate the losses, we need to now the internal RON/ROFF (Source/sink output resistance). Could you please let us know the values. (We tied it using Table 7, but it leads to different values)
We can only use one DESAT channel. What do we do with the unused DESDAT Pin?
Kind Regards
Rolf
Solved! Go to Solution.
 Labels:

ispn:4068:1:0

l1:144:1:0

l2:1244:1:0
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Hello @RonnieRolfes
Yes, you can use 8 Ohms if Vincotech topology and input side driver voltage is 5V is used @25kHz.
Cooling option for Driver IC:
The thermal coupling to the chips is given by the pins GND1 (for the primary) and VEE2LS&VEE2HS (for
the secondary) since these are also connected to the lead frame that is below the chip. In order to optimize
thermal performance, these pins should be connected to the bigger copper area (such as a polygon) with a
couple of vias.
Thanks
Guru
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Hello @RonnieRolfes
Thanks for posting your question in the Infineon community.
Q1 the maximal ambient temperature could not be higher than 65+100/12 = 73.3°C ? but the device is rated for 40…125°.Do I understand this rating in a wrong manner?
Ans: Yes your approach is not in the right direction. Your calculation represents internal temperature, not ambient temperature. The maximum ambient temperature can be calculated by using this formula
Tamb_max=TjPloss*Rth.
Q2To calculate the losses, we need to now the internal RON/ROFF (Source/sink output resistance). Could you please let us know the values? (We tied it using Table 7, but it leads to different values)
Ans: The RON and Roff are not fixed resistor they are nonlinear.
Ron=VCC_MAX/Source current(Given in Table2)
Roff=VEE_MAX/Sink current(Given in Table2)
Q3 We can only use one DESAT channel. What do we do with the unused DESDAT Pin?
Ans: It is advisable to connect one capacitor to that pin the value of nearly 100pF.(no specific calculation is required. here just the DESAT pin response will be slowly no effect in driver IC )
Thanks
Guru
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ANS/Q1:
Thank you for correcting me
ANS/Q2:
I already thought that RON/ROFF will be the "Rdson" of the corresponding internal switch. Will taking the Absolute Maximum Ratings in Table 2 be the "worstcase losses" for the RON/ROFF?
e.g. RON = 20/2.4 = 8.33 Ohm and ROFF = 12/2.4 = 5 Ohm
Are the formulas fot the power dissiapation correct? (see attached picture)
ANS/Q3:
We are using 100pF or 150pF for the desaturation at Pin 30 DESATHS. Won't the desaturation detection at Pin 20 DESATLS react quicker than DESATHS if we use 100pF?
Would it make sense to use a capacitance e.g 100x higher than the one on DESATHS? So the blanking time is 100x longer and will not trigger a FAULT before DESATHS
See attached schematic
Best Regards
Rolf
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Hello @RonnieRolfes
The Q2 answer is Correct.
I have some points for Q2 mentioned below.
Q1. By observing the schematic, I assume it is for driving the top & bottom IGBT Leg. If yes, then it is recommended to
use Desat for both the IGBTS to protect the system in fault events.
Q2. are you using R1701, R1703, R1706 & R1712, R1714, R1717 for both Turn ON & Turn OFF?
Q3. may I know what topology you are using here (IGBT connections)?
Q4. if you want to use Desat protection, then use calculation C = i*dt/dv where dt is the required saturation response
time, dv is (desat pin voltage) 9V typical given in datasheet and I is internal current 500uA given in datasheet to
charge the desat pin capacitor.
Q5. both the channels used the same power supply with the ground, so isolation has been bypassed. Please check
once again.
Could your share your application details
Thanks
Guru
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See my answers below in green
Best Regards
Rolf
The Q2 answer is Correct.
I have some points for Q2 mentioned below.
Q1. By observing the schematic, I assume it is for driving the top & bottom IGBT Leg. If yes, then it is recommended to
use Desat for both the IGBTS to protect the system in fault events.
I know it, but the threelevel module we like to use does not have an access to the common collector of the inner switches
[cid:image001.png@01D8563A.976B6780]
Q2. are you using R1701, R1703, R1706 & R1712, R1714, R1717 for both Turn ON & Turn OFF?
Only if the power dissipation of the 2ED is to high. But we would like to avoid the external booster if not necessary. That’s the main reason why I started this discussion 😊
Q3. may I know what topology you are using here (IGBT connections)?
3 phase threelevel with active neutral point clamp (ttype). See picture above
Q4. if you want to use Desat protection, then use calculation C = i*dt/dv where dt is the required saturation response
time, dv is (desat pin voltage) 9V typical given in datasheet and I is internal current 500uA given in datasheet to
charge the desat pin capacitor.
That’s exactly how we did it for ONE channel of the 2ED. The question is hoe to “disable” the second one
Q5. both the channels used the same power supply with the ground, so isolation has been bypassed. Please check
once again.
You observed it well! But this is made on purpose. Guess the picture below will explain it
To reduce costs, some of the supplies and “Eice 2ED’s” needed to switch the IGBTs T1, T2, T3 and T4 are combined (as they share the same reference potential)
· 1 x Supply for all T1 and T2 Phase U
· 1 x Supply for all T1 and T2 Phase V
· 1 x Supply for all T1 and T2 Phase W
· [DC+] 1 x Supply for all T3
· 1 x Supply for all T4
[DC+][DC/2][Phase][cid:image014.png@01D8563F.7C665800][cid:image015.png@01D8563F.7C665800][cid:image016.png@01D8563F.7C665800][cid:image017.jpg@01D8563F.7C665800]
[T1 / T2 Phase U,T4 Phase U/V/W] [T1 / T2 Phase W,T1 / T2 Phase V,T3 Phase U/V/W]
[cid:image026.png@01D8563F.7C665800][2ED allows the needed 2mm Creepage between DC and DC/2][cid:image029.png@01D8563F.7C665800]
Could your share your application details
The inverter shall be use for high speed (>120’000rpm) machines
Thanks
Guru
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Hello @RonnieRolfes
I have reviewed your schematics and power supply connections looks all are correct. except below points
are you using R1701, R1703, R1706 & R1712, R1714, R1717 for both Turn ON & Turn OFF?
Only if the power dissipation of the 2ED is to high. But we would like to avoid the external booster if not necessary. That’s the main reason why I started this discussion
the next question is: Which Pins do have the least thermal path? Or in other words: To which potential shall we attach a "cooling" plane? VCC2, GND2, VEE2 or OUT?
Since you wanted to use only a driver instead of an external booster, need to evaluate the power loss and temperature rise of the Driver IC. To calculate the same can you provide IGBT details, and switching frequency?
Thanks
Guru
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Hello @Guru_Prasad
Yes, we will use R1701, R1703, R1706 & R1712, R1714, R1717 for both Turn ON & Turn OFF. Having no option to set different switching speeds is fine for us
It will be a SemiCustom 3 Phase 3 Level Module
 1200V /40A IGBT4 of Infineon, for the "outer" Switches T1/T4, together with a parallel SiC Diode
 650V /40A Trenchstop 5 IGBT of Infineon, S5 for the "inner" Switches T2/T3
Our, and also your best Datasheet is Datasheet 10PY12M3A040SH09M749F38Y (vincotech.com)
The switching frequency is 25 to 50kHz, Ambient < 90°C
Best Regards
Rolf
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Hello @RonnieRolfes
The power loss for Gate driver IC was calculated based below formula
Where
2 is for twochannel
∆Vge is Gate VCC_Max
Ipeak is the Peak current(Calculated based on VCC_max/Rg_total)
trRaise time of your IGBT
tfFall time of your IGBT
fs is switching frequency
Iq_in is the Input Quiescent current of Gate Driver IC
VCC1 is the input voltage of Driver IC
Iq_Out is the Input Quiescent current of Gate Driver IC
VCC2 is the Out voltage of Driver IC
The losses for your case are given below
PICloss=2*(20*2.5A*(14.7ns+46.6ns)*25kHz)+(15*6mA)+(5*9mA)=153.2mW+90mW+45mW=288.2mW
The derating capacity of IC at 90 deg is 1000mW300mW=700mW
Note:
1)The peak current 2.5A is calculated from 20V/8ohm. The 8ohm considered from your Vincotech Datasheet
2)The tr,tf calculated from Vincotech Datasheet at your power supply conditions
3) switching frequency I have taken 25kHz from your given range
Rg_total=Rg_on+Rg_int_Igbt(Rg_int_Igbt=0 in Vincotech Datasheet)
Conclusions:
1)The IC can drive rated current with Rg_total of 8 ohm. If you want to use of Rg_total <8ohm kindly use external booster.
2)Desat disabling can be done by connecting resistor of 6kohm(3V/500uA) OR you can connect diode from Desat pin to ground in Forward direction.
I have one small question may I know what is semi custom 3 Phase 3 Level Module topology? How are you connecting Vincotech module, IGBT4+Sic diode and Trench stop 5 IGBT for 3 level NPC?
Thanks
Guru
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Hi @Guru_Prasad
So the fact is, that we can drive the module with 25kHz and 8Ohm without Booster. For lower Rgon/off or higher switching frequencies, we will need the booster.
Thank you so far for that!
So would the best cooling option now be to connect the planes to VCC2, GND2, VEE2 or OUT?
For disabling DESAT i'll connect a 3.3k (> max 1.815@DESAT) resistor just because it is already on the BoM
It will be a pressfit module. That is all I know
Greetings
Rolf
We will also start with 8Ohm.
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Hello @RonnieRolfes
Yes, you can use 8 Ohms if Vincotech topology and input side driver voltage is 5V is used @25kHz.
Cooling option for Driver IC:
The thermal coupling to the chips is given by the pins GND1 (for the primary) and VEE2LS&VEE2HS (for
the secondary) since these are also connected to the lead frame that is below the chip. In order to optimize
thermal performance, these pins should be connected to the bigger copper area (such as a polygon) with a
couple of vias.
Thanks
Guru