The circuit shown below is the schematic of infineon PIM module. I am specifically interested in FP15R12W1T4. From the schematic I could understand all elements other than the 4th leg which has an igbt in series with a diode. I am new to this field. I assume, that leg and terminals are provided to feedback the present voltage level under a specified switching frequency.
Electronic – the use of the fourth leg, the one with igbt in series with the diode, in the infineon power module
igbt
Related Solutions
Fundamentally yes but not quite
You have captured the basic concept
\$ P_{total} = P_{cond} + P_{sw} + P_{leak} \approx P_{cond} + P_{sw} \$
I have added in leakage losses for completeness but this only becomes of significance one you get to series stacked devices.
Conduction loss
This is the subtle difference. Conceptually it is \$ P_{cond} = V_{ce} \cdot I_c\$ however, \$V_{ce}\$ might not be good enough. This needs to be the saturation voltage at your operating point. Now you may have read the datasheet and chosen \$V_{ce} = 2.1V\$ when \$I_c\$ equals 130A but there is no named part to cross-reference.
Assuming you have done this correctly then yes \$ P_{cond} = V_{ce} \cdot I_c\$ scaled by duty.
However, there is a more flexible method and that is approximating the IGBT operating point as a series connected DC source \$V_{ce0}\$ and a collector-emitter resistance \$r_c\$
In this example I an interested in an operating point around 20A.
1V = 18.942mm
zero @ 17.071mm
therefore Vce0 = 0.901V
dI = 20A
dv = 12.527mm = 0.661V
therefore rc = 0.00331R
The instantanious power eqates to
\$P_{cont}(t) = V_{ce0}(t)\cdot I_c(t) + r_c\cdot i_c^2(t)\$
The average loss is therefore
\$P_{cont}(t) = \frac{1}{T_{sw}}\int_{0}^{T_{sw}} V_{ce0}(t)\cdot I_c(t) + r_c\cdot i_c^2(t) dt = V_{ce0}\cdot I_{c,avg} + r_c\cdot I_{c,rms}^2\$
You can then derive the average and rms currents based upon your waveform.
A similar process can be done with the diode IF appropriate. NOTE: thermally the co-packaged diode typically becomes the limiting factor in a half-bridge
Switching loss
AS you have captured... Switching loss is the accumulation of switching energy with respect to the switching frequency
\$P_{sw} = (E_{on} + E_{off})\cdot f_{sw} \$
However... the Eon and Eoff are typically stated at a given operating point
This means you need to scale the switching energy with respect to your operating point AND the test point. a linear rescaling is usually good enough BUT please check the Eon,off curves to see if a linear interpolation is good enough
\$P_{sw} = (E_{on,t} + E_{off,t})\cdot \frac{V_{DC}}{V_{DC,test}}\cdot \frac{I_{peak}}{I_{I,test}}\cdot f_{sw} \$
Likewise there is is something comparable with the diode as you accumulate the reverse recovery charge.
The real fun is deriving the average and rms current waveforms and there are given equations for the profile all related to modulation depth
With these types of things the devil is in the detail.
1EDI60I12AF : Very nice little driver chip, I am making use of a selection of these for a SiC inverter (except the wide body variant due to a few hundred volts )
SIM-0512D : Not used this exact DC:DC but SIL DC:DC with suitable rating and rails are my goto for lab lash-ups (newport: bad experience, Traco: robust but noisy, Murata: very nice) otherwise it is a flyback due to more specific needs.
So conceptually everything looks fine... The 8V regulator to generate -8V from -12V is odd, especially as the Gate-emitter can accept +-20V.
This will work, but it is lossy... I personally would have just powered the EiceDriver from the full 24V.
The turn-on voltage (+12V) is a bit low for efficient driving and the collector-emitter characteristics are down from the more preferred drive case of +15V... but this is just a loss problem. Could you be driving this such that the junction temperature is exceeding 175C and causing thermal runaway? possibly... it really depends on the heatsinking (The video does indicate a hsnk is used)
I suspect the real problem is the actual routing of the circuit. Your visual sketch and equally the linkwire version you are showing is very ... messy... The 1EDI60I12AF is a very fast driver, 20ns risetime (nice for SiC...) and equally your turn-off resistor (3r3) is very low. I appreciate the entire drive circuit is copied from the 1EDI60I12AF (the -8V, 10R:3r3) but you are meant to tailor for your specific needs. In this case an Roff closer to 15-20R would be better.
Back to the messy part and the modcard. They key here is the Emitter return
You have about 30cm of gate-emitter with not the greatest of twist nor routing, switching at 20ns and then finally the actual current loop from the driver, IGBT, psu is non-idea.
What I suspect is happening is, during turn-off, due to the additional emitter inductance, there is a lot of oscillation at the IGBT gate-emitter. This will result in turn-on events due to the miller capacitance. Because the ability of the gatedrive to hold the gate potential at teh desired voltage is now impeded, the IGBT is oscillating around the active region resulting in thermal runaway. Equally this gate oscillation could burn the gate region out.
Here is the output stage of a gatedrive I have been working on.
This is also using an EiceDRIVER (but the wide-body, higher drive version). The Ve is a plane on this card and the twisted-pair leads to the IGBT are 2cm in length.
Using a lamp would be very forgiving as this presents a fixed resistive load to the circuit so any oscillations would still result in limited current.
So in summary
1) There is a setup issue - tab of the IGBT isn't actually isolated causing an additional ground loop, the FWD across the inductive load is actually open resulting in a large voltage that then kills the IGBT.
2) the drive capability of the gatedrive is compounded by the gate leads resulting in.
a) thermal runaway due to inductive load operation.
b) gate region burnout due to excessive gate current.
3) Your pulsewidth code has an oversight and is holding the IGBT on for too long. With an inductive load this will produce large current and will kill the IGBT
Best Answer
By applying a DC shunt to the battery, the motor instead of free wheeling can regenerate and brake the motor. using the back EMF of the motor/generator.
There are different implementations including Battery Regen and single phase Remanence braking.
This is why there is a single B phase brake on a 3 phase UVW phase rectifier or bridge driver.