I am working on a 7-level power micro-inverter which composed mainly by power MOSFETs. These logic-level MOSFETs (IRLZ24NS) are driven by separate photovoltaic gate drivers with integrated fast turn-off (Broadcom ACPL-K30T). The problem comes when I try to simulate the circuit and compare the results with those obtained by experiments. The circuit topology with relevant signals looks like this:

As you can see, the gate-to-source voltage of the MOSFET M11 (V(g)-V(s)) at time starting from 11.68ms should be near to zero volts and not 2.31V, because the charge of the parasitic capacitors (Cgd and Cgs) will be discharged through the gate driver circuitry. This assertion was verified by breadboarding and gives the following result:

Noting that during my experiment I was using available Panasonic AVP1122 gate drivers instead of Broadcom ACPL-K30T which has comparative capabilities. However, I guess that problem comes from the ACPL-K30T SPICE model itself, I've tried to decipher it but didn't succeed, so I ask you SPICE gurus to help me understand the reasons behind this issue. The SPICE code is given below:

Avago ACPL-K30T Spice Macromodel:

```
* ACPL-K30T Spice Macromodel
.subckt ACPL-K30T AN CA VOUT- VOUT+
E1 N002 CA N014 N016 {CTR}
XX1 AN CA N016 N014 vbu
XX2 N002 CA N017 N015 pdnoc
XX3 N002 CA N008 N017 pdnoc
XX4 N002 CA N015 N013 pdnoc
XX5 N002 CA N013 N012 pdnoc
XX6 N002 CA N012 N010 pdnoc
XX7 N002 CA N010 N007 pdnoc
XX8 N002 CA N007 N004 pdnoc
XX9 N002 CA N004 N003 pdnoc
XX10 N002 CA N003 P001 pdnoc
XX11 N002 CA P002 P003 pdnoc
E2 N011 VOUT- N008 N001 1
R2 N009 N011 1k
D1 VOUT- N001 D
D2 N008 VOUT+ D
E3 N009 N006 N008 VOUT- 1
Q1 VOUT+ N006 N005 0 NPN
R3 VOUT- N005 3k
C1 VOUT- N009 37n
XX12 N002 CA P003 N001 pdnoc
XX13 N002 CA P001 P002 pdnoc
C2 VOUT- VOUT+ 100p
.param CTR=0.083
.ends ACPL-K30T
.subckt vbu AN CA LOPN LOPP
RSERIES AN 5 5
DELECT 5 CA VBUNOR
ELED 6 LOPN 5 CA 1
DOPTIC 6 8 VBUNORC
FPHOTO LOPN 3 VSENSE 1
VSENSE 8 LOPN 0
RL 3 LOPN 0.1
EOUT LOPP LOPN 3 LOPN 60
VSIM LOPN CA 0
Rnl 6 N001 5k
Vnl N002 LOPN 0
Fnl LOPN LOPN Vnl 1
Dsw N001 N002 DSW
.model DSW D Is=1e-4
.model VBUNOR D IS=330E-21 N=1.5 XTI=3 EG=1.52 BV=10.38 IBV=100u
+ CJO=60p VJ=.75 M=.3333 FC=.5 TT=20n
.model VBUNORC D IS=330E-21 N=1.5 XTI=3 EG=1.52 BV=10.38 IBV=100u
+ CJO=0 VJ=.75 M=.3333 FC=.5 TT=0
.ends vbu
.subckt pdnoc LOPP LOPN AN CA
D1 AN CA PDC
G1 CA AN LOPP LOPN 0.0010
.model PDC D IS=1E-14 N=1.5 CJO=0p M=0.95 VJ=0.75 ISR=100.0E-12 BV=100 TT=5E-9
.ends pdnoc
.model D D
.model NPN NPN
.model PNP PNP
```

Thank you