Hi there,

I have created a Zener diode model in LTSpice, and now I am trying to simulate it with different values of the saturation current (IS), emission coefficient (N) and breakdown voltage (BV) using Spice.
These terms are not entirely clear to me, so analyzing the results of my simulations would be problematic. This is my simple understanding of the above:

Saturation current - This is the maximum reverse current caused by the combination of the minority particles in the p and n junctions, beyond which the zener diode enters breakdown voltage and this current exceeds rapidly, likely to damage the diode.

Emission coefficient - This concept is not clear to me. From what I read online, it is a value between 1 and 2. Closer to 1 means greater forward bias. But how do I apply this to my Zener diode simulations?

Breakdown voltage - The voltage which is greater than the barrier potential, beyond which the reverse current is not longer negligible and increases greatly.

I would highly appreciate if the accuracy of my understanding of the terms could be confirmed. Also, how do I put all this together to analyze and conclude my results of varying IS, BV and N i.e what results would I expect when changing these values?

Many thanks!

 

See models:
.model DFLZ33 D(Is=12.5p Rs=.6 N=1.1 Cjo=220p VJ=1 M=.33 tt=50n BV=29 IBV=1e-10 NBV=8 Vpk=33 mfg=Diodes_Inc. type=Zener)
.model 1N746 D(Is=31.47f Rs=9.494 N=1 Xti=3 Eg=1.11 Cjo=220p M=.5959 Vj=.75 Fc=.5 Isr=2.035n Nr=2 Bv=3.3 Ibv=45.862m Nbv=3.0477 Ibvl=29.831m Nbvl=11.606 Tbv1=-636.4u Vpk=3.3 mfg=Motorola type=zener)
.model 1N747 D(Is=1.242f Rs=1.137 N=1 Xti=3 Eg=1.11 Cjo=210p M=.6063 Vj=.75 Fc=.5 Isr=1.922n Nr=2 Bv=3.6 Ibv=13.987m Nbv=3.031 Ibvl=10.212m Nbvl=12.73 Tbv1=-555.6u Vpk=3.6 mfg=Motorola type=zener)
.model 1N748 D(Is=1.252f Rs=1.156 N=1 Xti=3 Eg=1.11 Cjo=205p M=.6004 Vj=.75 Fc=.5 Isr=1.867n Nr=2 Bv=3.9 Ibv=17.244m Nbv=2.4016 Ibvl=8.619m Nbvl=13.283 Tbv1=-384.62u Vpk=3.9 mfg=Motorola type=zener)
.model 1N749 D(Is=880.5E-18 Rs=.25 N=1 Xti=3 Eg=1.11 Cjo=190p M=.6124 Vj=.75 Fc=.5 Isr=1.743n Nr=2 Bv=4.3 Ibv=16.748m Nbv=1.7936 Ibvl=5.0382m Nbvl=12.554 Tbv1=-232.56u Vpk=4.3 mfg=Motorola type=zener)
.model 1N750 D(Is=.88f Rs=.25 Cjo=175p M=.55 nbv=1.7 bv=4.7 Vj=.75 Isr=1.86n Nr=2 Ibv=20.245m Ibvl=1.96m Nbvl=15 Tbv1=-21.3u Vpk=4.7 mfg=OnSemi type=Zener)
.model 1N751 D(Is=1.004f Rs=.5875 N=1 Xti=3 Eg=1.11 Cjo=160p M=.5484 Vj=.75 Fc=.5 Isr=1.8n Nr=2 Bv=5.1 Ibv=27.721m Nbv=1.1779 Ibvl=1.1646m Nbvl=21.894 Tbv1=176.47u Vpk=5.1 mfg=Motorola type=zener)
.model 1N752 D(Is=1.154f Rs=.9471 N=1 Xti=3 Eg=1.11 Cjo=150p M=.5788 Vj=.75 Fc=.5 Isr=1.625n Nr=2 Bv=5.6 Ibv=62.583m Nbv=.62382 Ibvl=631.96u Nbvl=50 Tbv1=267.86u Vpk=5.6 mfg=Motorola type=zener)
.model 1N754 D(Is=1.616f Rs=1.818 N=1 Xti=3 Eg=1.11 Cjo=120p M=.5117 Vj=.75 Fc=.5 Isr=1.698n Nr=2 Bv=6.8 Ibv=2.8814 Nbv=.28248 Ibvl=1.9426u Nbvl=.27168 Tbv1=485.29u Vpk=6.8 mfg=Motorola type=zener)
.model 1N755 D(Is=2.077f Rs=2.467 N=1 Xti=3 Eg=1.11 Cjo=104p M=.5061 Vj=.75 Fc=.5 Isr=1.645n Nr=2 Bv=7.5 Ibv=2.5701 Nbv=.39227 Ibvl=4.0222e-5 Nbvl=.25042 Tbv1=533.33u Vpk=7.5 mfg=Motorola type=zener)
.model 1N757 D(Is=2.453f Rs=2.9 N=1 Xti=3 Eg=1.11 Cjo=78p M=.4399 Vj=.75 Fc=.5 Isr=1.762n Nr=2 Bv=9.1 Ibv=.48516 Nbv=.7022 Ibvl=1m Nbvl=.13785 Tbv1=604.396u Vpk=9.1 mfg=Motorola type=zener)

Help==>
BV - Reverse breakdown voltage
nbv - Reverse breakdown emission coefficient
Ibv - Current at breakdown voltage
Ibvl - Low-level reverse breakdown knee current
nbvl - Low-level reverse breakdown emission coefficient
Tbv1 - Breakdown voltage temp coeff
Tbv2 - Quadratic breakdown voltage temp coeff