[SI-LIST] Re: high-order coupling between inductors in hspice

Andy

In HSPICE manual, it is really printed. The following is the original text.

Best Regards

Zhangkun
2004.10.25

When using the mutual inductor element to calculate the coupling between more
than two inductors, Star-Hspice can automatically calculate an approximate
second-order coupling. See the third example below for a specific situation.

Examples
Inductors Lin and Lout are coupled with a coefficient of 0.9.
K1 Lin Lout 0.9
Inductors Lhigh and Llow are coupled with a coefficient equal to the value of 
the
parameter COUPLE.
Kxfmr Lhigh Llow K=COUPLE
The two mutual inductors K1 and K2 couple L1 and L2, and L2 and L3,
respectively. The coupling coefficients are 0.98 and 0.87. Star-Hspice
automatically calculates the mutual inductance between L1 and L3, with a
coefficient of 0.98*0.87=0.853.
K1 L1 L2 0.98
K2 L2 L3 0.87

Warning: The automatic inductance calculation is only an estimation and is
accurate for a subset of geometries. The second-order coupling
coefficient is simply the product of the two first-order coefficients,
which is not correct for many geometries.



----- Original Message ----- 
From: "Andrew Ingraham" <a.ingraham@xxxxxxxx>
To: "si-list" <si-list@xxxxxxxxxxxxx>
Sent: Monday, October 25, 2004 8:30 PM
Subject: [SI-LIST] Re: high-order coupling between inductors in hspice


> Zhangkun,
> 
> I don't know where K3=K1*K2 comes from, but I think it is just plain wrong.
> You could have positive coupling between L1 and L2 and between L2 and L3,
> but the coupling between L1 and L3 could be negative or zero.  You can't
> guess what it is from the other coupling coefficients.
> 
> Or consider the case where L1 and L3 are tightly coupled but L2 is far away.
> K1 and K2 are small, but K3 isn't.  K3!=K1*K2.
> 
> The same goes for any number of coupled inductors.  You need to specify
> every non-zero coefficient; you can't guess one based on the others.
> 
> Regards,
> Andy
> 
> 
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