LVD Release Glitches (Gingerich)

RMoore at corp.adaptec.com RMoore at corp.adaptec.com
Fri Jan 12 05:44:00 PST 1996


* From the SCSI Reflector, posted by:
* <RMoore at corp.adaptec.com>
Form: Memo
Text: (41 lines follow)
Regarding Kevin's post on the release glitch problem, the second equation:

> 2) Vp = 0.5 Ia Zt - Vbias - 0.5 (|Ia| + |In|) Zl, where Ia is the 
assertion
> current.
> 
> Solving for In, substituting the values above, using the minimum loaded 
media
> impedance of 85 ohms (Zl), the maximum Zt, and solving gives,
> 
> In = 2/Zl ((Ia/2(Zl-Zt)) - Vp + Vbias)
> In > 2/85 (Ia/2(85-115)) + 0.03 + 0.058)
> In > -0.353 Ia + 0.002.
> 
> Assume that a maximum assertion current of 10 mA is allowed, then the
> constraints become 1.5 < |In| < 1.6 ma.

I believe there is a sign error here. When Vp and Vbias appear on the same 
side of the equation, they would have to have the same sign. I get the 
following instead:

Vp = 0.5 Ia Zt - Vbias - 0.5 (|Ia| + |In|) Zl
|Ia| + |In| = - (Vp + Vbias - 0.5 Ia Zt) / (0.5 Zl)
In = 2/Zl (Ia/2 (Zt) - Vp - Vbias) - Ia
In = 2/Zl ((Ia/2(Zt-Zl)) - Vp - Vbias)
In > 2/85 (Ia/2(115-85)) + 0.03 - 0.058)
In > 0.353 Ia - 0.000659 

Again using 10 mA for Ia, we get about 2.9 mA for the lower bound of In.

NOTE: I hope I am correct in interpreting In and Ia to be defined such that 
they are positive quantities, so that the absolute value symbols in Eq. 2 
can be ignored.

I agree with Kevin, that the glitch will go away if Z0 and Zt are matched 
closely enough. But lowering Z0 means lowering Zl, so that the minimum In 
derived from equation 2 will get bigger. For example, if Zl is 50 ohms, then 
the minimum In would be nearly 12 mA, if Ia stays the same.

	Richard Moore
	Adaptec Irvine Technology Center
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