SPI-2 driver test conditions

Kevin Gingerich S=Kevin_Gingerich%S=Gingerich%G=Kevin%TI at mcimail.com
Tue Apr 30 14:47:00 PDT 1996

* From the SCSI Reflector, posted by:
* Kevin Gingerich <S=Kevin_Gingerich%S=Gingerich%G=Kevin%TI at mcimail.com>

  To  SCSI Reflector   X400
>From  Kevin Gingerich  GING
/pn=scsi/mbx1=scsi at symbios.com/ems=internet/admd=mci/c=us/
Subj  SPI-2 driver test conditions
">" my comments. /Kevin/
* From the SCSI Reflector, posted by:
* RMoore at corp.adaptec.com
Form: Memo
Text: (32 lines follow)
Following up my proposal at the Milpitas SPI-2 meeting:
I recalculated the test conditions under the new assumptions for terminator
characteristics, i.e., Rt ranges from 100 to 110 ohms and Vb from 100 to 125
I also added another limit as Bill Ham suggested to enforce asymmetric
driver strengths, by limiting In to 11 mA.
The new test conditions I am proposing for Section 10.1.1, Table 10, are as
follows. Replace the first condition:
        416 mV <= Va <= 706 mV
        249 mV + 0.3 * |Vn| <= Va <= 706 mV
> I calculate that this to be 170 mV + 0.47 * |Vn| <= Va <= 650 mV
>My derivation is rather lengthy so, I'll just give the result here as the
>constraints on the assertion current as a function of the negation current
>(In), cable impedance (Zo), the reflection coefficient at the load (pL),
>attenuation factor (A), termination impedance (Zt), and the bias voltage
>(Vos). The basis is to achieve at least 25% of the steady-state voltage on the
>first transition to the opposite state at any point on the bus.
>From steady-state negated to assertion;
> (In(1.25 Zt/2 - (1+pL)AZo/2) + 1.25Vos)/((1+pL)AZo/2) <= Ia
>From steady-state asserted to negated;
> Ia <= (In(1+pL)AZo/2 + 1.25Vos)/(1.25 Zt/2 - (1+pL)AZo/2)
>This can be converted to the form you give by inserting the test circuit
>values, rearranging terms, and multiplying through by 55 ohms to get voltages.
>(If you try it, watch your sign convention.)
Replace the second condition:
        381 mV <= |Vn| <= 977 mV
        255 mV + 0.3 * Va <= |Vn| <= 714 mV
> and this should be 170 mV + 0.47 * Va < |Vn| <= 700 mV.
>I got the new maximums from trial and error and the spreadsheet model. The
>constraint was to keep the differential voltages less than +/-1 V including
>reflected waves.
Remove the third condition:
        -277 mV < |Va| - |Vn| < 181 mV
It turns out that this condition is not necessary because restricting the
maximum In to 11 mA (or |Vn| <= 714 mV) keeps the voltage ratio at the
receiver within the 4:1 we are looking for.
I'd like to point out that my derivation assumes current mode drivers. Are
these test conditions valid for voltage mode? Unfortunately I haven't had a
chance to study that question.
>Either voltage- or current-mode drivers must deliver the first-step voltages
>required. The difference occurs in the polarity and magnitude of subsequent
>reflected waves. Within the defined bus parameters, the reflections should not
>cause problems.

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