# SPI-2 driver test conditions

Wed May 1 11:27:00 PDT 1996

```* From the SCSI Reflector, posted by:
*
Form: Memo
Text: (42 lines follow)
Kevin Gingerich wrote,

> (In(1.25 Zt/2 - (1+pL)AZo/2) + 1.25Vos)/((1+pL)AZo/2) <= Ia

I took a shot at deriving this and came up with something similar, except:

(1) Add N*IL to the left side for leakage current. Actually, it could be
argued that this should be 2*N*IL, since the worst-case leaking device could
be expected to switch from full leakage sinking to full leakage sourcing as
the line switches, or vice versa. For now I'll leave it at N*IL.

(2) Don't include the (1+pL) factor for the positive reflected wave. The
reason is: The device at the end of the bus does see the attenuated wave
boosted by the reflection, resulting in a combined factor of (1+pL)A, but a
device one or two feet from the end first sees only the attenuated wave,
followed by the reflection several ns later. For clean switching we should
rely on the incident waveform only.

(3) I figured in the source impedance, Rs. My full equation for the 4:1
ratio is

Ia >= N*IL + 1.25 * (In+It) * (Rs||(Rt/2)) / (A(Rs||(ZL/2))) - In

and of course a similar formula exists for In. Including Rs allows
calculations to be made for voltage mode.

(4) This formula covers the 4:1 ratio requirement, but not the minimum
signal voltage, which I'll call Vmin. To ensure V+ >= Vmin,

Ia >= N*IL + (Vmin + (It + In) * (Rs||(Rt/2)) / (A(Rs||(ZL/2))) - In

Since the equation now includes attenuation, we no longer need to budget the
additional 20 mV in Vmin, so we can let Vmin = 130 mV.

Note, in the above inequalities, It is the Norton equivalent current source
for the two terminators, equal to 2*Vb/Rt.

I'll follow up with coefficients derived from the above.

Richard Moore
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