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Regardless
of what was the status of the above comments, The proposed response is OBE 86
for all of them including 444 and change the response for 86 as follows: 1.
We want to work in terms of Power. 2. We decided that Ipeak/Iport =
0.4/0.35. We need to
transform this ratio Ki=0.4/0.35 in terms of current ratio to a value in terms
of Power ratio Kp=Ppeak/Pclass. (See detailed
analysis in darshan_3_0108 pages 22 and 23 done for 29.5W/0.72A PD
showing how we got Ki and Kp. Now we have
to update the numbers for 25.5W/0.6A PD.) In this case
a Ki=0.4/0.35=1.142857 is transformed to Kp:
Ki=0.4/0.35=Ipeak/Iclass=
={((Vpse-(Vpse^2-4*Rc*Pclass)^0.5)/(2Rch)} /
{((Vpse-(Vpse^2-4*Rc*Pclass*Kp)^0.5)/(2Rch)}. Solving for Kp, we
get Kp=1.114 in order to keep Ki=0.4/0.35. ------- PROPOSED
ACCEPT IN PRINCIPLE. 1. Table 33-17, p66. item
7, first row: Change to:
Peak operating power, Class 0, 3, and Type 1 Class 4 = 14.4 W 2. Table 33-17, p66. item
7, forth row: Change to:
Peak operating power, Type 2 Class 4 = 1.114 x Pclass 3. Add the following text to
33.3.7.4: Pclass is the
average power demand of the PD. Pclass = Vpd x Iport. The PD is
approximated as constant power load i.e. when the voltage increases Iport
decreases to maintain about constant Pclass. The PD load
behavior may be different then constant power load (e.g. resistor or linear
regulator) as long as the power demand from
the PD is not acceding Pclass for any Vport value. 4. Figure 33-18: 4.1 Change
(400/350)xPport_max/Vport to Ppd_peak/Voverload 4.2 Change Pport_max/Vport
to Pclass/Vport 4.3 Change Table 33-17
Ppeak to Ppd_peak (to match with Eq 33-1) 4.4 Add below Figure 33-18
the following text:
Ppd_peak is defined in Table 33-17 item 7. Voverload is calculated according to
Eq 33-1. Please
review and comment so we can be done with it tomorrow morning. Thanks Yair
Analog Mixed Signal Group Microsemi Corporation
Cell: +972-54-4893019 E-mail: <mailto:ydarshan@xxxxxxxxxxxxx>. |