职相同性质工作做了六七年,
内心已经厌倦且疲累不堪.
职已经不再是能充满活力与热诚,
也不再是能自动自发, 主动积极,
工作与错误皆需主管提醒与注意.
什么是快闪存储器？
快闪存储器(Flash Memory)，是非挥发性内存的一种，不需电力来维持数据的储存。
又可分为NOR Flash以及NAND Flash两种，前者用于储存程式码，后者用于储存数据资料。
快闪存储器用一个浮动闸(Floating Gate)电晶体来储存数据，
利用每个电晶体所能储存的资料数量来区分，
可以将NAND Flash区分为两类：单级储存(Sigle Level Cell，SLC)和多级储存
(Multi Level Cell，MLC)。
单级储存具有速度快，耗电量低的优点，但是多级储存的成本比较低。
先生尊鉴
面试时承蒙先生费心细析xx制度与xx作法,
受益匪浅, 不胜感激.
然世事多变化, 不能前往任职, 至以为歉;
烦劳先生无谓等候, 惭愧无已.
瞻望来日, 如有需为先生效力之时, 必不敢辞.
祈请谅恕, 是所至盼.
肃此奉陈 不尽下怀
The statement about the thermal dissipated of the major components in R12Y
1. CPU thermal dissipated is 11.36 W
In the calculation?we assume that the maximum value of the VID is
1.5V?the consumption of the CPU total 56.805W?and we assume
that 20 percent of the consumption transform to thermal?
so the thermal dissipated is 56.805 * 0.2 = 11.36 W
2. North Bridge thermal dissipated is 3.7436 W
In the calculation?the consumption of the North Bridge total 18.718W?
and we assume that 20 percent of the consumption transform to thermal?
so the thermal dissipated is 18.718 * 0.2 = 3.7436 W
3. South Bridge thermal dissipated is 1.02 W
In the calculation?the consumption of the South Bridge total 5.0933W?
and we assume that 20 percent of the consumption transform to thermal?
so the thermal dissipated is 5.0933 * 0.2 = 1.02 W
4. ATi Graphic Processor thermal dissipated is 1.95 W
In the calculation?we assume that the maximum value of the CORE voltage
is 1.15V?the consumption of the ATi Graphic Processor total 9.76W?
and we assume that 20 percent of the consumption transform to thermal?
so the thermal dissipated is 9.76 * 0.2 = 1.95 W
5. SODIMM DDR Memory thermal dissipated is 4.032 W
In the calculation?the consumption of the SODIMM DDR Memory total 20.16W?
and we assume that 20 percent of the consumption transform to thermal?
so the thermal dissipated is 20.16 * 0.2 = 4.032 W
.
6. Clock Generator thermal dissipated is 0.264 W
In the calculation?the consumption of the Clock Generator total 1.32W?
and we assume that 20 percent of the consumption transform to thermal?
so the thermal dissipated is 1.32 * 0.2 = 0.264 W
7. Clock Buffer thermal dissipated can be omitted
In the calculation?the consumption of the Clock Buffer total 0.66W?
and we assume that 20 percent of the consumption transform to thermal?
so the thermal dissipated is 0.66 * 0.2 = 0.132 W?and we can omit this value.
8. CardBus Controller thermal dissipated can be omitted
In the calculation?the consumption of the CardBus Controller total 0.099W?
and we assume that 20 percent of the consumption transform to thermal?
so the thermal dissipated is 0.099 * 0.2 = 0.0198 W?and we can omit this value.
9. LAN Controller thermal dissipated can be omitted
In the calculation?the consumption of the LAN Controller total 1.0694W?
and we assume that 20 percent of the consumption transform to thermal?
so the thermal dissipated is 1.0694 * 0.2 = 0.21388 W?
and we can omit this value.
10. Audio CODEC thermal dissipated can be omitted
In the calculation?the consumption of the Audio CODEC total 0.3W?
and we assume that 20 percent of the consumption transform to thermal?
so the thermal dissipated is 0.3 * 0.2 = 0.06 W?and we can omit this value.
11. Super I/O thermal dissipated can be omitted
In the calculation?the consumption of the Super I/O total 0.033W?
and we assume that 20 percent of the consumption transform to thermal?
so the thermal dissipated is 0.033 * 0.2 = 0.0066 W?
and we can omit this value.
According to numeric above and analysis?
We think that the thermal dissipated of the other chips can be omitted too.
12. Voltage Regulator VCORE thermal dissipated is 6 W
In the calculation?we assume that the maximum value of the VID is
1.5V?power supply is 20V adapter?
and the efficiency of the regulator is 90 % ?
that is?10 percent of the power supply transform to thermal?
so the thermal dissipated is [1.5 * 36 / ( 20 * 0.9 )] * 20 * 0.1 = 6 W.
13. Voltage Regulator 3.3STB / 5STB thermal dissipated is 6.9 W
In the calculation?we assume that power supply is 20V adapter?
and the efficiency of the regulator is 90 % ? that is?
10 percent of the power supply transform to thermal?
so the thermal dissipated is
[3.3 * 7.5 / ( 20 * 0.9 )] * 20 * 0.1 + [5 * 7.5 / ( 20 * 0.9 )] * 20 * 0.1 =
6.9 W.
14. Voltage Regulator 1.05VS / 1.5VS thermal dissipated is 2.7 W
In the calculation?we assume that power supply is 20V adapter?
and the efficiency of the regulator is 90 % ? that is?
10 percent of the power supply transform to thermal?
so the thermal dissipated is
[1.05 * 11 / ( 20 * 0.9 )] * 20 * 0.1 + [1.5 * 8.5 / ( 20 * 0.9 )] * 20 * 0.1 =
2.7 W.
15. Voltage Regulator AGP_CORE_SW / AGP_1.8VS
thermal dissipated is 1.39 W
In the calculation?we assume that the maximum value of the AGP_CORE_SW is
1.15V?power supply is 20V adapter?and the efficiency of the regulator is
90 % ? that is?10 percent of the power supply transform to thermal?
so the thermal dissipated is
[1.15 * 6 / ( 20 * 0.9 )] * 20 * 0.1 + [1.8 * 3.1 / ( 20 * 0.9 )] * 20 * 0.1
= 1.39 W.
16. Voltage Regulator 1.8V thermal dissipated is 2.3 W
In the calculation?we assume that power supply is 20V adapter?
and the efficiency of the regulator is 90 % ?
that is?10 percent of the power supply transform to thermal?
so the thermal dissipated is [1.8 * 11.5 / ( 20 * 0.9 )] * 20 * 0.1 = 2.3 W.
17. Charger thermal dissipated is 3.5 W
In the calculation?we assume that power supply is 20V adapter?
and the efficiency of the regulator is 90 % ?
that is?10 percent of the power supply transform to thermal?
so the thermal dissipated is [2.5 * 12.6 / ( 20 * 0.9 )] * 20 * 0.1 = 3.5 W.
The explanation for estimation of discharge time using BatteryMark
In the calculation?we assume that the battery of the various
manufacturers have the same chemistry?
the discharge rate of the different battery is equal?
and the discharge rate is uniform for the duration of the discharge
I
The power consumption of the major Component in T8D
Component Description Watt
CPU Dothan 5
NB Alviso 6
SB ICH6M 2.3
Memory 1024MB 5
CPU VR 87% efficiency 0.74
System VR 90% efficiency 3.8
The power consumption of the major Component in T8D is 22.84W
according to the table above
The battery pack in T8D is 3S2P and each cell of the battery pack
is 1800mAh?the capacity when discharge from full to empty is 3073mAh
by manufacturer specification?
and the battery terminal voltage is 12.6V?
so the energy release from battery pack when discharge
from full to empty is 3073mAh * 12.6V = 38720mAh-V
The actual discharge time is 3 hours 30 minutes
We can calculate to obtain its consumption is
8.072721208mAh-V per minute-watt
(38720 / (210 * 22.84) = 8.072721208)
II
The power consumption of the major Component in GoBook IV
(The power consumption of CPU is 31W)
Component Description TDP(Watt)
CPU Yonah 31
NB Calistoga 8.0 - 8.5
SB ICH7M 2
Memory 1024MB 5
CPU VR 87% efficiency 4.6
System VR 90% efficiency 3.8
The power consumption of the major Component
in GoBooK IV is 54.9W according to the table above
i
The battery pack in GoBook IV is 3S2P and each cell of
the battery pack is 2200mAh?the capacity
when discharge from full to empty is 3777mAh
by manufacturer specification?and the battery terminal voltage is 12.6V?
so the energy release from battery pack
when discharge from full to empty is 3777mAh * 12.6V = 47590mAh-V
We suppose that the discharge time using BatteryMark is X minutes
47590 / (X * 54.9) = 8.072721208
X = 107.38
The estimation of discharge time using BatteryMark is 1 hour 48 minutes
ii
The battery pack in GoBook IV is 3S3P
and each cell of the battery pack is 2200mAh?
the capacity when discharge from full to empty is 5665mAh
by manufacturer specification?and the battery terminal voltage is 12.6V?
so the energy release from battery pack
when discharge from full to empty is 5665mAh * 12.6V = 71379mAh-V
We suppose that the discharge time using BatteryMark is X minutes
71379 / (X * 54.9) = 8.072721208
X = 161.056
The estimation of discharge time using BatteryMark is 2 hour 41 minutes
iii
The battery pack in GoBook IV is 3S3P and each cell of the battery pack
is 2400mAh?the capacity when discharge from full
to empty is 6193mAh by manufacturer specification?
and the battery terminal voltage is 12.6V?
so the energy release from battery pack when discharge from full to empty
is 6193mAh * 12.6V = 78032mAh-V
We suppose that the discharge time using BatteryMark is X minutes
78032 / (X * 54.9) = 8.072721208
X = 176.068
The estimation of discharge time using BatteryMark is 2 hour 56 minutes
III
The power consumption of the major Component in GoBook IV
(The power consumption of CPU is 15W)
Component Description TDP(Watt)
CPU Yonah 15
NB Calistoga 8.0 - 8.5
SB ICH7M 2
Memory 1024MB 5
CPU VR 87% efficiency 2.23
System VR 90% efficiency 3.8
The power consumption of the major Component in GoBooK IV is 36.53W
according to the table above
i
The battery pack in GoBook IV is 3S2P and each cell of the battery pack
is 2200mAh?the capacity when discharge from full to empty is 3777mAh
by manufacturer specification?and the battery terminal voltage is 12.6V?
so the energy release from battery pack
when discharge from full to empty is 3777mAh * 12.6V = 47590mAh-V
We suppose that the discharge time using BatteryMark is X minutes
47590 / (X * 36.53) = 8.072721208
X = 161.38
The estimation of discharge time using BatteryMark is 2 hour 2 minutes
ii
The battery pack in GoBook IV is 3S3P and each cell of the battery pack is
2200mAh?the capacity when discharge from full to empty is 5665mAh
by manufacturer specification?and the battery terminal voltage is 12.6V?
so the energy release from battery pack when discharge from full to empty
is 5665mAh * 12.6V = 71379mAh-V
We suppose that the discharge time using BatteryMark is X minutes
71379 / (X * 36.53) = 8.072721208
X = 242.05
The estimation of discharge time using BatteryMark is 4 hour 2 minutes
iii
The battery pack in GoBook IV is 3S3P and each cell of the battery pack is
2400mAh?the capacity when discharge from full to empty is 6193mAh
by manufacturer specification?and the battery terminal voltage is 12.6V?
so the energy release from battery pack when discharge from full to empty
is 6193mAh * 12.6V = 78032mAh-V
We suppose that the discharge time using BatteryMark is X minutes
78032 / (X * 36.53) = 8.072721208
X = 264.61
The estimation of discharge time using BatteryMark is 4 hour 25 minutes
IV
The power consumption of the major Component in GoBook IV
(The power consumption of CPU is 5W)
Component Description TDP(Watt)
CPU Yonah 5
NB Calistoga 8.0 - 8.5
SB ICH7M 2
Memory 1024MB 5
CPU VR 87% efficiency 0.74
System VR 90% efficiency 3.8
The power consumption of the major Component in GoBooK IV is 25.04W
according to the table above
i
The battery pack in GoBook IV is 3S2P and each cell of the battery pack
is 1800mAh?the capacity when discharge from full to empty is 3073mAh
by manufacturer specification?and the battery terminal voltage is 12.6V?
so the energy release from battery pack when discharge from full to empty
is 3073mAh * 12.6V = 38720mAh-V
We suppose that the discharge time using BatteryMark is X minutes
38720 / (X * 25.04) = 8.072721208
X = 191.55
The estimation of discharge time using BatteryMark is 3 hour 12 minutes
ii
The battery pack in GoBook IV is 3S2P and each cell of the battery pack is
2200mAh?the capacity when discharge from full to empty is 3777mAh
by manufacturer specification?and the battery terminal voltage is 12.6V?
so the energy release from battery pack when discharge from full to empty is
3777mAh * 12.6V = 47590mAh-V
We suppose that the discharge time using BatteryMark is X minutes
47590 / (X * 25.04) = 8.072721208
X = 235.43
The estimation of discharge time using BatteryMark is 3 hour 56 minutes
iii
The battery pack in GoBook IV is 3S3P and each cell of the battery pack is
2200mAh?the capacity when discharge from full to empty is 5665mAh
by manufacturer specification?and the battery terminal voltage is 12.6V?
so the energy release from battery pack when discharge from full to empty is
5665mAh * 12.6V = 71379mAh-V
We suppose that the discharge time using BatteryMark is X minutes
71379 / (X * 25.04) = 8.072721208
X = 353.12
The estimation of discharge time using BatteryMark is 5 hour 53 minutes
iv
The battery pack in GoBook IV is 3S3P and each cell of the battery pack is
2400mAh?the capacity when discharge from full to empty is 6193mAh
by manufacturer specification?and the battery terminal voltage is 12.6V?
so the energy release from battery pack when discharge from full to empty
is 6193mAh * 12.6V = 78032mAh-V
We suppose that the discharge time using BatteryMark is X minutes
78032 / (X * 25.04) = 8.072721208
X = 386.03
The estimation of discharge time using BatteryMark is 6 hour 26 minutes