semi合集-English.pdf - 第4153页

SEMI F64-0701 © SEMI 2001 15 V eq is the equivalent internal control vol ume of the DUT. Note that it may not represent the actual internal vol ume of the DUT; i.e., ) min / sec 60 atm i/std. s p 14.7 ( x P Q = V 1 eq ∆ …

100%1 / 7923
SEMI F64-0701 © SEMI 2001 14
For example (see Table A1–1), if Q
N
= 50.01 sccm, Q
max
= 55.23 sccm, and Q
min
= 46.28 sccm, then
δ
Q
+
= 55.23 sccm - 50.01 sccm = 5.22 sccm
δ
Q
-
= 46.28 sccm - 50.01 sccm = -3.73 sccm
A1–1.1.2 t
s
is settling time to Q
N
; i.e.,
t
s
= elapsed time from initiation of pressure transient (t
o
) to when Q
A
is within 0.5% of
reading of Q
N
(t
f
) = t
f
- t
o
δ
M is deviation of mass of material delivered during the disturbance relative to steady state mass delivery; i.e.,
min/sec 60
)
t
-
t
()
Q
-
Q
(
= M
iih
NA
t
t
=
t
f
oi
δ
(2)
This quantity may be thought of as the net area between the Q
A
versus time curve and the Q
N
versus time line. See
Figure A1–1 below.
Figure A1–1
Q
N
versus t (sec)
NOTE 1:
δ
M is expressed in units of sccm.
A1–2 Data Analysis: Method A-2
Refer again to Table A1–1
A1–2.1 Data Interpretation, Method A-2
Q
R
is the steady state actual flow while the inlet pressure is being ramped. It is calculated by averaging Q
A
values from time t
o
+ 50 sec to t
o
+ 110 sec.
∆Q is the steady state deviation of actual flow during the inlet pressure ramp from that while inlet pressure
is constant; i.e.,
∆Q = Q
R
- Q
N
(3)
SEMI F64-0701 © SEMI 200115
V
eq
is the equivalent internal control volume of the DUT. Note that it may not represent the actual internal volume
of the DUT; i.e.,
)
min/sec 60
atm i/std. s p 14.7
(x
P
Q
=
V
1
eq
∆
(4)
For example, if ∆Q = 5.02 sccm and P1 = 0.7 kPa (0.1 psi)/sec, then
Veq = [(5.02 sccm)/( 0.7 kPa (0.1 psi)/sec)] × [(101.325 kPa (14.7 psi)/std. atm)/(60 sec/min)] = 12.3 ccm
A1–3 Data Analysis: Method B
Table A1–2 Sample Data Sheet: Method B
Date:________________ Time: __________
DUT #________________ Ambient Temp __________°C
Standard Flow Device ________________
Full Scale________________ sccm
P1 (psig) Setpoint Flow (sccm) Q
A
(sccm) Q
I
(sccm)
15.02 0 0.00 -0.01
15.00 50 50.01 50.01
15.01 100 100.12 100.01
50.06 0 0.00 -0.05
50.04 50 50.12 50.01
50.03 100 100.56 100.02
A1–3.1 Data Interpretation, Method B
PC
o
is the pressure coefficient of indicated flow per pressure change at zero flow; i.e.,
100
Q
P
Q
= (%)
PC
sp
1
A
A
x
∆
∆
(5)
x100
Q
)
P
-
P
(
Q
-
Q
= (%)
PC
FS
IiIf
IiIf
o
(6)
PC
A
is the pressure coefficient of actual flow per pressure change at a setpoint; i.e.,
For example, from Table A2.1,
PC
o
(%) = {[-0.05 sccm-(-0.01 sccm)]/[(50.06 psig-15.02 psig)(100 sccm)]} × 100
= -0.0011% of FS/psi
PC
A
(@ 50%) = [50.12 sccm-50.01 sccm]/[(50.04 psig-15.00 psig)(50 sccm)] × 100
= 0.0063% of reading/psi at 50 sccm
PC
A
(@ FS = [100.56 sccm-100.12 sccm]/[(50.03 psig-15.01 psig)(100 sccm)] × 100
= 0.013% of reading/psi at 100 sccm
SEMI F64-0701 © SEMI 2001 16
A–4 Data Analysis: Method C
Table A4–1 Sample Data Sheet: Method C
Date:________________ Time: __________
DUT #________________ Ambient Temp __________°C
Standard Flow Device ________________
Setpoint________________ sccm
Te = ________________ °C P1 = __________psig
Time (sec) P2 (kPa) Q
A
(sccm) Q
I
(sccm)
0 101.338 50.05 50.0
0.05 101.338 50.05 50.0
.
.
.
.
.
.
.
.
.
.
.
.
2.20 95.086 52.28 51.5
2.25 88.406 54.56 52.1
.
.
.
.
.
.
.
.
.
.
.
.
8.45 32.424 50.08 50.0
8.50 31.811 50.07 50.0
.
.
.
.
.
.
.
.
.
.
.
.
15.20 26.838 50.06 50.0
A1–4.1 Data Interpretation, Method C – See Appendix A1–1.1
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