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| 型號： | HEF4049BT |
| 廠商： | NXP Semiconductors N.V. |
| 元件分類： | 通用總線功能 |
| 英文描述： | Hex inverting buffers |
| 封裝： | HEF4049BP<SOT38-4 (DIP16)|<<http://www.nxp.com/packages/SOT38-4.html<1<Always Pb-free,;HEF4049BT<SOT109-1 (SO16)|<<http://www.nxp.com/packages/SOT109-1.html<1<week |
| 文件頁數(shù)： | 5/11頁 |
| 文件大?。?/td> | 334K |
| 代理商： | HEF4049BT |
HEF4049B_8
NXP B.V. 2009. All rights reserved.
Product data sheet
Rev. 08 — 2 December 2009
5 of 11
NXP Semiconductors
HEF4049B
Hex inverting buffers
11. Dynamic characteristics
[1]
The typical values of the propagation delay and transition times are calculated from the extrapolation formulas shown (C
L
in pF).
Table 7.
V
SS
= 0 V; C
L
= 50 pF; t
r
= t
f
≤
20 ns; T
amb
= 25
°
C; unless otherwise specified.
Symbol
Parameter
Conditions
t
PHL
HIGH to LOW
propagation delay
see
Figure 5
Dynamic characteristics
V
DD
5 V
10 V
15 V
5 V
10 V
15 V
5 V
10 V
15 V
5 V
10 V
15 V
Extrapolation formula
26 ns + (0.18 ns/pF)C
L
11 ns + (0.08 ns/pF)C
L
9 ns + (0.05 ns/pF)C
L
23 ns + (0.55 ns/pF)C
L
14 ns + (0.23 ns/pF)C
L
12 ns + (0.16 ns/pF)C
L
3 ns + (0.35 ns/pF)C
L
3 ns + (0.14 ns/pF)C
L
2 ns + (0.09 ns/pF)C
L
10 ns + (1.00 ns/pF)C
L
9 ns + (0.42 ns/pF)C
L
6 ns + (0.28 ns/pF)C
L
Min
-
-
-
-
-
-
-
-
-
-
-
-
Typ
35
15
12
50
25
20
20
10
7
60
30
20
Max
70
30
25
100
50
40
40
20
14
120
60
40
Unit
ns
ns
ns
ns
ns
ns
ns
ns
ns
ns
ns
ns
nA to nY;
[1]
t
PLH
LOW to HIGH
propagation delay
nA to nY;
see
Figure 5
[1]
t
THL
HIGH to LOW output
transition time
see
Figure 5
[1]
t
TLH
LOW to HIGH output
transition time
see
Figure 5
[1]
Table 8.
P
D
can be calculated from the formulas shown. V
SS
= 0 V; t
r
= t
f
≤
20 ns; T
amb
= 25
°
C.
Symbol
Parameter
V
DD
Typical formula for P
D
(
μ
W)
P
D
dynamic power
dissipation
10 V
P
D
= 11000
×
f
i
+
Σ
(f
o
×
C
L
)
×
V
DD2
15 V
P
D
= 35000
×
f
i
+
Σ
(f
o
×
C
L
)
×
V
DD2
Dynamic power dissipation P
D
where:
f
i
= input frequency in MHz;
f
o
= output frequency in MHz;
C
L
= output load capacitance in pF;
V
DD
= supply voltage in V;
Σ
(f
o
×
C
L
) = sum of the outputs.
5 V
P
D
= 2500
×
f
i
+
Σ
(f
o
×
C
L
)
×
V
DD2
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