參數(shù)資料
型號: M25PE40
廠商: 意法半導(dǎo)體
元件分類: DRAM
英文描述: 4 Mbit Uniform Sector, Serial Flash Memory
中文描述: 4兆位統(tǒng)一部門,串行閃存
文件頁數(shù): 50/60頁
文件大?。?/td> 315K
代理商: M25PE40
DC and AC parameters
M25PE40
50/60
AC characteristics (50 MHz operation,
T9HX (0.11μm) process
(1)
)
(2)
(3)
Table 20.
Test conditions specified in
Table 14
and
Table 15
Symbol
Alt.
Parameter
Min.
Typ.
Max.
Unit
f
C
f
C
Clock Frequency for the following instructions:
FAST_READ, RDLR, PW, PP, WRLR, PE, SE,
SSE, DP, RDP, WREN, WRDI, RDSR, WRSR
D.C.
50
MHz
f
R
Clock Frequency for READ instructions
D.C.
33
MHz
t
CH(4)
t
CL(4)
t
CLH
t
CLL
Clock High Time
9
ns
Clock Low Time
Clock Slew Rate
2
(peak to peak)
S Active Setup Time (relative to C)
9
ns
0.1
V/ns
t
SLCH
t
CHSL
t
DVCH
t
CHDX
t
CHSH
t
SHCH
t
SHSL
t
SHQZ(5)
t
CLQV
t
CLQX
t
WHSL(6)
t
SHWL(6)
t
DP(5)
t
RDP(5)
t
W
t
PW(7)
t
CSS
5
ns
S Not Active Hold Time (relative to C)
5
ns
t
DSU
t
DH
Data In Setup Time
2
ns
Data In Hold Time
5
ns
S Active Hold Time (relative to C)
5
ns
S Not Active Setup Time (relative to C)
5
ns
t
CSH
t
DIS
t
V
t
HO
S Deselect Time
100
ns
Output Disable Time
8
ns
Clock Low to Output Valid
8
ns
Output Hold Time
0
ns
Write Protect Setup Time
50
ns
Write Protect Hold Time
100
ns
S to Deep Power-down
3
μs
S High to Standby Mode
30
μs
Write Status Register Cycle Time
3
15
ms
Page Write Cycle Time (256 bytes)
11
23
ms
t
PP(7)
Page Program Cycle Time (256 bytes)
0.8
3
ms
Page Program Cycle Time (n bytes)
int(n/8) × 0.025
(8)
t
PE
t
SE
t
SSE
t
BE
Page Erase Cycle Time
10
20
ms
Sector Erase Cycle Time
1
5
s
SubSector Erase Cycle Time
40
150
ms
Bulk Erase Cycle Time
5
10
s
1.
See
Important note on page 6
.
2.
Preliminary data.
3.
Details of how to find the Technology Process in the marking are given in AN1995, see also
Section 13: Part numbering
.
4.
t
CH
+ t
CL
must be greater than or equal to 1/ f
C
Value guaranteed by characterization, not 100% tested in production.
5.
6.
Only applicable as a constraint for a WRSR instruction when SRWD is set at 1.
7.
When using PP and PW instructions to update consecutive bytes, optimized timings are obtained with one sequence
including all the bytes versus several sequences of only a few bytes (1
n
256).
int(A) corresponds to the upper integer part of A. E.g. int(12/8) = 2, int(32/8) = 4 int(15.3) =16.
8.
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