LM555CMX Fairchild Semiconductor, LM555CMX Datasheet - Page 7

• Manufacturer Part Number: LM555CMX
• Description: IC TIMER SINGLE PREC 8-SOP
• Manufacturer: Fairchild Semiconductor
• Type: 555 Type, Timer/Oscillator (Single)r

Specifications of LM555CMX

• Voltage - Supply: 4.5 V ~ 16 V
• Current - Supply: 7.5mA
• Operating Temperature: 0°C ~ 70°C
• Package / Case: 8-SOIC (3.9mm Width)
• Frequency: 100kHz
• Number Of Internal Timers: 1
• Supply Voltage (max): 4.5 V to 16 V
• Supply Voltage (min): 4.5 V
• Maximum Power Dissipation: 600 mW
• Maximum Operating Temperature: 85 C
• Minimum Operating Temperature: - 40 C
• Mounting Style: SMD/SMT
• # Internal Timers: 1
• Power Dissipation: 600mW
• Operating Supply Voltage (min): 4.5V
• Operating Supply Voltage (typ): 5/9/12/15V
• Operating Supply Voltage (max): 16V
• Package Type: SOIC N
• High Level Output Current: -200mA
• Low Level Output Current: 50mA
• Pin Count: 8
• Operating Temp Range: 0C to 70C
• Operating Temperature Classification: Commercial
• Mounting: Surface Mount
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• Count: -
• Lead Free Status / Rohs Status: Lead free / RoHS Compliant
• Other names: LM555CMXFS

The equivalent circuit for discharging capacitor C1, when timer output is low is, as follows:
Since the duration of the timer output low state(
Since R
t
Consequently, if the timer operates in astable, the period is the same with
'T=t
time. And since frequency is the reciprocal of the period, the following applies.
3. Frequency divider
3. Frequency divider
3. Frequency divider
3. Frequency divider
By adjusting the length of the timing cycle, the basic circuit of Figure 1 can be made to operate as a frequency divider. Figure
8. illustrates a divide-by-three circuit that makes use of the fact that retriggering cannot occur during the timing cycle.
L
=0.693R
t
V
L
frequency,
H
V C1 t
t
C 1
C1
H
+t
=
C1
L
D
dv
--------------
=
C
=0.693(RA+R
t
dt
1
is normally R
C
C1
B
1
-- - V
3
=
1
R
C
=
V
1
CC
B
+
2
-- - V
3
R
2
-- - V
3
C1
---------------------- - V
R
+
A
(0-)=2Vcc/3
CC
CC
A
R
+
=
D
1
+
R
2
-- - V
3
R
e
f
=
R
B
B
-
In2
=
------------------------------------ -
B
B
CC
V
B
R A
In2
)C
CC
>>R
-- -
T
1
=
C1
+
e
1
=
R D
-
t
+0.693R
0.693 R
=
------------------------------------ -
1
D
R A
=
--------------------------------------- -
0.693 R
R
–
C1
0
although related to the size of discharging Tr.,
+
A
2
-- - e
3
t L
R D
+
1.44
-
B
2R
B
–
C1
------------------------------------
C
A
+
R A
B
6
1
7
R
+
=0.693(R
+
D
C
R
t H
R B
1
B
C
8
C1
1
C
t
L
1
R
) is the amount of time it takes for the V
D
A
11
+2R
9
5
B
4
)C
(10)
1
' because the period is the sum of the charge time and discharge
C1
(t) to reach Vcc/3,
LM555/NE555/SA555
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