LM555CN Fairchild Semiconductor, LM555CN Datasheet - Page 7
LM555CN
Manufacturer Part Number
LM555CN
Description
IC TIMER SINGLE 0-70DEG C 8-DIP
Manufacturer
Fairchild Semiconductor
Type
555 Type, Timer/Oscillator (Single)r
Datasheet
1.LM555CM.pdf
(14 pages)
Specifications of LM555CN
Voltage - Supply
4.5 V ~ 16 V
Current - Supply
7.5mA
Operating Temperature
0°C ~ 70°C
Package / Case
8-DIP (0.300", 7.62mm)
Frequency
100kHz
Number Of Internal Timers
1
Supply Voltage (max)
16 V
Supply Voltage (min)
4.5 V
Maximum Power Dissipation
600 mW
Maximum Operating Temperature
+ 70 C
Minimum Operating Temperature
0 C
Mounting Style
Through Hole
Lead Free Status / RoHS Status
Lead free / RoHS Compliant
Count
-
Lead Free Status / Rohs Status
Lead free / RoHS Compliant
Other names
LM555CNFS
Available stocks
Company
Part Number
Manufacturer
Quantity
Price
Company:
Part Number:
LM555CN
Manufacturer:
Fairchild Semiconductor
Quantity:
30 401
Part Number:
LM555CN
Manufacturer:
NS/国半
Quantity:
20 000
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
7 7 7 7