An astable timer operation " />
參數(shù)資料
型號(hào)： NE555N
廠商： Fairchild Semiconductor
文件頁(yè)數(shù)： 10/13頁(yè)
文件大?。?/td> 0K
描述： IC OSC MONO TIMING 8-DIP
標(biāo)準(zhǔn)包裝： 50
類(lèi)型： 555 型，計(jì)時(shí)器/振蕩器（單路）
電源電壓： 4.5 V ~ 16 V
電流 - 電源： 7.5mA
工作溫度： 0°C ~ 70°C
封裝/外殼： 8-DIP（0.300"，7.62mm）
包裝： 管件
供應(yīng)商設(shè)備封裝： 8-DIP
安裝類(lèi)型： 通孔
其它名稱(chēng)： NE555NFS
LM555
—
Single
T
imer
2002 Fairchild Semiconductor Corporation
www.fairchildsemi.com
LM555 Rev. 1.1.0
6
An astable timer operation is achieved by adding resistor RB to Figure 2 and configuring as shown on Figure 6. In the
astable operation, the trigger terminal and the threshold terminal are connected so that a self-trigger is formed, operat-
ing as a multi-vibrator. When the timer output is high, its internal discharging transistor. turns off and the VC1 increases
by exponential function with the time constant (RA+RB)*C.
When the VC1, or the threshold voltage, reaches 2 VCC/3; the comparator output on the trigger terminal becomes
high, resetting the F/F and causing the timer output to become low. This turns on the discharging transistor and the C1
discharges through the discharging channel formed by RB and the discharging transistor. When the VC1 falls below
VCC/3, the comparator output on the trigger terminal becomes high and the timer output becomes high again. The dis-
charging transistor turns off and the VC1 rises again.
In the above process, the section where the timer output is high is the time it takes for the VC1 to rise from VCC/3 to 2
VCC/3, and the section where the timer output is low is the time it takes for the VC1 to drop from 2 VCC/3 to VCC/3.
When timer output is high, the equivalent circuit for charging capacitor C1 is as follows:
Since the duration of the timer output high state (tL) is the amount of time it takes for the VC1(t) to reach 2 VCC/3,
Figure 8. Waveforms of Astable Operation
Vcc
R
A
R
B
C1
Vc1(0-)=Vcc/3
C
1
dv
c1
dt
-------------
V
cc
V0-
()
–
R
A
R
B
+
-------------------------------
=1
()
V
C1
0+
()
V
CC
3
=2
()
V
C1
t
()
V
CC
1
2
3
---e
-
t
R
A
R
B
+
()C1
-------------------------------------
–
–
=3
()
V
C1
t
()
2
3
---V
CC
V
=
CC
1
2
3
---e
-
t
H
R
A
R
B
+
()C1
-------------------------------------
–
–
=4
()
t
H
C
1
R
A
R
B
+
()In2 0.693 R
A
R
B
+
()C
1
=
=5
()
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