NE555/SA555/SE555
LINEAR RAMP
When the pullup resistor, R
A
, in the monostable cir-
cuit is replaced by a constant current source, a linear
ramp is generated. Figure 17 shows a circuit con-
figuration that will perform this function.
Figure 17.
50% DUTY CYCLE OSCILLATOR
For a 50% duty cycle the resistors R
A
and R
E
may
beconnected as in figure19. The time preriod for the
output high is the same as previous,
t
1
= 0.693 R
A
C.
For the output low it is t
2
=
R
B
±
2R
A
[(R
A
R
B
) ⁄ (R
A
+
R
B
)
]
CLn
2R
B
±
R
A
Thus the frequency of oscillation is f
=
V
CC
R
E
4
Trigger
2
1
t
1
+
t
2
R1
Note that this circuit will not oscillate if R
B
is greater
Figure 19 :
50% Duty Cycle Oscillator.
8
7
NE555
Output
3
1
2N4250
or equiv.
6
C
5
V
CC
V
CC
0.01
µ
F
R2
4
555-21.EPS
R
A
51k
Ω
8
R
B
2
7
22k
Ω
NE55
Out
3
1
6
Figure 18 :
Linear Ramp.
than 1/2 R
A
because the junction of R
A
and R
B
can-
not bring pin 2 down to 1/3 V
CC
and trigger the lower
comparator.
ADDITIONAL INFORMATION
Adequate power supply bypassing is necessary to
protect associated circuitry. Minimum recom-
mended is 0.1µF in parallel with 1µF electrolytic.
V
CC
= 5V
Time = 20µs/DIV
R
1
= 47kΩ
R
2
= 100kΩ
R
E
= 2.7kΩ
C = 0.01µF
Top trace : input 3V/DIV
Middle trace : output 5V/DIV
Bottom trace : output 5V/DIV
Bottom trace : capacitor voltage
1V/DIV
8/10
555-23.EPS
555-22.EPS
Figure 18 shows waveforms generator by the linear
ramp.
The time interval is given by :
(
2
/3 V
CC
R
E
(R
1+
R
2)
C
T
=
V
BE
=
0.6V
R
1
V
CC
±
V
BE
(R
1+
R
2)
5
0.01
µ
F
C
0.01
µ
F
STMICROELECTRONICS [ STMICROELECTRONICS ]
