MultivibratorAn electronic circuit used to implement a variety of simple two-state systems such as oscillators, timers and flip-flops.
Note:• Astable
Not stable in either state, switch from one state to other state continuously
• MonostableWhich one of state s is stable , but other state is unstable
• BistableStable in either state, triggered by two external trigger
IC 555
Control Voltage (5)
Threshold Voltage (6)
Trigger Voltage (2)
Ground (1)
Vcc (8) Discharge (7)
Reset (4)
Output (3)
-+
-+
RESET
SET
Q
Q
COMP1
COMP2
Flip-Flop T1
555 Timer Design Equations
5
tHIGH : Calculations for the Oscillator’s HIGH Time
CRR0.693 BA HIGHt
5v
3.333 v
Vc1.666 v
0 v
tHIGH
Output
HIGH
LOW
THE OUTPUT IS HIGH WHILE THE CAPACITOR IS CHARGING THROUGH RA + RB.
555 Timer Design Equations
6 C0.693RBLOWt
tLOW : Calculations for the Oscillator’s LOW Time
5v
3.333 v
Vc1.666 v
0 v
tLOW
Output
HIGH
LOW
THE OUTPUT IS LOW WHILE THE CAPACITOR IS DISCHARGING THROUGH RB.
555 Timer – Period / Frequency / DC
7
C R2R 693.0T
CR 693.0C RR 693.0TttT
CR 693.0tC RR 693.0t
BA
BBA
LOWHIGH
BLOW
BAHIGH
Period:
C R2R 693.01F
T1F
BA
Frequency:
%100 R2R
RRDC
%100C R2R 693.0C RR 693.0DC
%100TtDC
BA
BA
BA
BA
HIGH
Duty Cycle:
Detail Analysis of a 555 Oscillator
8
5v
3.333 v
Vc1.666 v
0 v
RESET HIGHLOW
SET HIGHLOW
HIGHLOW
T1 ONOFF
Q HIGHLOW
Q
9
5v
3.333 v
Vc1.666 v
0 v
RESET HIGHLOW
SET HIGHLOW
HIGHLOW
T1 ONOFF
Q HIGHLOW
Q
Detail Analysis of a 555 Oscillator
10
5v
3.333 v
Vc1.666 v
0 v
RESET HIGHLOW
SET HIGHLOW
HIGHLOW
T1 ONOFF
Q HIGHLOW
Q
Detail Analysis of a 555 Oscillator
11
5v
3.333 v
Vc1.666 v
0 v
RESET HIGHLOW
SET HIGHLOW
HIGHLOW
T1 ONOFF
Q HIGHLOW
Q
OUTPUT IS LOW WHILE THE CAPACITOR IS DISCHARGING THROUGH RB.
OUTPUT IS HIGH WHILE THE CAPACITOR IS CHARGING THROUGH RA + RB.
Detail Analysis of a 555 Oscillator
Improved 555 Duty Cycle
By connecting this diode, D1 between the trigger input and the discharge input, the timing capacitor will now charge up directly through resistor R1 only, as resistor R2 is effectively shorted out by the diode. The capacitor discharges as normal through resistor, R2.
Now the previous charging time of t1 = 0.693(R1 + R2)C is modified to take account of this new charging circuit and is given as: 0.693(R1 x C). The duty cycle is therefore given as D = R1/(R1 + R2). Then to generate a duty cycle of less than 50%, resistor R1 needs to be less than resistor R2.
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