How to Design IC 555 Astable Circuit

 

The post explains how to design a IC 555 astable multivibrator circuit through formulas and calculations.

Free-running multivibrator or Astable multivibrator is a circuit that generates rectangular wave. This circuit in comparison to others can run by itself i.e. it does not require any external trigger to run.

However, in order to design the astable circuit the first care that should be taken is to ensure 555 IC is in a working condition.

To develop an astable multivibrator does not involve a complex process as it can be developed by adding a capacitor and resistors into the Timer IC. The capacitor and the resistors connected to the IC helps to determine the output [High or Low]. The diagram in below illustrates the circuit.

 

 

The Pin1 is laid in ground state. Pin 4 and Pin 8 are first shorted and is then connected to generate +Vcc.

The output i.e. V_OUT is adapted from Pin 3. Pin 2 and Pin 6, which are also shorted is connected to the ground via Capacitor C.

Pin 7 on the other hand supplies +V_CC via R_A resistor. R_B resistor is connected in between of Pin 6 and Pin 7. Furthermore, the Pin 5, there are two options: applying modulation input or connect F with a bypass capacitor 0.01.

 

How it operates?

 

In order to understand the way Timer 555 works as free-running multivibrator, here lies a circuit diagram as given below

 

Referring to Fig 1.1, when V_OUT output is in high state or Q in low state, the transistor in discharged state is disconnected.

The capacitor C then charges towards V_CC via R_A and R_B resistance, thus keeping the charge time C as constant (R_A + R_B). However, this makes the voltage threshold go up to +2/3 V_CC.

The Comparator 1 generating hi-output and also enabling the flip-flop, in order to keep Q high and output low. With Q in high state, there become saturation of discharge transistor and Pin 7 gets grounded so as to enable Capacitor C can discharge via R_B resistance keeping the discharged time constant to R_BC.

As the capacitor gets discharged the voltage triggered at the inverting input of Comparator 2 gets discharged. As the input goes down to 1/3V_CC, the Comparator 2 output increases.

This result reset of the flip-flop in order to keep Q low and output of the timer high and it further proves that the output of auto-transition goes from low to high and vice versa, further continuing the repetition.

 

Designing a 555 IC astable multivibrator circuit (calculations)

 

It is vital to note that the charging time of Capacitor C from 1/3V_CC to 2/3V_CC is equal to the output time when high, which is equated as t_c or T_HIGH = 0.693 (R_A + R_B). The following section shows the inference:

At any point of the charging period of the capacitor’s voltage is signified as V_C= V_CC(1– e^t/RC)

The charging time of the capacitor from 0 – 1/3 V_CC is calculated as: 1/3 V_CC = V_CC (1-e ^t/RC )

The charging time of the capacitor from 0 – 2/3 V_CC is calculated as: t₂ = RC log_e3 = 1.0986 RC.

Therefore, the time consumed by the capacitor for charging [from +1/3 V_CC - +2/3V_CC) is:
t_c = (t₂ – t₁) = (10986 – 0.405) RC = 0.693 RC

Further upon substitution of R = (R_A + R_B) as in the above mentioned equation we get:

T_HIGH = t_c = 0.693 (R_A + R_B) C [NOTE: R_A & R_B in ohms, C in farads)

The discharging time of capacitors from +2/3 V_CC - +1/3 V_CC equates to when the output time is low and is shown as:

t_d or T_LOW = 0.693 R_B C [Note: Here, R_B in ohms, C in farads]

However, following is the process of the equation:

At any point during discharge state, the voltage maintained across the capacitor is signified as: V_C = 2/3 V_CC e – t_d/ R_BC OR t_d = 0.693 R_BC

Oscillation period [overall], T = T_HIGH + T_LOW = 0.693 (R_A + 2R_B) C. The oscillation frequency as the reciprocal of T is signified as: f = 1/T = 1.44/ (R_A + 2R_B) C

The aforementioned calculation signifies that the oscillation frequency remains independent of + V_CC [Supply voltage collector]

 

NOTE: Duty cycle term is often used together with astable multi-vibrator.

The time ratio t_c, duty cycle when output in high state in respect to T [Total time period] is stated as:

 

% duty cycle, D = t_c / T * 100 = (R_A + R_B) / (R_A + 2_RB) * 100

 

Based upon the aforementioned equation it has become evident that the output of the square wave [Duty cycle 50%] cannot be achieved until and unless when R_A = 0.

Furthermore, there is also possible issue to short R_A resistance to 0. Keeping R_A = 0 ohm, terminal 7 has its direct connection to +V_CC followed by repetition of the cycle.

When capacitor is discharged from transistor and R_B the transistor receives extra current from V_CC via a short in between of +V_CC and Pin 7. This situation may put risk damaging the timer and the transistor.

If there is a need to acquire symmetrical square wave it can be done if a diode can be connected across R_B resistor.  Capacitor C get its charge Diode D and R_A to +2/3 V_CC (approx.) further discharging via the transistor [Terminal 7] and R_B [Resistor] till the voltage of the capacitor goes down to 1/3 V_CC and further repetition of cycle. In order to avail output in square wave, R_A must be combined of a pot and a fixed resistor R. This will help to acquire the correct square wave.

While 555 is used extensively these days, it is rather better to use 7555 [a CMOS version of 555]. This actually saves from high current input and noise control.