DESIGN OF ASTABLE MULTIVIBRATOR USING BJTs

EDITOR: B. SOMANATHAN NAR

           

I. DESIGN FOR SYMMETRICAL WAVEFORMS

1. SPECIFICATIONS

·         Free-running Frequency                :           5 kHz

·         Output-voltage Swing                    :            10 V

·         Rise time                                        :            10 ms

·         Duty cycle                                       :            50%

2. DESIGN PROCEDURE

The first step in our design of the astable multi begins with the design of a two-stage Standard Amplifier. We design the amplifiers such that their Q-points will be in the middle of their respective active regions. As the overall gain of the two stages is far higher than unity (approximately, more than 10⁴ for a two-stage amplifier using BC 107 transistors), the back-to-back connections convert the two-stage amplifier into an oscillator that will produce square waves. The design described here is contradictory to the usual design of the astable multi in the saturation and cut-off modes. The unique feature of the design in the active region is that it will ensure sure oscillation of the circuit as it is that of a standard positive-feedback oscillator design.

For active region operation, we must have voltage divider biasing resistors and emitter resistors, a common feature of the standard RC-coupled amplifier. However, since the gain is very large the transistors will be operating in their saturation and cut-off modes, which are stable states. Therefore, we can remove both the emitter resistors from the circuit, as it is used for temperature stability. Based on the same argument, we may also replace biasing resistors R₁ and R₂ with a single resistor R_B. Thus, the structure of the astable multi becomes highly simplified. The astable multi to be designed is in shown in Fig. 1.

3. DESIGN STEPS

Step 1: Fixing V_CC

From Fig. 1, we find

                                                            V_OP = V_CC − V_CESAT = V_CC      (1)

where V_CESAT = 0.2 V only and hence neglected. In our current design problem, we require V_OP = 10 V. Therefore we fix

            V_CC = 10 V                                                  

Step 2: Selection of Transistor

As in the design of the RC-coupled Standard Amplifier (for design of the standard amplifier, see the earlier blog on RC-coupled amplifier), transistors selected must be capable of handling 2I_CMAX and 2V_CEMAX where I_CMAX is the maximum collector current, and V_CEMAX is the maximum collector-to-emitter voltage. Select BC 107 which meets these specifications.

Step 3: Selection of I_C

Select any suitable value of I_C, say, between 1 mA and 10 mA, as this is a low-power design. In our problem, we choose

I_C = 1 mA

Step 4: Design of R_C

Since this a Standard-Amplifier design, for operation in the active region, from Fig. 1,

                                                            V_RC = V_CEQ = V_CC/2    (2)

where V_RC is the voltage across R_C and V_CEQ is the quiescent (steady-state) voltage across the collector and emitter of the transistor. Now, if I_C is the current through R_C, then we find

                                                           

                                                    R_C = V_RC/I_C = V_CC/2I_C    (3)

                                                                                                                                                      

Substituting values, we get

  

                                                    R_C = 10 V/2x1 mA = 5 kΩ

Choose the nearest standard resistor of 4.7 kΩ as R_C.

   

Step 5: Design of R_B

From Fig. 1, we observe that the voltage drop across R_B

                                   V_RB = V_CC −V_BEQ                                                          

where V_BEQ  is the base-emitter quiescent voltage (= 0.65 V). Also

I_BQ = I_C/β_min

And

 R_B = V_RB/I­_BQ = (V_CC −V_BEQ) β_min/I_C (5)

Substituting values, we get

        R_B = (10−0.65) V/1 mA = 1 MΩ

Step 6: Design of C

The specification demands 50% duty cycle. Duty cycle is defined as

                                                D = T_A/(T_A + T_B­) = T_A/T_o (6)

where T_A and T_B are the two half-cycles of oscillation and T_o = 1/f_o. The specification of 50% duty cycle indicates that the waveform is symmetrical. Hence, we use the expression for the free-running frequency of the multivibrator

                                                            f_o = 1/1.38R_BC  (7)

for symmetrical operation. Using Eq. (7), we get

C = 1/1.38R_Bf_o   (8)

                                                                                                                            

Substituting values yields

C = 1/1.38x10⁶x5x10³ = 0.15 nF

                                                          

Step 10: Design for the Desired Rise Time

The rise time is given by

                                                           

                                                            t_r = 1.6/f_oβ_min  (9)

From Eq. (9), we find that to get a shorter rise time, the transistor selected must have a large β. Substituting values, we get

t_r = 1.6/5x10³ x100 = 3.2 μ

This is less than the required value of 10 ms. Hence the selection of BC 107 is justified. The fully designed circuit is shown in Fig. 2.

II. DESIGN FOR UNSYMMETRICAL WAVEFORMS

1. INTRODUCTION

In Section I, we discussed the design of astable multivibrator that produces symmetrical square waveforms. In such waveforms, the duty cycle is 50%, i.e., the two half-cycles are of equal duration. Now we undertake the design of an astable multivibrator that produces unsymmetrical waveforms (i.e., waveforms having unequal half-cycles or different duty cycles). Equation (6) states that

                                                            D = T_A/(T_A + T_B­) = T_A/T_o

From which we find

 

                                                             T_A = DT_o = D/f_o                           (10)

                                                               

                                                T_B = T_o – T_A = T_o– DT_o = (1 – D)/f_o        (11)

We know that for a free-running multivibrator

                            T_A = 0.69R_BC₁  (12)

                             T_B = 0.69R_BC₂    (13)

where we keep R_B constant for biasing reasons and choose two different values for capacitors. From Eqs. (12) and (13), we find the values of the capacitors

C₁ = T_A/0.69R_B­ = D/f_ox0.69R_B   (14)

                                        C₂ = T_B/0.69R_B= (1–D)/f_ox0.69R_B  (15)

                         

Using these design equations, we can determine the values of the capacitors that produce the waveforms having desired duty cycle.

                                                                                 

2. SPECIFICATIONS

·         Free-running Frequency                :           5 kHz

·         Output-voltage Swing                    :            10 V

·         Duty cycle                                      :            25%

3. DESIGN PROCEDURE 

Step 1: Choice of C₁

Choosing R_B = 1 M, f_o = 5 kHz, and D = 0.25 and substituting in Eq. (14), we get

            

C₁ = D/f_ox0.69R_B = 0.25/0.69x10⁶x5x10³ = 72 pF

Step 2: Choice of C₂

Similarly, from Eq. (15), substituting values

C₂ = (1–D)/f_ox0.69R_B = (1–0.25)/0.69x10⁶x5x10³ = 216 pF

The fully designed circuit for unsymmetrical waveforms is shown in Fig. 3.

NOTE: In the design, we change only C’s, and do not touch R_B’s at all. This is because, if R_B’s are changed, the biasing will be affected and the circuit may not oscillate at all.