ASTABLE MULTIVIBRATOR USING BJTS- WORKING PRINCIPLES

EDITOR: B. SOMANATHAN NAIR

1.    INTRODUCTION

In our previous blog, we had given one method of explaining the operation of an astable multivibrator. We now discuss another method, which is more common in the scientific world. For easiness of understanding the working, we have developed two equivalent circuits of the astable multi.

2. PRINCIPLES OF WORKING – A SECOND THEORY

We now discuss principles of operation of the astable multi by considering the action of individual components in producing the oscillations. For this, we draw the circuit diagram in the form shown in Fig. 1. The circuit is assumed to be constructed by using identical components. For example, we use matched transistors, matched resistors, and matched capacitors. This means that transistor T₁≡ T₂, resistor R_C1≡ R_C2, capacitor C₁≡ C₂, and so on (the symbol “ ≡ ” means identical or matched). Under this condition, if we have designed the circuit to operate in the active region of the transistors, we find that the collector currents I_C1 ≡ I_C2, and the circuit will not produce any oscillation. However, in reality, the situation will not be like this.

We know that it is not possible to manufacture two components exactly identical to each other. At macro levels they may look identical; but at the micro or nano level, there will be some differences in the construction of the two devices because of the limitations in the manufacturing methods. This is especially true in the construction of active devices such as transistors.

            Based on the above argument, let us assume that the collector current I_C1 of transistor T₁ be slightly larger  than the collector current I_C2 of transistor T₂. In general, we have collector current

I_C = αI_E + I_CO    (1)

where α is the common-base current amplificatin factor, I_E is the emitter current, and I_CO is the collector-to-base reverse saturation current. Using this, we find the collector currents of T₁ and T₂ as

I_C1 = αI_E1 + I_CO1    (2)

I_C2 = αI_E2 + I_CO2    (3)

                                                                           

         If we use matched transistors, then α₁ = α₂, and I_E1= I_E2. However, I_CO1 ≠ I_CO2. This is because the minority-carrier generation depends on the structure of the two transistors and, as stated above, at atomic level there will be some minute structural differences in the construction of the two transistors. Further, when there is a slight change in the atmospheric temperature, corresponding changes in I_CO1 and I_CO2 will be slightly different. As a result of this mismatch in  I_CO1 and I_CO2, let I_C1 be conducting a little more than I_C2. This meas that V_RC1 (voltage drop across collector resistor R_C1 due to I_C1) will be slightly greater than V_RC2  (voltage drop across collector resistor R_C2 due to I_C2).

        Since V_RC1 = V_CC − V_CE1, V_RC2 = V_CC − V_CE2, and V_CC = a constant, we find that V_CE1 < V_CE2. Since the collector of T₁ is coupled to the base of T₂ through the coupling capcitor C₁, when V_CE1 desreases suddenly, V_BE2 also desreases suddenly, since a capcitor can not react to sudden changes in voltages.

        The sudden decrease in V_BE2 results in a corrensponding amplified sudden decrease in I_C2, since I_C2 = g_m2V_BE2, where g_m2 is the transconductance of T₂. It can be seen that this decrease in I_C2 is (numerically) g_m2 times larger than the increase in I_C1 mentioned above. And this produces a corresponding decrease in V_RC2, which in turn results in an increase in V_CE2.

         Since the collector of T₂ is coupled to the base of T₁ through the coupling capcitor C₂, when  V_CE2 increases suddenly, V_BE1 also results in an increase by the same amount as that in V_CE2. This increase in V_BE1 results in a further increase in I_C1 by a factor g_m1V_BE1, where g_m1 is the transconductance of T₁.

         This sudden (further) rise in I_C1 makes a further rise in V_RC1 and hence a further reduction in V_CE1. The process discussed above repeats in a cumulative fashion and finally we observe that T₁ goes to saturation and T₂ goes to cut-off. The equivalent circuit for this situation is shown in Fig. 2. as shown in the figure, the ON-transistor T_1­ has its base shorted to the collector and emitter (voltage drops neglected). However, in the OFF-transistor T₂, these terminals are open.

         From the figure, we notice that both the capacitors C₁ and C₂ (initially assumed to be uncharged) get charged through the ON-transistor T₁. It can be seen that C₁ gets charged through  R_B1 and T_1ON through its collector-emitter (c₁e₁) path to V_CC. During this same period, we find that C₂ also is getting charged to V_CC through R_C2 and T_1ON through its base-emitter (b₁e₁) path. The respective charging currents I_B1 is denoted by the red-colored thick line and I_C1 is denoted by the blue-colored thick line  in the figure. In Fig. 2, resistors R_C1 and R_B2 are shown in dotted lines to indicate that they are inactive at this moment.

We know that base resistor R_B1>> collector resistor R_C2 and hence I_B1 << I_C1. Hence the charging time-constant R_B1C₁>> R_C2C₂. Assuming C₁=C₂, typically, R_B1C₁ = 100 R_C2C₂. In that case, we notice that by the time C₂ has charged to V_CC, C₁ would have charged to only about V_CC/100. for example, assume that in a certain time period, C₂ got charged to V_CC = 10 V. then during the same time imterval, C₁ would have charged to only about 10/100 = 0.1 V. We also notice that when the capacitor voltage V_C2 and hence the base-emitter voltage V_BE2 (from the figure we find that V_BE2 = V_CE1) becomes equal to 0.5 V, the cut-in voltage of transistor T₂, it turns-on and  the collector current I_C2 starts flowing.

         The increse in I_C2 is followed by an increas in the drop I_C2R_C2, resulting in a decrease in V_CE2, which in turn produces a decrease in V_BE1 through the coupling of C₂. This decrease in V_BE1 results in a decrease in I_C1, which produces an increase in V_CE1 and hence in V_BE2. This increase in  V_BE2 results in a further increase in I_C2 and as discussed earlier, this is a cumulative process which ultimately turns-on T₂ and turns-off T₁. This situation is depicted in Fig. 3. In this state, we notice that C₁ remains charged to about 0.5 V and C₂ remains charged to V_CC.  

         Now, with T₂ ON (and T₁ OFF), as shown in Fig. 3, we find that C₁ discharges through T_2ON and R_C1. Simultaneously, C₂ discharges through T_2ON and R_B2. Once, C₁ and C₂ discharge fully, and they start recharging in the opposite directions, as shown in Fig. 3.

         The discharging and recharging paths are clearly indicated by using blue and red-colored thick line in Fig. 3. As shown, C₂ first discharges to zero and then recharges in the opposite direction towards V_CC. Similarly, C₁ first discharges to zero and then recharges in the opposite direction towards V_CC. These arguments clearly indicate that both the capacitors are discharging first to zero and then recharging towards V_CC at the same time. The main difference is that by the time one capacitor gets fully charged to V_CC, the other capacitor would have charged to only a small percentage of V_CC. From Fig. 3, we also notice that during this interval, resistors R_B1 and R_C2 are inoperative and are indicated by dotted lines. 

         To explain further, we notice that C₁ discharges much faster than C₂ because, in this condition, the time constant R_C1C₁<< R_B2C₂. As a result of this, C₁ gets discharged from 0.5 V to 0 V and recharged to +V_CC at a fast rate. However, during this period, C₂ gets discharged from V_CC (since it was fully charged to V_CC in the previous charging) to zero; but gets recharged to 0.5 V only in the opposite direction, as shown. Finally, when V_BE1 (=V_CE2) = 0.5 V, T₁ again turns-on, and by regeneration, T₂ again gets turned-off. This process of switching-on and switching-off of the two transistors alternately generates square-waves. The associated waveforms will be explained in the next blog.