Topic Covered

`color{blue}{star}` JUNCTION TRANSISTOR
`color{blue}{star}` LOGIC GATES
`color{blue}{star}` INTEGRATED CIRCUITS


`color{blue} ✍️`The credit of inventing the transistor in the year 1947 goes to J. Bardeen and W.H. Brattain of Bell Telephone Laboratories, U.S.A. That transistor was a point-contact transistor.
The first junction transistor consisting of two back-to-back p-n junctions was invented by William Schockley in 1951.

`color{blue} ✍️`As long as only the junction transistor was known, it was known simply as transistor. But over the years new types of transistors were invented and to differentiate it from the new ones it is now called the `"Bipolar Junction Transistor (BJT)."`

`color{blue} ✍️` Even now, often the word transistor is used to mean BJT when there is no confusion. Since our study is limited to only BJT, we shall use the word transistor for BJT without any ambiguity.

`color{brown}bbul("Transistor: structure and action")`
`color{blue} ✍️`A transistor has three doped regions forming two p-n junctions between them. Obviously, there are two types of transistors, as shown in Fig. 14.27.

`color{purple}bb("(i) n-p-n transistor :")` Here two segments of n-type semiconductor (emitter and collector) are separated by a segment of p-type semiconductor (base).

`color{purple}bb("(ii) p-n-p transistor:")` Here two segments of p-type semiconductor (termed as emitter and collector) are separated by a segment of n-type semiconductor (termed as base).

`color{blue} ✍️`The schematic representations of an `n-p-n` and a `p-n-p` configuration are shown in Fig. 14.27(a). All the three segments of a transistor have different thickness and their doping levels are also different. In the schematic symbols used for representing p-n-p and n-p-n transistors [Fig. 14.27(b)] the arrowhead shows the direction of conventional current in the transistor. A brief description of the three segments of a transistor is given below:

`color{blue} {•}` Emitter: This is the segment on one side of the transistor shown in Fig. 14.27(a). It is of moderate size and heavily doped. It supplies a large number of majority carriers for the current flow through the transistor.

`color{blue} {•}` Base: This is the central segment. It is very thin and lightly doped.

`color{blue} {•}` Collector: This segment collects a major portion of the majority carriers supplied by the emitter. The collector side is moderately doped and larger in size as compared to the emitter.

`color{blue} ✍️`We have seen earlier in the case of a p-n junction, that there is a formation of depletion region acorss the junction. In case of a transistor depletion regions are formed at the emitter base-junction and the basecollector junction.

`color{blue} ✍️`For understanding the action of a transistor, we have to consider the nature of depletion regions formed at these junctions. The charge carriers move across different regions of the transistor when proper voltages are applied across its terminals. The biasing of the transistor is done differently for different uses.

`color{blue} ✍️`The transistor can be used in two distinct ways. Basically, it was invented to function as an amplifier, a device which produces a enlarged copy of a signal. But later its use as a switch acquired equal importance.
We shall study both these functions and the ways the transistor is biased to achieve these mutually exclusive functions.

`color{blue} ✍️`First we shall see what gives the transistor its amplifying capabilities. The transistor works as an amplifier, with its emitter -base junction forward biased and the base-collector junction reverse biased.

`color{blue} ✍️`This situation is shown in Fig. 14.28, where `V_(C C)` and `V_(E E)` are used for creating the respective biasing. When the transistor is biased in this way it is said to be in active state.We represent the voltage between emitter and base as `V_(E B)` and that between the collector and the base as `V_(C B).`

`color{blue} ✍️`In Fig. 14.28, base is a common terminal for the two power supplies whose other terminals are connected to emitter and collector, respectively. So the two power supplies are represented as `V_(E E)`, and `V_(C C),` respectively.

`color{blue} ✍️`In circuits, where emitter is the common terminal, the power supply between the base and the emitter is represented as `V_(B B)` and that between collector and emitter as `V_(C C.)`

`color{blue} ✍️`Let us see now the paths of current carriers in the transistor with emitter-base junction forward biased and base-collector junction reverse biased. The heavily doped emitter has a high concentration of majority carriers, which will be holes in a p-n-p transistor and electrons in an n-p-n transistor.

`color{blue} ✍️`These majority carriers enter the base region in large numbers. The base is thin and lightly doped. So the majority carriers there would be few. In a p-n-p transistor the majority carriers in the base are electrons since base is of n-type semiconductor.

`color{blue} ✍️`The large number of holes entering the base from the emitter swamps the small number of electrons there. As the base collector-junction is reversebiased, these holes, which appear as minority carriers at the junction, can easily cross the junction and enter the collector.

`color{blue} ✍️`The holes in the base could move either towards the base terminal to combine with the electrons entering from outside or cross the junction to enter into the collector and reach the collector terminal. The base is made thin so that most of the holes find themselves near the reverse-biased base-collector junction and so cross the junction instead of moving to the base terminal.

`color{blue} ✍️`It is interesting to note that due to forward bias a large current enters the emitter-base junction, but most of it is diverted to adjacent reverse-biased base-collector junction and the current coming out of the base becomes a very small fraction of the current that entered the junction.

`color{blue} ✍️` If we represent the hole current and the electron current crossing the forward biased junction by `I_h` and `I_e` respectively then the total current in a forward biased diode is the sum `I_h + I_e`.

`color{blue} ✍️`We see that the emitter current `I_E = I_h + I_e` but the base current `I_B << I_h + I+e,` because a major part of `I_E` goes to collector instead of coming out of the base terminal. The base current is thus a small fraction of the emitter current.

`color{blue} ✍️`The current entering into the emitter from outside is equal to the emitter current `I_E.` Similarly the current emerging from the base terminal is `I_B` and that from collector terminal is `I_C.` It is obvious from the above description and also from a straight forward application of Kirchhoff’s law to Fig. 14.28(a) that the emitter current is the sum of collector current and base current:

`color{blue}(I_E = I_C+I_B)`


`color{blue} ✍️`We also see that `I_C approx I_E.`
Our description of the direction of motion of the holes is identical with the direction of the conventional current. But the direction of motion of electrons is just opposite to that of the current.

`color{blue} ✍️`Thus in a `p-n-p` transistor the current enters from emitter into base whereas in a `n-p-n` transistor it enters from the base into the emitter. The arrowhead in the emitter shows the direction of the conventional current.

`color{blue} ✍️`The description about the paths followed by the majority and minority carriers in a n-p-n is exactly the same as that for the p-n-p transistor. But the current paths are exactly opposite, as shown in Fig. 14.28.

`color{blue} ✍️` In Fig. 14.28(b) the electrons are the majority carriers supplied by the n-type emitter region. They cross the thin p-base region and are able to reach the collector to give the collector current, `I_C` . From the above description we can conclude that in the active state of the transistor the emitter-base junction acts as a low resistance while the base collector acts as a high resistance.

Basic transistor circuit configurations and transistor characteristics

`color{blue} ✍️`In a transistor, only three terminals are available, viz., Emitter (E), Base (B) and Collector (C). Therefore, in a circuit the input/output connections have to be such that one of these (E, B or C) is common to both the input and the output.

`color{blue} ✍️`Accordingly, the transistor can be connected in either of the following three configurations: Common Emitter (CE), Common Base (CB), Common Collector (CC).

`color{blue} ✍️`The transistor is most widely used in the CE configuration and we shall restrict our discussion to only this configuration. Since more commonly used transistors are n-p-n Si transistors, we shall confine our discussion to such transistors only. With p-n-p transistors the polarities of the external power supplies are to be inverted.

`color{blue} ✍️`When a transistor is used in CE configuration, the input is between the base and the emitter and the output is between the collector and the emitter. The variation of the base current `I_B` with the base-emitter voltage `V_(B E)` is called the input characteristic.

`color{blue} ✍️`Similarly, the variation of the collector current `I_C` with the collector-emitter voltage `V_(C E)` is called the output characteristic. You will see that the output characteristics are controlled by the input characteristics. This implies that the collector current changes with the base current.

`color{blue} ✍️`The input and the output characteristics of an n-p-n transistors can be studied by using the circuit shown in Fig. 14.29.

`color{blue} ✍️`collector-emitter voltage `V_(C E)` is kept fixed while studying the dependence of `I_B` on `V_(B E).` We are interested to obtain the input characteristic when the transistor is in active state.

`color{blue} ✍️`So the collector -emitter voltage `V_(C E)` is kept large enough to make the base collector junction reverse biased. Since `V_(C E) = V_(C B)+V_C E` and for Si `V_(B E)` is 0.6 to `0.7 V, V_(C E)` must be transistor sufficiently larger than 0.7 V.

`color{blue} ✍️`Since the transistor is operated as an amplifier over large range of `V_(C E,)` the reverse bias across the base collector junction is high most of the time. Therefore, the input characteristics may be obtained for `V_(C E)` somewhere in the range of `3 V` to `20 V.`

`color{blue} ✍️`Since the increase in `V_(C E)` appears as increase in `V_(C B,)` its effect on `I_B` is negligible. As a consequence, input characteristics for various values of `V_(C E)` will give almost identical curves Hence, it is enough to determine only one input characteristics. The input characteristics of a transistor is as shown in Fig. 14.30(a).

`color{blue} ✍️`The linear segments of both the input and output characteristics can be used to calculate some important ac parameters of transistors as shown below.

`color{green} {(i)bbul "Input resistance" (r_i ) : }` This is_ defined as the ratio of change in baseemitter voltage (DeltaV_(B E)) to the resulting change in base current (DeltaI_B) at constant collector-emitter voltage `(V_(C E)).`

`color{blue} ✍️`This is dynamic (ac resistance) and as can be seen from the input characteristic, its value varies with the operating current in the transistor:

`color{blue}(r_i= (DeltaV_(E E))/(DeltaI_B) V_(C E))`


`color{blue} ✍️`The value of `r_i` can be anything from a few hundreds to a few thousand ohms.

`color{green}((ii) bbul"Output resistance" (r_o) : )` This is defined as the ratio of change in collector-emitter voltage `(DrltaV_(C E))` to the change in collector current `(DeltaI_C)` at a constant base current `I_B.`

`color{blue}((r_0= (DeltaV_(CE))/(DeltaI_C) I_B))`


`color{blue} ✍️`The output characteristics show that initially for very small values of `V_(C E) , I_C` increases almost linearly. This happens because the base-collector junction is not reverse biased and the transistor is not in active state.

`color{blue} ✍️`In fact, the transistor is in the saturation state and the current is controlled by the supply voltage `V_(C C) (=V_(C E))` in this part of the characteristic. When `V_(C E)` is more than that required to reverse bias the base-collector junction, `I_C` increases very little with `V_(C E).`

`color{blue} ✍️`The reciprocal of the slope of the linear part of the output characteristic gives the values of `r_o`. The output resistance of the transistor is mainly controlled by the bias of the base collector junction.

`color{blue} ✍️`The high magnitude of the output resistance (of the order of 100 kW) is due to the reverse-biased state of this diode. This also explains why the resistance at the initial part of the characteristic, when the transistor is in saturation state, is very low.

`color{green} {(iii)bbul "Current amplification factor" (beta):}` This is defined as the ratio of the change in collector current to the change in base current at a constant collector-emitter voltage `(V_CE)` when the transistor is in active state.

`color{blue}(beta_(ac)= (DeltaV_C)/(DeltaI_B) V_(C E))`


`color{blue} ✍️`This is also known as small signal current gain and its value is very large. If we simply find the ratio of `I_C` and `I_B` we get what is called dc b of the transistor. Hence,

`color{blue} {beta_(dc)= (I_C)/(I_B)}`


`color{blue} ✍️`Since `I_C` increases with `I_B` almost linearly and `I_C = 0` when `I_B = 0,` the values of both `beta_(dc)` and `beta_(ac)` are nearly equal. So, for most calculations `beta_(dc` can be used. Both `beta_(ac)` and `beta_(dc)` vary with `V_(C E)` and `I_B` (or `I_C`) slightly.

Q 3119480310

From the output characteristics shown in Fig. 14.30(b), calculate the values of `beta_(ac)` and `beta_(dc)` of the transistor when `V_(CE)` is `10V` and `I_C` = 4.0 mA.
Class 12 Chapter 14 Example 8

`beta_(ac)= (DeltaI_C)/(Delta_B) V_(CE), beta_(dc) = (I_C)/(I_B)`

For determining `b_ac` and `b_dc` at the stated values of `V_(CE)` and IC one can proceed as follows. Consider any two characteristics for two values of `I_B` which lie above and below the given value of `I_C` . Here `I_C = 4.0 mA.` (Choose characteristics for `I_B= 30` and `20 μA.`) At `V_(CE) = 10 V` we read the two values of `I_C` from the graph. Then `DeltaI_B= (30-20)muA= 10muA, DeltaI_C = (4.5-30)mA=1.5mA`

Therefore, `beta_(ac) = 1.5 mA// 10 μA = 150`
For determining `beta_(dc)`, either estimate the value of `I_B` corresponding to `IC = 4.0 mA at V_(CE) = 10 V` or calculate the two values of `b_dc` for the two characteristics chosen and find their mean.

Therefore, for `I_C= 4.5 mA` and `I_B = 30 μA,`
`beta_(dc)= 4.5mA//30muA= 150`
and for `I_C = 3.0 mA` and `I_B = 20 μA`
`beta_(dc) = 3.0 mA//20muA=150`
Hence, `beta_(dc) = (150+150)//2=150`

Transistor as a device

`color{blue} ✍️`The transistor can be used as a device application depending on the configuration used (namely C B, CC and CE), the biasing of the E-B and B-C junction and the operation region namely cutoff, active region and saturation.

`color{blue} ✍️`As mentioned earlier we have confined only to the CE configuration and will be concentrating on the biasing and the operation region to understand the working of a device.

`color{blue} ✍️`When the transistor is used in the cutoff or saturation state it acts as a switch. On the other hand for using the transistor as an amplifier, it has to operate in the active region.

`color{green}ulbb("(i) Transistor as a switch")` We shall try to understand the operation of the transistor as a switch by analysing the behaviour of the base-biased transistor in C E configuration as shown in Fig. 14.31(a).

`color{blue} ✍️`Applying Kirchhoff’s voltage rule to the input and output sides of this circuit, we get

`color{blue}(V_(B B) = I_BR_B + V_(B E))`

..........(14.12) and

`color{blue}(V_(C E) = V_(C C) – I_CR_C.)`


`color{blue} ✍️`We shall treat VBB as the dc input voltage `V_i` and `V_(C E)` as the dc output voltage `V_O.` So, we have

`color{purple}(V_i = I_BR_B + V_(BE))` and `color{purple}(V_o = V_(C C) – I_CR_C.)`

`color{blue} ✍️` Let us see how `V_o` changes as `V_i` increases from zero onwards. In the case of Si transistor, as long as input Vi is less than 0.6 V, the transistor will be in cut off state and current `I_C` will be zero.
Hence `V_o = V_(C C)`

`color{blue} ✍️`When `V_i` becomes greater than 0.6 V the transistor is in active state with some current `I_C` in the output path and the output `V_o` decrease as the term `I_CR_C` increases. With increase of `V_i , I_C` increases almost linearly and so `V_o` decreases linearly till its value becomes less than about 1.0 V.

`color{blue} ✍️`Beyond this, the change becomes non linear and transistor goes into saturation state. With further increase in `V_i` the output voltage is found to decrease further towards zero though it may never become zero.

`color{blue} ✍️`If we plot the `V_o` vs Vi curve, [also called the transfer characteristics of the base-biased transistor (Fig. 14.31(b)], we see that between cut off state and active state and also between active state and saturation state there are regions of non-linearity showing that the transition from cutoff state to active state and from active state to saturation state are not sharply defined.

`color{blue} ✍️`Let us see now how the transistor is operated as a switch. As long as `V_i` is low and unable to forward-bias the transistor, `V_o` is high (at `V_(C C)` ). If `V_i` is high enough to drive the transistor into saturation, then Vo is low, very near to zero.

`color{blue} ✍️`When the transistor is not conducting it is said to be switched off and when it is driven into saturation it is said to be switched on. This shows that if we define low and high states as below and above certain voltage levels corresponding to cutoff and saturation of the transistor, then we can say that a low input switches the transistor off and a high input switches it on.

`color{blue} ✍️`Alternatively, we can say that a low input to the transistor gives a high output and a high input gives a low output. The switching circuits are designed in such a way that the transistor does not remain in active state.

Transistor as an amplifier

`color{blue} ✍️`For using the transistor as an amplifier we will use the active region of the `V_o` versus `V_i` curve. The slope of the linear part of the curve represents the rate of change of the output with the input.

`color{blue} ✍️`It is negative because the output is `V_(C C) – I_CR_C` and not `I_CR_C.` That is why as input voltage of the `C_E` amplifier increases its output voltage decreases and the output is said to be out of phase with the input. If we consider `DeltaV_o` and `DeltaV_i` as small changes in the output and input voltages then `DeltaV_o//DeltaV_i` is called the small signal voltage gain `A_V` of the amplifier.

`color{blue} ✍️`If the `V_(BB)` voltage has a fixed value corresponding to the mid point of the active region, the circuit will behave as a CE amplifier with voltage gain `DeltaV_o// DeltaV_i`. We can express the voltage gain AV in terms of the resistors in the circuit and the current gain of the transistor as follows.

`color{blue} ✍️`We have, `color{purple}(V_o = V_(C C) – I_CR_C)`

Therefore, `color{purple}(DeltaV_o = 0 – R_C Delta I_C)`

`color{blue} ✍️`Similarly, from `V_i = I_BR_B + V_(BE)`

`color{purple}(DeltaV_i = R_B DeltaI_B + DeltaV_(BE))`

`color{blue} ✍️`But `DeltaV_(BE)` is negligibly small in comparison to `DI_BR_B` in this circuit.

`color{blue} ✍️`So, the voltage gain of this `C_E` amplifier (Fig. 14.32) is given by `A_V`

`A_V = – R_C Delta I_C // RB DeltaI_B`

`color{blue}(= –beta_(ac)(R_C //R_B ))`........(14.14)

`color{blue} ✍️`where `beta_(ac)` is equal to `Delta I_C/DeltaI_B` from Eq. (14.10). Thus the linear portion of the active region of the transistor can be exploited for the use in amplifiers. Transistor as an amplifier (CE configuration) is discussed in detail in the next section.

`color{brown}bbul("Transistor as an Amplifier" ("CE-Configuration"))`
`color{blue} ✍️`To operate the transistor as an amplifier it is necessary to fix its operating point somewhere in the middle of its active region. If we fix the value of `V_(BB)` corresponding to a point in the middle of the linear part of the transfer curve then the dc base current `I_B` would be constant and corresponding collector current `I_C` will also be constant. The dc voltage `V_(CE) = V_(C C) - I_CR_C` would also remain constant. The operating values of `V_(C E)` and `I_B` determine the operating point, of the amplifier.

`color{blue} ✍️`If a small sinusoidal voltage with amplitude vs is superposed on the dc base bias by connecting the source of that signal in series with the `V_(BB)` supply, then the base current will have sinusoidal variations superimposed on the value of `I_B.` As a consequence the collector current also will have sinusoidal variations superimposed on the value of `I_C`, producing in turn corresponding change in the value of `V_O.`
We can measure the ac variations across the input and output terminals by blocking the dc voltages by large capacitors.

`color{blue} ✍️`In the discription of the amplifier given above we have not considered any ac signal. In general, amplifiers are used to amplify alternating signals. Now let us superimpose an ac input signal vi (to be amplified) on the bias `V_(B B )(dc)` as shown in Fig. 14.32.
The output is taken between the collector and the ground.

`color{blue} ✍️`The working of an amplifier can be easily understood, if we first assume that `v_i = 0.` Then applying Kirchhoff’s law to the output loop, we get

`color{blue}(V_(c c) = V_(C E) + I_cR_L)`


`color{blue} ✍️`Likewise, the input loop gives

`color{blue}{V_(B B) = V_(B E) + I B * RB}`


When `v_i` is not zero, we get `color{purple}(V_(B E) + v_i = V_(B E) + I_B R_B + DeltaI_B (R_B + r_i))`

`color{blue} ✍️`The change in `V_(B E)` can be related to the input resistance `r_i` [see Eq. (14.8)] and the change in `I_B.` Hence
`v_i = DeltaI_B (R_B + r_i )`
`= r DeltaI_B`

`color{blue} ✍️`The change in `I_B` causes a change in `I_c`. We define a parameter `beta_(ac),` which is similar to the beta_(dc) defined in Eq. (14.11), as

`color{blue}(beta_(ac)=(DeltaI_c)/(DeltaI_B)= (i_C)/(I_b))`


`color{blue} ✍️`which is also known as the ac current gain `A_i.` Usually `beta_(ac)` is close to `beta_(dc)` in the linear region of the output characteristics.
The change in `I_c` due to a change in `I_B` causes a change in `V_(C E)` and the voltage drop across the resistor `R_L` because `(V_c c)` is fixed.

`color{blue} ✍️`These changes can be given by Eq. (14.15) as `color{purple} {DeltaV_(C C) = DeltaV_(C E) + R_L DeltaI_C = 0}`
or `color{purple} {DeltaV_(C E) = - R_LDeltaI_C}`

`color{blue} ✍️`The change in `V_(CE)` is the output voltage `v_0.` From Eq. (14.10), we get

`color{purple} {v_0 = DeltaV_(C E) = –beta_(ac) R_L DeltaI_B}`

The voltage gain of the amplifier is
`color{purple} {A_v = (v_0)/(v_i) = ((DeltaV_(CE))/(rDeltai_B))}`

`color{blue} {= - (beta_(ac)R_L)/r}`


`color{blue} ✍️`The negative sign represents that output voltage is opposite with phase with the input voltage.

`color{blue} ✍️`From the discussion of the transistor characteristics you have seen that there is a current gain `beta_(ac)` in the CE configuration. Here we have also seen the voltage gain `A_v` Therefore the power gain `A_p` can be expressed as the product of the current gain and voltage gain. Mathematically `A_P = beta_(ac)xxA_v` ...(14.19)

`color{blue} ✍️`Since `beta_(ac)` and `A_v` are greater than 1, we get ac power gain. However it should be realised that transistor is not a power generating device. The energy for the higher ac power at the output is supplied by the battery.

Feedback amplifier and transistor oscillator

`color{blue} ✍️`In an amplifier, we have seen that a sinusoidal input is given which appears as an amplified signal in the output. This means that an external input is necessary to sustain ac signal in the output for an amplifier. In an oscillator, we get ac output without any external input signal. In other words, the output in an oscillator is self-sustained. To attain this, an amplifier is taken.

`color{blue} ✍️`A portion of the output power is returned back (feedback) to the input in phase with the starting power (this process is termed positive feedback) as shown in Fig. 14.33(a).

`color{blue} ✍️`The feedback can be achieved by inductive coupling (through mutual inductance) or LC or RC networks. Different types of oscillators essentially use different methods of coupling the output to the input (feedback network), apart from the resonant circuit for obtaining oscillation at a particular frequency.

`color{blue} ✍️` For understanding the oscillator action, we consider the circuit shown in Fig. 14.33(b) in which the feedback is accomplished by inductive coupling from one coil winding `(T_1)` to another coil winding `(T_2)`.

`color{brown} {"Note"}` that the coils `T_2` and `T_1` are wound on the same core and hence are inductively coupled through their mutual inductance. As in an amplifier, the base-emitter junction is forward biased while the base-collector junction is reverse biased. Detailed biasing circuits actually used have been omitted for simplicity.

apply proper bias for the first time. Obviously, a surge of collector current flows in the transistor. This current flows through the coil `T_2` where terminals are numbered 3 and 4 [Fig. 14.33(b)].

`color{blue} ✍️`This current does not reach full amplitude instantaneously but increases from X to Y, as shown in Fig. [14.33(c)(i)]. The inductive coupling between coil `T_2` and coil `T_1` now causes a current to flow in the emitter circuit (note that this actually is the ‘feedback’ from input to output). As a result of this positive feedback, this current (in `T_1`; emitter current) also increases from X´ to Y´ [Fig. 14.33(c)(ii)].

`color{blue} ✍️`The current in `T_2` (collector current) connected in the collector circuit acquires the value Y when the transistor becomes saturated. This means that maximum collector current is flowing and can increase no further. Since there is no further change in collector current, the magnetic field around `T_2` ceases to grow.

`color{blue} ✍️`As soon as the field becomes static, there will be no further feedback from `T_2` to `T_1`. Without continued feedback, the emitter current begins to fall. Consequently, collector current decreases from Y towards Z [Fig. 14.33(c)(i)]. However, a decrease of collector current causes the magnetic field to decay around the coil `T_2`.

`color{blue} ✍️` Thus, `T_1` is now seeing a decaying field in `T_2` (opposite from what it saw when the field was growing at the initial start operation). This causes a further decrease in the emitter current till it reaches Z when the transistor is cut-off. This means that both `I_E` and `I_C` cease to flow.

`color{blue} ✍️`Therefore, the transistor has reverted back to its original state (when the power was first switched on). The whole process now repeats itself. That is, the transistor is driven to saturation, then to cut-off, and then back to saturation.

`color{blue} ✍️`The time for change from saturation to cut-off and back is determined by the constants of the tank circuit or tuned circuit (inductance L of coil `T_2` and C connected in parallel to it). The resonance frequency (n ) of this tuned circuit determines the frequency at which the oscillator will oscillate.

`color{blue} {v = 1/2Esqrt(LC)}`


`color{blue} ✍️`In the circuit of Fig. 14.33(b), the tank or tuned circuit is connected in the collector side.

`color{blue} ✍️`Hence, it is known as tuned collector oscillator. If the tuned circuit is on the base side, it will be known as tuned base oscillator. There are many other types of tank circuits (say RC) or feedback circuits giving different types of oscillators like Colpitt’s oscillator, Hartley oscillator, RC-oscillator.

Q 3169580415

In Fig. 14.31(a), the `V_(BB)` supply can be varied from 0V to 5.0 V. The Si transistor has `beta_(dc) = 250` and `R_B = 100 K Omega, R_C = 1K Omega V_(CC)= 5.0V.` Assume that when the transistor is saturated `V_(CE) = 0V` and `V_(BE) = 0.8V`. Calculate (a) the minimum base the transistor will reach saturation. Hence, (b) determine `V_1` whencurrent, for which the transistor is ‘switched on’. (c) find the ranges of `V_1` for which the transistor is ‘switched off’ and ‘switched on’.
Class 12 Chapter 14 Example 9

Given at saturation `V_(CE) = 0V, V_(BE)= 0.8V`
`V_(CE) = V_(C C) – I_CR_C`
`I_C = V_(C C)//R_C = 5.0V//1.0kOmega = 5.0 mA`
Therefore `I_B = I_C//beta=5.0 mA//250=20muA`
The input voltage at which the transistor will go into saturation is given by
`V_(IH) = V_(BB) = I_BR_B +V_(BE)`
`= 20μA × 100 kOmega + 0.8V = 2.8V`
The value of input voltage below which the transistor remains cutoff is given by
`V_(IL) = 0.6V, V_(IH) = 2.8V`
Between 0.0V and 0.6V, the transistor will be in the ‘switched off state. Between 2.8V and 5.0V, it will be in ‘switched on’ state.

Note that the transistor is in active state when IB varies from 0.0mA to 20mA. In this range, `I_C = betaI_B` is valid. In the saturation range,
`I_C =< betaI_B.`
Q 3189580417

For a CE transistor amplifier, the audio signal voltage across the collector resistance of `2.0 kOmega` is `2.0 V.` Suppose the current amplification factor of the transistor is 100, What should be the value of `R_B` in series with `V_(BB)`supply of `2.0 V` if the dc base current has to be 10 times the signal current. Also calculate the dc drop across the collector resistance. (Refer to Fig. 14.33).
Class 12 Chapter 14 Example 10

The output ac voltage is 2.0 V. So, the ac collector current `i_C = 2.0//2000 = 1.0 mA.` The signal current through the base is, therefore given by `i_B=i_C//beta=1.0mA//100= 0.1010mA` The dc base current has to be `10× 0.010 = 0.10 mA.`
From Eq. 14.16, `R_B = (V_(BB)-V_(BE))//I_B` Assuming `V_(BE) = 0.6V`
`R_B = (2.0 – 0.6 )//0.10 = 14 kOmega.`
The dc collector current `I_C= 100x0.10=10mA.`