BJT  Amplifier | BJT full form in electronics | bjt amplifier problems and solutions | circuit analysis

circuit analysis , BJT  Amplifier | BJT full form in electronics | bjt amplifier problems and solutions configurations and characteristics.
BJT  Amplifier
BJT  The BJT is a main building block of all modern electronic systems. it is three- teminal device whose output current, voltage, and/or power are controlled by its input current.
Basically, a transistor consists of two back- to- back p-n junctions manufactured in a single piece of a semiconductor crystal.
These two junctions give rise to three regions called emitter, base and collector. the emitter, base and collector and C. The two terminals which are labelled as E,B junction and collector- Base (CB) junction.

1. Emitter It forms the left hand section of region of the transistor as shown in fig. 1. It is more heavily doped than any of the other region because its main function is to supply majority charge carriers (either electrons or holes )to the base.
2. Base It forms the middle section of the transistor. it is very thin (10 m)as compared to either the emitter or collector and is very lightly doped.
3. Collector It forms the right hand side section of the is to collect majority carriers through the base.

In most transistors, collector region is made physically larger than the emitter region because, it has to dissipate much greater power. because of this difference, there is no possibility of inverting the transistor,i.e., making its collector the emitter and its emitter the collector.
Transistor Circuit Configuration
Basically, there are three types of circuit connections (called configurations) for operating a transistor.

1. Common- Base(CB)
2. Common – Emitter(CE)
3. Common- Collector(CC)
4. The Common – Base Configuration

If the base is common to the input and output circuits, it is known as common – base configuration as shown in Fig. 2.
For a p-n-p transistor, the largest current components are due to holes. Holes flow from emitter to collector and base terminal,the current directions are shown in fig. 2.
I_{e =} I_{c +} I_B
For a forward biased junction, V_ed is positive and for a reverse biased junction V_cb is negeative. the complete transistor can be described by the following two relations, which give the input voltage V_EB and output current I_C in terms of the output voltage (V_CB) and input current (I_E).
V_EB  = F₁ (V_CB,I_E)
I_C = F₂ (V_CB, I_E)
The Output Characteristics
The collector current I_C is completely determined by the input current I_E and the V_CB voltage. the relationship is given in fig. 3. It is plot of I_C versus V_CB, with emitter current I_E as parameter. the curves are known as the output or collector or static characteristics. the transistor consists of two diodes placed in series back- to- back (with two cathodes connected together). the complete characteristics can be divided in three regions

1. Active region

In this region, the collector diode is reverse biased and the emitter diode is forwaed biased. consider first that the emitter diode is forward biased biased. consider first that the emitter current is zero. then the collector current is small and equals the reverse saturation current I_CO of the collector junction considered as a diode.
If the forward current I_B is increased, then a fraction of I_E i.e., a_dc I_E will reach the collector. in the active region, the collector current is essentially independent of collector voltage and depends only upon the emitter current. because a_DC is less than one but almost equal to unity, the magnitude of the collector current is slightly less than that of emitter current. the collector current is almost constant and work as a current source.
The collector current slightly increases with voltage. this is due to early effect. at higher voltage collector gathers in a fem more electrons. this reduces the base current. the difference is so small, that it is usually neglected. if the collector voltage is increased, then space charge width increases; this decreased the effective base width. then there is less chance for recombination within the base region.

2. Saturation region

The region to the left of the ordinate V_CB = 0 and above the I_E = 0, characteristics in which both emitter and collector junctions are forward biased, is called saturation region.
When collector diode is forwaed biased, there is large change in collector current with small change in collector voltage. a forward bias means that p is made positive with respect to n, there is a flow of holes from p to n. this changes the collector current direction. if diode is sufficiently forward biased, the current changes rapidly. it does not depend upon emitter current.

3. Cut – off region

The region below I_E = 0 and to the right of V_CB for which emitter and collector junctions are both reversed biased is referred to cut- off region. the characteristic I_E = 0, is similar to other characteristics but not coincident with horizontal axis. the collector current is same as I_CO. I_CBO is frequently used for I_CO. It means collector to base current with emitter is open. this is also temperature dependent.
The Input Characteristics
In the active region, the input diode is forward biased biased therefore, input characteristic is simply the forward biased cheracteristics of the emitter to base diode for various collector voltages (fig. 4). below cut – in voltage (0.7 or 0.3) the emitter current is very small. the curve with the collector open represents the forward biased emitter diode. because of the early effect the emitter current increases for same V_EB ( the diode becomes better diode).when the collector is shorted to base, the emitter current increases for a given V_EB. since, the collector now removes minority carriers from the base and hence base can attract more holes from the emitter. this means that the curve V_CB = 0, is shifted from the character when V_CB = open.
Equivalent Circuit of a Transistor (Common – Base)
In an ideal transistor a_DC = 1. This means all emitter electrons entering the base region go on to the collector.therefore, collector current equals emitter current. for transistor action, emitter diode acts like a forward bias diode and collector diode acts like a current source. the equivalent circuits of n-p-n and p-n-p transistors are shown in fig. 5. the current source arrow points for conventional current the current source is controlled by emitter current.
Common – Base Amplifier
The common – base amplifier circuit is shown in fig. 6. tha V_EE source forward biased the emitter diode and V_CC source reverse biased collector diode. the AC source V_in is connected to emitter through a coupling capacitor so that it blocks DC. this AC voltage produces small fluctuation in current and voltages. the load resistance R_L is also connected to collector through coupling capacitor so the fluctuation in collector- base voltage will be observed across R_L.
The DC equivalent circuit is obtained by reducing all AC sources to zero and opening all capacitors. the DC collector current is same as I_E and V_CB is given by
V_CB = V_CC – I_CR_C
These current and voltage fix the Q point. the AC equivalent circuit is obtained by reducing all DC sources to zero and shorting all coupling capacitors.R’_E represents the AC resistance of the diode as shown in fig. 7(a).
fig. 7(b) shows the diode curve relating I_E and V_BE. In the absence of AC signal, the transistor operates at Q point (point of intersection of load line and input characteristic). when the AC signal is applied the emitter current and voltage also change. If the signal is small, the operating point swings sinusoidally about Q point (A to B).
by a straight line and diode appears to be a resistance given by
R’_B = V_BE/ I_E  small change
If the input signal is small, input voltage and current will be sinusoidal but if the input voltage is large then, current will no longer be sinusoidal because of the non-linearity of diode curve. the emitter current is elongated on the poditive half cycle and compressed on negative half cycle. therefore, the output will also be distorted.
R’_E is the ration of V_BE and I_E and its value depends upon the location of Q. higher up the Q point samll will be the value of R’_E because the same change in V_BE produces large change in I_E . the slop of the curve at Q determines the value of R’_E. From calculation, it can be proved that
R’_E = 25/I_E mV
Proof    In general, the current through a diode is given by
I = I_CO qv(e^kt – 1)
where, q is the charge on electron, V is the drop across diode, T is the temperature and K is a constant.
On differentiating w.r.t V, we get
dI/dV = I_CO qv/e^kt  q/KT
The value of (q/KT) at 25^.c is approximately 40.
dI/dV = 40I_CO qv/e^kt
= 40 (I+I_CO)
Therefore,        dv/dI    = 1/ 40(I + I_CO) ~ 1/ 40 i
Therefore, AC resistance of the emitter diode
dv/di = 25/I mV
To a close approximation, the small change in collector current is equal to the small changes in emitter in emitter current. in the AC equivalent circuit, the current I_C is shown upward because if I_e increases, then I_C also increases, then I_C also increases in the same direction.
Voltage gain
Since the AC input voltage source is connected across R’_E therefore, the AC emitter current is given by
I_E = V_in/R’_E
or        V_in = I_E R’_E
The output voltage is given by
V_out = i_c (R_C || R_L)
Therefore, voltage gain A_V = V_out/ V_in = (R_C || R_E) /R’_E
= R_C /R
Under open circuit condition, V_out = I_CR_C
Therefore, voltage gain in open circuit condition
= A_V = R_C/R’_E
Example 1. find the voltage gain and output of the amplifier shown is figure, if the input voltage is 1.5 mV
Sol. The emitter DC current I_E is given by
I_E = 10 – 0.7 /6.8 = 1.37 mA
Therefore, emitter AC resistance = A_V = R_C/ R’_E
= 3.3 || 1.5 / 18.2
A_V  = 56.6
V_out = 1.5 x56.6
= 84.9 mV
Example 2.  Repeat example 1, if AC source has resistance R = 100 .
Sol.The AC equivalent circuit with AC source resistance is shown is figure.
The emitter AC current is given by
I_E = V_in  R_S + (R_E || r’_e) X R_E/R_E + r’_e
V_in / (R_S + r’_e) R_E + R_Er’_E   R_E: V_in/ R_s + R’_E
Therefore, voltage gain of the amplifier = A_v = V_out/v_in
I_oR_o / I_E(R_S + R’_E) = R_o/R_S + R’_E
A_V = 3.3 || 1.5 /(100 + 18.2) = 8.71
V_out = 1.5 x 8.71 = 13.1mV

2. Common – Emitter Configuration

The common – emitter configuration of BJT is shown is fig. 8.
in CE configuration, the emitter is made common to the input and output. it is also referred to as grounded emitter configuration. it is most commonly used configuration. in this base current and output voltage are taken as independent parameters.
V_BE = F₁ (I_B, V_CE)
I_C = F₂ (I_B, V_CE)
Input Characteristics
The curve between I_B and V_BE for different values of V_CE are shown in fig. 8 (b). since, the base-emitter junction of a transistor is a diode, therefore the characteristic is similar to diode one. with higher values of V_CE collector gathers slightly more electrons and therefore, base current reduces. normally, this effect is neglected (early effect).when collector is shorted with emitter then, the input characteristic is the characteristic of a forward biased diode when V_BE is zero and I_B is also zero.
Output Characteristic
The output characteristics are the curve between V_CE and I_C for various values of I_B. for fixed value of I_B,I_C is shown in fig. 9. for fixed value of I_B , I_C is not varying much dependent on V_CE but slopes are greater than CE characteristics. the output characteristics can again be devided into three parts

1. Active region

In this region, collector junction is reverse biased and emitter junction is forward biased. it is the area to the right of V_CE = 0.5 V and above I_B = 0. in this region transistor current responds most sensitivity to I_B. if transistor is to be used as an amplifier. it must operate in this region.
I_E = I_C + I_B
Since,  I_C  = I_CO + a_DCI_E
I_C = I_CO + a_DC (I_C + I_B)
(I – a_DC) I_C = a_DCI_B + I_CO
I_C = (a_dc/1 – a_dc) I_B + (1/1-a_dc) I_co
Lat         B_DC = a_dc / 1 – a_dc
I_C = (1 + B_DC) I_CO + B_DC I_B
B_DC  is defined as current gain of the transistor is given by
B_DC = I_C – I_CO / I_B + I_CO
If a_dc is truly constant then, I_C would be independent of V_CE but because of early effect, a_dc increases by 0.1% (0.001) e.g., from 0.995 to 0.996 as V_CE increases from few volts to 10 v. then, B_DC increases from 0.995 /(1-0.995) = 200 to 0.996 / (1-0.996) = 250 or about 25%. this shoes that small change in a reflects large change in B. therefore. the curves are subjected to large variations for the same type of transistors.

2. Cut-off region

Cut-off in a transistor is given by I_B = 0, I_C = I_CO. A transistor is not at cut-off, if the base current is simply reduced to zero (open circuited). under this condition
I_C = I_E = I_CO/ (1 – a_DC) = I_CEO
The actual collector current with base open is designed as I_CEO. Since, even as 0.9 for ge then, I_C = 10I_CO (approximately), at zero base current. accordingly in order to cut-off transistor it is not reverse bias the emitter junction slightly. it is found that reverse voltage of 0.1 V is sufficient for cut-off a transistor. in si,the a_dc is very nearly equal to zero therefore, I_C = I_CO. Hence, even with I_B = 0, I_C = I_E = I_CO, So that transistor is very close to cut-off.
In summaey, cut-off means I_E = 0, I_C = I_CO, I_B = – I_CO and V_BE is a reverse voltage whose magnitude is the order of 0.1 V for ge and 0 V for si.
Reverse Collector saturation current I_CBO
When in a physical transistor emitter-current is reduced to zero,then the collector current is known as I_CBO (approximately equal to I_CO). Reverse collector saturation current I_CBO also varies with temperature, avalanche multiplication and variabiliy from sample to sample. consider the circuit shown in fig. 10. V_BB is the reverse voltage applied to reduce the emitter-current to zero.
I_E = 0, I_B = – I_CBO
If we require,               V_BE = 0.1 V
Then                  – V_BB + I_CBOR_B < – 0.1 V
If R_B = 100 K, I_CBO = 100 mA, then V_BB must be 10.1 V. Hence, transistor must be capable to withstand this reverse voltage before breakdown voltage exceeds.

3. Saturation region

In this region, both the diodes are forwaed biased by at least cut in voltage. since,the voltage V_BE and V_CE across a forward is approximately 0.7 V therefore,
V_CE  = V_CB + V_BE = – V_BC + V_BE is also few tenths of volts. hence, saturation region is very close to zero voltage axis, where all the current repidly reduce to zero. in this region the transistor collector current is approximately given by V_CC /R_C and independent of base current. normally, transistor action is last and it acts like a small ohmic resistance.
Large signal current gain B_DC
The ratio I_C /I_B is defined as transfer ratio or large signal current gain B_DC.
B_DC = I_C / I_B
Where I_C is the collector current and I_B is the base current. the B_DC is an indication of how well the transistor works. the typical value of B_DC varies from 50 to 300.
In terms of h-parameters, B_DC is known as DC current gain and in designated h_FE (B_DC = h_FE). Knowing the maximum collector current and B_DC the maximum base current can be found which will be needed to saturate the transistor.
I_{C (max)} = V_cc – V_ce(sat) /R_C = I_{C (Sat)}
I_{B (min)} = I_{C (Sat)} /B_DC
This expression of B_DC is defined neglecting reverse leakage current (I_CO).
Taking reverse leakage current (I_CO) into account, the expression for the B_DC can be obtained as follows:
B_DC in terms of a_dc is given by
B_DC = a_DC / 1 – a_dc
I_C – I_CO
I_E / I_C – I_CO = I_C – I_CO / I_E – I_{C +} I_CO
= I_C – I_CO / I_B + I_CO
I_CO = I_CBO
B_DC = I_C – I_CBO / I_B + I_CBO
Cut-off a transistor means I_E = 0, then I_C = I_CBO and I_B = – I_CBO. Therefore, the above expression B_DC gives the collector current increment to the base current increment, it represents the large signal current gain of all common emitter transistors.
Biasing Circuit Techniques or Locating the Q Point
Fixed bias or base bias
In order for a transistor to amplify, it has to be properly biased. this means forward biasing the base-emitter junction and reverse biasing collector base junction. for linear amplification, the transistor should operate in active region (if I_E increases, I_C increases, V_CE decreases proportionally).
The source V_BB, through a current limits resistor R_B forward biases the emitter diode and V_CC through R_C (load resistance) reverse biases the collector junction as shown in fig. 11(a).
The DC base current through R_B is given by
I_B = (V_BB – V_BE) /R_B
Or              V_BE = V_BB – I_BR_B
Normally, V_BE is taken 0.7 V or 0.3 V if exact voltage is required, then the input characteristic (I_B vs V_BE) of the transistor should be used to solve the above equation. the load line for the input circuit is drawn on input characteristic. the two points of the load line can be obtained as given below.
For I_B = 0,                     V_BE = V_BB
and for    V_BE = 0,         I_B = V_BB /R_B
The intersection of this line with input characteristic gives the operating point Q as shown in fig. 11(b). if an AC signal is connected to the base of transistor, then variation in V_BE is about Q points. this gives variation in I_B and hence I_C.
In the output circuit, the load equation can be written as
V_CE = V_CC – I_CR_C
This equation involves two unknown V_CE and I_C and therefore, cannot be solved. to solve this equation output characteristic (I_{C VS} V_CE) is used.
The load equation is the equation of a straight line and given by two points
I_C = 0, V_CE = V_CC
and       V_CE = 0, I_C = V_CC/R_C
The intersection of this line which is also called DC load line and the characteristic gives the operating point Q as shown in fig. 11(c).
The point at which the load line intersects with I_B = 0 characteristic is known as cut-off point. at this point base current is zero and collector current is almost negligbly small. at cut-off, the emitter diode comes out of forward bias and normal transistor action is last. to a close approximation.
V_CE (cut-off) >>V_CC         (approximately)
The intersection of the load line and I_B = I_B (_max) characteristic is known as saturation point. At this point I_B = I_{B (MAX),} I_C = I_{C (sat) .} At this point collector diode comes out of reverse bias and again transidtor action is last. to a close approximation,
I_{C (sat)} >> V_CC /R_C               (approximately)
The I_{B (sat)} is the minimum current required to operate the transistor in saturation region. if the I_B is less than I_{B (sat)}, the transistor will operate in active region. if I_B > I_{B (sat)}, it always operates in saturation region.
If the transistor operates at saturation or cut-off points and no where else then, it is operating as switch shown in fig. 11(d)
If I_B > I_{B (sat),}then it operates at saturation in I_B = 0 then, it operates at cut-off.
If a transistor is operating as an amplifier then Q point must be selscted carefully. although, we can select the operating point anywhere in the active region by choosing different values of R_B and R_C but the various transistor ratings such as maximum collector dissipation P_{C (max)} maximum collector voltage V_{C (max)} and I_{C (max)} and V_{BE (max)} limit the operating range.
Once the Q point is established an AC input is connected. due to this, the AC source the base current varies, AS a resut of this collector currect and collector voltage also varies and the amplified output is obtained.
If the Q point is not selected properly then, the output waveform will not be exactly the input waveform i.e., it may be dipped from one side or both sides or it may be distored one.
Example 3. Find the transistor current in the circuit shown in figure, if I_CO = 20 nA, B = 100.
Sol.  For the base circuit,
5 = 200 X I_B + 0.7
I_B = 5 -0.7 / 200
= 0.0215 mA
Since, I_CO << I_B therefore, I_C = BI_B = 2.15 mA
From the collector circuit V_CE = 10 – 3 x 2.15 = 3.55 V
Since, V_CE = V_CB + V_BE
Thus    V_CB = 3.55 – 0.7 = 2.55 V
Therefore, collector junction is reverse biased and transistor is operating in its above region.
Example 4. If a resistor of 2 k is connected in series with emitter in the circuit as shown in figure, find the currents. Given, I_CO = 20 mA. B = 100.
Sol.     I_E  = I_B  + I_C = I_B + 100I_B = 101 I_B
For the base circuit,
5 = 200 x I_B + 0.7 + 2 x 101 I_B
I_B = 5 – 0.7 / 402 = 0.0107 mA
Since I_CO << I_B therefore, I_C  = BI_B = 1.07 mA
From the collector circuit,
V_CB = 10 – 3 x 1.07 – 0.7 – 2 x 101 x 0.0107 = 3.93 V
Therefore, the collector junction is reverse biased and transistor is operating in its active region.
Stability of Operating point
Let us consider three operating points of transistor operating in common – emitter amplifier.

1. Near cut-off
2. Near saturation
3. In the middle of active region

If the operating point is selected near cut-off region, the output is clipped in negative half cycle as shown in fig. 12(a).
If the operating point is selected near saturation region, then the no is clipped and the output follows input faithfully as shown in fig. 12 (c) if input is large then clipping at both sides will take place. the first circuit for biasing the transistor in CE configuration is fixed bias.
In biasing circutit shown in fig. 12(d), two different power supplies are required. to avoid the use of two supplies the base resistance R_b is connected to V_CC as shown in fig.12(d).
Now, V_CC is still forward biassed emitter diode. in this eircut Q point is very unstable. the bias resistance R_B is selected by notiong the required base current I_B for operating point Q.
I_B  = (V_CC – V_BE) /R_B
Voltage across base emitter junction is approximately 0.7 V. since, V_CC is usally very high.
i.e.,              I_B = V_CC / R_B
since, I_B is constant therefore, it is called fixed bias circuit.