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

By   June 28, 2020

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.

Ie = Ic + IB

For a forward biased junction, Ved is positive and for a reverse biased junction Vcb is negeative. the complete transistor can be described by the following two relations, which give the input voltage VEB and output current IC in terms of the output voltage (VCB) and input current (IE).

VEB  = F1 (VCB,IE)

IC = F2 (VCB, IE)

The Output Characteristics

The collector current IC is completely determined by the input current IE and the VCB voltage. the relationship is given in fig. 3. It is plot of IC versus VCB, with emitter current IE 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 ICO of the collector junction considered as a diode.

If the forward current IB is increased, then a fraction of IE i.e., adc IE 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 aDC 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.

  1. Saturation region

The region to the left of the ordinate VCB = 0 and above the IE = 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.

  1. Cut – off region

The region below IE = 0 and to the right of VCB for which emitter and collector junctions are both reversed biased is referred to cut- off region. the characteristic IE = 0, is similar to other characteristics but not coincident with horizontal axis. the collector current is same as ICO. ICBO is frequently used for ICO. 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 VEB ( the diode becomes better diode).when the collector is shorted to base, the emitter current increases for a given VEB. 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 VCB = 0, is shifted from the character when VCB = open.

Equivalent Circuit of a Transistor (Common – Base)

In an ideal transistor aDC = 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 VEE source forward biased the emitter diode and VCC source reverse biased collector diode. the AC source Vin 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 RL is also connected to collector through coupling capacitor so the fluctuation in collector- base voltage will be observed across RL.

The DC equivalent circuit is obtained by reducing all AC sources to zero and opening all capacitors. the DC collector current is same as IE and VCB is given by

VCB = VCC – ICRC

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 IE and VBE. 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 = VBE/ IE  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 VBE and IE 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 VBE produces large change in IE . the slop of the curve at Q determines the value of R’E. From calculation, it can be proved that

R’E = 25/IE mV

Proof    In general, the current through a diode is given by

I = ICO qv(ekt – 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 = ICO qv/ekt  q/KT

The value of (q/KT) at 25.c is approximately 40.

dI/dV = 40ICO qv/ekt

= 40 (I+ICO)

Therefore,        dv/dI    = 1/ 40(I + ICO) ~ 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 IC is shown upward because if Ie increases, then IC also increases, then IC 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

IE = Vin/R’E

or        Vin = IE R’E

The output voltage is given by

Vout = ic (RC || RL)

Therefore, voltage gain AV = Vout/ Vin = (RC || RE) /R’E

= RC /R

Under open circuit condition, Vout = ICRC

Therefore, voltage gain in open circuit condition

= AV = RC/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 IE is given by

IE = 10 – 0.7 /6.8 = 1.37 mA

Therefore, emitter AC resistance = AV = RC/ R’E

= 3.3 || 1.5 / 18.2

A= 56.6

Vout = 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

IE = Vin  RS + (RE || r’e) X RE/RE + r’e

Vin / (RS + r’e) RE + REr’E   RE: Vin/ Rs + R’E

Therefore, voltage gain of the amplifier = Av = Vout/vin

IoRo / IE(RS + R’E) = Ro/RS + R’E

AV = 3.3 || 1.5 /(100 + 18.2) = 8.71

Vout = 1.5 x 8.71 = 13.1mV

  1. 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.

VBE = F1 (IB, VCE)

IC = F2 (IB, VCE)

Input Characteristics

The curve between IB and VBE for different values of VCE 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 VCE 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 VBE is zero and IB is also zero.

Output Characteristic

The output characteristics are the curve between VCE and IC for various values of IB. for fixed value of IB,IC is shown in fig. 9. for fixed value of IB , IC is not varying much dependent on VCE 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 VCE = 0.5 V and above IB = 0. in this region transistor current responds most sensitivity to IB. if transistor is to be used as an amplifier. it must operate in this region.

IE = IC + IB

Since,  IC  = ICO + aDCIE

IC = ICO + aDC (IC + IB)

(I – aDC) IC = aDCIB + ICO

IC = (adc/1 – adc) IB + (1/1-adc) Ico

Lat         BDC = adc / 1 – adc

IC = (1 + BDC) ICO + BDC IB

BDC  is defined as current gain of the transistor is given by

BDC = IC – ICO / IB + ICO

If adc is truly constant then, IC would be independent of VCE but because of early effect, adc increases by 0.1% (0.001) e.g., from 0.995 to 0.996 as VCE increases from few volts to 10 v. then, BDC 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.

  1. Cut-off region

Cut-off in a transistor is given by IB = 0, IC = ICO. A transistor is not at cut-off, if the base current is simply reduced to zero (open circuited). under this condition

IC = IE = ICO/ (1 – aDC) = ICEO

The actual collector current with base open is designed as ICEO. Since, even as 0.9 for ge then, IC = 10ICO (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 adc is very nearly equal to zero therefore, IC = ICO. Hence, even with IB = 0, IC = IE = ICO, So that transistor is very close to cut-off.

In summaey, cut-off means IE = 0, IC = ICO, IB = – ICO and VBE is a reverse voltage whose magnitude is the order of 0.1 V for ge and 0 V for si.

Reverse Collector saturation current ICBO

When in a physical transistor emitter-current is reduced to zero,then the collector current is known as ICBO (approximately equal to ICO). Reverse collector saturation current ICBO also varies with temperature, avalanche multiplication and variabiliy from sample to sample. consider the circuit shown in fig. 10. VBB is the reverse voltage applied to reduce the emitter-current to zero.

IE = 0, IB = – ICBO

If we require,               VBE = 0.1 V

Then                  – VBB + ICBORB < – 0.1 V

If RB = 100 K, ICBO = 100 mA, then VBB must be 10.1 V. Hence, transistor must be capable to withstand this reverse voltage before breakdown voltage exceeds.

  1. Saturation region

In this region, both the diodes are forwaed biased by at least cut in voltage. since,the voltage VBE and VCE across a forward is approximately 0.7 V therefore,

VCE  = VCB + VBE = – VBC + VBE 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 VCC /RC and independent of base current. normally, transistor action is last and it acts like a small ohmic resistance.

Large signal current gain BDC

The ratio IC /IB is defined as transfer ratio or large signal current gain BDC.

BDC = IC / IB

Where IC is the collector current and IB is the base current. the BDC is an indication of how well the transistor works. the typical value of BDC varies from 50 to 300.

In terms of h-parameters, BDC is known as DC current gain and in designated hFE (BDC = hFE). Knowing the maximum collector current and BDC the maximum base current can be found which will be needed to saturate the transistor.

IC (max) = Vcc – Vce(sat) /RC = IC (Sat)

IB (min) = IC (Sat) /BDC

This expression of BDC is defined neglecting reverse leakage current (ICO).

Taking reverse leakage current (ICO) into account, the expression for the BDC can be obtained as follows:

BDC in terms of adc is given by

BDC = aDC / 1 – adc

IC – ICO

IE / IC – ICO = IC – ICO / IE – IC + ICO

= IC – ICO / IB + ICO

ICO = ICBO

BDC = IC – ICBO / IB + ICBO

Cut-off a transistor means IE = 0, then IC = ICBO and IB = – ICBO. Therefore, the above expression BDC 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 IE increases, IC increases, VCE decreases proportionally).

The source VBB, through a current limits resistor RB forward biases the emitter diode and VCC through RC (load resistance) reverse biases the collector junction as shown in fig. 11(a).

The DC base current through RB is given by

IB = (VBB – VBE) /RB

Or              VBE = VBB – IBRB

Normally, VBE is taken 0.7 V or 0.3 V if exact voltage is required, then the input characteristic (IB vs VBE) 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 IB = 0,                     VBE = VBB

and for    VBE = 0,         IB = VBB /RB

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 VBE is about Q points. this gives variation in IB and hence IC.

In the output circuit, the load equation can be written as

VCE = VCC – ICRC

This equation involves two unknown VCE and IC and therefore, cannot be solved. to solve this equation output characteristic (IC VS VCE) is used.

The load equation is the equation of a straight line and given by two points

IC = 0, VCE = VCC

and       VCE = 0, IC = VCC/RC

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 IB = 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.

VCE (cut-off) >>VCC         (approximately)

The intersection of the load line and IB = IB (max) characteristic is known as saturation point. At this point IB = IB (MAX), IC = IC (sat) . At this point collector diode comes out of reverse bias and again transidtor action is last. to a close approximation,

IC (sat) >> VCC /RC               (approximately)

The IB (sat) is the minimum current required to operate the transistor in saturation region. if the IB is less than IB (sat), the transistor will operate in active region. if IB > IB (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 IB > IB (sat),then it operates at saturation in IB = 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 RB and RC but the various transistor ratings such as maximum collector dissipation PC (max) maximum collector voltage VC (max) and IC (max) and VBE (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 ICO = 20 nA, B = 100.

Sol.  For the base circuit,

5 = 200 X IB + 0.7

IB = 5 -0.7 / 200

= 0.0215 mA

Since, ICO << IB therefore, IC = BIB = 2.15 mA

From the collector circuit VCE = 10 – 3 x 2.15 = 3.55 V

Since, VCE = VCB + VBE

Thus    VCB = 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, ICO = 20 mA. B = 100.

Sol.     IE  = IB  + IC = IB + 100IB = 101 IB

For the base circuit,

5 = 200 x IB + 0.7 + 2 x 101 IB

IB = 5 – 0.7 / 402 = 0.0107 mA

Since ICO << IB therefore, IC  = BIB = 1.07 mA

From the collector circuit,

VCB = 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 Rb is connected to VCC as shown in fig.12(d).

Now, VCC is still forward biassed emitter diode. in this eircut Q point is very unstable. the bias resistance RB is selected by notiong the required base current IB for operating point Q.

IB  = (VCC – VBE) /RB

Voltage across base emitter junction is approximately 0.7 V. since, VCC is usally very high.

i.e.,              IB = VCC / RB

since, IB is constant therefore, it is called fixed bias circuit.

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