Question Bank

Unit-3

Q1) Determine the values of VCC RC and RB of a fixed bias configuration for a given load line and defined Q point.

Solution:

Q2) Explain input characteristics of CB ?

Sol: Input Characteristic: is always a graph of input current verses input vtg. For common base (CB) configuration input current is the emitter current (IE) & I/p vtg. Is the emitter  to the base vtg (VEB)

The I/p Characteristic is plotted at a constant O/p vtg. VCB

Q3) What is BJT explain its construction?
Sol: *BJT:-The BJT is constructed with the three draped semiconductors region’s separate by two Pn junctions as shown below

-Basic epitaxial planner structures.

-Three terminal with region’s are called emitter, base and collector.

-The physical representation of the two types of BJT’s,

One type consists between two regions separated by a P region (npn) and other type consists of two p regions separated by an n region (pnp).

-The Pn junction joining the base region and the emitter region is called the base emitter junction.

-The Pn junction joining the base region and the collector region is called the base collector junction.

-The base region is lightly doped and very thin compared to the heavily doped emitter and the moderately doped collector regions.

Q4) How do the BJT work?

Sol: Base Transistor Operation:-

        In order for the transistor to operate properly as an amplifier the two pn junction must be correctly biased with the external D.C vtg.

-The next figure shows the proper bias arrangement for both the npn and pnp transistors for active operation as an amplifier.

-In both the cases the base emitter

(BE) junction is forward biased & the base collector junction (BC) junction is reverse biased

• As from above figure consider n-p-n transistor. The forward bias from base to emitter narrow’s the BE depletion region and the reverse bias from base to collector widens the BC depletion region shown in figure.
• The heavily doped N-TYPE emitter region is full with conduction band(frep)  electron’s that easily diffuse through the forward biased BE junction into the p-type base region where they become minority carrier’s same as forward biased diode region
• The base region is lightly doped & very thin so that it has a very limited number of holes.
• Those only a small percentage of all the e-flowing the BE junction can combine with the available holes in the base.
• The relatively few recombined flow out of the base lead as valance electrons, forming as small base current.
• Most of the e flowing from the emitter into the thin lightly dooped base region do not recombine but diffuse into the BC depletion region.
• The BC depletion region diffuse e is being pulled across the reverse biased BC junction by the attraction of the collector supply vtg.
• The electrons now move through the collector region, out through the collector lead into the +ve terminal of the collector vtg source. This forms the collector electrons current.
• The collector current is much larger than the base current.
• This is the reason transistor exhibit current gain.

Q5) Derive the relation for base current and emitter current in CB mode?

Sol:The collector current is IC of the common base configuration is given by

Ic=Ic(INI)+ICBO

-Where the Ic(INI) called the injected collectors current and it is due to the number of electrons crossing the collectors base junction.

-ICBO :- This is the reverse saturation current flowing due to the minority carrier’s between the collector and base when the emitter is open

-ICBO flow’s due to the reverse biased collector’s base junction. As ICBO negligible as compared to Ic(INI) we can neglect it in practice.

.’. Ic=Ic(INI)………………………practically

Ic=ICBO………………with emitter open

Emitter is open

ICBO

Collector is to base control

-Since the ICBO flow’s due to terminally generated minority carrier’s it increases with increase the temperature.

-It doubles it’s value for every 100c rise in temperature.

-Current amplification factor or current gain (ddc):-

Current amplification factor or current gain is the ratio of collector current due to the injection to the total emitter current

αd.c = Ic(INI)

-The value of the ddc for CB configuration will always be less than 1.This is because

Ic(INI)<IE.

-Typically the value of d.d.c ranges between 0.95 to 0.995 depending upon the thickness of the base region.

-Larger the thickness of the base region smaller the value of the d.d.c

Ic(INI)=d.d.c.IE

Hence the expression for IC is given by

IC=αd.c IE + ICBo--------------------I

But the ICBo is negligibly small

ICd.d.c IE

* Expression for IB:-

IE=IB+IC

IE=αd.cIE+ICBo+IB…………………from I

IB=(1-αd.c)IE-ICBo

Neglecting ICBo

IB=(1-αd.c)IE

Q6) Explain Eber’s moll model?

Sol: This model of transistor can operate in forward active mode. The circuit has two diodes with two current sources connected as shown below. The diodes represent base-emitter and base collector. The directions of current are as shown in the figure below. The minority carriers flow through the base region.

                   Fig: Eber’s Moll model for NPN bipolar junction transistor

IE = IF – αR IR

IB = (1- αF) IF + (1- αR) IR

IC = -IR + αF IF

The Eber’s Moll model parameters are related as

IE,sαF = IC,sαR

The above relation is called as reciprocity relation. But for case when there is no minority carrier diffusion due to uniform base doping and equal voltages of base-emitter and base-collector junctions the relation becomes

IF(VBE) αF = IR (VBC+VBE) αR

Where

IF = Forward current of diode

IR = reverse current of diode

αF = Forward transport factor

αR = Reverse transport factor

IE,s = Saturation current of base-emitter diode

IC,s = Saturation current of base-collector diode

Q7) what are the region of operations of BJT?

Sol:

Q8) Explain output characteristics of CB?
Sol: Output Characteristic is always a graph of O/p current versus O/p Vtg.

For the CB configuration the O/p current is collector current (IC) of the output voltage is collector to base vtg. (VCB)

Output Characteristic is plotted for a constant value of I/p current (IE)

O/p Characteristic of a n-p-n transistor in CB Configuration

Dynamic O/p resistance (ro)

Ro = / I constant

In the active region Ic does not depend on VCB. It depends only on the I/p current IE. That is why the transistor is called as a current controlled or current operated device.

Q9) What is load line in transistor characteristics?

• Sol: The load line solution was found by superimposing the actual characteristics on a plot of network equation involving the same network variables.
• It is known as load line analysis as load of network defined the slope of the straight line connecting the points defined by network parameters.
• The smaller the load resistance, the steeper the slope of the network load line.

               Fig.: DC Load Line (Ref. 2)

• The output equation is given by,

• The characteristic curve is drawn as

                           Fig.: DC Load Line characteristic curve (Ref. 2)

• When Ic = 0mA then

• If we choose VCE = 0V then,

Q10) Draw npn CB transistor?

Sol

:

The I/p is applied between the emitter and the base. The base acts as a common terminal between the I/p and o/p.

-The input vtg is therefore VEB and the input current is IE.

-The output is taken between the collector and the base therefore the output vtg is VCB and the output current is IC.