TRANSISTORS by Pranghi Miu

Assignment -2 NAME :

PRASANTH KRISHNA

SECTION:

B-10

ROLLNO :

1210316059

ANSWERS

1. Explain the structure of PNP and NPN transistor in detail? NPN TRANSISTOR STRUCTURE: This NPN transistor is made with two layers of N material and sandwiched layer of P. Transistor is composed three regions:  emitter  base  collector Emitter:  It is a heavily doped and it is an n type material and having more free electrons.  It supplies charge carriers (electrons and holes).  It is always forward biased with respective to base to supply large no of majority carriers. Base:

 It is a another region composed of lightly doped n type material  It has a few holes and having capable of carrying electrons.  It is lightly doped and very thin.  It is reverse bias & provide high resistance in collector circuit. Collector:  It is the n type material but not heavily doped and has free electrons.  It is moderately doped.

 Most of emitter carriers diffuse with the n base region and are attracting with the collector region  Some of the electrons are attracted by the holes in the base region and move to the collector side.

 In an NPN transistor the emitter current is equal to sum of base current and collector current.  In the NPN transistors current flows from collector to emitter.  The working principle of an NPN transistor is such that when you increase current to the base terminal, then the transistor turns ON and it conducts fully from the collector to emitter. When you decrease the current to the base terminal, the transistor turns ON less and until the current is so low, the transistor no longer conducts across the collector to emitter, and shuts off.

PNP TRANSISTOR STRUCTURE:  It is exact opposite to the NPN transistor.  It is made up with two layers P type materials and sandwiched with one N type material.  The direction of current flow from emitter to current.  It is always reverse biased.  The working principle of a PNP transistor is such that when the current exists at the base terminal of the transistor, then the transistor is shut off. When there is no current at the base terminal of the PNP transistor, then the transistor turns on.

 Number of electrons in base region is very small hence no of holes combined with electrons in n type base is very small, to constitute base current (IB).  Remaining holes (more than 95%) cross over collector region to generate collector current (IC).

 The relation of PNP transistor is IE=IC+IB

2) What are the different modes of operation of a transistor and explain the operation of n-p-n transistor under active mode? The three different modes of operation of a transistor are  Saturated  Active region  Cut-off When a transistor is in the fully-off state (like an open switch), it is said to be cutof. When it is fully conductive between emitter and collector (passing as much current through the collector as the collector power supply and load will allow), it is said to be saturated. Bipolar transistors don’t have to be restricted to these two extreme modes of operation. As we learned in the previous section, base current “opens a gate” for a limited amount of current through the collector. If this limit for the controlled current is greater than zero but less than the maximum allowed by the power supply and load circuit, the transistor will “throttle” the collector current in a mode somewhere between cut-off and saturation. This mode of operation is called the active mode. In Active region emitter base junction is forward biased and collector base junction is reverse biased. In saturation region emitter base junction is forward biased and collector base junction is forward biased In cut-off region emitter base junction is reverse biased and collector base junction is reverse biased. In this mode transistor has zero current. Transistor

3) Explain the input and output characteristics of a transistor in CE Configuration with a neat graph.

Input Characteristics:  Keep output voltage VCE = 0V by varying VCC.  Varying VBB gradually, note down base current IB and base-emitter voltage VBE.  Step size is not fixed because of nonlinear curve. Initially vary VBB in steps of 0.1V. Once the current starts increasing vary VBB in steps of 1V up to 12V.  Repeat above procedure (step 3) for VCE = 5V. Output Characteristics:  Connect the circuit as shown in the circuit diagram.  Keep emitter current IB = 20 A by varying VBB.  Varying VCC gradually in steps of 1V up to 12V and note down collector current IC and Collector-Emitter Voltage (VCE).  Repeat above procedure (step 3) for IB = 60µA, 0µA.

Plot the input characteristics by taking VBE on X-axis and IB on Y-axis at a constant VCE as a constant parameter. Plot the output characteristics by taking VCE on X-axis and taking IC on Y-axis taking IB as a constant parameter.

4) Briefly explain Common Base & Common collector Configuration of a Transistor. COMMON BASE:  Common-base transistor amplifiers are so-called because the input and output voltage points share the base lead of the transistor in common with each other, not considering any power supplies.  The current gain of a common-base amplifier is always less than 1.  The voltage gain is a function of input and output resistances, and also the internal resistance of the emitter-base junction, which is subject to change with variations in DC bias voltage.  Suffice to say that the voltage gain of a common-base amplifier can be very high.  The ratio of a transistor’s collector current to emitter current is called α.  The value for any transistor is always less than unity, or in other words, less than 1.

5) Explain the operation of a transistor as a switch?  Generally switch is used to on or off the current.  Transistor can act as a switch in active region.  Transistor can act as switch when it is given a pulse to the transistor then the currnet passes through it.  When lcd is passed on a circuit, when pulse given to it the led glows.  When pulse is absent then led will be in off. It will not glow.  There are three types of modes they are  Active mode-amplifier  Cutoff mode----------|  Saturation mode--- | Act as aswitch active mode: the value of voltages resistances are adjusted so that active region. the input is controlled by the input pulse. Off state: No connective current Input pulse is zero. Both the base and emitter at zero volts so zero forward bias to base emitter junction. Ib=0

Vbe<0.7(silicon transistor) It is not forward bias It is a diode When base current is zero Ic=0 In this it a reverse bias. By Kirchhoff law VCE=VCC-ICRC. Ic=0 Vec=VCC In this led will not glow when these are and true then it is in off state. On state: Pulse is present here and be voltage increases and base current is starts following Vbe>0.7V Ib increases then Ic is also increases.

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Vce=vcc-IcRc Vce=0 Ic=Vcc/Rc

6. Draw the circuit diagram of CE amplifier and explain its working?

 In electronics, a common emitter amplifier is one of three basic single-stage bipolarjunction-transistor (BJT) amplifier topologies, typically used as a amplifier.  In this circuit the base terminal of the transistor serves as the input, the collector is the output, and the emitter is common to both, hence its name. The analogous field-effect transistor circuit is the common source amplifier, and the analogous tube circuit is the common cathode amplifier.  Common emitter amplifiers give the amplifier an inverted output and can have a very high gain that may vary widely from one transistor to the next.  The gain is a strong function of both temperature and bias current, and so the actual gain is somewhat unpredictable.  Stability is another problem associated with such high gain circuits due to any unintentional positive feedback that may be present.  Other problems associated with the circuit are the low input dynamic range imposed by the small-signal limit; there is high distortion if this limit is exceeded and the transistor ceases to behave like its small-signal model.  One common way of alleviating these issues is with emitter degeneration. This refers to the addition of a small resistor (or any impedance) between the emitter and the common signal source. This impedance reduces the overall trans conductance of the circuit by a factor of, which makes the voltage gain.  The voltage gain depends almost exclusively on the ratio of the resistors rather than the transistor's intrinsic and unpredictable characteristics. The distortion and stability characteristics of the circuit are thus improved at the expense of a reduction in gain.

7) Write the differences between the BJT and FET?  Transistors can be categorized according to their structure, and two of the more commonly known transistor structures, are the BJT and FET.  BJT, or Bipolar Junction Transistor, was the first kind to be commercially mass-produced.  BJTs conduct using both minority and majority carriers, and its three terminals have corresponding names “ the base, emitter, and collector.  It basically consists of two P-N junctions – the base-collector and the base-emitter junctions. A material called the base region, which is a thin intervening semiconductor, separates these two junctions.  BJTs basically function as regulators of currents.  A small current is regulating a larger current. However, for them to properly operate as current regulators, the base currents and the collector currents must be moving in the right directions.

 FET, or Field-effect Transistor, also controls the current between two points, but it uses a different method to the BJT.  As the name suggests, FETs’ function is dependent on the effects of electric fields, and on the flow, or movement, of electrons in the course of a particular type of semi-conductor material.  FETs are sometimes referred to as unipolar transistors, based on this fact.  FET uses either holes (P channel), or electrons (N channel), for conduction, and it has three terminals – source, drain, and gate – with the body connected to the source in most cases.  In many applications, FET is basically a voltage controlled device, due to the fact that its output attributes are established by the field that is dependent on the applied voltage.

8) Explain how transistor is used as an amplifier? Transistor as an amplifier  A transistor raises the strength of a weak signal and thus acts as an amplifier.  Fig.

 The weak signal is applied between emitter-base junction and output is taken across the load Rc connected in the collector circuit.  The input circuit is always forward biased. To do so, a dc voltage Vee is applied in the input circuit in additional to the signal. The dc voltage is known as bias voltage.  As the input circuit has low resistance therefore a small change in signal voltage causes an appropriate change in emitter current.  This causes the same change in collector current due to transistor action.  The collector current flowing through a high load resistance Rc produces a large voltage across it.  Thus weak signal applied in the input circuit appears in the amplified form in the collector circuit. so it acts as a amplifier.

9. Explain the DC analysis of a transistor?

D.C Analysis of a Bipolar Junction Transistor:  A typical bipolar junction transistor circuit.  DC Analysis When doing DC analysis, all AC voltage sources are taken out of the circuit because they're AC sources.  DC analysis is concerned only with DC sources. We also take out all capacitors because in DC, capacitors function as open circuits.  For this reason, everything before and after capacitors are removed, which in this circuit includes resistor, Rs.  Below is the schematic of the circuit above with respective to DC analysis:  Now let's do the calculations to find the Vbb, Rb, Ieq, and Vceq. From this then, we can find the quiescent or just simply Q-point of this transistor circuit.

10) Draw and explain the small signal equivalent circuit of a transistor?

There are three transistor configurations CB,CE,CC. Each configuration will have its separate hybrid or h parameter equivalent circuit but we will develop a general equivalent circuit which can be for any of three configurations. The equivalent circuits thus obtained for any type of configuration is also called as small signal, low frequency hybrid model of a transistor Thus h parameters equations are

   

V1=h11+I1+h12V2 I2=h21I1+h22V2 V1=hi.I1+hr.V2 I2=hf.I1+ho.V2

 Since, each term of equation has the units of volts.  we can use Kirchhoff’s voltage law to find a circuit that fits this equation.  Figure shows this circuit.  Similarly each term of equation has the units of currents we can use Kirchhoff’s current law to find a circuit that fits this equation fig shows this circuit

11) Explain the different amplification factors αβƳ and derive the relation between them? The ratio of charge in output current to change in input current is known as CAF.(current amplification factor). In common base (CB) configuration CAF, α=Δ IC /Δ IE In common emitter (CB) configuration CAF, β=Δ IC /Δ IB In common collector configuration CAF Ƴ=Δ IE /Δ IB Relationship between α and β: We have

Δ IE=ΔIC+ΔIB ΔIC = α Δ IE Δ IE= α Δ IE + ΔIB ΔIB= Δ IE(1- α)

Dividing both sides by ΔIC, we get Δ IB /Δ IC = Δ IE/ Δ IC (1- α) 1/ β=1/ α (1- α) β = α/ (1- α) β = α/ (1- α) for Ƴ=Δ IE /Δ IB Substituting Δ IB= Δ IE-

The above equation is the relation of β, α, Ƴ.

13) A common base transistor amplifier has an input resistance of 20Ω and output resistance of 100kΩ. if a signal of 400 mV is applied between emitter and base, find voltage amplification. Assumed α=1? Given, Signal voltage =400mV Input resistance=20 ohms IE = signal voltage/input resistance =400/20 = 20mA Ic=αIE =1*20mA =20mA Output voltage = vo=IC*RL=20mA*1kΩ = 20V Voltage amplification(A) = output voltage/signal voltage = 20V/400mV = 50

14) The emitter current Ie in a transistor is 3mA. If the leakage current Icbo=5µA and α=0.98, Calculate the collector and base current.

Ans.

Ie = 3 mA ICBO = 50 x 10-6 A

A = 0.98 IC = ? IB = ?

= a.IE + ICBO = 0.98 x 3 x 10-6 + 5 x 10-6 = 2.94 x 10-3 + 0.005 x 10-3 = 2.945 x 10-3 A

I e = I B + IC

= I e – IC = 3 x 10-3 – 2.945 x 10-3 = 0.055 x 10-3 A

15) Determine the values of emitter current and Collector Current of a transistor having α=0.98 and collect to base leakage current (Icbo) =4µA. The base current is 50 µA

Ans.

a = 0.98 Icbo = 4 x 10-6 A Ib = 50 x 10-6 A Ie =? Ic =?

= (a.Ib/1-a) + (ICBO/1-a)

= (0.98 x 50 x 10-6/1-0.98) + (4 x 10-6/1-0.98) =24.5 x 10-4 + 2 x 10-4 =26.5 x 10-4 A

= Ib + I c = 50 x 10-6 + 26.5 x 10-4 = 27 x 10-11 A