Bipolar Junction Transistor(part 1)

3.0 BIPOLAR JUNCTION TRANSISTORS (BJT)

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3.1 The basic of Bipolar Junction Transistor (BJT) 3.1.1 Physical structure and schematic symbols for BJT

Symbol

Physical Structure

Symbol

Physical Structure

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3.1.2 Basic transistor operation Two conditions to make the transistor operate is: (a) (b)

E - B junction must get the forward bias voltage C - B junction must get the reverse bias voltage

3.1.2 Basic transistor operation

1. 2.

3. 4. 5.

The Emitter(E) in NPN device is made of N-type material, where the majority carriers are electrons. When the Base-Emitter junction is forward biased, the electron move from the N-type region towards the P-type region, and the holes move towards the N-type region. When they reach each other, they combine and enabling a current to flow across the junction. When a current flows between Base-Emitter, electron leave the Emitter and flow into the base. Normally the electrons would combine when they reach this area. However the doping level in this region is very low and base is also very thin.

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3.1.2 Basic transistor operation

6. 7.

This means the most of the electrons are able to travel across this region without recombining with the holes. As a result, the electrons migrate towards the Collector, because they are attracted by the positive potential.

Electron Direction Flow in Transistor

3.1.2 Basic transistor operation The diagram above shows the flow of electrons in NPN bipolar transistor. By Kirchhoff's Current Law, we have :-

I E = IC + I B 3.1.2 BJT Operation BJT can operates as a (a) Switch (b) Amplifier

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3.2 Characteristic and Operations of BJT 3.2.1 Characteristic Curve and the Operating Regions of BJT. These are called ‘transistor-characteristic’ curves. When base current ( IB ) increases, the collector current ( IC ) increases. In the linear or amplification area, Beta, β , the gain, is constant.

The direct current gain, βdc = hFE = IC / IB = 1mA/20µA = 50 The parameter hFE indicates what is known as the DC Forward current transfer ratio of a transistor in the common-Emitter mode.

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1. Cutoff region: (i) B-E junction is reverse biased. (ii) No current flow , high voltage (iii) when BJT is used it will act as a open switch) 2. Saturation region: (i) B-E junction forward biased C-B junction is forward biased. (ii) High current, low voltage.(Ic reaches a maximum which is independent of IB and β. (iii) when BJT is used it will act as closed switch) VCE common (neutral) IE value(Emitter current) depends on value of forward bias.(VEE ). When VEE value increased or decreased, IE value also will increase or decrease. So, IC also lower or higher because it’s always depends on IE but the value is a little bit lower than IE.

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Common Base (CB)

Input signal

Equation at Emitter:

Output signal

Equation at Collector:

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Refer to Figure 6.5 below, if transistor type is Silicone and value of IC = IE. Calculate value of Vo.

(B) Common Emitter Configuration (CE) Emitter become common (neutral) to base and collector circuit.

RB resistor - to limit the IB current so not to high and can destroy B-E junction. RC resistor is connected at Collector and known as 'Load resistor'. The function is to drop the Collecter voltage, Vc

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Based on Kirchoff's law and Ohm law:-

Equation at Base:

Equation at Collector:

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Example 3.1 : Refer to figure 6.15 , if the Beta factor transistor is 120 and the transistor type is silicone, Calculate IC value and VC. Solution 3.1 :

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3.3.1

Circuit of three basic transistor configuration

RE resistor - to limit the IE current so not to high and can destroy E-B junction. RL resistor -known as 'Load resistor'. Function - to drop the Collecter voltage, Vc when IC went through it.

3.3.2

Electrical Characteristic for Each Configuration

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3.4 DC Operation of BJT 3.4.1 β DC and β AC Beta Factor can be divided to 2 situations :-

β DC : Value of collector current(IC) that flow according to value Base current ( IB) but Vc is permanent.

β AC

: Changes happened at collector current when Base current change but Vc is permanent.

Example 3.2 : A 20µA base current flows through an emmiter-base transistor, with collector current become 2mA. After that,the base current was changing to 40µA, while the collector current also changing to 4mA. Determine the β d.c and βa.c.

Solution 3.2 :

β DC = IC/IB = 2mA/20µA = 100

β AC = ∆IC/∆IB = (4mA – 2mA) / (40µA - 20µA) = 100

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3.4.2

DC Operating Point (Q-Point) The point which the base current intersects the load line is referred to as the operating point or the quiescent point. A line which connecting the cutoff point, q-point and saturation point called d.c. load line

DC saturation point or IC(sat) is level where IC at maximum value and Vc=0.

IC(D.C. SAT.) = VCC

DC cut-off point is at ideal cut-off where Ic =0 in the circuit and value VC=VCC

VCE(D.C. CUT-OFF) = Vcc

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3.4.3 Current & Voltage Calculation Of Common-emitter Amplifier Equation :-

Equation :-

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Example 3.3 :

Refer to circuit at figure 6.25, Determine :(i) Base current; IB (ii) Collector current; IC (iii) Emitter Voltage Solution 3.3 : i. Base current; IB

Type of transistor not given, so VBE = 0

Given β =50 (refer to figure) Equation :-

Example 3.4 : Refer to circuit below, if transistor used is Silicone, Determine the value of: (i ) Base current, IB (ii) Collector current, IC (iii) Collector Voltage, VCE

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Solution 3.4 : i. Base current; IB

(ii) Collector current, IC Given β =100

Given type of transistor is Silicone, so VBE = 0.7 (iii) Collector Voltage, VCE

3.4.4 : 3.4.5 : 3.4.6

DC Saturation Point & DC cut-off point

Example 3.5 : Draw DC loadline for emitter common circuit and show the location of Q-point. Assume β=100

Solution 3.5 : Operational point (Q-Point):

Type of transistor not given, so VBE = 0

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D.C. SATURATION POINT

D.C. CUT-OFF POINT

EXERCISE Example 3.6 : Based on figure below, find out DC load line of the circuit given

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Solution 3.6 : Operational point (Q-point):-

Type of transistor not given, so VBE = 0

D.C. SATURATION POINT

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D.C. CUT-OFF POINT

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3.4.7

Operation of CE Amplifier With AC input signal 1. If Q-point during IB=60µA, IC=6mA dan VC=8V. Assuming IB swing from 40µA to 80µA during a.c. input signal, meanwhile IC swing from 4mA to 8mA, therefore causes VC swing between 4V to 12V.

2. i. VC is a.c. output voltage, and swing at 1 complete cycle at 8Vp-p. Note: Output voltage is 180° out of phase with the input voltage.

3. If a.c. input signal too large, IB also too large (refer to dash) -IC also too large and followed by VC. -We found that IC and VC swing over than saturation point. At this point, gain will not happen. So, output will be clipped and distortion.

3.4.7

Operation of CE Amplifier With AC input signal 1) When AC. signal is applied, load at output loop sometimes different compared DC. signal analysis. 2) Capacitor will allowed AC signal flow through it. So, load that are facing at output is rL=RC//RL. 3) Collector current value will not drop at DC load line again, so in AC analysis it will have a new load line. It also have its own saturation point and cut off. This line known as ‘AC load line’. 4) Formula to get AC load line

VCQ

(a)

I C ( AC .SAT ) = I CQ +

(b)

VC ( AC .CUT −OFF ) = VCQ + I CQ .rL

(c)

rL = R C // R L =

rL

RC × R L RC + RL

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Example 3.6 : Base on Figure 6.35, draw DC. and AC load line for the circuit. Locate the location of Q-point.

Solution 3.6 : Operational point (Qpoint):-

D.C. LOADLINE

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Solution 3.6 :

3.4.8

3.4.9

Comparison AC and DC Load line in terms of amplitude and phase shift

Voltage Gain Based, AV = VOUTPUT / VINPUT 1) Definition : ratio between output voltage and input voltage

Another formula can be used is :-

Where

Whereas

I E = IC

re’ – resistance internal of transistor(ac emitter resistance)

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3.4.10

Maximum output voltage value without distortion, Vo(max)(without distortion). 1) Maximum output voltage without distortion means output signal which is swing at operation point with symetry and without limited/clipped.

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3.4.11

Undistorted output Signal Amplitude to Maximum Input Signal 1) If we know the value of maximum output voltage without distortion (Vo(max)(without distortion)), so we can determine the value of input voltage without distortion (Vin(max)(witout distortion)). But, the value of voltage gain for the circuit need to know first. Example 3.7 : Refer to the maximum output voltage without distortion (Vo(max)(without distortion)) at figure 6.37, get the value of input voltage without distortion (Vin(max)(without distortion)) for amplifier if voltage gain, AV circuit is 100.

Solution 3.6 :

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Determine the value of Q-point, IC(DC sat), VC(DC cut-off), IC(AC sat), Vc (AC cut-off), and draw DC. and AC load line at the same graph. From the graph, locate maximum output voltage without distortion and get maximum input voltage without distortion

A.C. LOADLINE

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