Bipolar Junction Transistor(part 2)

3.0 BIPOLAR JUNCTION TRANSISTORS(BJT)-PART 2

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3.5 Frequency response curve 3.5.1 Frequency response curve of an amplifier 3.5.2:3.5.3 Important parameters of a frequency response curve: a. Maximum voltage gain in decibel unit, Avmax(dB) b. Cut-off frequencies / f(-3b) c. Frequency bandwidth A frequency-response curve is a graphical representation of the relationship between amplifier gain and operating frequency. A generic frequency response curve is shown in figure below. This particular curve illustrates the relationship between power gain and frequency.

figure 3.5: Frequency response curve

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Which Response Curve is Best? •When a system or component reproduces all desired input signals with no emphasis or attenuation of a particular frequency band, the system or component is said to be "flat", or to have a flat frequency response curve. •For an example, an ideal "flat" frequency response means that the microphone is equally sensitive to all frequencies. •In this case, no frequencies would reduced (the chart would show a flat line = B is better than A), resulting in a more accurate representation of the original sound. •We therefore say that a flat frequency response produces the purest audio.

3.5.4 The term decibel, AV max and Avmax(dB) Av (max) is the actual power gain, Vout / Vin Av max (dB) is a logarithmic measurement of the ratio of output voltage to input voltage Av max (dB) = 20 log Av max Example : Express each of the following ratios in dB a. Vout / Vin = 250 b. Av = 10 Solution : a. Av (dB) = 20 log 250 = 48 dB b. Av (dB) = 20 log 10 = 20 dB

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3.6 Frequency response characteristics of an amplifier. 3.6.1 Calculate the maximum voltage gain in decibel, Avmax(db), using the formula Avmax(dB)=20 log Avmax.

1.5 V

5V

Amplifier In

Out

Avmax = Vout Vin = 5V 1.5V = 3.33

Avmax(db) =20 log Avmax = 20 log 3.33 = 10dB

3.6.2 Explain cut off frequencies or frequency at -3b gain •

Cut-off frequency is the frequency at which the response of an amplifier is 3 dB less than at midrange

•

The voltage gain is down 3dB or is 70.7% of its midrange value : 20 log 0.707 = -3dB

•

These cut-off frequencies are referred to as the lower cutoff frequency,fc1 and upper cut-off frequency, fc2

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3.6.3 Example : Determine the cut-off frequencies (fc1 and fc2) from the frequency response diagram.

Solution : The two points in red on the response curve mark where the output of the amplifier has fallen to 70.7 % of the maximum output. This means that that the 100mV output has fallen to 70.7 mV at these frequencies. These are called the -3 dB points.

Therefore, fc1 = 5Hz and fc2 = 900 kHz.

3.64:3.65 Determine frequency bandwidth from the formula BW = fc2 – fc1 • The bandwidth of the amplifier is the range (band) of frequencies lying between fc1 and fc2 • The frequency bandwidth is expressed in units of Hertz (Hz) as BW = fc2 – fc1 Example : Determine the frequency bandwidth from the frequency response curve

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Exercise: In an amplifier, the maximum voltage gain is 2000 and occurs at 2 kHz. It falls to 1414 at 10 kHz and 51 Hz. Find: i) Bandwidth ii) Lower cut-off frequency iii) Upper cut off frequency

Solution: i) Referring to the frequency response in figure below, the maximum gain is 2000. Then 70.7% of the gain is 0.707 x 2000=1414.It is given that gain is 1414 at 50Hz and 10kHz. Bandwidth = 10kHz – 50Hz = 9950Hz

ii)

Lower cut-off frequency = 50 Hz

ii)

Upper cut-off frequency = 10kHz

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3.7 Classification of amplifier 3.7.1 Distinguish A, B, AB and C class Amplifiers. Amplifiers are classified according to their circuit configurations and methods of operation into different classes such as A, B, AB and C.

Class A Amplifier Class A Amplifier operation is where the entire input signal waveform is faithfully reproduced at the amplifiers output as the transistor is perfectly biased within its active region, thereby never reaching either of its Cut-off or Saturation regions. The output characteristics with operating point Q are also shown. ICQ and VCEQ represent no signal collector current and collector-emitter voltage respectively. When ac input signal is applied, the operating point Q shifts up and down causing output current and voltage to vary about it. The output current increases to Ic max and falls to Ic min. Similarly, the collector-emitter voltage increases to Vce max and falls to Vce min

This then results in the AC input signal being perfectly "centered" between the amplifiers upper and lower signal limits as shown below.

Class A Amplifier output characteristic – AC load line

Class A output waveform

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Class B Amplifier In class B operation the transistor is so biased that zero-signal collector current is zero. Hence class B operation does not need any biasing system. The operating point is set at cut-off. It remains forward biased for only half cycle of the input signal i.e its conduction angle is 180 degree. As illustrated in figure, during the positive half cycle of the input ac signal, the circuit is forward biased and, therefore, collector current flows. On the other hand, during negative half cycle of the input ac “signal, the circuit is reverse biased and no collector current flows.

Therefore, at zero input there is zero output. This then results in only half the input signal being presented at the amplifiers output giving a greater efficiency as shown below.

Class B Amplifier output characteristic – AC load line

Class B output waveform

Class AB Amplifier

The Class AB Amplifier is a compromise between the Class A and the Class B configurations above. While Class AB operation still uses two complementary transistors in its output stage a very small biasing voltage is applied to the Base of the transistor to bias it close to the Cut-off region when no input signal is present. An input signal will cause the transistor to operate as normal in its Active region thereby eliminating any crossover distortion which is present in class B configurations. A small Collector current will flow when there is no input signal but it is much less than that for the Class A amplifier configuration. This means then that the transistor will be "ON" for more than half a cycle of the waveform. This type of amplifier configuration improves both the efficiency and linearity of the amplifier circuit compared to a pure Class A configuration.

Class AB output waveform

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Class C Amplifier The operating point of the class "C" amplification is not on the load line. The part of the input signal is amplified. So, it isn't possible to use for the amplification such as the sound. It is used for the high frequency multiplying circuit and so on.

3.7.2 : 3.7.3 Compare the different classes Class

A

B

AB

C

Conduction Angle

360o

180o

180 to 360o

Less than 90o

Position of the Q-point

Centre Point of the Load Line

Exactly on the X-axis

In between the X-axis and the Centre Load Line

Below the X-axis

Overall Efficiency

Poor, 25 to 30%

Better, 70 to 80%

Better than A but less than B 50 to 70%

Higher than 80%

Signal Distortion

None if Correctly Biased

At the X-axis Crossover Point

Small Amounts

Large Amounts

Signal amplifier

Audio amplifier

Audio amplifier

Radio RF transmitter tuned amplifier

Application

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“Why need to avoid Beta factor?” This is due to Beta factor have relationship with leakage current ICBO in circuit . Leakage current occur because changes of the environment temperature. So, we can call that Beta value is always change depends on environment temperature.

3.8 Other biasing techniques of Common Emitter Amplifier 3.8.1 One of the bias technique which is stable and can avoid Beta factor changes with components connection characteristics in circuits are : i. Base bias technique with emitter feedback. ii. Biased voltage divider technique . i. Common-emitter Amplifier Using Base Bias With Emitter Feedback.

The amplifiers bias voltage can be stabilized by placing a single resistor, RE in the transistors emitter circuit as shown. This resistance is known as the Emitter Resistance. This resistor introduces negative feedback that stabilizes the Q-point. From Kirchhoff's voltage law, the voltage across the base resistor is VRb = VCC - IeRe - Vbe. From Ohm's law, the base current is Ib = VRb / Rb.

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The function of emitter resistance and shunt capacitor

In the basic series feedback circuit above, the emitter resistor (RE) performs two functions: DC negative feedback for stable biasing and AC negative feedback for signal transconductance and voltage gain specification. But as the emitter resistance RE is a feedback resistor, it will reduce the amplifiers gain due to fluctuations in the emitter current IE owing to the AC input signal. This is overcome by placing “a shunt capacitor” CE across the emitter resistance. Over the operating range of frequencies, the capacitors reactance will be extremely low giving little AC feedback effect but good AC gains.

Existing of resistor, RE at emitter division will cause voltage gain formula changed:

⇒ AV =

rL re'+ R E

Therefore, definitely voltage gain will decrease as usual due to additional of RE value in formula. This is the weakness when using the biasing technics. To avoid this problem, one capacitor will be joint parallel together with RE in the circuit. Vcc RC

RB

Q1

RE

CE

CE capacaitor usually known as shunt capacitor because the function is to shunt RE from pass through by a.c. signal. On the other hand, its reduced input circuit resistance. Formula for voltage gain will be change as before :

⇒ ΑV =

rL re'

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Example: Base on figure below, calculate IC, VC, IC(SAT), voltage gain. If one shunt capacitor, CE is connecting parallel with RE, calculates new voltage gain for the circuit. Given β = 100.

Vcc 30V RC 2k

RB 300k

Q1

RE 1k

Solution :

V CC

⇒ IC = R

E

=

+

R

B

β

30 V 300 100

= 7 . 5 mA

1k Ω +

⇒ V C = V CC − I C . R

L

= 30V - (7.5mA

⇒ I C(sat) =

⇒ re' =

VCC RE + RL 30V = = 10mA 2kΩ + 1kΩ

25mV = 3.33Ω 7.5mA

⇒ ΑV =

rL rL = ri re' + R E 2kΩ = = 1.99 3.33Ω + 1kΩ

)(2 k Ω ) =

15 V

If CE is added in the circuit:

rL re' 2kΩ = 3.33Ω = 600

⇒ AV =

By observation, circuit during using shunt capacitor and without using the capacitor is different.

Advantages :

Then to summarize, the bias of a common emitter amplifier circuit on the transistor input can be stabilized against changes in supply voltage by the addition of an Emitter Resistance, RE. In order to overcome any loss of gain with AC signals, a shunt capacitor, CE is connected in parallel with the emitter resistor.

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ii. Biased voltage divider technique Vcc R1

RL

Q1

R2

RE

CE

Figure above shows the voltage divider bias circuit. R1 and R2 network actually voltage divider network. Therefore, this bias technique called voltage divider bias technique. Voltage drop at R2 resistor (VB) is accumulated voltage at base.

Let’s observe formula derivation to get circuit operation current as shown in figure below: Input Loop:

Vcc R1

Q1

R2

RE

CE

⇒ VCC = VR1 + VR2 ⇒ VR2 =

R2 × VCC R1 + R 2

⇒ VR2 = VB ⇒ VB = VE + VBE ........Ignore..VBE ⇒ VE = IE .R E ⇒ IE =

VB RE

⇒ I E = IC ⇒ IC =

R2 V × CC R1 + R2 RE

To calculate operation current ( IC ) above, totally not involve Beta factor (β β ). So, the bias technique is good

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Output loop: Vcc RL

Q1

CE

RE

⇒ V C = V CC − I C R ⇒ Current ⇒ I C (sat)

operation V CC = RE + RL

L

equation

for saturation

is :

Input, output resistance and Voltage Gain circuit: When a.c. signal inject to the circuit, we can draw a.c. equation circuit as below: ib

Vm

ic

R1

R2

RL

βre'

rm

VK

rl

By a.c. equation circuit, overall resistance at input and output stage can be calculated: rm = R1 // R2 // βre'

rl = RL

Circuit voltage gain : Av =

rl re'

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Example:

Base on figure 7.10 below, if R1=R2=6.2kΩ, RL=1kΩ, β=100 RE = 2kΩ and Vcc=12V ;

Vcc RL

R1

C2

C1 VC R2

VB

RE

CE

Ignore VBE, calculate VB, IE current, VC and gain voltage gain if Vi = 1.0mVp-p. Solution:

VB =

=

R2 xVcc R1 + R2

6.2kΩ x12 6.2kΩ + 6.2kΩ

=

6.2kΩ x12 12.4kΩ

= 6Volt

Ignore VBE , VE=VB : VE = I E .RE = 6

IE =

6 2 kΩ

I E = 3mA

Equation at output loop, Vcc = VRC + VC 12V = I C .RC + VC VC = 12 − ((3mA)( . 1kΩ )) = 12 − 3 = 9V

Voltage gain if Vi = 1mVp-p (Change to a.c circuit equation)

rm = R1// R 2 //

β .25mV

rL = 1kΩ

3mA

= 6.2 kΩ // 6.2 kΩ // 833Ω

= 657Ω

Ib =

Vi 0.5mVp = = 0.79µA ri 657Ω

I C = β .ib = 100 x 0.76 µA = 0.076mA

VC = Ic.RL = 0.076x10−3 x1KΩ = 0.076Vp Av= Vo = 0.076Vp = 152 Vi 0.5mVp

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Exercise: For the transistor amplifier shown in figure 1, R1= 10KΩ, R2=5KΩ, RC=1KΩ, RE=2KΩ and RL=1KΩ. Assume VBE= 0.7V

i. ii. iii.

Draw d.c load line Determine the operating point Draw a.c load line

Solution : i.

VCE cut off = VCC = 15V

IC sat = VCC / (RC+RE) = 15 / (1KΩ + 2KΩ) = 5mA Dc loadline: ii.

V2 = VBE + IERE

IE = (V2 – VBE)/RE = (5 – 0.7)/ 2K = 2.15mA.

VCE = VCE – IC (RC + RE) = 15 -2.15mA(3K) = 8.55V

Operating point, VCEQ = 8.55,

ICQ = 2.15mA.

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iii. a.c load, RAC = RC//RL =(1K x 1K) / (1K + 1K) =0.5 KΩ ICsat = VCE + IC. RAC = 8.55mA + 2.15mA. 0.5KΩ = 9.62V. VCcut off = IC + VCE/RAC =8.55 + 8.55/0.5KΩ = 19.25mA.

a.c Loadline :

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