Lecture-15 BJT

The inner junction capacitances are shown in Fig. They are not physically connected but inherently present with the device.

1

The capacitor Cbc between the base and the collector connects the output with the input. Because of this, negative feedback takes place in the

circuit and the gain decreases. This feedback effect is more, when Cbc provides a path for higher

frequency ac currents1

Xc =2πfC 2

The capacitance Cbe between base

and Emitteroffers a low input impedance at higher frequency thus reduces the effective input signal and so the gain falls.

3

The capacitance Cce between

collector and Emitter

Similarly, Cce provides a shunting effect at high frequencies in the output side and reduces gain of the amplifier.

4

Besides these junction capacitances there are wiring capacitance CW1 & CW2.

These are relatively very small but at high frequencies they become 5 to 20 p.f. For a multistage amplifier, the effect of the capacitances Cce, CW1 and CW2 can be represented by single shunt capacitance.

CS = CW1 + CW2 +Cce.5

At higher frequency, the capacitor CS offers low input impedance and thus reduces the output.

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The gain is constant over a frequency

range. The frequencies at which the gain reduces to 70.7% of the maximum gain are known as cut off frequencies, there are upper cut off frequency and lower cut off frequency.

Bandwidth of an amplifier (BW)

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Fig. shows these two frequencies. The difference of these two frequencies is called Bandwidth (BW )of an amplifier.

Therefore BW = f2 – f1. 9

At f1 and f2, the gain voltage becomes 70.7% i.e. (Am / 2). The output voltage reduces to (1 / 2 )of maximum output voltage.

Since the power is proportional to voltage square the output power at these frequencies becomes half maximum power.

The gain on dB scale is given by

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20log10(V2 / V1)

= 20 log10(0.707)

= 10 log10 (1 / 2)2

= 10 log10(1/2)

= - 3dB.If the difference in gain is more than 3 dB, then it can be detected by human. If it is less than 3 dB it cannot be detected.

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Finding the two Cutoff Frequencies

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For a CE Amplifier

Zin = R1 || R2 || Zin(base)

The input coupling capacitor is in series with RS and Zin

From the circuit theory at XC = R the current reduces to 0.707 point where the R is total series resistance

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I 2 = V2 / ( R2 + Xc 2 )

At R = Xc

I2 = V2 / ( 2.R2 )

= 0.707 (V/R)

Therefore

(1/2πfCin) = (RS + Zin)

Solving for frequency we get

fin = 1/(2π(RS + Zin). Cin)

this is the cutoff frequency introduced by input coupling capacitor

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on the output side the capacitor Cout is in series with the output impedance RC and load resistance RL

Similarly the output ac current is down to 0.707 point when

XC = R

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Or

(1/2πfCout) = (Rc + RL)

Solving for frequency we get

fout = 1/(2π(Rc + RL). Cout)

this is the cutoff frequency introduced by output coupling capacitor

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The Bypass capacitor sees a resistance of RE in parallel with the output impedance of the emitter

Zout = re’ + (RS || R1 || R2)/β

Applying the same condition at XC = R

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We get

(1/2πfCb) = (RE||Zout)

Solving for frequency we get

fE = 1/(2π(RE||Zout). Cb)

this is the cutoff frequency introduced by output bypass capacitor

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The three capacitors produce three different cut off frequencies

fin = 1/(2π(RS + Zin). Cin)

fout = 1/(2π(Rc + RL). Cout)

fE = 1/(2π(RE||Zout). Cb)

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The highest one is most criticalBecause that is the one where the voltage gain first reduces from the pass band valuesUsually these three are different

While analyzing the amplifier highest of the Lower Cut off frequencies is chosen for lower cut off

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The ac α of the transistor is ratio of ac collector current and ac emitter current

α = iC / iE

at low frequency α approaches unity but as frequency increases after a point alpha start decreasing

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Alpha cutoff frequency fα

defined as the frequency at which alpha drops to 0.707 of its lower frequency value

fα is one of the limitation in CB amplifier 23

Beta cut off frequency fβ

is defined as the frequency at which β of the transistor is reduced to 0.707 of its low frequency value

Current bandwidth product fγ

is defined as frequency at which β equals unity24

Further

fβ = fγ / β and

fγ = fα / 1.2 approx

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fαIs Limitation of CB Amplifier

, are Limitations of CE Amplifier

fβ fγ

h- Parameters

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All the transistor amplifiers are two port networks having two voltages and two currents. The positive directions of voltages and currents are shown in fig..

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Out of four quantities two are independent and two are dependent. .

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If the input current i1 and output voltage v2 are taken independent then –

other two quantities i2 and v1 can be expressed in terms of i1 and v2.

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1 1 1 2

2 2 1 2

v = f (i ,v )

i = f (i , v )31

The equations can be written as

1 11 1 12 2

2 21 1 22 2

v = h i + h v

i = h i + h v

where h11 , h12, h21 and h22 are called h-parameters.

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Defining

h11

111

1v =02

vh =

i

h11 designated as hi & is

input impedance with output short circuit to ac.

1 11 1 12 2

2 21 1 22 2

v = h i + h v

i = h i + h v

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Defining

h12

h12 designated as hr is fraction of output voltage at input with input open circuited or

reverse voltage gain with input open circuited to ac (dimensionless)

1 11 1 12 2

2 21 1 22 2

v = h i + h v

i = h i + h v

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Defining

h21

221

1 v =02

ih =

i

designated as hf And is Negative of current gain with output short circuited to ac.The current entering the load is negative of I2. This is also known as

forward short circuit current gain.

1 11 1 12 2

2 21 1 22 2

v = h i + h v

i = h i + h v

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Defining

h22

designated as ho & is

Output admittance with input open circuited to ac

1 11 1 12 2

2 21 1 22 2

v = h i + h v

i = h i + h v

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If these parameters are specified for a particular configuration, then suffixes e, b or c are also included corresponding to CE , CB and CC Configurations respectively

hie ,hre ,hfe ,hoe …….. CE hib ,hrb ,hfb ,hob …….. CB

hic ,hre ,hfc ,hoc …………. CC41

Using two equations the generalized model of the amplifier can be drawn as shown in fig.

1 11 1 12 2

2 21 1 22 2

v = h i + h v

i = h i + h v

Reverting back to two equations

h11 hie Input impedance with a shorted output

h12 hre Reverse voltage gain with a open input

h21 hfe Current gain with a shorted output

h22 hoe Output admittance with an open input 42

hybrid model for a transistor amplifier

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Let us consider CE configuration as show in fig.. The variables iB, iC,vC & vB represent total instantaneous values iB and vC can be taken as independent variables and vB, IC as dependent variables

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ThereforeVB = f1(iB, Vc) & Ic = f2 (iB, Vc).

1 1B B C

B CC B

2 2C

B CC B

f fΔv = Δi + Δv

i vv i

f fΔi = Δi + Δvc Bi vv i

Using Taylor’s series expression, and neglecting higher order terms we obtain

1 1 1 2

2 2 1 2

v = f (i ,v )

i = f (i , v )

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The partial derivatives are taken keeping the collector voltage or base current constant.

The ΔVB, Δ Vc, Δ iB, Δ ic represent the small signal (incremental) base and collector current and voltage and can be represented as Vb ,ib, Vc ,ic.

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1 1B B C

B CC B

2 2C

B CC B

f fΔv = Δi + Δv

i vv i

f fΔi = Δi + Δvc Bi vv i

VB = f1(iB, Vc) & Ic = f2 (iB, Vc).

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The Hybrid model for CE configuration

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Determination of h - parameters

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To determine the four parameters input and output characteristic are used.

Input characteristic depicts the relationship of between input voltage and input current with output voltage as parameter.

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The output characteristic depicts the relationship between output voltage and output current with input current as parameter. From the output characteristic shown.

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C C2 C1fe

B b2 b1C

i - iih = =

i i - iV

The current increments are taken around the quiescent point Q which corresponds to iB = IB and to the collector voltage VCE = VC

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The value of hoe at the quiescent operating point is given by the slope of the output characteristic at the operating point (i.e. slope of tangent AB)

Coe

CB

ih =

V i

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hie is the slope of the appropriate input Chs on fig. at the operating point (slope of tangent EF at Q).

Bie

BBC

V VBh =i i V

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A vertical line on the input characteristic represents constant base current. The parameter h re can be obtained from the ratio (VB2 – VB1) and (VC2 – VC1) for at Q.

B B B2 B1re

CC C2 C1B

V V - VVh =

V V V - V I

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Typical CE h-parameters:

(Transistor 2N1573).

hie = 1000 ohm.

hre = 2.5 * 10 –4

hfe = 50

hoe = 25 mA/V

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r parameter h parameter

β hfe

α hfe /( hfe+1)

re’ hie /hfe

rc’ hfe/ hoe

rb’ hrb/ hob

Approximate relations

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Typical CE h-parameters:

(Transistor 2N1573).

hie = 1000 ohm.

hre = 2.5 * 10 –4

hfe = 50

hoe = 25 mA/V

What is the value of α, β, re ’

& rc’

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std CE CC CB meaning

h11 hie hic hib Input impedance with a shorted output

h12 hre hrc hrb Reverse voltage gain with a open input

h21 hfe hfc hfb Current gain with a shorted output

h22 hoe hoc hob Output admittance with an open input

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Consider the CE amplifier as two-port network. Rs is the source resistance and ZL is the load impedance. h-parameters are assumed to be constant over the operating range. The ac equivalent circuit is shown

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The quantities of interest are the 1.current gain, 2.input impedance, 3.voltage gain, and 4.output impedance

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Current gain:

For the transistor amplifier stage,Gain Ai is defined as the ratio of output

to input currents.

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CLi L C L C

b b

C fe b

L L C L

C fe b oe C L

C fe

b oe L

fei

oe L

-IIA = = (I +I = 0. I = -I )

I I

I = h I + h Voe cV = I Z = -I ZcI = h I + h (-I Z )

I hor =

I 1+h Z

hA = -

1+h Z

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Input Impedance:The impedance looking into the

amplifier input terminals (1,1’) is the input impedance Zi

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Voltage gain:

The ratio of output voltage to input voltage gives the gain of the transistors

C C Lv

b b

B i L i Lv

b i

V I ZA = = -

V V

I .A .Z A ZA = =

V Z

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Output Admittance:

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Voltage amplification taking into account source impedance (Rs) is given by

C C b s = *b iVS +s is b S

iV

i s

i L

i s

V V V VA = = * V ZR ZV V V

Z= A .

Z +R

A Z=

Z +R

AV is the voltage gain for an ideal voltage

source (Rs = 0). 71

parameter Exact Approximate

AVh21.ZL/((h11.(1+h22.ZL)-h12.h21.ZL) h21.ZL/h11

Aih21/(1+ h22. ZL) h21

Zinh11 – (h12.h21/(h22+1/ZL)) h11

Zout(RS+h11)/(h22.( RS+h11)- h12.h21 1/h22

Summary72