www.bookspar.com | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS | FORUMS www.bookspar.com | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS | FORUMS 1 The Transistor At Low frequencies Graphical Analysis of CE configuration Consider a CE amplifier as shown below • V CC and V BB provide the biasing for the transistor • R L is the load resistor • R S is the source resistance The input and output characteristics are shown below B Q A i C v BE V i B v CE R L V CC R B B Collector voltage v CE , V Collector current i C , mA I B3 I B2 I B I B1 I B0
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The intersection of input characteristics, corresponding to quiescent base current IB , and the load line results in the operating point Q. corresponding to this operating point, the base current, collector current, base voltage and collector voltage are indicated by the symbols IB, IC, VB and VC respectively
Due to the application of ac signal, the base current is subjected to variation around the q point. The instantaneous value of current and voltage are indicated by lower case letter with an upper case subscript ( for ex: iB, iC, vb, vc).
The instantaneous value of the varying component from the quiescent value is indicated by lower case letter with a lower case subscript (for ex: ib, ic, vb, vc). The following relations can be established ic=ic-Ic = Δ ic Ib = iB-IB = Δ iB Vc =vc-Vc= Δ Vc Vb = vB - VB = Δ vB
Thus, variation of collector current and voltage can be seen as alternating current and voltages. The waveforms When an AC signal is applied, the instantaneous base current will fluctuate between two extreme values iB1 and iB3 as indicated in the output characteristics . The corresponding Q point will fluctuate between two points A and B on the DC load line the corresponding collector current and voltage variations are as indicated in the output characteristics. Due to the nonlinearity of the output characteristics, the output characteristics are not parallel lines.
Also these characteristics are not at equally spaced for equal increments in the base current. This results in the collector current and voltage waveforms being non sinusoidal. This called output linear distortion. The input characteristics is non linear, therefore the base voltage Vb is not identical to the base current ib, which is sinusoidal. This change in waveform is known as input nonlinear distortion. Due to the non-linearity for large voltage , the base current swing is not symmetrical around the Q point. The base current swing and hence collector current swing is larger in BQ region then in QA region. Thus if the operating point Q is situated in linear region, then for small changes , the distortion would be less. Under this condition of small amplitude of the input signal, linear circuit model can be used. NOTE: Refer to PPT slides for waveforms and diagrams
SMALL SIGNAL MID FREQUENCY ANALYSIS OF AMPLIFIER
MODELING OF TRANSISTOR AMPLIFIER
To analyze and assess the performance of transistor amplifiers, equivalent circuits are used. The behavior and performance of the transistor can be expressed by a set of mathematical equations. These equations are called a mathematical model. These equations are based on suitable theory. For example, considering the transistor to be operating in linear region it can be modeled with help of two-port network theory. The solution of these mathematical equations results in the response of the amplifier for a given excitation.
These mathematical equations can be synthesized in to an electrical network called equivalent circuit. This equivalent circuit along with the external components conned to the network represents the transistor amplifier. The solution of the equivalent circuit provides to the response of the system.
The elements of the equivalent circuit are called the parameters of the transistor. Various parameters can be defined based on the requirements of the designer. For example h-parameters are used for mid and low frequency analysis and hybrid-pi parameters are used for high frequency analysis.
Besides the above two models, Z-parameters, Y-parameters or A-B-C-_D parameters can also be used. Identification of proper independent and dependent variables results in suitable model.
THE TRANSISTOR HYBRID MODEL:
Consider the amplifier circuit as shown in the fig. A common emitter configuration is considered. The transistor can be considered as a linear device in the active region for small signal amplitude. Then, the two-port network theory can be applied to the transistor. According to this theory, the four variables of the transistor can be related by the following set of equations. For defining h-parameters, the quantities i b
(input current) and vce (output voltage) are taken as independent variables and the remaining two quantities are represented in terms of independent variables Vobe= f1(ib , v ce ) ic = f2(ib ,vce )
ΔvBE = δf1 Δ iB + δ f1 ΔvCE δ iB δ vCE ΔiC = δf2 Δ iB + δ f2 ΔvCE δ iB δ vCE ΔvBE, Δ iC, Δ iB , ΔvCE represent small change in the base and collector voltage and currents. Rewriting the above equation vbe = δf1 ib + δ f1 vce δ iB δ vCE iC = δf1 iB + δ f2 vce δ iB δ vCE
All the terms shown in the parenthesis which are the co-efficients of the equations also called transistor parameter. The above equations are rewritten as vbe = hie ib + hre vce --------------------------------(1) ic = hfe ib + hoe vce---------------------------------- (2) Put vce =0 in the equation (1), vbe = ib * hieor hie = vbe/ib Ohm = ΔvBE = δvBE ΔiB δ iB Thus, hie has the unit of resistance. The terms vbe and ib represents input voltage and input current. Therefore, hie is defined as the input resistance of the transistor when the output terminals are short-circuited. Put ib = 0 in equation (2), ic = vce * hoe or hoe = ic/vce mho ΔiC = with ΔvCE =0 ΔvCE δ iC = with ΔvCE =0 or at Vc δvCE Thus, hoe has the unit of conductance. The terms ic and vce represents the output current and voltage of the transistor. Therefore, hoe is called output conductance of the transistor when the input terminals are open circuited. Put vce = 0 in equation (2), ic = ib * hfe or hoe = ic/ib no-unit ΔiC = with vce = 0 Δ ib Thus, the parameter hfe is dimensionless and is equal to ratio output current to input current. Therefore, hfe is called current gain of the transistor when the output terminals are short-circuited. It is also called simply the short-circuit current gain. Put ib =0 in equation (1), vbe= vce * hre or hoe=vbe/vce no-unit. ΔvBE = with iB = constant Δ vCE
δ vBE = at IB = constant δ vCE Thus, hoe is also dimensionless. The above ratio represents the reciprocal of voltage gain (voltage gain is the ratio of output voltage to input voltage). Therefore, hoe is called reverse voltage ratio of reciprocal of voltage gain when the input terminals are open-circuited. Since the units of all parameters are different, they are said to be hybrid in nature or simply h-parameters. To obtain the equivalent circuit, the equation (1) and (2) can be used to write the following circuit. The equation (1) is a KVL, hence it can be realized with a series circuit consisting of a voltage source vs , a resistance hie and a voltage source hre* vce. The equation (2) represents a KCL, hence it can be realized by a parallel circuit with three branches. One branch has a dependent current source hfe* ib , the second branch is resistor of value 1/hoe and the third branch is the total current ic.
Analysis of CE amplifier
h- parameters equivalent circuit To find current gain
Applying KVL to input circuit Vs = hie ib + hre vce Vs = ib hie + hre iL RL from equation (1) Vs = ib hie + hre Ai ib RL ( iL =Ai ib ) Substituting in equation (2)
By inspection, RS is in series with the entire circuit whose resistance on the input side is Ri Ris = RS + Ri Ris = 1.98KΩ To find current gain Ais Converting the voltage source in to current source
Question:In a transistor amplifier, explain the effect of internal resistance of ac voltage source on the performance of the amplifier. Derive the necessary equations in support of the answer. CKT DIAGRAM Let Ri is the input resistance of the amplifier without Rs- internal resistance.
Let Ris is the input resistance of the amplifier with RS
From the above circuit b
sis i
vR =
Ris = RS + Ri ( Since Rs is in series with input circuit) Thus due to the internal resistance Rs, the input resistance of the amplifier increases. It is desirable since the input resistance of the amplifier should be high to avoid loading effect on the previous stage of the amplifier, the input resistance of the amplifier should be high. Then the amplifier will draw less input current from the previous stage. Loading effect is the drop in the input voltage of the amplifier due to the large current drawn by the amplifier. The drop in the voltage is due to the voltage drop in the internal resistance of the previous amplifier. Current gain:
Let Ai is the current gain without Rs. b
Li i
iA =
Let Ais be the current gain with Rs
iS
Liis RR
RAA+
=
From the above equation, the current gain decreases with Rs. This is not desirable Voltage gain Let AV is the voltage gain without Rs.
Comparing equations (1) and (2), voltage reduces with Rs. Output resistance
ieS
fereoe
O
hRhh
hR
+−
=1
Due to Rs present in the denominator, the output resistance decreases. This is desirable to reduce the loading effect. 3.For the following circuit, find ac performance quantities. hie =1KΩ,hre=10-4, hfe =100, hoe =12µmho.
Considering only Ac signal and taking the frequency of ac signal to be high
fcX C ∏
=2
1
becomes very low. It is approximated to zero, or all capacitors are replaced by short circuit. AC equivalent circuit. h- parameters equivalent circuit without RB
Since RB is in parallel with ideal voltage source the resistance RB will not affect the input voltage.∴ voltage gain remains the same. To find output resistance
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1
=−×−==
Ω==
Ω=−
=
gainPowerAAgainpower
KRRR
K
hhh
hR
ibv
LOO
ie
efreoe
O
4.With the help of suitable circuit and mathematical proof, explain the effect of biasing resistance on the performance of the amplifier.
Let Aib be the current gain with RB Ri is the input resistance of the amplifier without RS
S
Lib i
iA =
S
bi
b
b
S
Lib i
iA
ii
ii
A ×=×=
Applying current divider rule,
iB
Biib
iB
BSb
RRRA
A
RRRi
i
+=
+=
From the above equation current gain decreases with RB. This is undesirable. The voltage gain, without RS ( AV) and with RB ( Avb) are equal since the biasing resistor RB is in parallel with the ideal voltage source.
The equation for RO is independent of RB. Hence the output resistance of the amplifier is unaffected due to the redundancy of RB Input resistance Let Ri and Rib be the input resistance of the amplifier without and with RB respectively.
Approximate h-parameter equivalent circuit. hre is the reverse voltage gain or reciprocal of voltage gain. For an amplifier, voltage gain is high. There fore hre is very low. For approximation, take hre Vce because zero or the voltage source is replaced by short circuit.
In the h-parameter equivalent circuit, hoe
1 is the internal resistance of the current
source hfe ib. For an ideal condition, the internal resistance must be infinity. Because hoe
1 =
infinity or hoe =0Ω or the corresponding branch is replaced by open circuit. With the above approximations, the circuit is called approximate h-parameter equilent circuit .
8. Repeat the previous problem using approximate h-parameter equilent circuit .
Since hre, hoe values are not given for approximation. Assume the data as zero that is use approximate h-parameter equivalent circuit Ac equivalent circuit,
Since there is no source on the input side , ib =0.
∞==
==∴
C
OO
Cbef
ivR
iih 0
In the above problem, what is the effect of connecting a bypass capacitor across RE ?? With emitter bypass capacitor, RE becomes redundant or mathematically RE = 0. Substituting RE = 0 in the above equations.
Ai= -hfe not affected since it does not depend on the RE value or emitter bypass capacitor does not affect current gain.
Ri = hie + ( 1 + hfe ) RE Ri = hie Ri = 1KΩ With the emitter bypass capacitor, input resistance decreases. An amplifier should have high input resistance. But with bypass capacitor, Ri decreases. ∴Bypass capacitor should not be connected to get high input resistance.
2001
)2(100
−=
−=
=
V
i
CiV
A
RR
AA
The voltage gain increases due to reduction in the input resistance. ∴ Emitter bypass capacitor should be connected to get high voltage gain. ∞=0R not affected. ∴Bypass capacitor will not alter the output resistance. Thus the function of RE and CE is to provide high dc stability ( -||- behaves like open circuit for dc) and to get high voltage gain. For ac signal of mid frequency, reactance of the -||- becomes very low and capacitor behaves like short circuit.
1. EMITTER FOLLOWER Draw the circuit of emitter follower. Find ac performance quantities. What are the limitations of emitter follower circuit?? Take R1=100K R2 = 10K RE = 1K hie=1K hfe=100
Applying KCL to output terminal (E) io+ hfe ib + ib = 0 Apliying KVL to input circuit -ib hie -Vo= 0 Vo = -ib hie
Ω=+
=+−
−= 9.9
1)1( ie
ie
feb
iebo h
hhi
hiR
Limitations of Emitter Follower
1. Input resistance Ri (102 KΩ) is very high for an emitter follower circuit. But when biasing network is considered (RE), input resistance reduces. .|||
ii RRR B= Thus the input resistance is
limited by BR , for any value of .iR Therefore the biasing network defeats the purpose of using emitter follower.
2. Current gain of emitter follower is very high [Ai = 1+ hfe]. But due to biasing network, the overall current gain |
iA is reduced.
iB
Bii RR
RAA
+=|
The above 2 limitations can be over come by using Boot strapping Technique.
This is a voltage divider circuit. With an additional resistor R. assuming IB and R to be very small, the voltage drop across R can be neglected and circuit resembles standard voltage divider bias. Hence the biasing condition is unaltered. For ac condition
The capacitor behave like short circuit. fc
X C π21
= , if the frequency is in mid frequency range.
The boot strap resistor R is directly connected to output terminal E. therefore, the input VS is connected to output terminal through the resistance R. This method of connecting input and output terminals only under ac condition is called boot strapping STATE AND PROVE MILLER’S THEOREM
Consider an amplifier with a resistance ( or capacitance ) connected across input and output terminals. Statement: Miller’s theorem states that when a resistance ( or capacitance ) is connected across input and output terminals, the same can be replaced by two independent resistances ( or capacitances ) connected one across the input terminals and the other across output terminals. These are called Miller equivalent resistances ( or capacitances).
comparing equations (1) and (2), the same current i will flow through a resistance MIR when it is between input terminal and ground. Therefore, the KCL at the input terminal is not affected by replacing the resistance R by MIR . MIR is called Miller equivalent resistance on the input side. To find Miller equivalent resistance on the output side.
MOR is the Miller equivalent resistance on the output side.
The same current I will flow through the resistance MOR if R is replaced by MOR which is connected
across output terminal and ground. ∴Replacing R by MOR will not affect KCL at the output terminal. ∴Replacing R by Miller equivalent resistances, the circuit is as follows.
Draw the circuit of Boot strapped Darlington Emitter Follower and find ac performance quantities.
R1=100K R2 = 10K RE = 1K hie=1K hfe=100
Dual of Miller’s Theorem Consider a general amplifier as shown below. In this amplifier, the resistance R is common to the input and output circuits do not have a common resistance. This resistance R can be removed so that the input and output circuits do not have a common resistance. Removing of the resistance R affects the KVL equation on input and output side. The circuit must be modified as explained follows so that the removal of resistance R should not affect the KVL equations. The purpose of the following analysis is to remove the inter dependence of input and output circuits. So that either input circuits or output circuits can be solved independently.
From the above equation, RAi )1( − is the resistance which when connected in series with the input circuit instead of R will not affect the input circuit. Repeating the above analysis for the output circuit.
RA
AR
RiVV
RA
iVV
RA
iVV
RiAiVV
RiiVV
iiButRiiVV
i
iMO
MOLOO
iLOO
iLOO
Li
LOO
LOO
L
OO
−=
=+−
=−+−
=−−−
=−−−
=−−−
−==+−−
1
0
0)11(
0)11(
0)(
0)(
0)(
|
|
|
|
1|
2
21|
We know that
i
LL
Li
Aii
iiA
=
=1
From the above equation, the resistance MOR when connected in the output circuit alone does not alter the KVL equation. Thus, without affecting the electrical nature of the circuit, the circuit can be redrawn a
HYBRID-π EQUIVALENT CIRCUIT To obtain Hybrid-π Equivalent circuit Consider a PNP transistor as shown above. The emitter current IE is divided in to base current IB and a component αIE of the collector current. This division of current takes place in the entire base layer at infinite number of points. For mathematical convenience, it is assumed that the division of current takes place at an imaginary terminal B1. rb
1e: It is the resistance of forward biased base to emitter junction and it is the resistance offered to the flow
of the current IE. rb
1c: It is the resistance of reverse biased collector to base junction. The flow of current in this resistance
represents the reverse saturation current Ico due to flow of minority charge carriers. rbb
1: It is the resistance of the base layer for the flow of the current IB. This is called base spreading resistance because the division of emitter current is spread across the entire region.
αIE: This is the current in the collector due to transistor action. When charge carriers reach the base layer from emitter, the potential gradient at the collector junction will result in the movement of the charge carriers in to the collector. This forms the current. αIE depends on the emitter current IE which inturn depends upon the voltage across base to emitter junction.
]1[1
−=T
EVB
OE VeIIη
Therefore, the voltage VB1
E controls αIE. VB1
E is the independent variable. This depends on charge carrier concentration and temperature. cb
1e and cb
1c: This is the stray capacitance across the two P-N junction. The reactance of the capacitor is
very high at mid-frequency. Hence approximately, capacitors are replaced by open circuit (not considered).
But for high frequency, fc
X C ∏=
21
the reactance becomes finite. Hence considered in the analysis.
All the above terms are called Hybrid-π parameters. These parameters can be represented by the following circuit and it is called Hybrid-π equivalent circuit or Giacollette equivalent circuit. gm vb
1e is the component of collector current(αie) expressed as a function of independent variable vb
1e. gm
is the Transconductance of the transistor. This represents ability of the transistor in transforming the input voltage vb
1e in to output current.
rce: rce is the internal resistance of the current source.
To find Hybrid-π parameters. Hybrid -π equivalent circuit: Let the output terminals be short circuited . Considering mid- frequency, reactance of all capacitors becomes infinite. Therefore, all capacitors can be replaced by open circuit. rb
1c is the resistance of reverse biased collector junction whose value is very high. Therefore it can be
approximated to open circuit. rce is short circuited, becomes redundant. Hence can be removed.
In the above equation, IC represents the dc collector current or quiescent current. Its value can be found graphically by drawing the dc load line, locating the Q point on the load line and then measuring IC. OR if know the biasing arrangement of the transistor, then the circuit can be solved using biasing technique and then IC can be calculated. To find rb
1e
From the two port network theory, we know that
vse = ib hie + hre vce -----(4) ic = ibhfe + hoe vce -----------(5) OR
C is the junction capacitance of reverse biased collector to base junction. When a PN junction is reverse biased, the width of the depletion layer increases and capacitance decreases. Therefore Cb
1C is very low of
the order of few pico farads. To find Cb
1e
This is the capacitance of forward biased PN junction. When a PN junction is forward biased, width of the depletion layer decreases and capacitance increases.
Cb1
e + Cb1
C =T
m
fg∏2
Where fT is called the transition frequency.
fT = hfe fβ
fβ is called upper cutoff frequency.
fβ = ][21
111 cbebeb ccr +∏
Problem: A transistor amplifier is operating with a dc condition of (10V,10mA). The operating temperature is 300C. The H-parameters of the transistor are hie =1Ko, hre =2.5X10-4, hfe=100, hoe=25X10-5mho. Calculate hybrid-π parameters given that CC=3PF. Take fT=1MHz.