PGDST 07 Small-signal BJT Amplifiers
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Transcript of PGDST 07 Small-signal BJT Amplifiers
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Features of an Amplifier
A device is said to be an ac voltage amplifier whenthe total change in the output voltage from the deviceis greater than the total change in the input voltagethat caused it.
An amplifier amplifies the level of a signal voltageor current, retaining the original waveform of theinput signal.
Under small signal conditions, the input variationsare small enough to confine the variation in theoutput voltage and current to the active portion of atransistors output characteristics.
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Amplifier Gain
Input signal power Pi = Vi. Ii , Output signal power P0 = V0 .I0
@ Power gain ( AP ) = P0 /Pi = (V0 /Vi ). (I0/Ii) = AV . AI
An amplifier may amplify either voltage or current or both.
An amplifier must exhibit power gain ( AP > 1 ) and voltage and /or current gain.
A step-up transformer amplifies ac voltage, but it cannot becalled an amplifier, because it does not provide power gain.
The gain of an amplifier is often expressed in decibel form:
dB = 10 log10 P2/P1 = 10 log10 [ V2/V1]2 = 20 log10 (V2/V1) {if R1=R2 }
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Some Parameters of an Amplifier
The input resistance ( Rin
, rin
) to an amplifier is the totalequivalent resistance at its input terminals and can becomputed as the ratio of input voltage to input current.
The output resistance (R0, r0) of an amplifier is the totalequivalent resistance at its output terminals.
Every signal source has internal resistance which is refer to assource resistance (rs). When a signal source is connected to theinput of an amplifier, the source resistance is in series with theinput resistance (rin).
An ac amplifier is always used to supply voltage, current and/orpower to some kind ofload connected to its output. The load
may be a speaker, an antenna, a siren, an indicating instrument,an electric motor, or any one of large number of other devices.Often the load is the input to another ac amplifier. Amplifierperformance is analyzed by representing its load as anequivalent load resistance (or impedance ).
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The frequency response of an electronic device or system isthe variation it causes, if any, in the level of its output signalwhen the frequency of the input signal is changed. It is themanner in which the device responds to changes in signalfrequency.
Frequency response also refers to phase shift as a functionof frequency, calledphase response.
The frequency response of an amplifier is usually presentedin the form of a graph that shows output amplitude ( orvoltage gain ) plotted versus frequency.
The frequency range over which the gain is more or lessconstant [Am] (flat) is called the midband range.
Some Parameters of an Amplifier
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Frequency Response of an Amplifier
The low frequency at which the gain equals 0.707 Am iscalled the lower cutoff frequency([1). The high frequency atwhich the gain once again drops to 0.707Am is called the
upper cutoff frequency([2). The bandwidth of the amplifier is defined to be the
difference between the upper and lower cutoff frequencies:BW ( bandwidth) = [2 [1
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Amplifier Classifications
An amplifier is a two port network that produces at its output
port an amplified version y(t) of an electrical signal x(t) applied
across its input port.
The concept of viewing an electrical signal as a voltage or a
current need not necessarily remain confined to the transducers.
The output signal of any electrical network, when appliedacross a load, can also modeled as voltage or a current signal.These models are acceptable to the extent the voltage(or current)signal remains insensitive to the variation of its load.
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Public Address System
The task of a small signal amplifier is to amplify a weak
electrical signal ( voltage or current ) into a larger voltage or
current swing with an amplitude that is adequate for
subsequent signal processing.
Power amplifier produces a large current-voltage product at
the output and takes care to efficiently transfer this power to
the next stage.S. Kal, IIT-Kharagpur
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Voltage Source
A practical voltage source
model is conceived byplacing a small internal
resistance (RS) in series with
an ideal voltage source (vS).
The voltage across the load
(vL) approaches vS with RL >>RS
A practical current source
can be conceived by placing
an internal resistance RS in
shunt with an ideal currentsource iS, wherein the
maximum current iS from the
source can be entirely flown
through RL with RS >> RL.
Current Source
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Classification of small signal amplifiers corresponding to the four
possible combinations of input and output signals as given below:
Amplifier
type
Input Signal Output Signal Transfer Ratio
Voltage
Amplifier
Voltage Voltage Voltage gain
(no dimension)
Trans-
impedance
Amplifier
Current Voltage Trans- impedance
(dimension :Ohm)
Trans-
AdmittanceAmplifier
Voltage Current Trans- admittance
(dimension :mho)
Current
Amplifier
Current Current Current gain
(no dimension)
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Effect of source and load resistance on amplifier gain
A voltage source vS with an internal resistance RS is connected at
the input for a necessary voltage amplification Av = v2/vS.
vS through RS is expected to provide a current i1, the finite valueof which visualizes the existence of a series input resistance Ri .
Thus the input would be reduced version of vS by a factor Ri / (RS+ Ri ). v1 will achieve its maximum value vS with Ri >> RS
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Output port of the voltage amplifier should behave as a practical
voltage source to its load and it is effected by configuring theoutput port as an ideal voltage source v0 followed by an internal
resistance R0 which looks into the load resistance RL
RL would be able to extract the largest output v0 provided that R0
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vS
Aint
Ri
RL
v2 = = vS Aint E = vS AV
Ri RL [(1 + RS/Ri ) (1+ R0/RL)] ~ (3)
The voltage amplifier with finite Ri and non-negligible R0 offers an
external voltage gain AV
( = v2
/vS
) which is less than the
internal voltage gain Aint by a factor ofE p strongly determined
by a relative magnitude of Ri and R0 with respect to RS and RLrespectively.
AV approaches to its maximum possible value Aint as closely as
one can satisfy the conditions on Ri and R0 given by,
Ri >> RS , R0
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RC Coupled CE Amplifier
C1 and C2 are coupling ca-
pacitors. They allow AC to
go through and block DC.
CE is called the emitter by pass
capacitor. It acts as a shortcircuit to AC components
while blocking DC.
For medium frequency range, called mid-frequency band, the
capacitors C1, C2, CE behave as short-circuits, i.e.,
1/([C1)
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Amplifier Analysis Using Small Signal Models
Small-signal parameters:
To analyze transistor circuits using algebraic method, transistors
are substituted by an equivalent circuit. The form of the
equivalent circuit depends on transistor parameters.
Small-signal parameters are parameters whose values aredetermined under small-signal (ac) operating conditions.
Small-signal value ofF is defined to beF = ic /ib VCE= const.
Small-signal E is defined in terms of ac currents-
E = ic/ie VCB const.S. Kal, IIT-Kharagpur
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An important physical parameter of a transistor is its small-signal resistance from emitter to base, called emitter resistanceand designated as re .
Emitter resistanceis defined as re= vbe /ie VCEconst.
Recall thatin case ofdiode, rD } VT/I }0.026/I at roomtemp. I isthe dc current ofthe diode.
Similarly, re} 0.026 /IE
ohms, where IEisthe dc current.
Thesmall-signal collector resistance ( rc )
rc = vcb/icIE const. ohms
Small Signal Parameters
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Small signal CE Amplifier Model
An equivalent circuit for an electronic device is called a
model.
The re model employs a diode and controlled current
source to duplicate the behavior of a transistor in the
region of interest. Here all the voltages and currents are ac
quantities.
The re model does not show the feed back effect, i.e. ie is
assumed to be independent of VCB. Sophisticated hybrid
model includes the feedback effect. For most practicaldesign and analysis problems, the feedback relationship
can be ignored.
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In CE configuration, input terminals are base, emitter and outputterminals are collector and emitter.
Small signal CE Amplifier Model
Here ib and ic are input and output current respectively.
The ac input resistance, rin = vi /ii = vbe /ibS. Kal, IIT-Kharagpur
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BJT re Model
and vi = vbe = iere, ie = ic + ib = F ib + ib = (1+F) ib
As vbe = ie re, rin = ie re/ib = (1 +F) ib re / ibThus, r in = (1 +F) re } F re
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BJT re Model
The input resistance in CE is approximately F times greaterthan that of CB configuration ( } re)
The output resistance,
Thus, r0 = vce / ic } vcb / F ib } rc / F[ for r0, input is short circuit ie vbe = 0 and vce } vcb ]
Because the CE input resistance is F times greater and outputresistance is F times smaller than the corresponding values in
CB configuration, CE amplifier is inherently better suited forvoltage amplification than its CB counterpart
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Mid Frequency Analysis of CE Amplifier
The DC ground of a circuit means a point where the DC or total
potential is zero. The AC ground means a point where the AC
potential with respect to the DC ground is zero.
In RC coupled CE amplifier circuit, the DC ground is shown by
the symbol ( ). The power supply Vcc is an AC ground.
At mid- and high- frequencies, the emitter point is also ground.
The coupling capacitances ( C1, C2 ) are also assumed to be
short circuited at these frequency ranges.
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Small Signal rE model of CE Amplifier
Approximate small-signal model for a transistor in CE configuration:
Assume a load Rc is connected at the output and the voltage
across it is v0. The output resistance of the stage is
rc/F Rc } Rc as rc/F "" Rcv0 = - ic Rc = - F ibRc and vi = ib F re
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Voltage and Current Gain of CE Amplifier
Thus voltage gain,
Av = v0 /vI = - F ib Rc / F ib re = - Rc / re
The minus sign reveals that the output andinput voltages are 180o out of phase.
Current gain, Ai = ic / ib } F
Thus, the CE amplifier can provide voltage andcurrent gains that are both greater than unity.
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AC Equivalent Circuit of CE Amplifier Using re Model
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Analysis of CE Amplifier
Case
I, S
ource resistance (RS) and
load resistance(RL) are not considered.
The output voltage ,
V0 = i0 R0 = - Ic R0 = - F ib R0 or, V0 } - F ib Rc (r0>>RC)
Input voltage, Vi = F re ib
Thus voltage gain = Av = V0 /Vi = - Rc / re
The minus sign is to denote phase inversion between
input and output.
The current gain, Ai= i0/ ii = i0 /ib y ib /iiii = i2 + ib = Vi /RB + Vi / Fre
Input Impedance, Ri = Vi /ii = 1/ [ 1/RB + 1/Fre ]or Ri = FRB re / [ RB +Fre ]
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Analysis of CE Amplifier
Now Ib = Vi /Fre = ii RB/ (RB +Fre), i0 = FibThus, Ai = F RB / ( RB +Fre)
Case 2, When source (RS) andload (RL) resistances are
Considered
The overall voltage gain from source to load is given by-
AVs = vL / vs = Av [ rin / (Rs + rin)] y [ RL / (R0 +RL ) ]
orAVs = - Rc/re [(RB``Fre) / (Rs + RB``Fre )]y[ RL /(r0+RL)]
The overall current gain is given by
Ais = iL /iS = Ai [Rs / (Rs + rin) ] y [ R0 / (R0 + RL)]
orAis = Ai [ (Rs``RB) / ( Rs``RB + rin )] y [ R0 / (R0 + RL)]
Ais = Ai [ (Rs``RB) / ( Rs``RB +Fre )] y [ R0 / (R0 + RL)]S. Kal, IIT-Kharagpur
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Low Frequency Response of CE Amplifier
At low frequencies below the mid-frequency range, the capacitorsC1, C2, and CE no longer behave as short circuit. So the gainbecomes a function of frequency.
Note that the mid-freq gain is independent of frequency.
The contribution from C1, C2 and CE will be dealt with separately.
Effect of C1 assuming C2 = CE = w
In the above circuit effective Ri = hie R1 R2 = hieRB / ( hie +RB )
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Low Frequency Behavior
where [c = 1/RC = 1/X and phase angle of ( v0/vi ) = tan-1 ([c/[) = U
[c = cut-off frequency at which `vo/vi` is 1/2 times or 70.7% of themaximum value.
At [ ! [c , ` vo/vi ` in dB = 20 log10 1/2 = - 3.0103 dB
Thus at [ ! [c , the value of`vo/vi ` is 3 dB less than the maximumvalue.
2
0
0
0
1
1
1
1
1
)/({V
V
/jV/V
)t(V.C/j
R)t(V
ci
ci
i
[[
[[
[
!@
!@
!
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Low Frequency Behavior
c
c
c
i
for
log
log
dBinV
V
[[
[
[
[
[
$
!
10
2
10
0
20
120
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High Frequency Analysis ( Hybrid- T Model )
The re model / h-parameter model of transistors becomesinconvenient at high frequencies above the mid-frequency
range, because the h-parameters become complex and
frequency dependent at high frequencies. A model appropriate
for high frequency analysis, derived from physical principles, is
called hybrid T model.
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rb b = base spreading resistance between the extrinsic base
terminal (B) and the intrinsic base region ( B ) of the shallow
base layer.
rbe = resistance of the forward biased B-E junction
Cb
e = Capacitance of the forward biased B-E junction
( diffusion plus transition capacitance)
rb
c = resistance of the reverse biased C-B junction
Cb
c= Capacitance of the reverse biased C-B junction
( transition/ jn. depletion capacitance)
gm = transconductance $ ic / vb
e
High Frequency Analysis ( Hybrid- T Model )
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gm vb e = output current source responsible for transfer ofpower for input to output giving rise to amplification.
rce = Leakage resistance between C-E.
High-frequency equivalent circuit of CE amplifier can berepresented by a simple RC network (low pass) and higher cut-offfrequency is obtained.
Hybrid- T Parameters
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