UNIT-2
ELECTRONIC CIRCUITS-ANALYSIS
AND DESIGN
Topics Covered:
• Introduction to Frequency Response
• Different Frequency Ranges
• Series & Parallel Circuit Transfer fucntions
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UNIT-2
ELECTRONIC CIRCUITS-ANALYSIS
AND DESIGN
Frequency Response
JNTUA R-19
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Topics to be Covered:
• Frequency Response:
• Amplifier frequency response-different ranges, short
circuit and open circuit time constants, time response
• Transistor amplifiers with circuit capacitors
–
–
–
–
coupling capacitor effects
load capacitor effects
Bypass capacitor effects
Problem solving
• Combined effects of coupling and bypass capacitor,
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Topics to be Covered:
• High-frequency response model for BJT and
MOSFETs
• short circuit current gain
• Miller effect and its applications
• Unity-gain bandwidth in BJT and FET amplifiers
• CE and CS circuits, CB and CG circuits
• Cascode amplifier analysis
• Emitter and source follower circuits
• High frequency response- design application.
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Frequency Response:
• All amplifier gain factors are functions of signal frequency.
• These gain factors include voltage, current, transconductance, and
transresistance.
• The curve drawn b/w frequency vs Gain is called Frequency response curve.
• The frequency ranges are divided as:
– Low frequency (f < fL)
– High Frequency (f > fH)
– Medium Frequency
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Frequency Ranges:
• Mid band Range:
• The coupling and bypass capacitors in this region are treated as short
circuits.
• The stray and transistor capacitances are treated as open circuits.
• In this frequency range, there are no capacitances in the equivalent
circuit.
• These circuits are referred to as midband equivalent circuits.
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Frequency Ranges:
• Low-Frequency Range:
• In this frequency range, we use a low-frequency equivalent circuit.
• In this region, coupling and bypass capacitors must be included in the
equivalent circuit and in the amplification factor equations.
• The stray and transistor capacitances are treated as open circuits.
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Frequency Ranges:
• High-Frequency Range:
• In the high-frequency range, we use a high-frequency equivalent circuit.
• In this region, coupling and bypass capacitors are treated as short circuits.
The transistor and any parasitic or load capacitances must be taken into
account in this equivalent circuit.
• The mathematical expressions obtained for the amplification factor in this
frequency range must approach the midband results as f approaches the
midband frequency range, since in this limit the capacitors approach open-
circuit conditions.
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Voltage Transfer Functions: (Series)
•
The voltage transfer function for the circuit in Figure
can be expressed in a voltage divider format, as follows:
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Voltage Transfer Functions: (Parallel)
• Writing a Kirchhoff current law (KCL) equation at
the output node, we can determine the voltage
transfer function for the circuit, as follows:
In this case, the element RS is in series between the input and
output signals, and the elements RP and CP are in parallel with
the output signal. Rearranging the terms in Equation
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Voltage Transfer Functions: (Parallel)
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UNIT-2
ELECTRONIC CIRCUITS-ANALYSIS
AND DESIGN
Topics Covered:
• Short Circuit & Open Circuit Time Constants
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Short-Circuit and Open-Circuit Time Constants:
• Capacitor CS is the coupling capacitor
• CP is the load capacitor and is in parallel
with the output and ground.
• Applying KCL at output node,
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Short-Circuit and Open-Circuit Time Constants: (Contd..)
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Short-Circuit and Open-Circuit Time Constants: (Contd..)
where τS and τP are the same time constants as previously defined.
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Short-Circuit and Open-Circuit Time Constants: (Contd..)
•
•
•
•
•
CS affects the low frequency response and CP affects the high-frequency
response.
At low frequencies, we can treat the load capacitor CP as an open circuit.
To find the equivalent resistance seen by a capacitor, set all independent
sources equal to zero.
Therefore, the effective resistance seen by CS is the series combination of RS
and RP.
The time constant associated with CS is
τS = (RS + RP)CS
• Since CP was made an open circuit, τS is called an open-circuit time
constant
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Short-Circuit and Open-Circuit Time Constants: (Contd..)
• At high frequencies, we can treat the coupling capacitor CS as
a short circuit.
• The effective resistance seen by CP is the parallel combination
of RS and RP, and the associated time constant is
τP = (RS || RP)CP
• which is called the short-circuit time constant.
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Short-Circuit and Open-Circuit Time Constants: (Contd..)
• The lower corner, or 3 dB frequency, which
is at the low end of the frequency scale, is a
function of the open-circuit time constant
and is defined as
Where, τS = (RS + RP)CS
• The upper corner, or 3 dB, frequency,
which is at the high end of the frequency
scale, is a function of the short-circuit time
constant and is defined as
Fig: Bode plot of the voltage
transfer function magnitude
Where, τP = (RS || RP)CP
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Short-Circuit and Open-Circuit Time Constants: (Contd..)
• The amplifier gain is constant over a wide frequency range, called the
midband (all capacitance effects are negligible)
•
At the high end of the frequency spectrum, the gain drops as a result of
the load capacitance.
•
At the low end of the frequency spectrum, the gain decreases because
coupling capacitors and bypass capacitors do not act as perfect short
circuits.
•
The midband range, or bandwidth, is defined by the corner frequencies fL
and fH, as follows:
•
fBW = fH − fL
Since fL , value is low , the bandwidth is essentially given by
fBW ∼= VIVTRONICS
fH
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Problem:
• Determine the corner frequencies and bandwidth of a
passive circuit containing two capacitors.
• Consider the circuit shown in Figure with parameters
• RS = 1 k, RP =10 k, CS = 1 μF, and CP = 3 pF.
The open-circuit time constant is
The short-circuit time constant is
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The maximum magnitude of the voltage function is again
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UNIT-2
ELECTRONIC CIRCUITS-ANALYSIS
AND DESIGN
Topics Covered:
• Time Response
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Time Response:
• Some times it we need to amplify
non sinusoidal signals ( Square waves)
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Time Response: (Contd..)
• If the input voltage is a step function, then Vi (s) = 1/s.
The output voltage can be written as
• Taking the inverse Laplace transform, we find the output voltage
time response as
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Time Response: (Contd..)
• If we are trying to amplify an input voltage pulse using a coupling
capacitor, the voltage applied to the amplifier (load) will begin to droop
• So we need to take the τs > T
• Where, T = input pulse width of input signal.
• A large time constant implies a large coupling capacitor.
Fig(a): Output response of circuit for a squarewave input signal for large time constant
Fig(b): Steady-state output response for
a square-wave input response (coupling
capacitor) and a large
time constant
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Time Response: (Contd..)
•
•
The capacitor CP may represent the input capacitance of an amplifier.
The transfer function was given as
• Again, if the input signal is a step function, then Vi (s) = 1/s. The output
voltage can then be written as
Taking the inverse Laplace transform, we find the output voltage time
response as
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Time Response: (Contd..)
•
•
If we are trying to amplify an input voltage pulse, we need to ensure that
the time constant τP is short compared to the pulse width T, so that the
signal v0 (t) reaches a steady-state value.
A short time constant implies a very small capacitor CP as an input
capacitance to an amplifier.
Fig(a): Output response of circuit for
a square-wave input signal and for a
short time constant
Fig(b): Steady-state output response for a
square-wave input response (load capacitor) in
short time constant YouTube: VIVTRONICS
UNIT-2
ELECTRONIC CIRCUITS-ANALYSIS
AND DESIGN
Topics Covered:
• Transistor Amplifiers with Circuit Capacitors
• Coupling Capacitor Effect (CE Amplifier)
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Transistor Amplifiers with Circuit Capacitors:
• In a Single stage Amplifier section includes
three types of capacitors namely:
– Coupling Capacitor
– Load capacitor
– Bypass capacitor
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Coupling Capacitor Effects:
• Consider a Single stage CE Amplifier with input coupling capacitor CC.
• At high frequencies, the capacitor CC acts as a short circuit, and the input
signal is coupled through the transistor to the output.
• At low frequencies, the impedance of CC becomes large and the output
approaches zero.
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Effects of coupling capacitor: (Contd..)
• The input current can be written as
---(1)
---(2)
To determine the input resistance to the base of
the transistor, we multiplied the emitter resistance by the factor (1 + β).
---(3)
---(4)
---(5)
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Effects of coupling capacitor: (Contd..)
• Combining equations (1) through (5)
---(6)
---(7)
---(8)
---(9) YouTube: VIVTRONICS
Effects of coupling capacitor: (Contd..)
• The equation (7) is in the form of series coupling capacitor circuit voltage
transfer function. & The corner frequency is
---(10)
---(11)
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Ex: Problem:
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Ex: Problem:
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UNIT-2
ELECTRONIC CIRCUITS-ANALYSIS
AND DESIGN
Topics Covered:
• Load Capacitor Effects
• Bypass Capacitor Effects
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Load Capacitor Effects:
• An amplifier output may be connected to a load or to the input or another
amplifier.
• The model of the load circuit input impedance is generally a capacitance in
parallel with a resistance.
• In addition, there is a parasitic capacitance between ground and the line
that connects the amplifier output to the load circuit.
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Load Capacitor Effects: (Contd..)
•
Consider a MOSFET common-source amplifier with a load resistance RL and a
load capacitance CL connected to the output. Which forms a Low pass network.
•
At high frequencies, the impedance of CL decreases and acts as a shunt between
the output and ground, and the output voltage tends toward zero.
•
The equivalent resistance seen by the load capacitor CL is RD ||RL . Since we set
•
Vi = 0, then gmVsg = 0, which means that the dependent current source does not
affect the equivalent resistance.
•
The time constant for this circuit is
• The maximum gain, which is found by
assuming CL is an open circuit.
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Load Capacitor Effects: (Contd..)
• When CL is an open circuit.
• KVL at i/p.
• From the output side circuit,
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Bypass Capacitor Effects:
• In Amplifiers the emitter and source bypass capacitors included so that
the emitter and source resistors can be used to stabilize the Q-point
without sacrificing the small-signal gain.
• The bypass capacitors are assumed to act as short circuits at the signal
frequency.
• To choose bypass capacitor, determines the circuit response in the
frequency range where these capacitors are neither open nor short
circuits.
• Consider a CE amplifier with bypass capacitor CE
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Bypass Capacitor Effects: (Contd..)
• The small signal voltage gain as a function
of frequency.
• Using the impedance reflection rule, the
small-signal input current is
• The total impedance in the emitter is multiplied by the factor (1 + β).
The control voltage is
Combining equations produces the small-signal voltage gain, as follows:
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Bypass Capacitor Effects: (Contd..)
• Take Parallel Combination of RE and CE
• The gain equation can be written as,
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Bypass Capacitor Effects: (Contd..)
• Assuming,
in terms of time constants
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Bypass Capacitor Effects: (Contd..)
• The Bode plot of the voltage gain magnitude has two limiting horizontal asymptotes.
• If we set s = jω, we can then consider the limit as ω →0 and ω→∞.
• For ω →0, CE acts as an open circuit; for ω→∞, CE acts as a short circuit.
Fig: Bode plot of the voltage gain magnitude
for the circuit with an emitter bypass
capacitor
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Bypass Capacitor Effects: (Contd..)
The corner frequency due to τB is
Fig: Bode plot of the voltage gain magnitude for the circuit with an emitter bypass capacitor
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UNIT-2
ELECTRONIC CIRCUITS-ANALYSIS
AND DESIGN
Topics Covered:
• Combined Effects of Capacitors
•
•
Coupling & Load Capacitors
Coupling & Bypass Capacitor effect
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Coupling & Load Capacitor Effect:
• A circuit with both a coupling
capacitor and a load capacitor is
shown in Figure.
• The small-signal equivalent
circuit is shown in below fig.
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Coupling & Load Capacitor Effect: (Contd..)
• The lower corner frequency fL is given by
• where τS is the time constant associated with the coupling
capacitor CC.
• The upper corner frequency fH is given by
• where τP is the time constant associated with the load capacitor CL
• From the eq circuit, to find the
equivalent resistance associated with
the CC . By setting vi = 0;
• Where,
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Coupling & Load Capacitor Effect: (Contd..)
• The related time constant is
• Similarly, the time constant related to CL is
• The two corner frequencies are far apart,
the gain will reach a maximum value in
the frequency range between fL and fH,
which is the midband.
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Coupling & Load Capacitor Effect: (Contd..)
• We can calculate the midband gain by assuming that the
coupling capacitor is a short circuit and the load capacitor is
an open circuit.
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Combined Effects: Coupling and Bypass Capacitors:
• When a circuit contains multiple
capacitors, the frequency response
analysis becomes more complex.
• Consider a circuit with two coupling
capacitors and an emitter bypass
capacitor, all of which affect the
circuit response at low frequencies.
• The transfer function includes all the
components.
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Combined Effects: Coupling and Bypass Capacitors:
Case-1 :
•
In this case, the bypass capacitor is assumed to be a short circuit.
•
The plots consider C1 and C2 individually, as well as together.
•
As expected, with two capacitors both acting at the same time, the slope is 40 dB/decade or 12
dB/octave.
•
Since the poles are not far apart, in the actual circuit, we cannot consider the effect of each
capacitor individually.
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Combined Effects: Coupling and Bypass Capacitors:
Case-2 :
•
Consider the emitter bypass capacitor and the two coupling capacitors.
•
The plot shows the effect of the bypass capacitor, the effect of the two coupling capacitors, and the
net effect of the three capacitors together.
•
When all three capacitors are taken into account, the slope is continually changing; there is no
definitive corner frequency.
•
However, at approximately f = 150 Hz, the curve is 3 dB below the maximum asymptotic value, and
this frequency is defined as the lower corner frequency, or lower cutoff frequency.
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UNIT-2
ELECTRONIC CIRCUITS-ANALYSIS AND
DESIGN
Topics Covered:
• Expanded Hybrid-π Equivalent Circuit
• Short-Circuit Current Gain of CE Amplifier
• Unity Gain Bandwidth (Figure of Merit )
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Expanded Hybrid-π Equivalent Circuit:
•
•
•
Let us consider the cross section of the npn transistor in a classic integrated
circuit configuration.
The C, B, and E terminals are the external connections to the transistor, and
the C’, B’, and E’ points are the idealized internal collector, base, and emitter
regions.
Expanded hybrid – pi equivalent circuit was divided into 3 parts(B-E, C-E, C-B)
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Expanded Hybrid-π Equivalent Circuit: (Contd..)
• Resistance rb is the base series resistance between the external base
terminal B and the internal base region B’.
• The B’–E’ junction is forward biased; therefore,
Cπ is the forward biased junction capacitance
and rπ is the forward-biased junction diffusion
resistance.(Both functions of the junction current)
• Finally, rex is the emitter series resistance between
the external emitter terminal and the internal
emitter region.
• This resistance is usually very small, on the order of
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1 to 2 Ω .
Expanded Hybrid-π Equivalent Circuit: (Contd..)
•
The dependent current source, gmVπ , is the transistor collector current
controlled by the internal base–emitter voltage.
• Resistance ro is the inverse of the output conductance go and is primarily due
to the Early effect.
• Cs --junction capacitance of the reverse
biased collector–substrate junction.
• rc is the collector series resistance
• ro is the output resistance
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Expanded Hybrid-π Equivalent Circuit: (Contd..)
• B’–C’ junction reverse biased.
• Capacitance Cμ is the reverse-biased junction capacitance, and rμ is the
reverse-biased diffusion resistance.
• Normally, rμ is on the order of megohms and can be neglected.
• The value of Cμ is usually much smaller than Cπ ; however,
• Because of the Miller effect, Cμ usually cannot be neglected.
• rμ -- reverse-biased diffusion resistance.
• Cμ -- reverse-biased junction capacitance.
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Expanded Hybrid-π Equivalent Circuit: (Contd..)
• The complete Hybrid-π Equivalent Circuit is given by:
• rb - base series resistance
• rπ - FB junction diffusion
resistance
• Cπ -FB junction capacitance
• rex - emitter series
resistance
• rc -collector series
resistance
• Cs - junction capacitance
• rμ -- RB diffusion resistance.
• Cμ - RB junction capacitance
• ro - output
resistance
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Short-Circuit Current Gain:
• The frequency effects of the bipolar transistor can be
determining the short-circuit current gain, after simplifying
the hybrid-π model.
• Neglecting the parasitic resistances rb, rc, and rex , rμ and the
substrate capacitance Cs.
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Short-Circuit Current Gain (Contd..)
• The small-signal current gain Ai = Ic / Ib.
• Writing a KCL equation at the input node,
•
From a KCL equation at the output node,
Fig: Simplified hybrid-π equivalent circuit for
determining the short-circuit current gain
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Short-Circuit Current Gain (Contd..)
The input voltage Vπ can then be written as
Substituting this Vπ in Ib equation, Ib
• The small-signal current gain usually
designated as h f e,
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Short-Circuit Current Gain (Contd..)
• If we assume typical circuit parameter values of Cμ = 0.05 pF, gm = 50 mA/V,
and a maximum frequency of f = 500 MHz, then we see that ωCμ << gm.
• Therefore, to a good approximation, the small-signal current gain is
•
•
W.k.t gmrπ = β, then the low frequency current gain is just β,
The corner frequency, which is also the beta cutoff frequency fβ in this
case, is given by
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Short-Circuit Current Gain (Contd..)
Bode plots for the short-circuit current gain: (a) magnitude and (b) phase
• As the frequency increases, the small-signal collector current is no longer
in phase with the small-signal base current.
• At high frequencies, the collector current lags the input current by 90
degrees.
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Problem:
•
•
•
Determine the 3 dB frequency of the short-circuit current gain of a bipolar
transistor. Consider a bipolar transistor with parameters rπ = 2.6k, Cπ = 0.5
pF, and Cμ = 0.025 pF.
Solution:
We know that,
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Cutoff Frequency: (Unity Gain Bandwidth)
• Cutoff Frequency of a short circuit CE amplifier is
defined as,
• It is the frequency at which the gain of an amplifier
becomes unity.
• In the magnitude plot, the small-signal current gain
decreases with increasing frequency.
• At frequency fT , the gain goes to 1. Which is called
as cutoff frequency.
• The cutoff frequency is also called “Figure of Merit” of an amplifier.
•
Which characterizes the performance of an amplifier.
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Cutoff Frequency: (Contd..)
• W.k.t SC Current Gain of CE amplifier, hfe
• Assuming, cutoff frequency
• The gain equation can be written as
• The magnitude of h fe is ,
•
At the cutoff frequency fT , |h f e| = 1
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Cutoff Frequency: (Contd..)
• Normally, βo >> 1, which implies that fT >>fβ
• Frequency fβ is also called the bandwidth of the transistor.
• The cutoff frequency fT is the gain–bandwidth product of the transistor, or more
commonly the unity-gain bandwidth.
•
fT is a a function of IC & gm α IC,
Since;
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Problem:
• Calculate the bandwidth fβ and capacitance Cπ of a bipolar transistor.
• Consider a bipolar transistor that has parameters fT = 20 GHz at IC = 1 mA,
βo = 120, and Cμ = 0.08 pF.
• Solution: we know that,
• The transconductance is
• The Cπ capacitance is determined from Equation
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UNIT-2
ELECTRONIC CIRCUITS-ANALYSIS AND
DESIGN
Topics Covered:
• Miller Effect and Miller Capacitance (BJT)
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Miller Effect and Miller Capacitance:
• The Miller effect, or feedback effect, is a multiplication effect of Cμ in
circuit applications.
• The presence of Cμ complicates the analysis.
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Miller Effect and Miller Capacitance: (Contd..)
•
•
•
Small-signal current gain Ai = io / is
Assuming the coupling and bypass capacitors are short circuit and also
assuming capacitor Cμ as a two-port network.
Writing KVL equations at the input and output terminals,
• Using above Equations, we can form a two-port equivalent circuit
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Miller Effect and Miller Capacitance: (Contd..)
Fig: Small-signal equivalent circuit, including the two-port equivalent model of capacitor Cμ
•
•
To evaluate this circuit, we will make some simplifying approximations. Typical
values of gm and Cμ are, gm = 50 mA/V and Cμ = 0.05 pF.
For these values, we can assume the frequency at which the magnitudes of the
two dependent current sources are equal.
−→
2𝜋𝑓 𝐶𝜇 = 𝑔𝑚
• Since the frequency of operation of bipolar transistors is far less than 159 GHz,
the current source Isc = jωCμVπ is negligible compared to the gmVπYouTube:
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Miller Effect and Miller Capacitance: (Contd..)
•
We can now calculate the frequency at which the magnitude of the impedance of
Cμ is equal to RC||RL .
• If we assume RC = RL = 4 kΩ & typical values for discrete bipolar circuits, then
• If the frequency of operation of the BJT is very much smaller than 1.59 GHz, then the
impedance of Cμ will be much greater than RC || RL and Cμ can be considered an open
circuit.
Fig: Small-signal equivalent circuit, including approximations
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Miller Effect and Miller Capacitance: (Contd..)
• From the circuit segment between
the dotted lines in the fig,
• The output voltage is
• Substituting Vo in Ii equation,
• The circuit segment between the dotted
lines can be replaced by an equivalent
capacitance given by
Fig: Small-signal equivalent circuit, including the equivalent
Miller capacitance
• Where, Capacitance CM is called the Miller capacitance, and the multiplication effect of
Cμ is the Miller effect.
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UNIT-2
ELECTRONIC CIRCUITS-ANALYSIS AND
DESIGN
Topics Covered:
•
•
•
•
FET - High Frequency Model
Gain Bandwidth Product (Figure of Merit)
Miller Effect & Capacitance (FET)
Problems
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High-Frequency Equivalent Circuit (FET):
•
Figure shows a model based on the inherent capacitances
and resistances in an n-channel MOSFET.
•
Assuming that, the source and substrate are both tied to
ground.
•
Two capacitances connected to the gate are inherent in the
transistor.
• These capacitances, Cgs and Cgd , represent the interaction between the gate and the
channel inversion charge near the source and drain terminals.
•
Cgsp and Cgdp, are parasitic or overlap capacitances.
•
Cds is the drain-to-substrate pn junction capacitance.
•
rs and rd are the series resistances of the source and drain terminals
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High-Frequency Equivalent Circuit (FET): (Contd..)
• Voltage V’gs is the internal gate-to-source
voltage that controls the channel current.
• The resistance r0 is associated with the slope
of ID versus VDS.
• In the ideal MOSFET biased in the saturation
region, ID is independent of VDS, which means
that ro is infinite.
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High-Frequency Equivalent Circuit (FET): (Contd..)
• The Simplified low-frequency equivalent circuit for NMOSFET including rs
but not ro Shown in the above fig.
• From the circuit, the drain current is
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High-Frequency Equivalent Circuit (FET): (Contd..)
•
The relationship between Vgs and V’gs is
• The drain current can now be written as
• Above equation Shows that the source resistance reduces the effective
transconductance, or the transistor gain.
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Unity-Gain Bandwidth: ( Cut-off Frequency)
• The unity-gain frequency or bandwidth is a figure of merit for the FETs.
•
Neglect rs , rd , ro, and Cds , and connect the drain to signal ground, the
resulting equivalent small-signal circuit is (S.C Drain)
• We can derive the short-circuit current gain.
From that we can define and calculate the
unity-gain bandwidth.
• Writing a KCL equation at the input node,
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Unity-Gain Bandwidth: ( Cut-off Frequency) (Contd..)
• From a KCL equation at the output node,
• Substituting vgs value in Ii Equation, Ii =
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Unity-Gain Bandwidth: ( Cut-off Frequency) (Contd..)
• Therefore, the small-signal current gain is
Assuming, ωCgd << gm (for Typical values )
• The unity-gain frequency fT is defined as the frequency at which the magnitude of the
short-circuit current gain goes to 1.
• The unity-gain frequency or bandwidth is a parameter of the transistor and is
independent of the circuit.
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Problem:
• Determine the unity-gain bandwidth of an FET.
Consider an n-channel MOSFET with parameters Kn = 1.5 mA/V2, VT N = 0.4 V,
λ = 0, Cgd = 10 fF, and Cgs = 50 fF. Assume the transistor is biased at VGS = 0.8 V.
• Solution: The transconductance is
• The unity-gain bandwidth or frequency, is
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Miller Effect and Miller Capacitance: (FET/MOSFET)
• The Miller effect and Miller capacitance are factors in the high-frequency
characteristics of FET circuits.
• Figure shows a simplified high frequency
transistor model, with a load resistor RL
connected to the output.
• Applying KCL at input side node,
• Likewise, summing currents at the output drain node
Equivalent high-frequency small-signal
circuit of a MOSFET with a load
resistance RL
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Miller Effect and Miller Capacitance: (Contd..)
• Combine above two equations to eliminate voltage Vds.
• The input current is then
• (ω RL Cgd )is much less than 1; therefore , neglect
(jω RL Cgd) above equation becomes
Equivalent high-frequency small-signal
circuit of a MOSFET with a load
resistance RL
• The parameter CM is the Miller capacitance and is given by
• Equation shows the effect of the parasitic drain overlap capacitance.
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Cut-off Frequency :
•
When the transistor is biased in the saturation region, the total gate-to-drain
capacitance Cgd is the overlap capacitance.
• This overlap capacitance is multiplied because of the Miller effect and may
become a significant factor in the bandwidth of an amplifier.
• From the circuit,
and the ideal short-circuit output current is
MOSFET high-frequency circuit, including
the equivalent Miller capacitance
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Cut-off Frequency : (Contd..)
• The cutoff frequency fT of a MOSFET is defined
as the frequency at which the short circuit
current gain magnitude is 1, or the magnitude
of the input current Ii is equal to the ideal
current Id .
MOSFET high-frequency circuit, including
the equivalent Miller capacitance
• where CG is the equivalent input gate capacitance.
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UNIT-2
ELECTRONIC CIRCUITS-ANALYSIS AND
DESIGN
Topics Covered:
• High Frequency Response of Transistor Circuits
• CE & CS Amplifier Circuits (Common Emitter)
• CB & CG Amplifier Circuits (Common Base)
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High Frequency Response of Transistor Circuits:
• Common-Emitter and Common-Source Circuits:
• In this analysis, we will assume that CC and CE are
short circuits, and CL is an open circuit.
• The High Frequency Equivalent circuit is given by,
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Common-Emitter and Common-Source Circuits: (Contd..)
•
We can replace the capacitor Cμ with the equivalent Miller capacitance CM
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Common-Emitter and Common-Source Circuits: (Contd..)
• The upper 3 dB frequency can be determined by
where τP = ReqCeq .
Ceq = Cπ + CM
& Req = rπ || RB || RS
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Common-Emitter and Common-Source Circuits: (Contd..)
Voltage Gain:
• The midband voltage gain magnitude can be calculated
by assuming Cπ and CM are open circuits.
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Common-Base, Common-Gate : (High-Frequency Analysis)
•
The common-base circuit is show in the figure.
•
The circuit configuration is the same as the
common-emitter circuit, except a bypass capacitor
is added to the base and the input is capacitively
coupled to the emitter.
•
The coupling and bypass capacitors are replaced by
short circuits.
•
Neglecting R1 & R2 also the resistance ro is
assumed to be infinite
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Common-Base, Common-Gate : (High-Frequency Analysis)
High-frequency common-base equivalent circuit
• Capacitance Cμ, which led to the multiplication effect, is no longer between
the input and output terminals.
• One side of capacitor Cμ is tied to signal ground.
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Common-Base, Common-Gate : (High-Frequency Analysis)
Apply KCL at the emitter
; 𝑠𝑖𝑛𝑐𝑒 𝑉𝜋 = −𝑉𝑒
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Common-Base, Common-Gate : (High-Frequency Analysis)
•
The equivalent input portion of the circuit is shown in Figure (b)
• Figure (c) shows the equivalent output portion of the circuit.
• As, one side of Cμ is tied to ground, which eliminates the feedback or Miller
multiplication effect. Therefore, fH may larger than CE configuration.
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Common-Base, Common-Gate : (High-Frequency Analysis)
•
For the input portion of the circuit, the upper 3 dB frequency is given by
Where the time constant is
• Assume that CL is an open circuit. Capacitance Cμ
will also produce an upper 3 dB frequency, given by
Where the time constant is
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Common-Base, Common-Gate : (High-Frequency Analysis)
•
If Cμ << Cπ  fHπ due to Cπ to dominate the high-frequency response.
•
However, the factor rπ /(1 + β) in the time constant τPπ is small; therefore, the
two time constants may be the same order of magnitude.
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UNIT-2
ELECTRONIC CIRCUITS-ANALYSIS AND
DESIGN
Topics Covered:
• High Frequency Response of Transistor Circuits
• Cascode Amplifier Circuit
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Cascode Circuit: (High Frequency Analysis)
•
Cascode amplifier is a combination of CE-CB.
•
The input is applied to CE Amplifier (Q1), and the output of
the CE is fed to the CB Amplifier (Q2).
•
The input impedance to the CE circuit (Q1) is relatively large,
and the load resistance seen by Q1 is the input impedance of
Q2 and is fairly small.
•
The low output resistance of CE reduces the Miller
multiplication factor on Cμ1 and therefore extends the
bandwidth of the circuit.
•
To draw Small signal equivalent circuit assuming the coupling
and bypass capacitors are short circuits, and resistance ro2 to
be infinite.
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Cascode Circuit (Contd..)
• The high-frequency small-signal equivalent circuit is given by,
Fig: High-frequency equivalent
•
The input impedance to the emitter of Q2 is, can be given
From the CB Amplifier analysis,
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Cascode Circuit (Contd..)
• The input portion of the circuit shown in
Figure can be transformed to a circuit with
Miller capacitance as,
•
The Miller capacitance CM1 is included in
the input, and capacitance Cμ1 is included
in the output portion of the Q1 model.
• In the center of this equivalent circuit, ro1 is
in parallel with rπ2/(1 + β).
Fig: Rearranged high-frequency equivalent circuit
• Since ro1 is usually large, it can be
approximated as an open circuit.
• The Miller capacitance is then
Fig: Variation of the high-frequency circuit,
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Miller capacitance
Cascode Circuit (Contd..)
•
Transistors Q1 and Q2 are biased with essentially the same current;
therefore,
Then
• Above equation shows that this cascode circuit greatly reduces the Miller
multiplication factor.
• The time constant related to Cπ2 involves resistance rπ2/(1 + β).
• Since this resistance is small, the time constant is small, and the corner frequency
related to Cπ2 is very large.
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Cascode Circuit (Contd..)
• Therefore neglect the effects of Cμ1 and Cπ2 in the center portion of the
circuit.
• The time constant for the input portion of the circuit is
• where CM1 = 2Cμ1. The corresponding 3 dB frequency is
Assuming CL acts as an open circuit, the time constant of the output portion of the circuit,
and the corresponding corner frequency is
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Cascode Circuit (Contd..)
•
•
To determine the midband voltage gain we assume that all capacitances in the
circuit are open circuits.
The output voltage is then
• Neglect the effect of ro1 compared to rπ2/(1 + β). Also, since gm1rπ2 = β, above
equation becomes
and, from the input portion of the circuit,
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Cascode Circuit (Contd..)
• The midband voltage gain is
• The expression for the midband gain of the cascode circuit is identical to that of
the common-emitter circuit.
• The cascode circuit achieves a relatively large voltage gain, while extending the
bandwidth.
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UNIT-2
ELECTRONIC CIRCUITS-ANALYSIS AND
DESIGN
Topics Covered:
• High Frequency Response of Transistor Circuits
• Emitter & Source Follower Circuits
• Design Application(CE)
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Emitter- and Source-Follower Circuits:
• Figure shows an emitterfollower circuit with the output
signal at the emitter
capacitively coupled to a load.
(a) High-frequency equivalent circuit of emitter
follower
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Emitter- and Source-Follower Circuits: (Contd..)
(a) High-frequency equivalent circuit of emitter follower
•
(b) rearranged high frequency equivalent circuit
Figure (b) shows the rearrange the circuit. Cμ is tied to ground potential and also
that ro is in parallel with RE and RL.
• In this analysis neglect CL.
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Emitter- and Source-Follower Circuits: (Contd..)
• We can find the impedance Z’b looking into the
base without capacitance Cμ.
• The current I’b entering the parallel combination
of rπ and Cπ is the same as that coming out of the
combination.
• The output voltage is then
(b) rearranged high frequency equivalent circuit
• Voltage Vπ is given by
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Emitter- and Source-Follower Circuits: (Contd..)
• Combining Equations (1),(2) and (3), we obtain
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Emitter- and Source-Follower Circuits: (Contd..)
• Substituting the expression for yπ, we find
• Impedance Z’b is shown in the equivalent
circuit in Figure (c).
• Equation (4) shows that the effect of
capacitance Cπ is reduced in the emitterfollower configuration.
(c) high-frequency equivalent circuit with
effective input base impedance
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Emitter- and Source-Follower Circuits: (Contd..)
From Equations (1) and (2), we have
• which yields a zero when yπ + gm = 0.
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Emitter- and Source-Follower Circuits: (Contd..)
• Using the definition of yπ, the zero occurs at
• Since rπ/(1 + β) is small, frequency fo is usually very high.
• In many applications, the impedance of rπ(1 + gm R’L) in parallel with Cπ/(1 + gm R’L) is
large compared to R’L. If we neglect R’L, then the time constant is
and the 3 dB frequency is
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Design Application:
• TWO Stage Amplifier with Coupling Capacitors
• Design a two-stage BJT amplifier with coupling capacitors, such that the 3 dB
frequencies associated with each stage are equal.
Specifications: The first two stages of a
multistage BJT amplifier are to be capacitively
coupled and the 3 dB frequency of each stage is
to be 20 Hz.
Choices: Assume the BJTs have parameters
VBE(on) = 0.7 V, β = 200, and VA =∞.
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TWO Stage Amplifier with Coupling Capacitors –Design (Contd..)
Solution (DC Analysis): We find, for each stage,
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TWO Stage Amplifier with Coupling Capacitors –Design (Contd..)
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TWO Stage Amplifier with Coupling Capacitors –Design (Contd..)
• The small-signal equivalent circuit is shown in Figure.
• The time constant of the first stage is
and the time constant of the second stage is
If the 3 dB frequency of each stage is to be 20 Hz, then
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TWO Stage Amplifier with Coupling Capacitors –Design (Contd..)
Comment:
• This circuit design using two coupling capacitors is a brute-force approach.
• Since the 3 dB frequency for each capacitor is 20 Hz, this circuit is referred to as a
two-pole high-pass filter.
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