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Frequency Response
ROCHESTER INSTITUTE OF TECHNOLOGYMICROELECTRONIC ENGINEERING
Frequency Response of the CE Amplifier
Dr. Lynn FullerWebpage: http://people.rit.edu/lffeee/
Microelectronic EngineeringRochester Institute of Technology
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 1
3-1-11 Frequency_Response.ppt
Rochester Institute of Technology82 Lomb Memorial DriveRochester, NY 14623-5604
Tel (585) 475-2035
Email: [email protected] webpage: http://www.microe.rit.edu
Frequency Response
OUTLINE
Introduction
Gain Function and Bode Plots
Low Frequency Response of CE Amplifier
Millers Theorem
High Frequency Response of CE Amplifier
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 2
High Frequency Response of CE Amplifier
References
Homework Questions
Frequency Response
INTRODUCTION
We will be interested in the voltage gain of an electronic circuit as a function of frequency.
VoutVin Av = Vout/Vin
Decibel: the gain of some network can be expressed in logarithmic units. When this is done the overall gain of cascaded networks can be found by simple addition of the individual network gains.
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 3
be found by simple addition of the individual network gains.
The decibel is defined as:Ap = 10 log (Po/Pin) dBwhere Ap is the power gain in decibels
Po is the power out and Pin is the power in
The decibel has also been used as a unit for voltage gain.Po = Vout2/RL and Pin = Vin2/Rin
and if Rin=RLAp = 20 log (Vout/Vin) dB
Frequency Response
INTRODUCTION
Thus the decibel is often used to express voltage gains. (Really only correct if RL=Rin but many people are not precise about this point)
-
+VoVin
R2R1 If R1=2K and R2=47K
Vo/Vin = - 47K/2K = -23.5
Vo/Vin = 23.5 or 27.4 dB
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 4
Vo/Vin = 23.5 or 27.4 dB
1 10 100 1k 10k 100k
Gain vs Frequency
0
10
20
30
dB
Frequency Response
THE GAIN FUNCTION
The gain function, A(s): an expression for Vo/Vin which is found in a straight forward manor from the ac equivalent circuit.
Vo/Vin = A(s) or in particular s=jω thus A(jω)
a0 + a1 s + a2 s2 + a3 s3 ….b0 + b1 s + b2 s2 + b3 s3 ….
A(s) =
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 5
K (s-z1)(s-z2)(s-z3)…(s-p1)(s-p2)(s-p3)…
A(s) =
Where z1, z2, z3 are zeros, p1, p2, p3 are poles
K (jω-z1)(jω-z2)(jω-z3)…(jω-p1)(jω-p2)(jω-p3)…
A(jω) =
A0 (jω/ω1)Ν(jω/ω3+1)(jω/ω5+1)…(jω/ω2+1)(jω/ω4+1)(jω/ω6+1)…
A(jω) =
Frequency Response
GOALS
1. Obtain the gain function from the ac equivalent circuit.
2. Predict the frequency response of the gain function.
3. Use graphical techniques to sketch the frequency response
3. Introduce a new model for transistors at high frequencies.
5. Analyze and predict the frequency response of a common
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 6
5. Analyze and predict the frequency response of a common
emitter amplifier stage
Frequency Response
GRAPHICAL TECHNIQUES AND BODE PLOTS
Vo = Vin1/sC
R + 1/sC
Vo/Vin = 1/jωC
R + 1/jωC1
jωRC + 1=
1jω/ω1+ 1
=
Where ω1 = 1/RC and f1 = 1 / 2 π RC
Vin +-
R
CVout
Gain Function:
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 7
Where ω1 = 1/RC and f1 = 1 / 2 π RC
Bode Plot: a plot of the gain function versus frequency (ω or f). Note: both magnitude and phase are a function of frequency. The Bode Plot plots this information separately.
ω
ω
Vo/V
s (d
B)
Log10 scale
Log10 scale Phas
e (D
egre
es)
Frequency Response
CONTINUE PREVIOUS EXAMPLE 1
Av = Vo/Vin =jω/ω1 + 1
1
At low ω Vo/Vin = 1 0° Vo/Vin dB = 0 dB and; Θ = 0°
At high ω Vo/Vin = 1/(jω/ω1) -90°
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 8
At high ω Vo/Vin = 1/(jω/ω1) -90°
Vo/Vin dB = ω1/ω dB and Θ = -90°
Note: at ω = 10 ω1 Vo/Vin dB = -20 dB
Note: at ω = 100 ω1 Vo/Vin dB = -40 dB
Thus we see at high frequencies the gain decreases by -20 dB / decade
Frequency Response
CONTINUE EXAMPLE 1
ω
Vo/V
s (d
B)
ω1
0dB-3dB
-20dB
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 9
ω
Log10 scale
Phas
e (D
egre
es)
ω1
-45
-90
Frequency Response
EXAMPLE 2
Vin +-
RC Vout
Obtain the gain function for the network shown. Sketch the magnitude part of the Bode Plot.
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 10
Frequency Response
POLES AND ZEROS
Poles and Zeros: the complex frequency at which the gain function goes to infinity in the case of poles or to zero in the case of zeros.
Vo/Vin1
sCR + 1Example 1:
jωωωω
σσσσ
Pole at s1 = - 1/RC
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 11
Example 2:
Pole at s1 = - 1/RC
Vo/VinsCR
sCR + 1
jωωωω
σσσσ
Which has a Zero at zero and a Pole at s1 = - 1/RC
s-plane
s2
Frequency Response
CORNER FREQUENCY
Vo/Vin =1
jωCR + 1Example 1:
Has a corner at ω1 = 1/RC or f = 1/2πRC
Corner frequencies: that frequency (f or ω) at which the real and imaginary parts of one term of the gain function are equal/
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 12
Example 2:Vo/Vin =
jωCRjωCR + 1
Has a corner ω1 = 1/RC
Frequency Response
EXAMPLE 3
Find the gain function, poles, zeros and corner frequencies for the network shown, sketch the Bode plot.
VinR1
C
VoutR2
+
-
+
-
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 13
R2- -
Frequency Response
LOW FREQUENCY MODEL OF CE AMPLIFIER
Effect of the Coupling Capacitor, Cc assume Ce, and Cc2 act like a short.
Obtain the gain function from the ac equivalent circuit:
Vcc
Rc
vs+
R1
voRL
Rs+
Cc
Cc2
vo = -gm vbe Ryvs Rx
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 14
vs+
-
vo
ReR2 -
Cc
Ce
Rs
RL
rogmvbe
or
βibrπvbe
+
-
vo
+
-
vs +
-Rc
RthCc
Ry = ro//Rc//RL
vbe =
Rx = Rth//rπ
(Rs+1/sCc + Rx)vs Rx
vo/vs =-gm Rx Ry
(Rs+1/sCc + Rx)
s = jω
Frequency Response
EFFECT OF COUPLING CAPACITOR Cc
Manipulate the gain function until we have a form from which we can easily obtain the bode plot.
vo/vs =s Cc
(sCc(Rs+Rx)+1)-gm RyRx
vo/vs =s Cc
(sCc(Rs+Rx)+1)-gm RyRx
(Rs+Rx)
(Rs+Rx)
-gm RyRx
Vo/Vs (dB)
20Log10 |Avmid|
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 15
vo/vs =s Cc
(sCc(Rs+Rx)+1)
-gm RyRx
(Rs+Rx)
(Rs+Rx)
vo/vs =j ω/ω1
(j ω/ω1+1)Avmid
Where ω1 = 1/Cc(Rs+Rx) ω
Log10 scale
ω1
0dB
-3dB
-20dB/Dec
Frequency Response
SUMMARY FOR EFFECT OF Cc
1. At low frequencies the coupling capacitor “opens” up and the voltage gain drops as the frequency decreases.
2. The corner frequency ω1 equals the inverse of the product ReqCc where Req is the resistance “seen” looking from the capacitor terminals with Vin = zero in the ac equivalent circuit.
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 16
3. At mid frequencies the voltage gain is the expected gain.
Avmid =
4. Summary 1, 2, and 3 above are true but the results are slightly different if the emitter bypass capacitor acts like an open near where Cc begins to open. (start with new ac equivalent circuit)
-gm RyRx
(Rs+Rx)
-β RyRx
rπ(Rs+Rx)=
Frequency Response
EFFECT OF Ce ON FREQUENCY RESPONSE
The ac equivalent circuit of the CE amplifier on page 14 above is shown. Here we assume Cc is a short (note: it is possible that Cc acts like an open rather than a short)
Rs rogmvbe+
vo
+
vs
Let Rs = 0 and RL = ro = infinity to simplify the algebra
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 17
Rs
RL
roor
βibrπvbe
-
vo
-
vs +
-Rc
Rth
CeRe
Frequency Response
EFFECT OF Ce ON FREQUENCY RESPONSE
vo = - β ib Rcvs = ib rπ + (β+1) ib Re//(1/sCe)
vo/vs = -β Rc1
rπ + (β+1) Re//(1/sCe)
The gain function:
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 18
rπ + (β+1) Re//(1/sCe)
Manipulate the gain function:
vo/vs = -β Rc1
rπ + (β+1) Re/sCeRe + 1/sCe
= -β Rc1
rπ + (β+1) ResCeRe + 1
Frequency Response
EFECT OF Ce ON FREQUENCY RESPONSE
vo/vs = = -β Rcrπ + (β+1) Re(sCeRe + 1)
sCeRe + 1
= -β Rc+ (β+1) ResCeRe rπ + rπ
sCeRe + 1
rπ + (β+1) Re-β Rc sCeRe + 1
sCeRe rπrπ + (β+1) Re
+ 1
=
(jω/ωe + 1)
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 19
rπ + (β+1) Re-β Rc (jω/ωe + 1)
(jω/ωe1+ 1)
Where: ωe = 1/Ce Re
ωe1 = 1/(Ce Re//(rπ/(β+1)))
vo/vs =
k=AvlowNote: ωe1 is always > ωe
Note: Avmid = Avlow ωe1/ωe
Frequency Response
EFFECT OF Ce ON FREQUENCY RESPONSE
Vo/Vs (dB)
Avmid
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 20
ω
Log10 scale
ωe1 ωe1
Avlow
Frequency Response
SUMMARY FOR EFFECT OF Ce
1. At low frequencies the bypass capacitor, Ce, opens up and the voltage gain becomes that of an unbypassed CE amplifier, Avlow
2. At high frequencies the gain is Avmid
3. Because of 1 and 2 we see that there are two corner frequencies. They are:
ωe = 1/ReCe
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 21
They are: ωe = 1/ReCe
and ωe1 = 1/ReqCe where Req is the resistance seen from the terminals of Ce
Req = Re//rπ/(β+1) if Rs = 0 and Cc “short”
Req = Re//(rπ+ R1//R2//Rs )/(β+1) if Rs not 0 and Cc “short”
Req = Re//(rπ+ R1//R2 )/(β+1) if Cc “open”
Frequency Response
COMPLETE CE AMPLIFIER LOW FREQUENCY RESPONSE
Vcc
Rc
vs+
R1
voRL
Rs+
Cc
Cc2
Rs = 2KR1 = 40KR2 = 10KRC = 4KRe = 1KRL = 2K
β = 100Vcc = 20Cc = Ce = Cc2 = 10µf
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 22
vs+
-
vo
ReR2 -
Cc
Ce
vo/vs = K (jω/ω1) (jω/ω2) (jω/ω3+1)
(jω/ω1+1) (jω/ω2+1) (jω/ω4+1)
Find k, ω1, ω2, ω3, ω4
Frequency Response
EXAMPLE: SOLUTION
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 23
Frequency Response
HIGH FREQUENCY BJT TRANSISTOR MODEL
β ibrπ
Rbb’
Cb’e
CD
Cb’ccb b’
e
roib
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 24
Rbb’ is the series base resistancerπ is the base emitter small signal junction resistanceCb’e is the base emitter junction capacitanceCb’c is the base collector junction capacitanceCD is the diffusion capacitance, represents the change in charge
stored in the base caused by a change in base emitter voltage
ro is the small signal output resistance = VA/ICβ is the short circuit common emitter current gain
e
Frequency Response
MILLERS THEOREM
To predict the high frequency response of a common emitter amplifier we want to do some quick calculations. We would like to simplify the model given on the previous page. We can do this with the aid of Miller’s theorem. The resulting model is approximate and might not give good results above the upper corner frequency where the voltage gain begins to fall off.
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 25
Millers Theorem: Consider a linear network with N nodes. An impedance, Z, between any two nodes, N1 and N2, can be removed and another impedance Z1 placed from N1 to reference and impedance Z2 placed from N2 to reference. If Z1 = Z/(1-K) and Z2 = ZK/(K-1) where K=V2/V1, then the nodal equations will not be changed and the resulting circuit will yield equivalent node voltages, V1, V2, etc.
Frequency Response
MILLERS THEOREM
N1 N2 N2N1
Z
Z1 Z2Ref Ref
Z1 = Z/(1-K) Z2 = Z (K/(K-1))
V1 V2 V1 V2
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 26
Ref Ref
where K=V2/V1
at N1 term (V1-V2)/Z at N1 term V1/Z1 = V1/(Z/(1-K))
= V1/(Z/(1-V2/V1))
= (V1-V2)/Z
Frequency Response
HIGH FREQUENCY MODEL OF CE AMPLIFIER
β ibrπ
Rbb’
Cb’e
CD Cm’
cb b’
e
roib
Cm
1/sCb’c
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 27
e
From: Z1 = Z/(1-K) we have 1/sCm =1/sCb’c1-V2/V1
V2 = -β ib ro and V1 = ib rπ
Therefore: Cm = Cb’c (1- - β ro/rπ)
and : Cm’ = ~ Cb’c
From: Z2 = Z (K/(K-1)) Voltage gain
Frequency Response
EXAMPLE: HIGH FREQUENCY CE AMPLIFIER
Vcc
β ibrπvs +
- RB
Rbb’
CTiin ib
RSβ = 100rπ = 1K, Rb’b = 0Cb’c = 20pfCb’e + CD = 20pf + 1000pf
RL
Let CT = Cb’e + CD + Cm
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 28
Rc
vs +
-
Rb
IinRL
Let CT = Cb’e + CD + Cm
and Cm = Cb’c(1- - β RL/rπ)
To find the gain function:vo = - β ib RLib = Vb’e/rπVb’e = vs (RB//rπ//(1/sCT)
RS + (RB//rπ//(1/sCT)
Nextpg
Frequency Response
EXAMPLE: HIGH FREQUENCY CE AMPLIFIER
vo/vs = - β RL
RS + (RB//rπ//(1/sCT) rπ
(RB//rπ//(1/sCT)The gain function:
Manipulate the gain function: Let RB//rπ = R
vo/vs = - β RL
π
R(1/sCT)
R+ (1/sCT)
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 29
vo/vs = rπ
R+ (1/sCT)R(1/sCT)
R+ (1/sCT)RS +
vo/vs = - β RL
rπ
R
sCT R+ 1
R
s CT R+ 1RS +
vo/vs = - β RL
rπR
(s CT R+ 1)RS + R
vo/vs =- β RL
rπR
s CT R RS (RS + R)1
(RS + R)+1
Frequency Response
EXAMPLE: HIGH FREQUENCY CE AMPLIFIER
vo/vs =- β RL
rπR
jω CT R RS(RS + R)1
(RS + R)+1
ωh = 1/ CT (R//RS)Avmid
Vo/Vs (dB)
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 30
ω
Vo/Vs (dB)
Log10 scaleωh
20Log10 (Avmid)
Frequency Response
SUMMARY: HIGH FREQUENCY RESPONSE OF CE AMP
1. At high frequencies the internal capacitances in the transistor causes the voltage gain to decrease
2. At mid frequencies the gain is Avmid =
3. The corner frequency is ωh = 1/ (Req CT)
- β RLrπ
R
(RS + R)
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 31
where CT = Cb’e + CD + Cm
and Req = the equivalent resistance as “seen” from the terminals of the capacitor CT. (vs = zero)
4. There is a second corner due to the miller capacitance Cm’. Since ωh occurs first we are not normally interested in the corner due to Cm’
Frequency Response
HOW DOES MANUFACTURER SPECIFY CD, Cbe, Cbc
Cb’c is usually given by the manufacturer as the common base output capacitance which it is.
Cb’e and CD are given indirectly by the manufacturers specification of the transition frequency fT
fT is the frequency at which the CE short circuit current gain goes to 1Vcc
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 32
Vcc
Rc
vs +
-
RB
Iin Io β ibrπvs +
-
RB
Rbb’
CTiin
iout
ib
Frequency Response
2N3904
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 33
Rb = 10 ohms
Frequency Response
ANALYSIS OF SHORT CIRCUIT CURRENT GAIN TO EXTRACT Cb’e + CD
iout = β ib
ib = iin 1/sCT
rπ + 1/sCT
iout/iin = β
sCT rπ + 1
β
iout/iin (dB)
20Log10 (β)
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 34
iout/iin = β
jωCT rπ + 1
ωb = 1/(CT rπ)
ωωh
0 dB
ωΤ
ωΤ = 2π fT = transition freq in radians/s
iout/iin = β
jω/ωb + 1
Frequency Response
ANALYSIS OF SHORT CIRCUIT CURRENT GAIN TO EXTRACT Cb’e + CD
at ωT, iout/in = 1 = ~ β
jω/ωb
2 π fT =β
CT rπ
β
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 35
So CT = β
2 π fT rπ= Cb’e + CD + Cm
and Cm = Cb’c since Av = zero
Finally Cb’e + CD = β
2 π fT rπ- Cb’c
Frequency Response
EXAMPLE: DETERMINATION OF Cb’c, Cb’e +CD
Given:1. Common-Base Open Circuit Output Capacitace of 12 pf is measured at f = 1 Mhz, VCB = 10V and IE = zero.2. A transition frequency of 100 Mhz is measured using the following test conditions, VCE = 2V, IC = 50mA, β response with frequency is extrapolated at -20 dB/Dec to fT at which β = 1 from f = 20Mhz where β =100
12V
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 36
β =100
Find Cb’c,. Cb’e + CD12V
Rc=1K188K
vo
+
-
Frequency Response
SOLUTION TO EXAMPLE ON PREVIOUS PAGE
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 37
Frequency Response
ANOTHER EXAMPLE
Vcc
Rc
vs+
R1
vo
Rs+
Cc
Vcc = 15Rs = 100R1 = 150KRC = 500
Find rπ, Cm, CT and ωh
β = 100VA = infinityRb’b = 100Cb’c = 20pf at Vcb = 5fT = 100 Mhz
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 38
vs+
-
vo
-
CcFind rπ, Cm, CT and ωh
Frequency Response
EXAMPLE FROM OLD EXAM
Assume β = 150VA = 100Cb’c = 10pF @ 10 VfT = 200 MHzRb’b = 100 ohms
12V
4K
vs+
40K
vo2K+
10uf
1uf
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 39
Find k, ω1, ω2, ω3, ω4, each 5 pts and ωh 10 pts
vo/vs = k (jω/ω1) (jω/ω2) (jω/ω3+1)
(jω/ω1+1) (jω/ω2+1) (jω/ω4+1) (jω/ωh+1)
vs+
-1K10K -
4uf
Frequency Response
REFERENCES
1. Sedra and Smith, chapter 5.
2. Device Electronics for Integrated Circuits, 2nd Edition, Kamins
and Muller, John Wiley and Sons, 1986.
3. The Bipolar Junction Transistor, 2nd Edition, Gerald Neudeck,
Addison-Wesley, 1989.
4. Data sheets for 2N3904
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 40
4. Data sheets for 2N3904
Frequency Response
HOMEWORK – FREQUENCY RESPONSE OF CE AMP
1. For the circuit on page 22 find Avmid, k, ω1, ω2, ω3, ω4 given:Rs = 1KR1 = 50KR2 = 10KRC = 5KRe = 1KRL = 5K
β = 150Vcc = 24Cc = 1ufCe = 2ufCc2 = 10µf
2. Create a spread sheet to analyze CE circuits like that in problem 1 to
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 41
2. Create a spread sheet to analyze CE circuits like that in problem 1 to find Avlow, Avmid, k, low frequency corners. Extra points if you also do high frequency analysis?3. If Cb’c is measured at Vcb = 5 what is it at Vcb=10?4. Find Cb’e + CD for fT = 200 Mhz and IC = 5mA, β = 150 and Cb’c = 10pf5. Create a spread sheet to calculate and graph the magnitude part of the Bode Plot given k, ω1, ω2, ω3, ω4 and ωh
Frequency Response
EXAMPLE pg 22: SOLUTION
DC analysis: Rth = R1//R2 = (10)(40)/(10+40) = 8K
Vth= Vcc R2/(R1+R2) = 20 (10)/(10+40) = 4V
KVL: IB Rth +0.7 +(B+1)IB Re – Vth = 0
IB = 4 - 0.7 / (Rth +101K) = 30.3uA
IC = B IB = 100 (30.3uA) = 3.03 mA
gm = IC/VT = 3.03/0.026 = 117 mS
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 42
gm = IC/VT = 3.03/0.026 = 117 mS
rπ = Vt/IB = 0.026/30.3uA = 858 ohms
ro = VA/IC = assume large
Avmid = Voltage gain including RS and RL assume all C’s shorts
Vo = gm Rc//RL Vin
Vin = Vs Rin/(Rin + Rs)
Vo/Vs = Vo/Vin x Vin/Vs = -(gmRC//RL ){Rin/(Rin+Rs)}
Vo/Vs = -117m (4K//2K) (Rth//rπ)/((Rth//rπ)+2K)
= - 43.6
Frequency Response
EXAMPLE pg 22: SOLUTION(continued)
w1 = 1/Req Cc1 assume Ce is open unless it is 10X Cc1
Req = Rs+Rth// (rπ +(B+1)Re) = 2K+ 8K // (0.858 +101K)
= 9.42K
w1 = 1/ (9.42K 10 uF) = 10.6 r/s or 1.69 Hz
w2 = 1/Req Cc2
Req = RL + Rc = 6K
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 43
Req = RL + Rc = 6K
w2 = 1 / (6K 10uF) = 16.7 r/s or 2.65 Hz
w3 = 1/ReCe = 1/(1K 10uF) = 100 r/s or 15.9 Hz
W4 =1/ReqCe
Req = Re//((rπ+Rth//Rs)/(B+1)) = 1K//((0.858K+8K)/101)
= 80.6 ohms
w4 = 1/(80.6 10uF) = 1240 r/s or 198 Hz
K =Avlow = Avmid w3/w4 = -43.6 (100/1240) = -3.52
Frequency Response
SPREAD SHEET SOLUTION
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 44
Frequency Response
SPREAD SHEET SOLUTION
© March 1, 2011 Dr. Lynn Fuller, Professor
Rochester Institute of Technology
Microelectronic Engineering
Page 45