Low Frequency Response of CE Amplifier

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

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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




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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.




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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




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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) =




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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




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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:




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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°




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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




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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.




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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




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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/




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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

+

-

+

-




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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




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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|




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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.




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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




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Rs

RL

roor

βibrπvbe

-

vo

-

vs +

-Rc

Rth

CeRe

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:




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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)




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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


Vo/Vs (dB)

Avmid




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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




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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




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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




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Frequency Response

HIGH FREQUENCY BJT TRANSISTOR MODEL

β ibrπ

Rbb’

Cb’e

CD

Cb’ccb b’

e

roib




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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.




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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




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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




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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




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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


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)




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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


vo/vs =- β RL

rπR

jω CT R RS(RS + R)1

(RS + R)+1

ωh = 1/ CT (R//RS)Avmid

Vo/Vs (dB)




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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)




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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




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Vcc

Rc

vs +

-

RB

Iin Io β ibrπvs +

-

RB

Rbb’

CTiin

iout

ib

Frequency Response

2N3904




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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)




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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




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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




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β =100

Find Cb’c,. Cb’e + CD12V

Rc=1K188K

vo

+

-

Frequency Response

SOLUTION TO EXAMPLE ON PREVIOUS PAGE




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




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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




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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




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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




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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




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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




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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




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Frequency Response

SPREAD SHEET SOLUTION




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Low Frequency Response of CE Amplifier

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