chapter 9 frequency response - weber state university · 2017-01-04 · chapter 9 frequency...

1

ECE 3120 Microelectronics II Dr. Suketu Naik

Chapter 9

Frequency Response

2


Operational Amplifier Circuit Components

1. Ch 7: Current Mirrors and Biasing

2. Ch 9: Frequency Response

3. Ch 8: Active-Loaded Differential Pair

4. Ch 10: Feedback

5. Ch 11: Output Stages

3


Op Amp Circuit Components

Two Stage

Op Amp

(MOSFET)

4


Learning Objectives

1) How coupling and bypass capacitors cause the gain of discrete circuit amplifiers to fall off at low frequencies, and how to obtain an estimate of the frequency fL at which the gain decreases by 3dB below its value at midband.

2) The internal capacitive effects present in the MOSFET and the BJT and how to model these effects by adding capacitances to the hybrid-p model of each of the two transistor types.

3) The high-frequency limitation on the gain of the CS and CE amplifiers and how the gain falloff and the upper 3-dBfrequency fH are mostly determined by the small capacitances between the drain and gate (collector and base).

5


Learning Objectives

3) Methods for the analysis of the high-frequency response of amplifier circuits of varying complexity.

4) How the cascode amplifier studied in Ch 7 can be designed to obtain wider bandwidth than is possible with CS and CE amplifiers.

5) The high-frequency performance of the source and emitter followers.

6) The high-frequency performance of differential amplifiers (Note: this topic will be covered after Ch8: Differential Pair)

7) Circuit configurations for obtaining wideband amplification.

6


PART A:

Low Frequency Response

7


Middle-frequency band (midband) is the range of frequencies over

which amplifier gain is constant

Frequency Response of an Amplifier

8


Low Frequency Response of CS Amplifier

Q: What is the effect of

the coupling capacitors

CC1 and CC2 and the

bypass capacitor CS on the

overall gain (Vo/Vsig) at

low frequencies?

A: At low frequencies,

their reactances (1/jωC)

are high and gain is low.

Common Source Amplifier

Note: IC amplifiers don't have

coupling and by pass caps: midband

extends to DC

9


9.1.1 The CS Amplifier

Determining Vo/Vsig

1) Circuit with dc sources eliminated

2) Small-signal analysis

3) Ignore ro as it has negligible effect at low frequencies

Goal: find the lower -3dB frequency or fL

10



11


2

3

2

1 2

(9.5) output voltage:

(9.6) break frequency #3:

(9.9) midband gain:

(9.7) transfer func

1

1

||

tion:

D Lo o L d

D L

C D L

P

C D L

GM m D L

G sig

oM

sig P P

R R sV I R I

R R sC R R

C R R

RA g R R

R R

V s sA

V s s

3P

s

s


12


Low Frequency Response of CS Amplifier

fp2 to fp3: for every factor of 10 in the frequency, the gain drops (increases) by 20 dB


fp1 to DC: for every factor of 10 in the frequency, the gain drops (increases) by 60 dB

13


Determining the Pole Frequencies by Inspection

Reduce VSig to zero

Consider each capacitor separately

Find the total resistance seen between terminals of each

capacitor Req

time constant τ = ReqC

fL is the highest of the three

pole frequencies


14


Place Lower frequency fL at a specified value and minimize capacitor values

Cs should be selected at its pole frequency: fL=fp2

Place fp3 and fp1 between fp2/5 and fp2/10

Lower fp1, fp3 as much as possible while

choosing Cc1 and Cc2 so that they are

not so big

Design for Low Frequency

15


Example 9.1

16


Low Frequency Response of CE Amplifier

Q: What is the effect of the

coupling capacitors CC1 and

CC2 and the bypass capacitor

CE on the overall gain

(Vo/Vsig) at low frequencies?

1) Complicated analysis

due to finite β making

effects of CC1 and CE

inseperable

2) Determine the amplifier

gain and the transfer function

Common Emitter Amplifier

17


9.1.2 The CE Amplifier

1

1

1

2

|| ||

||

1/

(9.12) midband gain:

(9.10) voltage gain:

(9.11) break frequency #1:

(9.14) break frequency #2

||

1

||

:

B

M m C L

B sig

oM

sig C B sig

P

C B sig

P

R rA g R R

R r R

VA s s

V C R r R

C R r R

1

||

1

B sig

E e

R RC r

18


2

3

2

||||

||

1

(9.15) low frequency gain:

(9.16) break freque

ncy #3:

/

1

o Bm C L

sig B sig

C C L

P

C C L

V R rg R R

V R r R

s sC R R

C R R

1 2 3

1 1 2 2

(9.17) transfer function:

(9.18) fr

eq1 1 1 1

uency: 2

oM

sig P P P

L

C C E E C C

V s s sA

V s s s

fC R C R C R


19



Low-frequency response of

the CE amplifier:

(a) the effect of CC1 is

determined with CE and CC2

assumed to be acting as

perfect short circuits

(b) the effect of CE is

determined with CC1 and CC2



(c) the effect of CC2 is

determined with CC1 and CE



(a)

(b)

(c)

20


Assumption: CC1, CE, and CC2 do not interact and that their pole

frequencies are widely separated

Low Frequency Response of CE Amplifier



fp3 to DC: for every factor of 2 in the frequency, the gain drops (increases) by 18 dB

21


Example 9.2

22


Problem 9.4

23


Problem 9.9

Use one significant digit

chapter 9 frequency response - weber state university · 2017-01-04 · chapter 9 frequency...

Documents