Chapter 6 Frequency response of circuits. Stability

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Transcription:

Chapter 6 Frequency response of circuits. Stability

6.. The frequency response of elementary functions

6... The frequency bandwidth

6... The frequency bandwidth /A/(dB ) A 0 3dB min max

6... The frequency response of elementary functions

6... The frequency response of elementary functions A constant /A 0 / (db) 0 lg(a 0 ) lg A 0 ct. ϕ (A 0 ) lg

A simple zero /A / (db) 0dB/dec A j 0 0 ϕ (A ) 90 o 45 o 0 /0 0 0 0 45 o /dec lg lg A 0 lg 0 << 0 A 0 >> 0 A 0 lg 0 ϕ 0 ( A ) arctg

A simple pole -0 ϕ (A ) /A / (db) 0 /0 0 0 0-0dB/dec lg lg A A 0 lg << j 0 A 0 0 >> 0 A 0 lg 0 0-45 o -90 o -45 o /dec ϕ 0 ( A ) arctg

A simple zero in origin /A 3 / (db) 0 0dB/dec A 3 j 0 ϕ (A 3 ) 90 o 0 lg A 3 0 lg 0 45 o lg ϕ o ( A ) 3 90

A simple pole in origin /A 4 / (db) lg A 4 j 0-0 ϕ (A 4 ) 0-0dB/dec A 4 0 lg 0-45 o lg ϕ o ( A ) 4 90-90 o

A multiple zero in origin /A 5 / (db) 0n 0n db/dec A 5 j 0 n ϕ (A 5 ) 90n o 45n o 0 lg lg A5 0 nlg 0 ϕ o ( A ) n 5 90

A multiple pole in origin /A 6 / (db) lg A 6 j 0 n -0n 0 ϕ (A 6 ) -0n db/dec A6 0 nlg 0-45n o lg ϕ o ( A ) n 6 90-90n o

A multiple zero /A 7 / (db) 0n db/dec A 7 j 0 n 0n ϕ (A 7 ) 90n o 45n o 0 /0 0 0 0 45n o /dec lg lg A 0 nlg 7 >> << 0 A 7 0 0 0 0 A7 n lg 0 ϕ 7 arctg 0 ( A ) n

A multiple pole 0n ϕ (A 8 ) -45n o -90n o /A 8 / (db) 0 /0 0 0 0-45n o /dec lg -0n db/dec lg A A 8 0 nlg 8 << j 0 0 A 8 n 0 >> 0 A8 0 n ϕ 0 lg 0 0 ( A ) n 8 arctg

Example /A/ (db) 80 60 40 A ( j) 0 4 j j 0 j 0 0 4 0 0-0 -40-60 ϕ (A) 80 o 35 o 90 o 45 o -45 o -90 o -35 o -80 o 0 0 0 3 0 4 0 5 0 6 0 7 0 8 0 9 0 0 0 0 0 3 0 4 0 5 0 6 0 7 0 8 0 9 0 0

Example /A/ (db) 80 60 40 A ( j ) 0 3 j 4 0 j j 0 3 0 0 0-0 -40-60 ϕ (A) 80 o 35 o 90 o 0 0 0 3 0 4 0 5 0 6 0 7 0 8 0 9 0 0 45 o -45 o -90 o -35 o -80 o 0 0 0 3 0 4 0 5 0 6 0 7 0 8 0 9 0 0

(a) (b) Z v a Z v v i Z v a Z v a v Z v v i v v v ) ( ) ( v Z v a Z v i Z v i Z Z Z -a v v i v v i -a v v i v v i 6..3. Miller theorem Z a a Z Z Z a Z Z v v v << << << << ;

6..3. Miller theorem

6.. Amplifiers with reaction

6... The block diagram of the amplifier with reaction

6... The block diagram of the amplifier with reaction X - X F F Y Y F XF X Y X F X X X F XF X Y X, Y are currents/voltages The global gain: Y X F F

6... Types of reaction - Positive reaction: > F F < - Negative reaction: < F > F Particular case: strong negative reaction X Defining loop transmission: T F >> X F it results T >> F ( << ) F - independent on amplifier F Conclusion: for strong negative reaction, the gain with reaction is function only on reaction

6... Types of reaction

6.3. eaction effects

6.3.. Amplifier de-sensitivity

6.3.. Amplifier de-sensitivity F F F F F d d d d ) ( F F F F F d F d d T F F (reaction factor)

6.3.. Distortion reduction

6.3.. Distortion reduction The reaction reduces the effect of distortions.

6.3.3. The improving of frequency response

6.3.3. The improving of frequency response For min Supposing that the direct amplifier is characterized by a first-order function: ( j) F min and that we have a constant negative reaction 0, it results: ( ) ( ) F j j j F F 0 j F min j F ( ) 0 / F / F0 /(db) Fmin 0dB/dec -0dB/dec lg resulting: ( j) j F min F 0 j F min F 0 0 j F min

equivalent with: ( j) F 0 F 0 0 j F min j F min ( ) F 0 ( ) F 0 0 0 It is possible to find the following form of (j): where: ( j) 0 j min j min 0 F 0 F 0 0 min F min F 0 0

Conclusion: The amplifier bandwidth is increased with the same factor of the gain decreasing. // 0 /(db) 0dB/dec min lg -0dB/dec

6.3.3. The improving of frequency response For max Supposing that the direct amplifier is characterized by a first-order function: F ( j ) F 0 j F max / F / F0 /(db) and that we have a constant negative reaction 0, it results: ( j ) F F ( j) ( j) 0 Fmax -0dB/dec lg resulting: ( j) F 0 F 0 0 j F max

equivalent with: ( j) F 0 F 0 0 F max F 0 j ( ) F 0 0 It is possible to find the following form of (j): where: 0 F 0 F 0 0 ( j ) 0 j max ( ) max F max F 0 0

Conclusion: The amplifier bandwidth is increased with the same factor of the gain decreasing. // 0 /(db) max lg -0dB/dec

6.3.3. The improving of frequency response / F (j)/ (db) /(j)/ (db) Conclusion: F0 / F (j)/ 0 lg ( F0 0 ) 0 /(j)/ min Fmin Fmax max lg

6.3.4. The impact on input/output resistances

6.3.4. The impact on input/output resistances The reaction changes input/output resistances in such a way that the amplifier with reaction simulates better an ideal amplifier. ( T ) i ' i for series reactions i ' i ( T ) for parallel reactions ( T ) o ' o for series reactions o ' o( T ) for parallel reactions

6.4. Circuits stability

6.4.. Algorithm for evaluating the stability of a circuit

6.4.. Algorithm for evaluating the stability of a circuit. Annulate the input voltage. Split the reaction loop in an arbitrary point 3. Apply a test voltage in this point, V test 4. Calculate the return voltage in the same point, V tr 5. Compute the eturn atio T V tr /V test 6. epresent the Bode diagrams for T 7. epresent an horizontal line at 80 o A. If the horizontal does not intersect the phase graphic, the circuit is stable B. If the horizontal intersects the phase graphic in a point A, from A represent a vertical axis which intersects the module diagram in point B a. if /T/ B > 0, the circuit is not stable b. if /T/ B 0 0, the circuit at the stability limit c. if /T/ B < 0, the circuit is stable. In this case it is possible to determine the phase margin: mark with C the point in which /T/ 0, represent a vertical axis from this point, which will intersect the phase diagram in point D. The phase margin is ϕ 80 o ϕ(d)

6.4.. Example

6.4.. Example Evaluate the stability of the following circuit v i C µf kω 0kΩ - 3 90kΩ v O a ( j ) 5 0 j 0 v 3 90kΩ V t kω 0kΩ a v V tr C µf

( ) ( ) ( ) ( ) ( ) ( ) ( ) [ ] 3 4 3 3 3 3 C C t t tr 0 j 0 j 0 j 0 T // C j C j a C j C j C j C j a T X // X // a V v a V V T

/A/ (db) 80 60 40 0 0-0 -40-60 ϕ (A) 80 o 35 o 90 o 45 o -45 o -90 o -35 o -80 o 0 0 0 3 0 4 0 5 0 6 0 7 0 8 0 9 0 0 0 0 0 3 0 4 0 5 0 6 0 7 0 8 0 9 0 0 The horizontal line does not intersect the phase diagram, so the circuit is stable.

6.4.3. Example

6.4.3. Example Evaluate the stability of the following circuit v i kω - 90kΩ C nf 3 9kΩ v O a ( j ) 5 0 j 0 v kω 90kΩ C nf 3 9kΩ V t a v V tr

( ) [ ] 3 3 3 3 3 C 3 t t tr 0 j 0 j 0 j 0 T // C j C j a C j a T X // a V v a V V T

/A/ (db) 80 60 40 0 0-0 -40-60 ϕ (A) 80 o 35 o 90 o 45 o -45 o -90 o -35 o -80 o 0 0 0 3 0 4 0 5 0 6 0 7 0 8 0 9 0 0 0 0 0 3 0 4 0 5 0 6 0 7 0 8 0 9 0 0 The horizontal line does not intersect the phase diagram, so the circuit is stable.

6.4.4. Exemple 3

6.4.4. Exemple 3 Evaluate the stability of the following circuit v i kω - 90kΩ C nf 3 9kΩ v O a ( j ) 5 0 j 0 j 5 0 v kω 90kΩ C nf 3 9kΩ V t a v V tr

( ) [ ] 5 3 3 3 3 3 C 3 t t tr 0 j 0 j 0 j 0 j 0 T // C j C j a C j a T X // a V v a V V T

/T/ (db) 80 60 40 0 0-0 -40-60 ϕ (T) 80 o 35 o 90 o 45 o -45 o -90 o -35 o -80 o -5 o -70 o B 0 0 0 3 0 4 0 5 0 6 0 7 0 8 0 9 0 0 0 0 0 3 0 4 0 5 0 6 0 7 0 8 0 9 0 0 ϕ 0 o A C D The horizontal line at 80 O intersects the phase diagram in A point, /T B / 0, so the circuit is at the stability limit ( ϕ 0).

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