Eercses r Frequency espnse EE 0, Wnter 0, F. Najabad
Eercse : A Mdy the crcut belw t nclude a dnant ple at 00 Mz ( 00 Ω, k, k, / 00 Ω, λ 0, and nre nternal capactances the MOS. pute the dnant ple n the crcut yu dened s reduced by a hal. pute the dnant ple n the crcut yu dened s reduced by a hal. There are tw ways t ntrduce a dnant ple n a S apler: A capactr between utput & rund A capactr between nput & utput
nant ple a a capactr between utput & rund λ 0 r esstance seen between ternals s: (r 0 r / π ( π ( Part A: Fr k, π ( k 0.667k π 667 00 0 6.4 pf Part : s ndependent s n chane n the dnant ple Part : Fr k, π ( k 0.5k π 500.4 0 33 Mz
nant ple a a capactr between nput & utput Mller s There λ 0 r r d A ( r,, ( A ( / A ( ( /
nant ple a a capactr between nput & utput / ( ( /(,, π r / ( / ( ( (,, A A. Set 0. esstance seen between ternals, s 3. esstance seen between ternals, s r
0.69 pf,30 30 767 534 / ( ( 6.67 and 0.667k k k, Fr Ω π Part A: nant ple a a capactr between nput & utput / ( ( /(,, π 50 Mz,534 534 767 767 / ( ( 6.67 and 0.667k k k, Fr Ω π Part : 8 Mz,800 800 600 00 / ( ( 5.0 and 0.5k k k, Fr Ω π Part :
Eercse : pute the upper cut- requency the S apler belw wthut usn Mller s There. We can use the te-cnstant ethd t cpute wthut usn Mller s There. weer, we wll need t cpute the resstance seen between ternals d by usn MOS sall nal del.
s db τ π /( p s τ /(π ( ( r p db
T nd d, we cnnect t the crcut and nd : s d s s s 0 ( ( ( τ /(π [ ( ] 3 p3 d d d
τ π τ /( p τ /(π [ ( ] 3 p3 d d d /(π τ τ τ 3 /(π s ( /(π ( s s /(π ( db d ( r d p db [ ( d ( db /(π [ s d ( ] ( db d ] d Mller capactr at the nput: d, Mller capactr at the nput: d, Nte: (-/A ter s ssn because Mller s appratn S&S, 6 th Edtn, cpares the requency respnse a S apler n Eaples 9.8 (a Mller s There, 9.9 (te cnstant ethd, and 9.0 (Eact. parn results r Eaples 9.8 and 9.9, S&S clas that Mller s There resulted n 50% errr n. Ths s ncrrect. The reasn r the 50% derence s that S&S uses square rt rula t nd n 9.8 nstead / / p / p. Mller s appratn s usually pretty d. Nte that Mller s appratn (and te-cnstant ethd sses the Mller Ple.
Eercse 3: Fnd the d-band an, lwer and upper cut- requency ths tw-stae apler (dentcal transstrs wth s 50 F, d 80 F, / 00 Ω, λ 0, and nre db. Eanatn the crcut shws that bth 00 and 500 pf capactrs result n lw-requency ples. h-requency respnse s set by the nternal capactances transstrs. Q S apler ncludes a eedback resstr (0k lcated between nput and utput.
Md-band an: The lw-requency capactrs shuld be shrt The hh-requency capactrs pen. Ths s a tw-stae apler:
T cpute the d-band an a ult-stae apler, we hae t start r the lad sde as the nput resstance each stae s the lad r the preus stae The 0 k eedback resstr s between the nput and the utput. Use Mller s There t replace wth tw resstrs:,, A / A
,,, A / A Q s a S apler: A ( r (k, We can substtute r, r abe n the epressn r A and sle r A. weer, r A >>,, 0k. Nte: ( k, (k A ( k, (k 0k 90 Ω / 9. 9 k / 9. and (k,, A Whch s a ery d appratn cpared t,, 0k 990 Ω A 9. (k 9k 900 Ω ( k, 90 Ω
Q s als a S apler: A A 5 ( r (k, ecause 0, ( 9. ( 5 ( 45.5
w-requency respnse: Ple r the 500 pf capactr: / esstance seen between ternals 500 pf capactr (/ π 00 00 00 67 Ω π 67 500 0 p - 4.77 Mz
w-requency respnse: Ple r the 500 pf capactr: (k p, π k 990 Ω.99k π,990 00 0-0.80 Mz p p p 4.77 0.80 5.57 Mz
h-requency respnse: ue t nternal capactances transstrs. w-requency capactrs are shrt.
Inre db (per prble nstructn. eplace d usn Mller s There. eplace 0k resstr between ate and dran Q usn Mller s There.
A 80( 5 ( d, d s 50 F d 480 F, d( / A 80( 0. 96 F d s, d ( A 80( 9. 808 F 50 F / A 80( / 9. ( d, d 89 F
bne parallel capactrs: 89 F,54 F 50 808 96 730 F 50 480,,,, d n s d d s d n Ω Ω 90 (k 0 k 990,,,
Te cnstant r n : τ n 730 F 5 0 00 n 00 730 0.46 0
Te cnstant r : τ,54 F r k, k 0.990 k 497 Ω 497,54 0 5 5.73 0 0
5, 3,,, 0 8.0 0 89 90 90 k k 89F Ω d d r τ Te cnstant r d, : 99Mz 0 8.0 0 8.0 0 5.73 0.46 0 0 0 3 πτ τ τ τ τ