THE JAMES-BAXANDALL PASSIVE TONE CONTROL NETWORK Vic iu chnh c lp tn s bass v treble trong b khuch i m thanh c trung thc cao thng c thc hin bi mch iu chnh tone c thit k c bit. C rt phin bn iu chnh tone ch da trn cc linh kin th ng, chng hn nh mch ground-reference James c minh ha hnh 1. Trong s nhng phin bn s dng linh kin tch cc chng ta phi k n P.J. Baxandall proposal y s iu chnh tone c sp t nh mt b khuch i hi tip. (minh ha hnh 1 v 2)

By gi ta bt u nghin cu vi phn iu chnh m Bass (fig 2). N c nh hng n ton di tn s pha di tn s trung tm c thit k trc ca mch Jame tng th.

y, R4 c k n nh mt s cch ly gia phn ny vi phn iu chnh m Treble ( di tn s pha trn tn s trung tm b nh hng bi phn ny). R5 i din cho tr khng ng vo ca mch khuch i c kt ni vi ng ra ca mch Jame v nn c chn sao cho n khng nh hng ng k n ti trong mch. y ta gi nh rng C3 v C4 lm h mch di tn s thp. Khi tng m bass ln cc i ( R2 trt n nh trn), mch iu chnh tone tng ng c m t nh sau:

T mch ny cng vi kin thc v laplace, ta c c:

Vi :

Th vo ta c:

Sau nhiu bc quy ng v n gin, ta i n cng thc tng qut:

(1)

H s khuch i i vi nhng m bass cao l:

H s khuch i ti nhng tn s thp:

im 0 ( zero) ca hm truyn tng qut (1):

im cc:

Vi mch iu chnh chnh maxium bass cut (R2 trt thp nht-m bass nh nht). mch tng ng thay i theo fig4. Mt ln na chng ra b qua nh hng ca ti R4 v R5.

Lm li cc bc trn 1 ln na:

Sau cc bc n gin, chng ta c:

(2)

H s khuch i ti m bass cao:

H s khuch i ti m bass thp:

im zero ca hm truyn:

im cc ca hm truyn:

Kt lun v h s khuch i di tn s thp: Vi mt mc iu chnh 40dB th theo quan h phi tha mn: (3)

Hay: Ngc li, ti im maximum ca s tng bass, thng s v h s khuch i ti tn s thp v cao c cho nh sau:

Theo m t biu thc (3): Th:

Ti ngng tng 20dB m bass: n gin ha biu thc cho ta quan h:

Chng ta c th kim tra li rng:

y l nhng thng s h s khuch i ca m bass cao v m bass thp cng ti 20dB ti ngng gimti a m bass. t c tnh i xng tt nht ca ng cong p ng ta phi chn s01=sp4. V vy:

Th nhng nhng quan h c gia cc thnh phn tr khng:

V cui cng: C2= 10C1 Biu Bode ca p ng bass ti full boost v full cut c m t bn di:

Tip theo ta phn tch mch iu chnh treble. Chng ta bit rng di tn treble l nhng m pha trn tn s trung tm. Chng ta c th xem rng C1 v C2 b ngn mch ti di tn ny. V vy R1 v R3 cng vi R4 l mt phn ca mch treble ny (fig 6).

thun li cho vic tnh ton, chng ta s tm thnh phn Thevenin cho Vg, R1 v R3. Trong fig 7, thnh phn in p Thevenin c cho nh sau:

(4) in tr Thevenin RTH :

Vi mch treble c chnh cc i (R6 cn trn) v gi nh rng dng chy qua R6 nh hn rt nhiu dng chy qua R4, chng ta c th a ra bi ton sau:

Nu R5 ln hn rt nhiu so vi R4 Thay th cc gi tr c m t bi biu thc (4) cho V1:

Sp xp li biu thc ny ta c:

V: (5)

Ti di tn Treble thp y : Ging nh -20dB Ti di tn Treble cao y :

Hay l 0dB Biu thc (5) c im zero: V im cc: Vimch Treble c chnh gim cc tiu, (R6 cn di) chng ta c:

Th: T phng trnh trn cng vi biu thc (4):

(6) Ti di tn treble thp:

Hay -20dB Ti di tn Treble cao:

Hm truyn (6) c im cc: ng cong p ng i xng tt nht ta phi chn Sp7=S05. V vy:

Kt qu C4=11C3 Hnh 8 m t biu Bode ca p ng Treble ti im tng v gim ti a.

Mt gi tr thun li cho R6 phi c tnh ton ngay. Chng ta cn nhiu kin thc v mch p t cng bc, trc khi chng ra b qua dng qua R6, Ta c th xem Ir61.1R4. Do , biu thc (7) c u tin p dng. Sau y ta s thc hin 1 v d vi cc tn s f01,fp2,f03,fp4,f05,fp6,fp7. u tin cho thun tin cc quan h c nhc li nh sau: R1=10R3 R2=99R3 C2=10C1 C4=11C3 f01 c tnh nh sau:

(8) f02 c tnh nh sau:

(9) f03 c cho nh sau:

fp4 = f01, yu cu mong mun ng cong p ng i xng: f05 c tnh nh sau:

(10) fp6 c tnh nh sau:

fp7 = f05, yu cu mong mun i xng ca ng cong p ng. Tn s trung tm ca mch Jame c ly t f01 v f05 (11) N trng vi tn s bin cc tiu ca ng cong p ng khi bass v treble c chnh tng ti a. ng thi cng l bin cc i ca ng cong p ng khi bass v treble gim nh nht. Thng th tn s trung tm ngh l 1KHz. Thit k minh ha: Hy gi thit rng chng ta mong mun thit mt mch iu chnh tone cho mt phn ca thit b. gi tr ca R1 l 10K. R3 s l 1K v R2 l 100K iu thun tin rng f01 v f05 c tch ra trn 1 decade trn di tn. theo m t (11) ta c:

Fc=1KHz th f01 phi l 316Hz. Gi tr cho f05 s l 3.6Khz Theo (8) th C2= 503.65nF. v C1=50.36nF. theo (9) th fp2 = 31.6Hz R5c ly bng 5 ln R2 trnh nh hng n ti ca mch. R5=500K. R4 phi c chn =5K, R6 >> 65K, chng ta chn 500K cho R6. Cui cng tr khng ngun phi c to khong 20 ln nh hn R1.