CTONIC 3 XCI OW AMII. xpla what are classes A, B, AB and C amplifiers terms f DC biasg usg a MOT dra characteristic.. efer t the graphs f page and the table at the tp f page 3 f the thery ntes t answer the fllwg questins: A) Hw des efficiency vary with signal level? B) In classes B and AB, des efficiency vary with the type f wavefrm used? xpla. C) What wavefrm prduces the wrst case pwer dissipatin f the pwer transistrs? xpla. D) r what signal level is the pwer supply stressed the mst? xpla ) r what signal level are the pwer transistrs stressed the mst? xpla 3. xpla what crssver distrtin is and what causes it class B amplifiers? 4. Hw des class AB biasg reduce crssver distrtin? 5. xpla what harmnic distrtin is. 6. When feedback is used with an p amp, expla hw crssver distrtin is reduced and hw the p amp frequency respnse will affect the amunt f distrtin. 7. xpla why dides r a B multiplier are better than a simple resistr t bias a (BJT) pushpull stage class AB. 8. efer t the circuit diagram n the next page and the MOT transfer curves shwn belw. A) Assumg that the MOT's have a G(TH) that ranges frm t, determe the values f,, and that are apprpriate allw fr safety marg, say G adjustment range f 0.5 t.5. Assume W pts are used B) If D (ON) 0,3Ω fr the MOT's when is imum, determe the peak AC values f, OA, v gs3 and v gs6 and I OA. Assume MOT s heat up t 5 C use 5 C MOT curve. C) rm values btaed part B, calculate and the imum efficiency f this amplifier and f each MOT assumg a sewave put. pecify at what put signal level the pwer transistrs will dissipate imum pwer. D) epeat questin C fr a square wave put. ) What is the effect f the. µ capacitr n the ga respnse f the amplifier? C. auril wer Amplifier xercise ev. 5/5/03 age
CTONIC 3 XCI 35 5 G k Q3 Q Q 35 0K 00 p I OA OA 4Ω.µ 0k 00k 5 G Q4 Q5 Q6 5 G k 35 9. ketch a bridge pwer amplifier, expla hw it wrks and als expla why it quadruples the imum O/ pwer. 0. ketch a bridgeparallel pwer amplifier, expla hw it wrks and expla what the advantage is ver the simple bridge amplifier. C. auril wer Amplifier xercise ev. 5/5/03 age
CTONIC 3. XCI 35 35 B B 35 35 ut 8 Ω 35 35 B B 35 35 A) rm the clippg vltage graph the data sheets, determe the imum unclipped O/ vltage acrss the lad and then cmpute the imum O/ pwer (assume sewave put) and cmpare it agast the typical value specified the data sheets. B) What is the pwer cnsumptin f the abve amplifier at the imum pwer level fund abve? Assume sewave put. Determe the imum heat sk resistance required at T A 30 C assumg that the fur amplifiers are munted n a sgle heat sk with 0. C/W sulatrs. C) If the put signal ranges frm 0 t rms, determe all resistrs required t prvide imum O/ pwer t the speaker. Assume sewave put and allw fr 0% mre ga than required mimum t ensure imum pwer is achieved. D) If all s and s have % tlerance, what is the imum current imbalance between tw parallel amplifiers? ) What is the imum current imbalance caused by DC ffsets f the s between tw parallel amplifiers? ) T slve the current imbalance caused by DC ffsets, add an AC cuplg capacitr t the put and als sert a capacitr series with all s. Calculate prper values fr an audi signal and expla hw that reduces the DC current imbalance parallel amplifiers. G) ketch the ga respnse f the vertg amplifiers and that f the nnvertg amplifier. Include the effect f the AC cuplg capacitr used with the frntend buffer bth ga respnses. C. auril wer Amplifier xercise ev. 5/5/03 age 3
CTONIC 3 XCI 8. A) 35 5 G.06 7 k 3. 0.7.5 Q Q3 Q 4.3 35 0K 00 p.067 0.7...088 I A 0.66 7 0 0 4.3 t 4.3.µ 0k 0.7.. 0.4 0.4 I B 4.3 00k 4 Ω Q4 35 5 G 0.4. T calculate and assume matched MOT s, equal G values, this entails I A I B and I 0 and a range f current f 0.4 t.088 thrugh and. 8.6 6.9k 67.9k 0.4.088 et 4k then 4k 74k 0.7 k Q5 0.5 Q6 67.9k 6.9k 4.6k 50k When MOT s are mismatched, G values are different, has t absrb the difference between I A and I B that is I A I B I. shuld be selected such as nt t set t clse t 35 and nt t clse t 0, the wrst case beg 0. et 0, therefre TA 4.3 ±0 4.3 t 4.3 which is a safe range. 0 5k 0.666 et 0k and verify that imum current is nt excessive. W W 0k 50k I I 7.07 fr and I 4. 47 fr 4.3 5k.38 35 3.3 0k.75 4.3.9 50k 4k 4.3 Wrst case scenaris are shwn beside. One must ensure I t be belw imum values calculated abve. Any part f the pt that carries mre than limit becmes a ht spt that may be damaged. C. auril wer Amplifier xercise ev. 5/5/03 age 4
CTONIC 3 4 3.56 35 4 0.3 4 8 B) 3.56 I 8.4 A XCI AC GOUND 5 G 0.667 p p k Q p O Q3 0K 00 p 0p 0 I OA Q 0.667 p 0 0.667 p 4.53p 0p 35 0A 3.56 p 8.4 3.56 p.µ 0k.33p 0 OA 0.667 p 0 0p 00k 8.4 4Ω 5 G 0p Q4 p 0.667 p k Q5 p ON Q6 5 G AC GOUND 8.4 DC bias level p C. auril wer Amplifier xercise ev. 5/5/03 age 5
CTONIC 3 8 C) r sewave put see equatins and graph n page f thery ntes. XCI π 35 π 4 94.96W dri 3.56 0.9303 35 π π dri 0.9303 94.96 3.5W 4 4 Q Q 94.96 6.06W 3.03W per MOT π Q ccurs at 35.8 and.8 π π π η dri 0.9303 0.7306 73.06% 4 4 peak 8 D) r squarewave put see equatins and graph n page f thery ntes. Q 35 4 306.5W ccurs at Q Q dri 0.5 35 7.5 η dri 0.9303 93.03% dri 3.56 0.9303 35 0.9303 306.5 65W 0.5 306.5 76.56W 38.3W peak and.75 per MOT 8 ) A C Z πc C 3.6 Hz C C 0 db/dec 9A) Use lwer saturatin vltage, that is AT m U CI 3.6 Hz 0 db.8.7 3.p 3.975 3.975 The data sheets state 50W typical O/ pwer fr an 8Ω lad with 35 supplies 63.6p fr THD < %. We calculated 63.6W (fr sewave signal) per amplifier at the lwest clippg level which prduces 3.p a higher THD. 3.p 3.975 3.975 B) 8 Ω 7.95 3.p sat 35.8 3. The lad vltage is OAD 64.4*8/(8) 63.6. The lad seen by each amplifier O/ is a grunded 8.Ω lad. 3. 63.6 Q 8.Ω 5. 8W I 3.975 8 This crrespnds t 63.W per amplifier delivered t the lad. C. auril wer Amplifier xercise ev. 5/5/03 age 6
CTONIC 3 XCI 35 35 3.p 3.975 8 Ω 3.8p The data sheets state 50W typical O/ pwer fr an 8Ω lad with 35 supplies fr THD < %. We calculated 63.W (fr sewave signal) per amplifier at the lwest clippg level which prduces a THD higher than %. π Q IC 35 96.8W π 8. Q dri dri peak 3. 0.9 at O / r drive level 35 96.8 30.65W per M 3886 ccurs at π 35.8 π 0.9 96.8 88.58W per M 3886 ccurs at O / r drive level The ttal pwer cnsumptin is therefre 4 * 88.58 354.3W 50 C J 30.65W C/W C 0. C/W 30.65W C/W INUATO 0. C/W 3. C The imum heat sk resistance is θ 83..6 A 0.68 C / W 30.65W 30.65W C/W C/W 30 C 0. C/W 0. C/W.6W HAT INK θ A 83. C The abve value was btaed assumg a sewave put, therefre practice the pwer dissipatin may exceed the value we have calculated. T be safe, we shuld use a heat that has less than half f the abve calculated thermal resistance. A B) A 3. rms Nn vertg amp A vertg amp A. 5.05 / 5.05 4.05 The ga value itself is nt critical, but matchg f the vertg and nnvertg gas is very critical, therefre let us use a ga f 5 / and % tlerance resistrs fr and. The balancg resistrs are nt critical, they are there just t mimize the O/ DC ffset. C. auril wer Amplifier xercise ev. 5/5/03 age 7
CTONIC 3 XCI k % 35 M3 886 35 k % 35 35 k 0. % 48k % 50k % k 0. % 8 Ω k % 35 35 k % M3 886 35 35 k 0. % 48k 0. % 50k 0. % k % D) Usg the frmula frm the ntes, we have TO 4 nm al 4 0.00 3. 64.4 8. 0 I 00.88A I I I I 7. 95 lvg fr the abve tw equatins we have I 4.6 A and I 3.33 A. Mismatch f the ga settg resistrs and causes AC current imbalance between parallel amplifiers. ± ± I 48k ± 0m ± 0. µ 48k ± 0. 596 k 50k ± 0m ± 0. µ 50k ± 0. 7 k ) i i fr balanced puts Nnvertg amp Invertg amp A 0.596 0.596A 0A.596A 0A.7A.7A 0 0.7 As yu can see, DC ffsets can cause a large DC current t flw between tw parallel amplifiers. T reduce the imum DC current ne can crese the W resistrs r resrt t the scheme used step. 0.596 0. 0.7 C. auril wer Amplifier xercise ev. 5/5/03 age 8
CTONIC 3 ) et s use a cutff frequency f 5 Hz s we d nt attenuate audi signals. All f the C s will trduce a cutff frequency C 5.9 µ 5 std C π C π π 5 k µ C XCI k % 35 % % 35 k % 0k % 5 u % 35 35 5 u %.5 u % 4 k % k % 48k % 35 % 8 Ω ut % 50k % 35 k % k % 5 u % 35 35 5 u % k % 48k % 50k % k % Nnvertg amp Invertg amp 9.6m 9.60m 48k ± 0m ± 0.µ 48 ± 9.6m 50k ± 0m ± 0.µ 50k ± 0m 0 96A 0A 96A 0.A 0A 0.A 0 0m 0m A picture is wrth a 000 wrds Nw wrst case DC currents are reasnable. NOT: One needs the put DC blckg capacitr shwn abve t remve any DC cmpnent frm the signal, therwise the lad may have t absrb a large DC current. ) Ga espnse f put amplifier ( ) ( ) C C C C 0 p ω C p 3.33 r / s C 5. 3 Hz C C 5µ 0k 0 db/dec 0 db/dec 0 db/dec C. auril wer Amplifier xercise ev. 5/5/03 age 9
CTONIC 3 Ga espnse f vertg amplifiers k, 50k, C 5 µ ( ) ( ) C C C 0 db/dec 0 db/dec 0 db/dec 0 OG XCI 7.96 db The H frequency ga will start drppg ff because f the GBW f the s. k HI β ( H ) GBW GBW 8M 307. 7 khz k 50k 0 db/dec 40 db/dec 7.96 db 307.7 khz with put amplifier respnse cmbed 0 db/dec Ga espnse f nnvertg amplifiers k, 48k, C 5 µ ( ) ( ) C C C C C C 0 db/dec 0 db/dec ( ) 0. Hz C ( ) 0 OG 0. Hz 0 db/dec 7.6 db The H frequency ga will start drppg ff because f the GBW f the s. k HI β ( H ) GBW GBW 8M 30 khz k 48k 7.6 db 0. Hz 0 db/dec 40 db/dec with put amplifier cluded 30 khz 0 db/dec 0 db/dec C. auril wer Amplifier xercise ev. 5/5/03 age 0