Lecture 30: Audio Amplifiers

Transcription

1 Whtes, EE 322 Lecture 30 Page 1 of 9 Lecture 30: Audo Amplfers Once the audo sgnal leaes the Product Detector, there are two more stages t passes through before beng output to the speaker (ref. Fg. 1.13): 1. Audo amplfcaton, 2. Automatc gan control (AGC). We ll dscuss each of these separately, begnnng wth audo amplfcaton n ths lecture. In the NorCal 40A, the Audo Amplfer s the LM386N-1 ntegrated crcut. The LM38x amplfer seres s qute popular. A smplfed equalent crcut for the LM386 s shown n Fg and n the data sheet begnnng on p. 399 of the text: 2006 Keth W. Whtes

2 Whtes, EE 322 Lecture 30 Page 2 of 9 We ll descrbe the operaton of ths crcut begnnng near the nput. (Note that Sedra and Smth, 5 th edton, Sec has a nce descrpton of a closely related crcut: the LM380 IC.) There are three stages of amplfcaton n the LM386: 1. pnp common-emtter amplfers (Q1 and Q2), 2. pnp dfferental amplfer (Q3 and Q4), 3. Class AB power amplfer (Q7 and Q8+Q9). For the remander of ths lecture, we ll step through the LM386 equalent crcut and explan the operaton of each part. Q1 and Q2: Q3 Q4 -Input Q1 Q2 +Input 50 kω 50 kω Q1 and Q2 are pnp emtter follower amplfers. These prode bufferng of the nput to the LM386. The 50-kΩ resstors prode dc paths to ground for the base currents of Q1 and Q2. Consequently, the nput should be capactely coupled so to not dsturb ths nternal basng.

3 Whtes, EE 322 Lecture 30 Page 3 of 9 Because of these resstors, the nput mpedance wll be domnated by these 50-kΩ resstors. Q3 and Q4 wth R e : Q3 and Q4 form a pnp dfferental amplfer: Ω 1.35 kω Q3 R e Q4 Q5 and Q6: The dfferental amplfer s based by the current mrror formed by Q5 and Q6: I 5 I 6 I 0 Q5 + V b - Q6 In the current mrror, I 6 I 5. To see ths, notce that V be = V b for both transstors. Wth Vb / Vt I = I e (13.1) and V b the same for both transstors, then I = I c cs c5 c6

4 Whtes, EE 322 Lecture 30 Page 4 of 9 proded the two transstors are matched. Ths mples that I5 I6, f we neglect the base currents wrt the collector currents. Ths s ald f the β s are large. Ths current-source basng prodes a relable bas and consderably smplfes the analyss of amplfer crcuts. Sgnal Gan of the LM386 We re now n a poston to compute the sgnal gan proded by the LM386. We ll see that the Audo Amplfer s prodng much of the total gan n the NorCal 40A receer. The current mrror forces the currents on both hales of the dfferental amplfer to be equal: both dc and ac components. Consequently, the currents at the emtters of Q3 and Q4 must be the same, as shown n Fg. 13.2(c): V cc Ω kω 1 Output R f Q3 R e Q4

5 Whtes, EE 322 Lecture 30 Page 5 of 9 From ths crcut, the small sgnal ac model s: R e d + R f 2 Equal because of current mrror. Notce that the oltage across R e s smply the dfferental nput oltage d. Why? Because the base-emtter oltage drops n the pnp transstors are the same on each sde of the dff amp! Therefore, the oltage across R e s d. Trcky. Due to the mrror, the current through Rf 2, neglectng the current n the two 15-kΩ resstors (whch are large mpedances relate to the other parts of the crcut). Therefore, d 2 (1) R f Now, the output oltage s produced by a so-called class AB power amplfer:

6 Whtes, EE 322 Lecture 30 Page 6 of 9 6 V cc Q7 R f e 5 Out b Q8 Compound pnp transstor Q10 Q9 c 4 Gnd The combnaton of Q8 and Q9 s called a compound pnp transstor : e b β β Q8 β Q9 c Notce that β βq8βq9, whch s easy to show startng wth c8 = βq8b8 and c9 = β Q9 b9. Compoundng pnp s was done n early IC s to mproe the tradtonally poor performance of pnp transstors wrt frequency response, etc. That s not much of a problem today. Secton 10.6 of the text has a dscusson on class AB (and class B) power amplfers. The result, n any eent, s that the output oltage wll be much larger than d. Therefore, from (1)

7 Whtes, EE 322 Lecture 30 Page 7 of 9 2 R (13.4) f Also, from the small-sgnal model shown aboe, we can see that d = (13.5) Re Combnng these last two results, we fnd that 2 = d Rf Re R or G 2 f = = (13.6) d R e Ths s the dfferental oltage gan of the LM386 audo amplfer. Notce that ths gan does not nole nternal dece parameters (such as the transstor β s) other than R f and R e. Nce. Hae you eer seen such a result as (13.6) before? Sure, wth smple operatonal amplfer crcuts such as: R f R The oltage gan s R f =. R 1

8 Whtes, EE 322 Lecture 30 Page 8 of 9 Smlar to an op amp, the LM386 has ncorporated feedback nternally through R f and R e, n a fashon smlar to ths nertng op amp crcut that s usng external components. Now, usng (13.6), the gan of the LM386 shown n Fg (.e., no other components attached between pns 1 and 8) s: 3 2R f G = = 2 = 20 3 R e As you ll dscoer n Prob. 31, a capactor can be placed (externally) across pns 1 and 8 of the LM386 to bypass R e at hgh frequences [X c = (ωc) -1 ]. In such a case, 2R f G = = 2 = 200 Re 150 Ths s a szeable gan at hgh frequences. LM386 Connecton n the NorCal 40A The NorCal 40A Audo Amplfer s bult n stages n Prob. 31. The frst stage of ths constructon s shown n Fg. 13.6:

9 Whtes, EE 322 Lecture 30 Page 9 of 9 The nput s taken dfferentally, as shown, and s capactely coupled by C20 and C21 for reasons we dscussed on p. 2. Note that wth the polarty of V shown aboe, we wll expect the gan of ths audo amplfer to be the negate of (13.6). The output s also capactely coupled. Why? It can be shown that the dc output oltage s V cc /2 at pn 5 of the LM386. So once agan, we need to capactely couple n order not to dsturb ths nternal basng.