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. 13.1 and n the data sheet begnnng on p. 399 of the text: 2006 Keth W. Whtes
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. 14.8 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.
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: 8 1 150 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
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 7 150 Ω 8 1.35 kω 1 Output R f Q3 R e Q4
Whtes, EE 322 Lecture 30 Page 5 of 9 From ths crcut, the small sgnal ac model s: 7 0 - 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:
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)
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 1 - + The oltage gan s R f =. R 1
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. 13.1 (.e., no other components attached between pns 1 and 8) s: 3 2R f 15 10 G = = 2 = 20 3 R 1.5 10 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 15 10 3 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:
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.