Whites, EE 30 Lecture 9 Page 1 of 8 Lecture 9: MOSFET Small-Sigal Amplifier Examples. We will illustrate the aalysis of small-sigal MOSFET amplifiers through two examples i this lecture. Example N9.1 (text Example 7.3). etermie A v (eglectig the effects of R G ), R i, ad R out for the circuit below give that V 1.5 V, k W L 0.5 ma/v, ad V A = 50 V. t (Fig. 7.15) The first step is to determie the C operatig poit. The C equivalet circuit is: 017 Keith W. Whites
Whites, EE 30 Lecture 9 Page of 8 IG 0 I Sice VG 0 Vt the MOSFET is operatig i the saturatio mode if I 0. Assumig operatio i the saturatio mode the C drai curret from (5.3) is 1 W I k VGS Vt 1 VS (5.3) L Notice i the circuit that V GS V S, so we will evetually create a triatic equatio i V GS for I. However, the last factor i (5.3) will be quite small for large V A (small ). So, for simplicity we will eglect r o i (5.3) gig 1 W I k VGS Vt (7.44) L For this C circuit 1 0.5 10 3 4 1.5 1.5 10 I VGS VGS 1.5 Notice i the circuit that V GS V S so that this last equatio becomes Also, by KVL I V 0.15 1.5 ma (1) S
Whites, EE 30 Lecture 9 Page 3 of 8 Substitutig () ito (1) V 15 R I 15 10,000I (7.43),() S I 4 I 1.510 15 10,000 1.5 Solg this equatio gives I 1.06 ma VS 4.4 V( VGS) or I 1.7 ma VS. V( VGS) This latter result is ot cosistet with the assumptio of operatio i the saturatio mode sice VGS Vt 1.5 V. So the proper solutio for I is the first ( I 1.06 ma). Next, we costruct the small-sigal equivalet circuit. We ll use the small-sigal model of the MOSFET with r o icluded: 3 g k V V r o (Fig. 7.15c) W 0.510 4.4 1.5 0.75 ms L VA 50 47. k I 1.06 ma m GS t Recall from the preous lecture (ad also Lecture 6) that the proper I i the r o calculatio is that with = 0, which is what we eded up calculatig earlier.
Whites, EE 30 Lecture 9 Page 4 of 8 To compute the small-sigal voltage gai, we start at the output (assumig R G is extremely large RG ro R RL) vo gmvgsro R RL At the iput otice that vgs. Therefore vo Av gmro R RLgm(4,51) 3.8 V/V Notice that the assumptio RG ro R RL is met ad hugely exceeded sice 10 M >> 4,51. For the iput resistace R i calculatio, we caot set vgs 0 ad subsequetly ope circuit the depedet curret source sice this would artificially force R i 0. Rather, we eed to determie i i as a fuctio of v i ad use this i the defiitio Ri ii The depedet curret source will remai i these calculatios. Proceedig, at the iput of the small-sigal equivalet circuit show above vo v i v o ii 1 1 Av RG R G v i RG Therefore, ii (1 3.8) RG Cosequetly, usig this expressio we fid that RG Ri. 34 M i 4. 8 i
Whites, EE 30 Lecture 9 Page 5 of 8 Lastly, to determie the output resistace, we ca set vgs 0 i the small-sigal equivalet circuit above, which will ope circuit the depedet curret source leadig to the equivalet circuit: from which we see that R R r R 4 k out G o 8. Example N9. (text Exercise 7.6). etermie the followig quatities for the coceptual MOSFET small-sigal amplifier of Fig. 7.4 give that V = 5 V, R = 10 k, ad V GS = V. 0.si t V V (Fig. 7.4) The MOSFET characteristics are Vt 1 V, k 0 A/V, W/L = 0, ad = 0.
Whites, EE 30 Lecture 9 Page 6 of 8 (a) etermie I ad V. We see from the circuit that V GS V t. Therefore, the MOSFET is operatig i the saturatio or triode mode. We ll assume saturatio. I that case 1 W 1 0 10 6 0( 1) I k VGS Vt 0. ma L ad V I R 3 V V Let s check if the MOSFET is operatig i the saturatio mode: VG 31 Vt Therefore, the MOSFET is ideed saturated, as assumed. (b) etermie g m. Usig (7.3) W 6 gm k VGS Vt010 0 ( 1) 4.0 ms L (c) etermie the voltage gai A v. We begi by first costructig the small-sigal equivalet circuit v gs g v m gs v o irectly from this circuit, v g v R o m gs
Whites, EE 30 Lecture 9 Page 7 of 8 so A v v o 3 3 gmr 0.410 1010 4 vgs V/V (d) If v 0.sit gs V, fid v d ad the max/mi v. vo Av vd Av v gs 40.si t v gs Therefore, v 0.8sit d V Hece, v V 3 0.8 3. 8 V max V d while v V 3 0.8. V mi V d Let s check that the MOSFET stays i saturatio at the two extremes of the drai voltage: VGS v 3.81.8V max t saturated VGS v.0.v mi t saturated Therefore the trasistor stays saturated for the etire cycle of the output voltage. (e) etermie the secod harmoic distortio. From (7.8) or (6) i the preous lecture otes, the total drai curret is give as W 1 W i I k VGS Vtvgs k vgs L L 6 1 6 or i I 0 10 0( 1) vgs 010 0vgs 3 3 I 0.410 v 0. 10 v gs gs
Whites, EE 30 Lecture 9 Page 8 of 8 Substitutig vgs 0.sit 6 6 i I 8010 sit8 10 si t ito this equatio gives Usig the trigoometry idetity si t1 1cos t this last expressio becomes i 00 80sit 4 4cost A or i 04 80sit 4cost A The first term i i is I, the C curret. We see that there is a slight shift upward i value by 4 A. The third term i i is the secod harmoic term because it varies with time at twice the frequecy of the iput sigal. The secod harmoic distortio is 4 100% 80 5 %