Lecture 23: Common Emitter Amplifier Frequency Response. Miller s Theorem.

Transcription

1 Whites, EE 320 ecture 23 Page 1 of 17 ecture 23: Common Emitter mplifier Frequency Response. Miller s Theorem. We ll use the high frequency model for the BJT we developed the previous lecture and compute the frequency response of a common emitter amplifier, as shown below Fig. 5.71a. (Fig. 5.71) s we discussed the previous lecture, there are three distct region of frequency operation for this and most transistor amplifier circuits. We ll exame the operation of this CE 2009 Keith W. Whites

2 Whites, EE 320 ecture 23 Page 2 of 17 amplifier more closely when operated three frequency regimes. Mid-band Frequency Response of the CE mplifier t the mid-band frequencies, the DC blockg capacitors are assumed to have very small impedances so they can be replaced by short circuits, while the impedances of C and C μ are very large so they can be replaced by open circuits. The equivalent small-nal model for the mid-band frequency response is then We ll defe so that at the output R = r R R (1) o C V = g R V (2) o m Usg Théven s theorem followed by voltage division at the put we fd

3 Whites, EE 320 ecture 23 Page 3 of 17 r r R B V = VTH = V r + rx + RTH r + rx + RB R RB + R (3) Substitutg (3) to (2) we fd the mid-band voltage ga m to be Vo gmr RB m = ( ro RC R) (4) V r + r + R R R + R x B B High Frequency Response of the CE mplifier For the high frequency response of the CE amplifier of Fig. 5.71a, the impedance of the blockg capacitors is still negligibly small, but now the ternal capacitances of the BJT are no longer effectively open circuits. Usg the high frequency small-nal model of the BJT discussed the previous lecture, the equivalent small-nal circuit of the CE amplifier now becomes: (Fig. 5.72a)

4 Whites, EE 320 ecture 23 Page 4 of 17 We ll simplify this circuit a little by calculatg a Théven equivalent circuit at the put and usg the defition for R (1): (Fig. 5.72b) where it can be easily shown that V is V given (3) r RB V = V (5.167),(5) r + r + R R R + R x B B while R = r rx + ( RB R ) (5.168),(6) Miller s Theorem We can analyze the circuit Fig. 5.72b through traditional methods, but if we apply Miller s theorem we can greatly simplify the effort. Plus, it will be easier to apply an approximation that will arise if we use Miller s theorem. You may have seen Miller s theorem previously circuit analysis. It is another equivalent circuit theorem for lear

5 Whites, EE 320 ecture 23 Page 5 of 17 circuits ak to Théven s and Norton s theorems. Miller s theorem applies to this circuit topology: The equivalent Miller s theorem circuit is (Fig. 1) (Fig. 2) where x = v 1 v B and B x = v 1 v B (7),(8) The equivalence of these two circuits can be easily verified. For example, usg KV Fig. 1 v = ix + vb v vb or i = (9) while usg KV the left-hand figure of Fig. 2 gives x

6 Whites, EE 320 ecture 23 Page 6 of 17 i v = (10) Now, for the left-hand figure to be equivalent to the circuit Fig. 1, then i (9) and i (10) must be equal. Therefore, v vb v = x The equivalent impedance can be obtaed from this equation as xv x = = v v vb B 1 v which is the same as (7). similar result verifies (8). So, for a resistive element R x, Miller s theorem states that Rx Rx R = and RB = (12),(13) vb v 1 1 v v while for a capacitive element C x, Miller s theorem states that v B v C = Cx 1 and CB = Cx 1 (14),(15) v vb B

7 Whites, EE 320 ecture 23 Page 7 of 17 High Frequency Response of the CE mplifier (cont.) Returng now to the CE amplifier equivalent small-nal circuit of Fig. 5.72b, we ll apply Miller s theorem of Figs. 1 and 2 to this circuit and the capacitor C μ to give (Fig. 3) where, usg (14) and (15), V o C = Cμ 1 V V and CB = Cμ 1 V o (16),(17) ctually, this equivalent circuit of Fig. 3 is no simpler to analyze than the one Fig. 5.72b because of the dependence of C and C B on the voltages V o and V. However, this equivalent circuit of Fig. 3 will prove valuable for the followg approximation. Note from Fig. 5.72b that I + I = μ gmv I = gv I (18) m μ Up to frequencies near f H and better, the current I μ the small capacitor C μ will be much smaller than gv m. Consequently, from (18)

8 Whites, EE 320 ecture 23 Page 8 of 17 I gv m (19) V I R = g R V (5.169),(20) and o m Usg this last result (16) and (17) we fd that g 1 ( 1 ) mr V C Cμ + = Cμ + gmr V V 1 and CB Cμ 1+ = Cμ 1+ gmrv gmr (21) (22) Most often for this type of amplifier, gmr 1 so that (22) CB C μ. But as we itially assumed, the current through C μ is much smaller than that through the dependent current source gv, which ultimately led to equation (19). m Consequently, we can ignore C B parallel with gv m and the fal high frequency small-nal equivalent circuit for the CE amplifier Fig. 5.71a is (Fig. 5.72c)

9 Whites, EE 320 ecture 23 Page 9 of 17 where C ( ) C C C C 1 gmr μ + = + + (5.173),(22) Based on this small-nal equivalent circuit, we ll derive the high-frequency response of this CE amplifier. t the put C V = V (23) + R while at the output V C = g R V (24) o m Substitutg (23) to (24) gives Sce ( jωc ) If we defe C o = m V g R V 1 C + R C = then (25) becomes 1 jωc gmr Vo = gmr V = V 1 + R 1+ jωc R jωc ω = H 1 C R then substitute this to (26) gives Vo gmr gmr = = V ω f 1+ j 1+ j ω f H H (25) (26) (27) (28)

10 Whites, EE 320 ecture 23 Page 10 of 17 where f H ωh 1 = = 2 2 C R (5.176),(29) You should recognize this transfer function (28) as that for a low pass circuit with a cut-off frequency (or 3-dB frequency) of ω H. This is the response of a sgle time constant circuit, which is what we have the circuit of Fig. 5.72c. What we re ultimately terested is the overall transfer function Vo V from put to output. This can be easily derived from the work we ve already done here. Sce Vo V V o = (30) V V V We can use (28) for the first term the RHS of (30), and use (5) for the second givg Vo gmr r RB = (31) V f 1+ j r + rx + RB R RB + R f H We can recognize m from (4) this expression givg Vo m = (5.175),(32) V f 1+ j f H Once aga, this is the frequency response of a low pass circuit, as shown below:

11 Whites, EE 320 ecture 23 Page 11 of 17 (Fig. 5.72d) Comments and the Miller Effect Equation (32) gives the mid-band and high frequency response of the CE amplifier circuit. It is not valid for the low frequency response near f and lower frequencies, as shown Fig. 5.71b. It turns out that C (22) is usually domated by C ( 1 ) = Cμ + gmr. Even though C μ is usually much smaller than C its effects at the put are accentuated by the factor + g R. 1 m The reason that C undergoes this multiplication is because it is connected between two nodes (B and C Fig. 5.72a) that experience a large voltage ga. This effect is called the

12 Whites, EE 320 ecture 23 Page 12 of 17 Miller effect and the multiplyg factor 1+ gmr (22) is called the Miller multiplier. Because of this Miller effect and the Miller multiplier, the put capacitance C of the CE amplifier is usually quite large. Consequently, from (20) the f H of this amplifier is reduced. In other words, this Miller effect limits the high frequency applications of the CE amplifier because the bandwidth and ga will be limited. ow Frequency Response of the CE mplifier On the other end of the spectrum, the low frequency response of the CE amplifier and all other capacitively coupled amplifiers is limited by the DC blockg and bypass capacitors. This type of low frequency response analysis is rather complicated because there is more than a sgle time constant response volved. In the circuit of Fig. 5.71a there are three capacitors volved, C C1, C C2, and C E. ll three of these greatly affect the low frequency response of the amplifier and can t be ignored.

13 Whites, EE 320 ecture 23 Page 13 of 17 The text presents an approximate solution which the low frequency response is modeled as the product of three high pass sgle time constant circuits cascaded together so that V o jω jω jω m (5.183),(33) V jω ω p1 jω ω p2 jω ω p3 (Fig. 5.73e) So there isn t a sgle f as suggested by Fig. 5.71b but rather a more complicated response at low frequencies as we see Fig. 5.73e above. Computer simulation is perhaps the best predictor for this complicated frequency response, but an approximate formula for f is given the text as f fp 1+ fp2 + fp3 = CC1RC1 CERE CC2RC2 (5.184),(5.185),(34) where R C1, R E, and R C 2 are the resistances seen by C C1, C E, and C C 2, respectively, with the nal source V = 0 and the other two capacitors replaced by short circuits.

14 Whites, EE 320 ecture 23 Page 14 of 17 Example N23.1. Compute the mid-band small-nal voltage ga and the upper 3-dB cutoff frequency of the small-nal voltage ga for the CE amplifier shown Fig. 5.71a. Use a 2N2222 transistor and the circuit element and DC source values listed Example 5.18 the text. Use 10 μf blockg and bypass capacitors. The circuit gilent dvanced Den System appears as: C C C1 Start=100 Hz Stop=1.0 MHz Step=100 Hz V_DC SRC2 Vdc=10.0 V R R2 R=8 kohm vo vi V_C SRC1 Vac=polar(1,0) V Freq=freq R R3 R=5 kohm C C1 C=10 uf C C2 C=10 uf R R1 R=100 kohm ap_npn_2n2222_ Q1 R R4 R=5 kohm I_DC SRC4 Idc=1 m C C3 C=10 uf V_DC SRC3 Vdc=-10.0 V From the results of the DS circuit simulation

15 Whites, EE 320 ecture 23 Page 15 of 17 ( ) ( ) V CB = 2.03 V 400 mv = 2.43 V V = 0.4 V 1.02 V = 0.62 V BE From Fig. 9 the Motorola 2N2222 datasheet (see the previous set of lecture notes) For V CB = 2.43 V Ccb = C μ 5.8 pf. For V = 0.62 V C = 20 pf. BE IC 1 m gm = = = 0.04 S VT 25 mv From (5.163), gm 0.04 ft = = MHz 2 ( C C ) 2 ( 20 pf 5.8 pf + + μ ) This value agrees fairly with the datasheet value of 300 MHz. β0 265 from the DS parts list for this 2N2222 transistor. Therefore, β0 265 r = = = 6,625 Ω g 0.04 m From the 2N2222 datasheet, the nomal output resistance at I C = 1 m is ro 50 kω. What about r x? It s so small value (~ 50 Ω) that we ll easily be able to ignore it for the m calculations compared to r (which is 6,625 Ω as we just calculated). From (4), eb C

16 Whites, EE 320 ecture 23 Page 16 of 17 Therefore, g r R = r R R ( ) m B m o C r rx RB R RB R 2,898.6= R V m = V m = 20log10 m = 36.2 db or decibels ( ) From DS: m3 freq= Hz db(vo)= m3 m1 freq= 6.300kHz m2 freq= 84.40kHz db(vo)= db(vo)= m1 m2 db(vo) E2 1E3 1E4 1E5 freq, Hz 1E6 From this plot, DS computes a mid-band ga of m = db, which agrees closely with the predicted value above. From (29), f H 1 2C R

17 Whites, EE 320 ecture 23 Page 17 of 17 where from (22) C = C + C ( + g R ) = + ( + ) μ 1 m ,898.6 pf = = pf while from (6) R = r rx + ( RB R ) Because RB R = 100 k 5 k = 4,761.9 Ω is so much larger than r x (on the order of 50 Ω), we can safely ignore r x. Then, R 6,625 4,762 = 2,771 Ω. Therefore, fh ,771 = khz This agrees very closely with the value of khz predicted by the DS simulation shown above. dd a short discussion on the ga-bandwidth product m f H.