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Whites, EE 320 Lecture 23 Page 1 of 18 Lecture 23: Common Emitter Amplifier Frequency Response. Miller’s Theorem. We’ll use the high frequency model for the BJT we developed in the previous lecture and compute the frequency response of a common emitter amplifier, as shown below in Fig. 10.9a. (Fig. 10.9a) (Fig. 10.1) © 2016 Keith W. Whites Whites, EE 320 Lecture 23 Page 2 of 18 As we discussed in the previous lecture, there are three distinct region of frequency operation for this – and most – transistor amplifier circuits. We’ll examine the operation of this CE amplifier more closely when operated in three frequency regimes. Mid-band Frequency Response of the CE Amplifier At the mid-band frequencies, the DC blocking capacitors are assumed to have very small impedances so they can be replaced by short circuits, while the impedances of C and C are very large so they can be replaced by open circuits: (Fig. 10.19a) where RB RB1 || RB 2 . The equivalent small-signal model for the mid-band frequency response is then Whites, EE 320 Lecture 23 Page 3 of 18 We’ll define RL ro || RC || RL (1) Vo g m RLV (2) so that at the output Using Thévenin’s theorem followed by voltage division at the input we find r r RB V VTH Vsig (3) r rx RTH r rx RB || Rsig RB Rsig Substituting (3) into (2) we find the mid-band voltage gain Am to be V g m r RB ro || RC || RL (10.54),(4) Am o Vsig r rx RB || Rsig RB Rsig High Frequency Response of the CE Amplifier For the high frequency response of the CE amplifier of Fig. 10.19a, the impedance of the blocking capacitors is still Whites, EE 320 Lecture 23 Page 4 of 18 negligibly small, but now the internal capacitances of the BJT are no longer effectively open circuits. Using the high frequency small-signal model of the BJT discussed in the previous lecture, the equivalent small-signal circuit of the CE amplifier now becomes: (Fig. 10.19a) We’ll simplify this circuit a little by calculating a Thévenin equivalent circuit at the input and using the definition for RL in (1): (Fig. 10.19b) where it can be easily shown that Vsig is V given in (3) r RB Vsig Vsig (Fig. 10.19b),(5) r rx RB || Rsig RB Rsig while Rsig r || rx RB || Rsig (Fig. 10.19b),(6) Whites, EE 320 Lecture 23 Page 5 of 18 Miller’s Theorem We can analyze the circuit in Fig. 10.19b 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 in circuit analysis. It is another equivalent circuit theorem for linear circuits akin to Thévenin’s and Norton’s theorems. Miller’s theorem applies to this circuit topology: (Fig. 1) The equivalent Miller’s theorem circuit is (Fig. 2) where Whites, EE 320 Lecture 23 ZA Zx v 1 B vA and Z B Page 6 of 18 Zx v 1 A vB (7),(8) The equivalence of these two circuits can be easily verified. For example, using KVL in Fig. 1 v A i A Z x vB v v iA A B or (9) Zx while using KVL in the left-hand figure of Fig. 2 gives v iA A (10) ZA Now, for the left-hand figure to be equivalent to the circuit in Fig. 1, then iA in (9) and iA in (10) must be equal. Therefore, v A vB v A Zx ZA The equivalent impedance ZA can be obtained from this equation as Zv Zx ZA x A v A vB 1 vB vA which is the same as (7). A similar result verifies (8). So, for a resistive element Rx, Miller’s theorem states that Whites, EE 320 Lecture 23 Page 7 of 18 Rx Rx and RB (12),(13) vB vA 1 1 vA vB while for a capacitive element Cx, Miller’s theorem states that v v C A C x 1 B and CB Cx 1 A (14),(15) vA vB RA High Frequency Response of the CE Amplifier (cont.) Returning now to the CE amplifier equivalent small-signal circuit of Fig. 10.19b, we’ll apply Miller’s theorem of Figs. 1 and 2 to this circuit and the capacitor C to give (Fig. 3) where, using (14) and (15), V V C A C 1 o and CB C 1 V Vo (16),(17) Whites, EE 320 Lecture 23 Page 8 of 18 Actually, this equivalent circuit of Fig. 3 is no simpler to analyze than the one in Fig. 10.19b because of the dependence of CA and CB on the voltages Vo and V. However, this equivalent circuit of Fig. 3 will prove valuable for the following approximation. Note from Fig. 10.19b that I L I g mV I L g mV I (18) Up to frequencies near fH and better, the current I in the small capacitor C will be much smaller than g mV . Consequently, from (18) (19) I L g mV and Vo I L RL g m RLV (20) Using this last result in (16) and (17) we find that g R V C A C 1 m L C 1 g m RL V V 1 and C B C 1 C 1 g R V g R m L m L (21) (22) Most often for this type of amplifier, g m RL 1 so that in (22) CB C . But as we initially assumed, the current through C is much smaller than that through the dependent current source g mV , which ultimately led to equation (19). Whites, EE 320 Lecture 23 Page 9 of 18 Consequently, we can ignore CB in parallel with g mV and the final high frequency small-signal equivalent circuit for the CE amplifier in Fig. 10.19a is (Fig. 10.19c) where Cin C C A C C 1 g m RL (Fig. 10.19c),(22) Based on this small-signal equivalent circuit, we’ll derive the high-frequency response of this CE amplifier. At the input Z Cin V Vsig (23) Z Cin Rsig while at the output Vo g m RLV (24) Substituting (23) into (24) gives Vo g m RL Z Cin Z Cin Rsig Since Z Cin jCin then (25) becomes 1 Vsig (25) Whites, EE 320 Vo g m RL Lecture 23 1 jCin 1 Rsig jCin Vsig Page 10 of 18 g m RL 1 jCin Rsig Vsig (26) If we define H 1 Cin Rsig then substitute this into (26) gives Vo g m RL g m RL f Vsig 1 j 1 j H fH 1 fH H where 2 2 Cin Rsig (27) (28) (10.57),(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 H. This is the response of a single time constant circuit, which is what we have at the input to the circuit of Fig. 10.19c. What we’re ultimately interested in is the overall transfer function Vo Vsig from input to output. This can be easily derived from the work we’ve already done here. Since Vo Vo Vsig (30) Vsig Vsig Vsig Whites, EE 320 Lecture 23 Page 11 of 18 We can use (28) for the first term in the RHS of (30), and use (5) for the second giving Vo r g m RL RB (31) f Vsig 1 j r rx RB || Rsig RB Rsig fH We can recognize Am from (4) in this expression giving Vo Am (10.56),(32) f Vsig 1 j fH Once again, this is the frequency response of a low pass circuit, as shown below: (Fig. 10.19d) 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 Whites, EE 320 Lecture 23 Page 12 of 18 frequency response near fL and lower frequencies, as shown in Fig. 5.71b. The high frequency, small-signal equivalent circuit models for the BJT in a common emitter amplifier are: It turns out that Cin in (22) is usually dominated by C A C 1 g m RL . Even though C is usually much smaller than C its effects at the input are accentuated by the factor 1 g m RL . The reason that C undergoes this multiplication is because it is connected between two nodes (B’ and C in Fig. 10.19a) that experience a large voltage gain. This effect is called the Miller effect and the multiplying factor 1 g m RL in (22) is called the Miller multiplier. Whites, EE 320 Lecture 23 Page 13 of 18 Because of this Miller effect and the Miller multiplier, the input capacitance Cin of the CE amplifier is usually quite large. Consequently, from (20) the fH of this amplifier is reduced. In other words, this Miller effect limits the high frequency applications of the CE amplifier because the bandwidth and gain will be limited. Low Frequency Response of the CE Amplifier 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 blocking and bypass capacitors. This type of low frequency response analysis is rather complicated because there is more than a single time constant response involved. In the circuit of Fig. 10.9a there are three capacitors involved, CC1, CC2, and CE. All three of these greatly affect the low frequency response of the amplifier and can’t be ignored. The text presents an approximate solution in which the low frequency response is modeled as the product of three high pass single time constant circuits cascaded together so that Vo j j j (33) Am Vsig j p1 j p 2 j p 3 Whites, EE 320 Lecture 23 Page 14 of 18 (Fig. 10.7) So there isn’t a single fL as suggested by Fig. 10.1 but rather a more complicated response at low frequencies as we see in Fig. 10.7 above. Computer simulation is perhaps the best predictor for this complicated frequency response, but an approximate formula for fL is given in the text as 1 1 1 1 f L f p1 f p 2 f p 3 2 CC1RC1 CE RE CC 2 RC 2 (10.20),(34) where RC1 , RE , and RC 2 are the resistances seen by CC1 , CE , and CC 2 , respectively, with the signal source Vsig 0 and the other two capacitors replaced by short circuits. Example N23.1. Compute the mid-band small-signal voltage gain and the upper 3-dB cutoff frequency of the small-signal voltage gain for the CE amplifier shown in Fig. 10.9a. Use a 2N2222A transistor and the circuit element and DC source values listed in text Example 10.4. Use 10 F blocking and bypass capacitors. Whites, EE 320 Lecture 23 Page 15 of 18 The circuit in Agilent Advanced Design System appears as: V_DC SRC2 Vdc=10.0 V AC AC AC1 Start=100 Hz Stop=1.0 MHz Step=100 Hz R R2 R=8 kOhm vo C C2 C=10 uF vi V_AC R SRC1 R3 Vac=polar(1,0) V R=5 kOhm Freq=freq C C1 C=10 uF R R4 R=5 kOhm R R1 R=100 kOhm ap_npn_2N2222A_19930601 Q1 C I_DC C3 SRC4 Idc=1 mA C=10 uF V_DC SRC3 Vdc=-10.0 V From the results of the ADS circuit analysis VCB 2.03 V 400 mV 2.43 V VBE 0.4 V 1.02 V 0.62 V From Fig. 9 in the Motorola 2N2222A datasheet (see the previous set of lecture notes) For VCB 2.43 V Ccb C 5.8 pF. For VBE 0.62 V Ceb C 20 pF. Whites, EE 320 gm Lecture 23 Page 16 of 18 IC 1 mA 0.04 S VT 25 mV The unity gain frequency, fT, is a hugely important high frequency specification for a transistor. fT (or T) is the frequency at which the gain of the transistor operating as an amplifier is one (0 dB): From (10.41), fT gm 0.04 246.8 MHz 2 C C 2 20 pF 5.8 pF This value agrees fairly well with the datasheet value of 300 MHz. 0 265 from the ADS parts list for this 2N2222A transistor. Therefore, r 0 gm 265 6,625 0.04 Whites, EE 320 Lecture 23 Page 17 of 18 From the 2N2222A datasheet, the nominal output resistance at IC = 1 mA is ro 50 k. What about rx? It’s so small in value (~ 50 ) that we’ll easily be able to ignore it for the Am calculations compared to r (which is 6,625 as we just calculated). From (4), g m r RB Am ro || RC || RL r rx RB || Rsig RB Rsig 2,898.6 RL 0.02327 V V Am 36.2 dB Am 64.24 Therefore, or in decibels 0.9524 Am 20 log10 From ADS: m3 freq= 400.0 Hz dB(vo)=33.632 40 dB(vo) 35 m3 m1 m2 freq= 6.300kHz freq= 84.40kHz dB(vo)=36.053 dB(vo)=33.046 m1 m2 30 25 20 15 10 1E2 1E3 1E4 freq, Hz 1E5 1E6 Whites, EE 320 Lecture 23 Page 18 of 18 From this plot, ADS computes a mid-band gain of Am 36.05 dB, which agrees closely with the predicted value above. fH From (29), where from (22) 1 2 Cin Rsig Cin C C 1 g m RL 20 5.8 1 0.04 2,898.6 pF 20 678.3 698.3 pF while from (6) Rsig r || rx RB || Rsig Because RB || Rsig 100 k || 5 k 4,761.9 is so much larger than rx (on the order of 50 ), we can safely ignore rx. Then, Rsig 6,625 || 4,762 2,771 Therefore, f H 2 698.3 1012 2,771 82.25 kHz This agrees very closely with the value of 84.40 kHz predicted by the ADS simulation shown above. Notice that this frequency is dramatically smaller than the unity-gain frequency of the transistor fT 250 MHz. [Add a short discussion on the gain-bandwidth product Am f H .]