6 ratio of large signal to small signal

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: few tens of mV. This analysis has assumed that the DC 432 APPENDIX A. CIRCUIT MODELS FOR BJT AND FET small-signal transconductance can be obtained in the limit as x approaches 0, i.e., gm = lim IDC x→0 2 I1 (x) IDC = v1 I0 (x) kT /q (A.19) At room temperature kT /q = 25 mV, so gm = IDC ≈ 40IDC 25 mV (A.20) The large signal transconductance is Gm (x) = IDC 2 I1 (x) q 2 I1 (x) 2 I1 (x) = IDC = gm v1 I0 (x) kT x I0 (x) x I0 (x) (A.21) The ratio of the large signal to small-signal transconductance is shown in Figure A.6. This 1 0.8 Gm (x) gm 0.6 0.4 0.2 2 4 6 8 10 x Figure A.6: Ratio of large signal to small-signal transconductance result shows that the large signal transconductance of the transistor decreases from the small-signal value as the base-emitter voltage swing (x) increases. Thus in an oscillator application, as the oscillation amplitude grows, the effective transconductance decreases. 2 I1 (x) q 2 I1 (x) 2 I1 (x) Gm (x) = IDC = IDC = gm v1 I0 (x) kT x I0 (x) x I0 (x) e large signal to Small-signal loop gain = 2.5 small-signal transconductance is shown in Figure A 1 0.8 Gm (x) gm 0.6 0.4 0.2 2 4 6 8 10 x gure A.6: Base-emitter voltage, x=Vbe/(25 mV), grows until transconductance is Ratio of by the factor 1/2.5=0.4, which occurs with x=4.5. Hence, reduced large signal to small-signal transconductance saturated base-emitter voltage amplitude is 4.5*25 mV, or about 112 mV. hat the large signal transconductance of the transistor decreases 5.4. EXAMPLE - COMMON-COLLECTOR COLPITTS OSCILLATOR 155 Table 5.2: Parameters used for oscillator simulations. Case 1 2 3 4 5 6 C1 2500 pF 1000 pF 500 pF 200 pF 100 pF 22.5 pF C2 20.4 pF 20.6 pF 21.1 pF 22.5 pF 25.4 pF 200 pF gm,ss 45.3 mS 18.1 mS 9.42 mS 3.77 mS 1.83 mS 2.23 mS gm /gm,ss 0.88 2.2 4.4 11.0 21.8 17.9 the period 0.9-1.0 µs after the initial transient. For the cases where oscillation occurs, the expanded plots show the current and voltage waveforms when the circuit is undergoing steady-state oscillation. Figure 5.12: Case 1 - loop gain is 0.88, which is less than one, so oscillation does not start. The loop gain (gm /gm,ss , column 4 in Table 5.2) is smaller than 1 for Case 1. Therefore the initial disturbance excites damped oscillations at a frequency approximately equal to fo . This case illustrates the fact that sustained oscillations cannot develop if the small-signal 156 CHAPTER 5. OSCILLATORS Figure 5.13: Case 2 - gm /gm,ss = 2.2. Figure 5.13: Case 2 - gm /gm,ss = 2.2. Figure 5.14: Case 3 - gm /gm,ss = 4.4. 5.4. EXAMPLE - COMMON-COLLECTOR COLPITTS OSCILLATOR Figure 5.15: Case 4 - gm /gm,ss = 11.0. 157 Figure 5.15: Case 4 - gm /gm,ss = 11.0. Figure 5.16: Case 5 - gm /gm,ss = 21.8. 158 CHAPTER 5. OSCILLATORS Figure 5.17: Case 6 - gm /gm,ss = 17.9. collector current (lower, left) waveform exhibiting increasingly narrow and increasingly large current “spikes”. For the larger loop gains, the transistor is essentially cut off for most of the oscillation cycle. The transistor injects a short current pulse into the resonator once each cycle, near the positive peak of the voltage swing. Thus, for most of the oscillation period, the circuit is un-excited and oscillation is maintained by the “flywheel effect” of the high-Q 160 CHAPTER 5. OSCILLATORS 2 numerically solving the equation 0.45 = x I1 (x)/Io (x) yields x ￿ 3.75. The predicted steadystate amplitude of |Vbe | is then (3.75)(25 mV) = 94 mV, which compares favorably with what is observed in the simulations. Table 5.3 compares the base-emitter swings observed in the simulations with values predicted in this way. In cases 2-6, where oscillation occurs, the Table 5.3: Comparison between predicted and simulated |Vbe | and |Ve |/|Vbe | ratio. Case 1 2 3 4 5 6 gm /gm,ss 0.88 2.2 4.4 11.0 21.8 17.9 |Vbe | pred. 94 mV 207 mV 537 mV 1.08 V 882 mV |Vbe | sim. 71 mV 185 mV 554 mV 1.29 V 483 mV |Ve | sim. 3.27 V 4.30 V 4.70 V 5.40 V * C1 /C2 122.5 48.5 23.7 8.9 3.9 0.11 |Ve |/|Vbe | 46.1 23.2 8.5 4.2 * prediction yields a reasonable approximation to the simulated values. Once |Vbe | is known, or has been predicted, the amplitude of the emitter voltage, |Ve |, can be estimated. A node equation at the junction between the current source, Z1 , and Z2 in Figure 5.22 yields Ve Vbe = = Z2 + Z2 gm,ss Z1 C1 + Z2 gm,ss . C2 (5.23) (5.24) Vout V1 U=1 V R1 R=10 Ohm C1 C=15.9 pF L1 L=159 nH R2 R=1000 Ohm Vout2 V2 U=1 V R3 R=10 Ohm ac simulation AC1 Type=lin Start=10 MHz Stop=200 MHz Points=101 C2 C=47.7 pF L2 L=47.7 nH L3 L=477 nH C3 C=4.77 pF R4 R=1000 Ohm Equation Equation Equation Eqn1 Pavs=1/80 Eqn2 Gt=0.5*abs(Vout.v)^2/1000/Pavs Eqn3 Gt2=0.5*abs(Vout2.v)^2/1000/Pavs Equation Equation Eqn4 dBGt=dB(Gt) Eqn5 dBGt2=dB(Gt2) 1 0.8 Two cascaded L-nets 0.6 0.4 0.2 Single L-net 0 2e7 4e7 6e7 8e7 1e8 1.2e8 acfrequency acfrequency 1.4e8 1.6e8 1.8e8 2e8 2e7 4e7 6e7 8e7 1e8 1.2e8 acfrequency acfrequency 1.4e8 1.6e8 1.8e8 2e8 0 -10 -20 -30 -40 -50 -60 -70 7.3. PARALLEL, SERIES, CASCADE CONNECTIONS OF 2-PORT NETWORKS 217 a I1 I1 + a a Z11 Z12 a a Z21 Z22 + V1...
View Full Document

Ask a homework question - tutors are online