Unformatted text preview: few tens of mV. This analysis has assumed that the DC 432 APPENDIX A. CIRCUIT MODELS FOR BJT AND FET smallsignal 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 smallsignal 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 smallsignal transconductance
result shows that the large signal transconductance of the transistor decreases from the
smallsignal value as the baseemitter 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 Smallsignal loop gain = 2.5
smallsignal 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: Baseemitter 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 smallsignal transconductance
saturated baseemitter voltage amplitude is 4.5*25 mV, or about 112 mV. hat the large signal transconductance of the transistor decreases 5.4. EXAMPLE  COMMONCOLLECTOR 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 0.91.0 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
steadystate 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 smallsignal 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  COMMONCOLLECTOR 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 unexcited and oscillation is maintained by the “ﬂywheel eﬀect” of the highQ 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 baseemitter swings observed in the
simulations with values predicted in this way. In cases 26, 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 Lnets 0.6 0.4 0.2 Single Lnet
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 2PORT NETWORKS 217
a
I1 I1
+ a
a
Z11 Z12
a
a
Z21 Z22 + V1...
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This test prep was uploaded on 03/13/2014 for the course ECE 453 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.
 Spring '08
 Staff

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