CMOS Optical Preamplifier Design Using Graphical Circuit Analysis

Since this SFG is rather involved, we will use Mason’s Direct Rule rather than
manually manipulating the SFG. We observe that there are four distinct feedback
loops in Figure A.3. In terms of their loop transmittances, they are
Notice how the loop transmittances give us a sense of the relative strengths of the
various feedback loops of the circuit; loops L 1 and L 2 represent the feedback paths
across the two gain stages and back through feedback resistor R f and they are the
strongest while loop L 3 is by far the weakest, representing the reflection of the out-
put signal through the feedback resistor R f . Since all loops include branch transmit-
tance i, there are no non-touching loops, thus
∆ = 1 – ( L1 + L2 + L3 + L4) = 22.472 .
For the voltage gain, we can identify three forward transmission paths through the
SFG:
Again, from the SFG, we gain a sense of the relative strengths of the forward trans-
mission, and confirm how insignificant the feedforward path through the feedback
resistor R f is relative to the main signal path through the two transistors. The final
transfer function is
L1 = bcdefgij = 21.8456
L2 = defgik = – 43.6913
L3 = fgil = 0.0096
L4 = bhij = 0.3641
P1 = abcdef = 1515.2 ∆1 = 1
P2 = abhikdef = – 1103.2 ∆2 = 1
P3 = abhilf = 0.2 ∆3 = 1
vo P1∆1 + P2∆2 + P3∆3 Gain = ---- = ----------------------------------------------------- = 18.3 .
vs ∆
Finding the input impedance is simple; having already determined ∆
, we only need
to determine the forward transmission path from i sc1 to v 1 :
170