When you calculate the dc bias voltages required in the prelab
calculations for experiment 3, assume that beta is infinite
so the base currents of all transistors are zero.
Furthermore, remember that the impedance of all capacitors is
infinite at dc, so you can ignore the capacitors during your
dc calculations.  Furthermore, the operation of the differential
pairs and current mirrors in these circuits will be presented
in class this term.

However, the operation of the detector schematic is new and
will not be covered in class.  To help you understand
its operation and to help you do the dc bias calculations,
here is some information.

About Fig. 3.2 in experiment 3:
1.  The collector current that flows in Q9 in Figure 3.2 is
    [VCC - VBE(on)]/[R12 + 500].
2.  The collector current that flows in Q9 in Figure 3.2 also
    flows in Q8 and Q7 because Q7-Q9 along with the 500 Ohm resistors
    form a current mirror with one input and two outputs.
3.  From a dc standpoint, the collector current in Q6 is equal
    to that in Q8 and the collector current in Q5 is equal to that
    in Q7 because the dc voltage from node 1 to ground is equal to the
    dc voltage from node 4 to ground.
4.  From a dc standpoint, the collector current in Q1 is equal
    to that in Q2, and each is equal to half the collector current
    in Q6 because the dc voltage from node 8 to ground is equal
    to the dc voltage from node 10 to ground.
5.  From a dc standpoint, the collector current in Q3 is equal
    to that in Q4, and each is equal to half the collector current
    in Q5 because the dc voltage from node 8 to ground is equal
    to the dc voltage from node 10 to ground.
6.  From this information, the dc current flowing in R11 and the
    dc voltage from node 6 to ground can be calculated.

About Fig. 3.4 in experiment 3:
1.  The collector current that flows in Q5 in Figure 3.4 is
    [VCC - VBE(on)]/R33.  This is (12-0.7)/20k = 565 uA.
2.  Since Q4 and Q5 form a current mirror, the collector current
    in Q4 is equal to the collector current in Q5.
3.  From a dc standpoint, the negative feedback loop forces the
    dc voltage from node 7 to ground to equal the dc input voltage
    from node 2 to ground, 6 Volts.
4.  Then Q3 is on and conducting a current of 6V/R35.
5.  Therefore, the voltage from node 5 to ground is equal to the
    voltage from node 7 to ground plus VBE(on), 6+0.7 = 6.7 Volts.
6.  Therefore, the collector current in Q2 is 
    [VCC - 6.7]/R34 = (12-6.7)/10k = 530 uA.
7.  Note that the collector current in Q2 is almost equal to the
    collector current in Q4.  Therefore the collector current in Q1
    is small.
8.  So the collector current in Q1 is the collector current in Q4
    minus the collector current in Q2, 565 - 530 = 35 uA.
9.  Now when the voltage between node 2 and ground rises,
    the collector current in Q1 increases.  Therefore, the collector
    current in Q2 decreases.  As a result, the voltage from node 5
    to ground increases, which increases the voltage from node 7
    to ground.  In other words, the output follows the input when
    the input is rising.
10. On the other hand, when the voltage between node 2 and ground falls,
    the collector current in Q1 decreases.  Therefore, the collector
    current in Q2 increases.  As a result, the voltage from node 5
    to ground decreases.  At the same time, C32 tends to hold the
    voltage from node 7 to ground high.  (C32 is only discharged by
    R35.)  Q3 turns off when the base-emitter voltage of Q3 drops
    below VBE(on), about 0.7 Volts.  Since Q1, Q2, and R34 provide
    gain between node 2 and node 5, the decrease in the input voltage
    required to turn off Q3 is small.
11. If the voltage between node 2 and ground decreases by a large amount,
    Q1 turns off, and the collector current in Q2 increases from its
    dc value of 530 uA to a maximum value of 565 uA, set by the collector
    current in Q4.  As a result, the lowest voltage between node 5
    and ground is VCC - 565uA(R34) = 6.35 Volts.  Notice that this
    lower limit is set without causing any transistor to saturate.
    As a result, when the voltage between node 2 and ground rises again,
    the time required to turn Q1 and Q3 back on is small; that is,
    the circuit is fast.
12. Since R35 discharges C32, the choice of element values here boils
    down to this tradeoff involving the product of R35 and C32.
    In simulation, you should try different values until you can
    understand what happens when the product is too small and when
    it is too big.