[LAU] Shielded electrical wiring for studio (or not)

[David Christensen]

LAU:

I had a realization -- while the low-pass filter idea may apply to 
amplifier power supply transformer and related, the Ground switch and 
capacitor require a different model and analysis: they implement a 
voltage divider to true earth ground via the parasitic capacitances of 
the amplifier:

    H o------+------------+
             |            |
             o            |
              \           |
        S  o - o-----+    |
              /      |    |
             o       |    |
             |       |    |
    N o---+--+       |    |
          |          |    |
          = C2       = C  = C1
          |          |    |
    A o---+----------+----+
                          |
                          = C3
                          |
    G o-------------------+

Where:

    H is the electrical supply Hot (120 VAC)

    N is the electrical supply Neutral (~0 V)

    A is the amplifier chassis (and connected equipment and people)

    G is the earth (0 V)

    S is the "Ground" switch

    C is the grounding capacitor (0.047 F, 56.4 kOhm at 60 Hz)

    C1 is the parasitic capacitance between H and A

    C2 is the parasitic capacitance between N and A

    C3 is the parasitic capacitance between A and G


I don't have data for the values of C1, C2, and C3, but will assume:

1.  The impedance of C is much smaller than the impedance of C1, C2, and 
C3 (at 60 Hz).

2.  C1 and C2 are approximately equal.

3.  The series combination of C1 and C3, and of C2 and C3, is 1 MOhm (or 
greater).


To simplify calculations, let:

1.  VN = VG = 0 V

2.  ZC1 = ZC2 = ZC3 = 500 kOhm


There are three cases for calculating the voltage on the amplifier 
chassis, VA:

1.  SW connected to H -- C and C1 are in parallel:

    H o---+-----+
          |     |
          = C   = C1
          |     |
    A o---+-----+
          |     |
          = C2  = C3
          |     |
    G o---+-----+

    VA = VH * ( ZC2 || ZC3) / (ZC || ZC1 + ZC2 || ZC3)

       = 120 VAC * (500 k || 500 k) / (56.4 k || 500 k + 500 k || 500 k)

       = 120 VAC * (250 k) / (50.7 k + 250 k)

       = 120 VAC * (250 k / 300.7 k)

       = 99.8 VAC

2.  SW in the center OFF position -- C is gone:

    H o---------+
                |
                = C1
                |
    A o---+-----+
          |     |
          = C2  = C3
          |     |
    G o---+-----+

    VA = H * ( ZC2 || ZC3) / (ZC1 + ZC2 || ZC3)

       = 120 VAC * (500 k || 500 k) / (500 k + 500 k || 500 k)

       = 120 VAC * (250 k) / (500 k + 250 k)

       = 120 VAC * (250 k / 750 k)

       = 40 VAC

3.  SW connected to N -- C is in parallel with C2 and C3:

    H o--------------+
                     |
                     = C1
                     |
    A o---+----+-----+
          |    |     |
          = C  = C2  = C3
          |    |     |
    G o---+----+-----+

    VA = H * ( ZC || ZC2 || ZC3) / (ZC1 + ZC || ZC2 || ZC3)

       = 120 VAC * (56.4k || 500 k || 500 k)
                  / (500 k + 56.4 k || 500 k || 500 k)

       = 120 VAC * (46.0 k) / (500 k + 46.0 k)

       = 120 VAC * (46.0 k) / (546.0 k)

       = 10.1 VAC


David