#!/usr/local/bin/calc -d -f
#
# A license is hereby granted to reproduce this design for personal,
# non-commercial use.
#
# THIS DESIGN IS PROVIDED "AS IS". THE AUTHOR PROVIDES NO WARRANTIES
# WHATSOEVER, EXPRESSED OR IMPLIED, INCLUDING WARRANTIES OF
# MERCHANTABILITY, TITLE, OR FITNESS FOR ANY PARTICULAR PURPOSE.  THE
# AUTHOR DOES NOT WARRANT THAT USE OF THIS DESIGN DOES NOT INFRINGE
# THE INTELLECTUAL PROPERTY RIGHTS OF ANY THIRD PARTY IN ANY COUNTRY.
#
# So there.
#
# Copyright (c) 1992-2006, John Conover, All Rights Reserved.
#
# Comments and/or problem reports should be addressed to:
#
#     john@email.johncon.com
#
#     http://localhost/www.johncon.com/john/
#     http://localhost/www.johncon.com/ntropix/
#     http://localhost/www.johncon.com/ndustrix/
#     http://localhost/www.johncon.com/nformatix/
#     http://localhost/www.johncon.com/ndex/
#
# A calc(1) script for designing series pass full-wave bridge power
# supplies. Design notes are included. Calc(1) is a C-style arbitrary
# precision calculator, (and also handles complex arithmetic,) and is
# available from http://sourceforge.net/projects/calc/.
#
# The capacitor RMS current, transformer secondary RMS current,
# capacitor peak current, transformer secondary peak current, filter
# capacitor ripple voltage, and capcitor RMS voltage are
# calculated/computed for a given output current and series pass
# component values.
#
# The script contains a binary search-for-solution algorithm that
# computes the intersection of a linear line, (the filter capacitor
# discharge time interval,) and the sine wave input voltage, (the
# filter capacitor charge time interval.)
#
# Note: if all that is necessary is to find the value of a filter
# capacitor, use:
#
#     C = ((i / (2 * pi * f)) * acos (- (vf / vi))) / (vi - vf)
#
# where C is the value of the capacitor, f is the line frequency, vi
# is the peak voltage on the capacitor, (e.g., the input voltage,) and
# vf is the minimum voltage on the capacitor, (probably the output
# voltage of a regulator plus the drop out voltage of the regulator;
# the capacitor ripple voltage is vi - vf.)
#
# For example, using calc(1):
#
#     ; i = 1.5
#     ; f = 60
#     ; vi = 20
#     ; vf = 19
#     ; ((i / (2 * pi * f)) * acos (- (vf / vi))) / (vi - vf)
#             0.01123646719869675163
#
# What this script does is compute the RMS capacitor current, peak
# capacitor current, RMS transformer current, peak transformer
# current, etc., after the capacitor value has been chosen.
#
# The development notes-derivation and verification-are left intact at
# the beginning of this file. The actual script is at the end of the
# file.
#
# The peak voltage, (i.e., half the peak-to-peak voltage,) of the
# input to the diode bridge rectifier:
#
# Vpp = 30.7;
#
# The value of the filter capacitor:
#
# C = 13600E-6;
#
# The constant current output of the regulator:
#
# Iout = 1.54;
#
# The line frequency:
#
# f = 60;
#
# Numerical values:
#
# pi = ln (-1) / (0 + 1i);
# w = 2 * pi * f;
#
# The voltage input to the diode bridge rectifier, Vpp is the
# peak-to-peak voltage:
#
#     v(t) = Vpp * cos (w * t)
#
# The derivative of the voltage input to the diode bridge rectifier:
#
#     dv(t) / dt = - Vpp * w * sin (w * t)
#
# The voltage across the filter capacitor during discharge time:
#
#     v(t) = - (Iout * t) / C
#
# The derivative of the voltage across the filter capacitor during
# discharge time:
#
#     dv(t) = - Iout / C
#
# When the derivative of the voltage input to the diode bridge
# rectifier and derivative of the voltage across the filter capacitor
# equate, the charge time interval ends, and the discharge time
# interval begins:
#
#     - Iout / C = - Vpp * w * sin (w * t)
#     Iout / C = Vpp * w * sin (w * t)
#     Iout / (C * Vpp * w) = sin (w * t)
#     sin (w * t) = Iout / (C * Vpp * w)
#     w * t = asin (Iout / (C * Vpp * w))
#
# T_begin is when the discharge time interval begins:
#
# T_begin = asin (Iout / (C * Vpp * w)) /  w;
#
# Vd is the voltage input to the diode bridge rectifier when the
# discharge time interval begins:
#
# Vd = Vpp * cos (w * T_begin);
#
# Vs is the "effective" starting voltage across the capacitor at t =
# 0:
#
# Vs = Vd + (T_begin * (Iout / C));
#
# The formula for the discharge time interval is:
#
#     v(t) = Vs - ((Iout * t) / C)
#
# Or, uncomment for a "pastable" command line into gnuplot(1) to view
# the discharge time interval:
#
#     printf ("plot [0:%f][0:] abs \(%f * cos \(%f * x\)\),
#         %f - \(\(%f * x\) / %f\)\n", 1 / (2 * f), Vpp, w, Vs, Iout,
#         C);
#
# Search for the time the capacitor discharge interval ends, i.e.,
# where Vs - ((Iout * t) / C) = - Vpp * cos (w * t); Note: abs (Vpp *
# cos (w * t)) is equal to - Vpp * cos (w * t) for the time interval 1
# / (4 * f) <= t <= 1 / (2 * f).
#
# The search is a recursive binary search on the interval 1 / (4 * f)
# <= t <= 1 / (2 * f), starting with the variable begin = 1 / (4 * f),
# and the variable end = 1 / (2 * f); the test for which way to direct
# the binary search is half way through the interval at t = begin +
# ((end - begin) / 2), and depending on whether the Vs - ((Iout * t) /
# C) or - Vpp * cos (w * t) is larger, the end, or begin variable is
# set to t, the search interval becoming smaller and smaller, at a
# binary rate, until the end of the discharge time interval is found
# within an error of erroreps.
#
# define search_end (begin, end, erroreps, Vs, Iout, C, Vpp, w)
# {
#     local t = begin + ((end - begin) / 2);
#
#     if ((- (Vpp * cos (w * t))) > (Vs - ((Iout * t) / C)))
#     {
#         end = t;
#     }
#
#     else
#     {
#         begin = t;
#     }
#
#     if (abs (end / begin) > (1 + erroreps))
#     {
#         t = search_end (begin, end, erroreps, Vs, Iout, C, Vpp, w);
#     }
#
#     return (t);
# }
#
# begin = 1 / (4 * f);
# end = 1 / (2 * f);
# erroreps = 1e-9;
#
# T_end = search_end (begin, end, erroreps, Vs, Iout, C, Vpp, w);
#
# T_begin is the beginning time of the discharge time interval, and
# T_end is the ending time of the discharge time interval. Uncomment
# for a "pastable"command line into gnuplot(1) to view discharge time
# interval:
#
#     printf ("plot [%f:%f][0:] abs \(%f * cos \(%f * x\)\),
#         %f - \(\(%f * x\) / %f\)\n", T_begin, T_end, Vpp, w, Vs,
#         Iout, C);
#
# The charge time interval begins at T_end, and ends at (T_begin + (1
# / (2 * f))) for - Vpp * cos (w * t)
#
# The peak-to-peak ripple voltage, Vr, is:
#
# Vr = Vpp - (-Vpp * cos (w * T_end));
#
# The capacitor current, Ic(t), during the charge time interval is the
# derivative of the voltage - Vpp * cos (w * t), multiplied by the
# capacitor value, C:
#
#     Ic(t) = Vpp * C * w * sin (w * t)
#
# Uncomment for a "pastable"command line into gnuplot(1) to view
# capacitor current, Ic(t), during the charge time interval:
#
#     printf ("plot [%f:%f] %f * sin \(%f * x\)\n", T_end, T_begin +
#         (1 / (2 * f)), Vpp * C * w, w);
#
# As a straight line approximation to the capacitor current during the
# charge time interval, use the two end points, say (37, 0.007700) and
# (0, 0.008333) for a time interval of 0.008333 - 0.007700 =
# 0.0006333, and using calc(1) to generate the straight line
# approximation from a file, myfile, that contains:
#
#     #!/usr/local/bin/calc -d -f
#     #
#     for (x = 0; x <= 6333; x ++)
#     {
#         printf ("%f\t%f\n", x / 10000000,
#             37 - ((37 / 0.0006333) * (x / 10000000)));
#     }
#
# and:
#
#     ./myfile | tsrms -p
#
# gives an approximate answer of about 21.36 A RMS.
#
# Or, a little more formal construct:
#
#     ./myfile | tsmath -S | tsintegrate | tsmath -d 6333 |
#         tsmath -R | tail -1
#
# which gives about the same answer.
#
# The tsrms(1), tsmath(1), and tsintegrate(1), programs are available
# from the http://www.johncon.com/ndustrix/ and/or
# http://www.johncon.com/ntropix/ sites, (and are usually used in
# analyzing financial time series-the "tsmath -d 8463" pipe was for
# time scaling-the time series programs do not have time scales.)
#
# However, formally, the RMS^2 of Ic(t) in the charge interval will be
# (1 / t) * integral of (Ic(t))^2 dt, evaluated at T_end and T_begin,
# where Ic(t) = Vpp * C * w * sin (w * t), which is:
#
#     RMS^2 = ((Vpp * C * w)^2 * ((t / 2) - ((1 / (4 * w)) *
#                  sin (2 * w * t)))) / t
#
# evaluated at T_end and T_begin + (1 / (2 * f)).
#
# And, the rather formidable formula for the RMS capacitor current
# during the charge interval, IcRMS:
#
# IcRMS = sqrt ((((Vpp * C * w)^2 * (((T_begin + (1 / (2 * f))) / 2) -
#             ((1 / (4 * w)) * sin (2 * w * (T_begin +
#             (1 / (2 * f))))))) - ((Vpp * C * w)^2 * ((T_end / 2) -
#             ((1 / (4 * w)) * sin (2 * w * T_end))))) /
#             (T_begin + (1 / (2 * f)) - T_end));
#
# which has an instantaneous peak value, IcP, at the time T_end:
#
# IcP = Vpp * C * w * sin (w * T_end);
#
# RMS^2 of the capacitor current during the discharge interval is 1 /
# t * integral of - Iout^2dt, evaluated at T_end and T_begin:
#
#     (1 / t) * (-Iout^2) t = Iout^2
#
# and the RMS current of the capacitor during the discharge interval
# is simply, Iout. The total capacitor current, Ic, is:
#
# Ic = (((T_end - T_begin) / (1 / (2 * f))) * Iout) + ((1 - ((T_end - T_begin) / (1 / (2 * f)))) * IcRMS);
#
# The transformer current during the charge interval, IlRMS, is the
# capacitor current during the charge interval plus Iout:
#
# IlRMS = IcRMS + Iout;
#
# The peak transformer current during the charge interval, IlP, is the
# peak capacitor current during the charge interval plus Iout:
#
# IlP = IcP + Iout;
#
# The transformer RMS current, Il, is:
#
# Il = ((1 - ((T_end - T_begin) / (1 / (2 * f))))) * IlRMS;
#
# The diode bridge rectifier current is the same as the transformer.
#
# printf ("Peak-to-peak capacitor ripple voltage = %.4f V,\nMaximum capacitor voltage = %.4f V,\nMinimum capacitor voltage = %.4f V,\nPeak capacitor charging current = %.4f A,\nRMS capacitor charging current = %.4f A,\nRMS capacitor current = %.4f A,\nPeak transformer charging current = %.4f A,\nRMS transformer charging current = %.4f A,\nRMS transformer current = %.4f A\n", Vr, Vpp, Vpp - Vr, IcP, IcRMS, Ic, IlP, IlRMS, Il);
#
# The charge time interval begins at T_end, and ends at (T_begin + (1
# / (2 * f))) for - Vpp * cos (w * t); the RMS capacitor voltage,
# VcRMS:
#
# VcRMS^2 = (1 / t) integral v^2(t) dt
#
# For  T_begin <= t <= T_end:
#
#    v(t) = Vs - ((Iout * t) / C)
#
#    VcRMS^2 = (1 / (T_end - T_begin)) *
#                  integral (Vs - ((Iout * t) / C))^2 dt
#
#    VcRMS^2 = (1 / (T_end - T_begin)) * integral (((Vs^2) -
#                  (2 * Vs * ((Iout * t) / C)) +
#                  ((((Iout * t) / C))^2))) dt
#
#    VcRMS^2 = (1 / (T_end - T_begin)) *
#                  (((Vs^2) * t) - (Vs * ((Iout * (t^2)) / C)) +
#                  (((Iout / C)^2) * (1 / 3) * (t^3)))
#
#    VcRMS^2 = (1 / (T_end - T_begin)) *
#                  ((((Vs^2) * T_end) - (Vs *
#                  ((Iout * (T_end^2)) / C)) + (((Iout / C)^2) *
#                  (1 / 3) * (T_end^3))) -
#                  (((Vs^2) * T_begin) - (Vs * ((Iout *
#                  (T_begin^2)) / C)) + (((Iout / C)^2) * (1 / 3) *
#                  (T_begin^3))))
#
# For T_end <= t <= T_begin + (1 / (2 * f)):
#
#    v(t) = -Vpp * cos (w * t)
#
#    VcRMS^2 = (1 / ((T_begin + (1 / (2 * f))) - T_end)) *
#                  integral (((-Vpp)^2) * (cos^2(w * t))) dt
#
#    VcRMS^2 = (1 / ((T_begin + (1 / (2 * f))) - T_end)) *
#                  (((-Vpp)^2) * ((t / 2) + ((1 / (4 * w)) *
#                  sin (2 * w * t))))
#
#    VcRMS^2 = (1 / ((T_begin + (1 / (2 * f))) - T_end)) *
#                  ((((-Vpp)^2) * (((T_begin + (1 / (2 * f))) / 2) +
#                  ((1 / (4 * w)) * sin (2 * w *
#                  (T_begin + (1 / (2 * f))))))) - (((-Vpp)^2) *
#                  ((T_end / 2) + ((1 / (4 * w)) * sin (2 * w * T_end)))))
#
# and the rather formidable formula for the total RMS voltage on the
# capacitor:
#
#    VcRMS = ((T_end - T_begin) / (1 / (2 * f))) *
#                (sqrt ((1 / (T_end - T_begin)) *
#                ((((Vs^2) * T_end) - (Vs *
#                ((Iout * (T_end^2)) / C)) + (((Iout / C)^2) *
#                (1 / 3) * (T_end^3))) -
#                (((Vs^2) * T_begin) - (Vs * ((Iout *
#                (T_begin^2)) / C)) + (((Iout / C)^2) * (1 / 3) *
#                (T_begin^3)))))) +
#                (((T_begin + (1 / (2 * f))) - T_end) / (1 / (2 * f))) *
#                (sqrt ((1 / ((T_begin + (1 / (2 * f))) - T_end)) *
#                ((((-Vpp)^2) * (((T_begin + (1 / (2 * f))) / 2) +
#                ((1 / (4 * w)) * sin (2 * w *
#                (T_begin + (1 / (2 * f))))))) - (((-Vpp)^2) *
#                ((T_end / 2) + ((1 / (4 * w)) * sin (2 * w * T_end)))))))
#
# printf ("Peak-to-peak capacitor ripple voltage = %.4f V,\nMaximum capacitor voltage = %.4f V,\nMinimum capacitor voltage = %.4f V,\nPeak capacitor charging current = %.4f A,\nRMS capacitor charging current = %.4f A,\nRMS capacitor current = %.4f A,\nRMS capacitor voltage = %.4f V,\nPeak transformer charging current = %.4f A,\nRMS transformer charging current = %.4f A,\nRMS transformer current = %.4f A\n", Vr, Vpp, Vpp - Vr, IcP, IcRMS, Ic, VcRMS, IlP, IlRMS, Il);
#
# define parameters (Vpp, C, Iout, f)
# {
#     local w = 2 * pi * f,
#           T_begin,
#           Vd,
#           Vs,
#           begin = 1 / (4 * f),
#           end = 1 / (2 * f),
#           erroreps = 1e-9,
#           T_end,
#           Vr,
#           IcRMS,
#           IcP,
#           Ic,
#           IlRMS,
#           IlP,
#           Il,
#           VcRMS;
#
#     T_begin = asin (Iout / (C * Vpp * w)) /  w;
#     Vd = Vpp * cos (w * T_begin);
#     Vs = Vd + (T_begin * (Iout / C));
#
#     T_end = search_end (begin, end, erroreps, Vs, Iout, C, Vpp, w);
#
#     Vr = Vpp - (-Vpp * cos (w * T_end));
#
#     IcRMS = sqrt ((((Vpp * C * w)^2 * (((T_begin + (1 / (2 * f))) / 2) -
#                 ((1 / (4 * w)) * sin (2 * w * (T_begin +
#                 (1 / (2 * f))))))) - ((Vpp * C * w)^2 * ((T_end / 2) -
#                 ((1 / (4 * w)) * sin (2 * w * T_end))))) /
#                 (T_begin + (1 / (2 * f)) - T_end));
#
#     IcP = Vpp * C * w * sin (w * T_end);
#     Ic = (((T_end - T_begin) / (1 / (2 * f))) * Iout) + ((1 - ((T_end - T_begin) / (1 / (2 * f)))) * IcRMS);
#     IlRMS = IcRMS + Iout;
#     IlP = IcP + Iout;
#     Il = ((1 - ((T_end - T_begin) / (1 / (2 * f))))) * IlRMS;
#
#     VcRMS = ((T_end - T_begin) / (1 / (2 * f))) *
#                 (sqrt ((1 / (T_end - T_begin)) *
#                 ((((Vs^2) * T_end) - (Vs *
#                 ((Iout * (T_end^2)) / C)) + (((Iout / C)^2) *
#                 (1 / 3) * (T_end^3))) -
#                 (((Vs^2) * T_begin) - (Vs * ((Iout *
#                 (T_begin^2)) / C)) + (((Iout / C)^2) * (1 / 3) *
#                 (T_begin^3)))))) +
#                 (((T_begin + (1 / (2 * f))) - T_end) / (1 / (2 * f))) *
#                 (sqrt ((1 / ((T_begin + (1 / (2 * f))) - T_end)) *
#                 ((((-Vpp)^2) * (((T_begin + (1 / (2 * f))) / 2) +
#                 ((1 / (4 * w)) * sin (2 * w *
#                 (T_begin + (1 / (2 * f))))))) - (((-Vpp)^2) *
#                 ((T_end / 2) + ((1 / (4 * w)) * sin (2 * w * T_end)))))));
#
#    printf ("Peak-to-peak capacitor ripple voltage = %.4f V,\nMaximum capacitor voltage = %.4f V,\nMinimum capacitor voltage = %.4f V,\nPeak capacitor charging current = %.4f A,\nRMS capacitor charging current = %.4f A,\nRMS capacitor current = %.4f A,\nRMS capacitor voltage = %.4f V,\nPeak transformer charging current = %.4f A,\nRMS transformer charging current = %.4f A,\nRMS transformer current = %.4f A\n", Vr, Vpp, Vpp - Vr, IcP, IcRMS, Ic, VcRMS, IlP, IlRMS, Il);
# }
#
# The value of Pi:
#
pi = ln (-1) / (0 + 1i);
#
# Compute the intersection of a linear line, (the filter capacitor
# discharge time interval,) and the sine wave input voltage, (the
# filter capacitor charge time interval.)
#
# The search is a recursive binary search on the interval 1 / (4 * f)
# <= t <= 1 / (2 * f), starting with the variable begin = 1 / (4 * f),
# and the variable end = 1 / (2 * f); the test for which way to direct
# the binary search is half way through the interval at t = begin +
# ((end - begin) / 2), and depending on whether the Vs - ((Iout * t) /
# C) or - Vpp * cos (w * t) is larger, the end, or begin variable is
# set to t, the search interval becoming smaller and smaller, at a
# binary rate, until the end of the discharge time interval is found
# within an error of erroreps:
#
define search_end (begin, end, erroreps, Vs, Iout, C, Vpp, w)
{
    local t = begin + ((end - begin) / 2);

    if ((- (Vpp * cos (w * t))) > (Vs - ((Iout * t) / C)))
    {
        end = t;
    }

    else
    {
        begin = t;
    }

    if (abs (end / begin) > (1 + erroreps))
    {
        t = search_end (begin, end, erroreps, Vs, Iout, C, Vpp, w);
    }

    return (t);
}
#
# Calculation function. Given the peak input voltage, (i.e., RMS *
# sqrt (2),) from the diode bridge, Vpp, the value of the filter
# capacitor, C, the output current from the regulator, Iout, and the
# line frequency, compute the variables of the power supply:
#
define parameters (Vpp, C, Iout, f)
{
    local w = 2 * pi * f,
          T_begin,
          Vd,
          Vs,
          begin = 1 / (4 * f),
          end = 1 / (2 * f),
          erroreps = 1e-9,
          T_end,
          Vr,
          IcRMS,
          IcP,
          Ic,
          IlRMS,
          IlP,
          Il,
          VcRMS;

    T_begin = asin (Iout / (C * Vpp * w)) /  w;
    Vd = Vpp * cos (w * T_begin);
    Vs = Vd + (T_begin * (Iout / C));

    T_end = search_end (begin, end, erroreps, Vs, Iout, C, Vpp, w);

    Vr = Vpp - (-Vpp * cos (w * T_end));

    IcRMS = sqrt ((((Vpp * C * w)^2 * (((T_begin + (1 / (2 * f))) / 2) -
                ((1 / (4 * w)) * sin (2 * w * (T_begin +
                (1 / (2 * f))))))) - ((Vpp * C * w)^2 * ((T_end / 2) -
                ((1 / (4 * w)) * sin (2 * w * T_end))))) /
                (T_begin + (1 / (2 * f)) - T_end));

    IcP = Vpp * C * w * sin (w * T_end);
    Ic = (((T_end - T_begin) / (1 / (2 * f))) * Iout) + ((1 - ((T_end - T_begin) / (1 / (2 * f)))) * IcRMS);
    IlRMS = IcRMS + Iout;
    IlP = IcP + Iout;
    Il = ((1 - ((T_end - T_begin) / (1 / (2 * f))))) * IlRMS;

    VcRMS = ((T_end - T_begin) / (1 / (2 * f))) *
                (sqrt ((1 / (T_end - T_begin)) *
                ((((Vs^2) * T_end) - (Vs *
                ((Iout * (T_end^2)) / C)) + (((Iout / C)^2) *
                (1 / 3) * (T_end^3))) -
                (((Vs^2) * T_begin) - (Vs * ((Iout *
                (T_begin^2)) / C)) + (((Iout / C)^2) * (1 / 3) *
                (T_begin^3)))))) +
                (((T_begin + (1 / (2 * f))) - T_end) / (1 / (2 * f))) *
                (sqrt ((1 / ((T_begin + (1 / (2 * f))) - T_end)) *
                ((((-Vpp)^2) * (((T_begin + (1 / (2 * f))) / 2) +
                ((1 / (4 * w)) * sin (2 * w *
                (T_begin + (1 / (2 * f))))))) - (((-Vpp)^2) *
                ((T_end / 2) + ((1 / (4 * w)) * sin (2 * w * T_end)))))));

    printf ("Peak-to-peak capacitor ripple voltage = %.4f V,\nMaximum capacitor voltage = %.4f V,\nMinimum capacitor voltage = %.4f V,\nPeak capacitor charging current = %.4f A,\nRMS capacitor charging current = %.4f A,\nRMS capacitor current = %.4f A,\nRMS capacitor voltage = %.4f V,\nPeak transformer charging current = %.4f A,\nRMS transformer charging current = %.4f A,\nRMS transformer current = %.4f A\n", Vr, Vpp, Vpp - Vr, IcP, IcRMS, Ic, VcRMS, IlP, IlRMS, Il);
}
#
# The line transformer under consideration, (Amveco 62084,
# http://www.amveco.com/,) is rated at a full load voltage of 18 V
# RMS, at 1.388 A RMS per secondary, and produces 20.7 V RMS at 0 A
# RMS, 60 C maximum ambient. The peak voltage would be (1.05 * 20.7) *
# sqrt (2), (for 120 V RMS +/- 5% line voltage,) or 30.74 V.  The
# maximum current output of the transformer, using the manufacturer's
# recommended form factor of 1.8, would be 1.388 A RMS / 1.8 = 0.771 A
# DC. (see http://www.amveco.com/Technical_Notes_3.htm, for
# particulars.)
#
# The filter capacitor under consideration, 4700 uF Electrolytic,
# (Panasonic ECO-S1JA472,) is rated at 4700 uF, +/- 20%, 63 V, 3.0 A
# ripple current, an ESR of 0.063, -4- C to +105 C, and a 3000
# Hr. life expectancy.
#
# The diode bridge under consideration, (International Rectifier
# GBPC3502A,) has a 1.1 V forward drop @ 35 A rating per leg, and 0.75
# V forward drop @ 1 A and 125 C per leg, and, 0.60 V forward drop @ 1
# A @ 25 C per leg.
#
# The line frequency is either 50 Hz. or 60 Hz.
#
# The output voltage from the diode bridge would be between the
# ranges:
#
#     ((0.95 * 18) * sqrt (2)) - 2.2 = 21.98 V, Minimum
#     ((1.00 * 18) * sqrt (2)) - 2.2 = 23.26 V, Typical
#     ((1.05 * 18) * sqrt (2)) - 2.2 = 24.53 V, Maximum
#
Iout = 0.6;
C = 4700E-6;
f = 60;
#
Vpp = 23.26;
Cval = C;
#
printf ("\nVpp = %f, C = %f, Iout = %f, f = %f:\n\n", Vpp, Cval, Iout, f);
parameters (Vpp, Cval, Iout, f)
#
Vpp = 21.98;
Cval = C * 0.8;
#
printf ("\nVpp = %f, C = %f, Iout = %f, f = %f:\n\n", Vpp, Cval, Iout, f);
parameters (Vpp, Cval, Iout, f)
#
Vpp = 24.53;
Cval = C * 1.2;
#
printf ("\nVpp = %f, C = %f, Iout = %f, f = %f:\n\n", Vpp, Cval, Iout, f);
parameters (Vpp, Cval, Iout, f)
#
Vpp = 24.53;
Cval = C * 0.8;
#
printf ("\nVpp = %f, C = %f, Iout = %f, f = %f:\n\n", Vpp, Cval, Iout, f);
parameters (Vpp, Cval, Iout, f)
#
Vpp = 21.98;
Cval = C * 1.2;
#
printf ("\nVpp = %f, C = %f, Iout = %f, f = %f:\n\n", Vpp, Cval, Iout, f);
parameters (Vpp, Cval, Iout, f)
#
f = 50;
#
Vpp = 23.26;
Cval = C;
#
printf ("\nVpp = %f, C = %f, Iout = %f, f = %f:\n\n", Vpp, Cval, Iout, f);
parameters (Vpp, Cval, Iout, f)
#
Vpp = 21.98;
Cval = C * 0.8;
#
printf ("\nVpp = %f, C = %f, Iout = %f, f = %f:\n\n", Vpp, Cval, Iout, f);
parameters (Vpp, Cval, Iout, f)
#
Vpp = 24.53;
Cval = C * 1.2;
#
printf ("\nVpp = %f, C = %f, Iout = %f, f = %f:\n\n", Vpp, Cval, Iout, f);
parameters (Vpp, Cval, Iout, f)
#
Vpp = 24.53;
Cval = C * 0.8;
#
printf ("\nVpp = %f, C = %f, Iout = %f, f = %f:\n\n", Vpp, Cval, Iout, f);
parameters (Vpp, Cval, Iout, f)
#
Vpp = 21.98;
Cval = C * 1.2;
#
printf ("\nVpp = %f, C = %f, Iout = %f, f = %f:\n\n", Vpp, Cval, Iout, f);
parameters (Vpp, Cval, Iout, f)
#
# The thermal shutdown characteristics of the LM317HVT, (a TO-220
# packaged device,) are; a minimum current limit of 1.5 A, typical of
# 2.2 A, and a maximum of 3.7 A, over 0 C to 125 C, with a maximum
# junction temperature of 125 C, and Vin - Vout = 5 V. The
# documentation states that thermal shutdown occurs when the device
# dissipation is 20 W, or above, (its in the foot note to the device's
# Absolute Maximum Ratings,) and an output current of 1.5 A.
#
# Additionally, there is a current limit graph, (which is assumed to
# be typical values,) vs the voltage across the device, which for less
# than 8 volts is 2.2 A @ -55 C, 2.25 A @ 25 C, and, 1.9 A @ 150 C.
# The short circuit current, (with about 20 V to 22 V across the
# device,) is about 1.5 A at all temperatures.
#
# The junction-to-ambient thermal resistance, Theta j-a, of the
# LM317HVT is 50 C / W, typical, and junction-to-case, Theta j-c, is 5
# C / W.
#
# The maximum power dissipation at fuse opening is (23.785 - 13.85) *
# 1.2 = 11.92 W @ Iout = 1.2 A. The junction temperature rise above
# the case would be 5 * 11.92 = 59.61 C. For the mica insulator with a
# thermal resistance of 1 C / W, add 11.92 C. For a 3 C thermal
# resistance heat sink, (3 = 50 / sqrt (A), A = 278 cm^2, or an
# aluminum sheet, 6.56 inches on an edge, single sided convection, per
# device,) add 3 * 11.92 = 35.77 C, or a total junction temperature
# rise of 59.61 + 11.92 + 35.77 = 107.3 C above ambient. If the
# ambient temperature is 25 C, then the junction temperature would be
# 132.3 C, at an output current, Iout, of 1.2 A.
#
# The efficiency of the power supply is (15 * 1.2) / ((15 * 1.2) +
# 11.92) = 60.02%.
#
# Estimated current consumption:
#
#     Iout = 2 * 81 mA, peak
#     LT1206CT7 = 2 * 35 mA
#     LT1115CN8 = 2 * 11.5 mA
#     LT1097CN8 = 2 * 0.56 mA
#     LM317HVT = 1 * (3 mA quiescent + 12 mA minimum output current)
#     --------------------
#     Total = 271 mA, (162 mA of DC 0 dbFS signal and, 109 mA quiescent)
#
# A conservative specification would be 0.6 / 1.25 = 0.48 A average
# continuous output current, with a non-repetitive peak instantaneous
# current of 0.6 A. The RMS capacitor voltage would be 23.8741 V, for
# a power dissipation of (23.8741 - 13.85) * 0.48 = 4.81 W. The
# junction temperature would be 4.81 * 5 = 24.1 C above the case
# temperature, which is approximately 4.81 C above the mica insulator
# temperature, for a total of 24.1 + 4.81 C above the chassis
# temperature. If the chassis is an aluminum sheet, 6.56 inches on an
# edge, single sided convection, per device, the thermal resistance
# would be 50 / sqrt (A) = 3 C / W, (A in cm^2); the chassis
# temperature would be 4.81 * 3 = 14.4 C above ambient. The junction
# temperature at an ambient temperature of 25 C would be 24.1 + 4.8 +
# 14.4 = 43.3 C. The maximum operating ambient temperature would be
# 125 - 14.4 = 110.6 C, (for a junction temperature of 125 C; this
# exceeds the transformer specification of 60 C ambient, meaning a
# maximum ambient operating temperature of 60 - 14.4 = 45.6 C = 114.1
# F.) 0.48 A would allow an additional (0.48 - 0.271) / 0.0115 = 18
# additional LT1115CN8 devices for extensibility.
#
# As a simple reliability model, assume electrical/electronic failure
# is a stochastic, (i.e., a random,) chemical process with a reactrion
# rate that doubles for every increase in temperature of 10 C.
#
# For 25 C ambient, and a junction temperature of 43.3 C, and assuming
# a 2000 Hr. MTBF at 125 C, the MTBF of each regulator would be about
# 2000 * (2^((125 - 43) / 10)) = 2000 * 294 = 588,134 Hr., or about
# 67.1 years of 24/7 operation; about 33.6 years since there are two
# regulators.
#
# For 25 C ambient, and a chassis temperature of 25 + 14.4 C = 39.4 C,
# and assuming a 2000 Hr. MTBF at 125 C, (Note:
# http://www.amveco.com/Technical_Notes_2.htm states that the
# transformer is designed for a 65 C rise at full power rating, and a
# 60 C maximum ambient temperature-from the transformer specification
# on http://www.amveco.com/Miniature_Low_Profile_Transformers.htm-or
# the maximum internal temperature is 125 C,) at an average secondary
# current of 0.5609 A RMS, the temperature rise would be 65 * ((0.5609
# / (1.388 / 1.8))^2) = 34.4 C, (since the power is proportional to
# the square of the current,) or the MTBF would be 2000 * (2^((125 -
#  34.4 - 14.4) / 10)) = 2000 * 196.7 = 393,440 Hrs., which is about
# 44.9 years of 24/7 operation. (It may be more since the
# manufacturer's form factor was used as the reference for 2000 Hrs.,
# and not the secondary current.)
#
# For 25 C ambient, and a chassis temperature of 25 + 14.4 = 39.4 C,
# and assuming a 3000 Hr. MTBF at 105 C, the MTBF of each filter
# capacitor would be about 3000 * (2^((105 - 39.4) / 10)) = 3000 *
# 94.4 = 383,060 Hrs., or about 32.3 years; 16.2 years since there are
# two filter capacitors.
#
# The total system MTBF, for the LM317HVT, the transformer, and the
# two filter capacitors, would be estimated at about 1 / ((1 / 33.6) +
# (1 / 44.9) + (1 / 16.2)) = 8.8 years at 25 C ambient, 24/7 usage,
# and Iout = 0.48 A average continuous current, each regulator mounted
# on a an aluminum sheet, at least 6.56 inches on an edge, single
# sided convection, per device. The diode bridge has a reverse bias
# rating that is at least twice what is applied, and a forward current
# that is 35 / 0.5609 = 62.4 times what is necessary to source 0.48 A.
#
# A remaining issue is fault/exception handling. The LM317HVT has a
# minimum current limit of 1.5 A, a typical of 2.2 A, and a maximum of
# 3.7 A, which is larger than the transformer and filter capacitor can
# handle for an indefinite time; an over load or shorted regulator
# output could damage collateral components. At Iout = 0.48 A, the
# power dissipation of the devices is 2.49 W to 4.97 W, or a junction
# temperature rise, with a 3 C / W, 6.54 inch sheet of single sided
# convection cooled aluminum, of (5 + 1 + 3) * 2.49 = 22.41 C to (5 +
# 1 + 3) * 4.97 = 44.7 C above ambient-probably well below the 125 C
# thermal shutdown of the devices.
#
# Worst case, for fault/exception fusing with a 0.6 A fast acting
# fuse, the power supply must be be capable of regulating at 0.48 *
# 1.25 * 2 = 1.2 A, which is the 200% rating for a 3AG 0.6 A, 250 V
# fuse, which will open in 5 seconds, maximum at 200% current rating
# for the fuse. (See http://www.littelfuse.com/, which recommends
# selecting a fuse rating at 125% of the current rating of the
# circuit, and specifies a 5 second maximum opening time at 200% of
# the current rating for the fuse.) Additionally, the fuse has an open
# time of 4 Hrs., minimum, at 100% of the current rating, 1 Hr.,
# maximum, at 135% of the current rating
#
# Under these conditions, (C = 4700 uF, 0.6 A fuse, Iout = 1.2 A,) the
# minimum filter capacitor voltage is 19.3172 V, (which is larger than
# the 18.86 V minimum to the LM317HVT,) the RMS capacitor current is
# 2.3123 A, (which is smaller than the 3.0 A capacitor ripple current
# maximum for the ECO-S1JA472,) and the RMS transformer current is
# 1.4118 A, (which is 1.017, or about 2%, larger than the 1.388 A
# secondary current maximum for the 62084, but is reasonably close to
# the maximum for a maximum duration of 5 seconds-this is a
# conservative estimation, see
# http://www.amveco.com/Technical_Notes.htm which allows about a 3X
# current rating for 15 seconds per minute.) The fuses are only rated
# at 200% current rating for a 5 second open time, maximum, (the
# maximum at other factors of current rating is unknown,) which means,
# worst case, that the power supply must operate at the fuse 5 second
# open current of 2 * 0.6 = 1.2 A, corresponding to an output current
# of 1.2 A indefinitely. The power dissipation of the LM317HVT would
# be between 1.2 * (19.3172 - 16.35) = 3.56 W and 1.2 * (22.9661 -
# 13.85) = 10.94 W, corresponding to a junction temperature rise of
# from 3.56 * (5 + 1 + 3) = 32.0 C to 10.94 * (5 + 1 + 3) = 98.5 C,
# which would almost be in thermal shutdown for an ambient temperature
# of 25 C, which is at the maximum of the LM317HVT rating. Note that
# the power supply components are all near maximally stressed for the
# 5 second open time of the fuse-a reasonable design under
# fault/exception handling.
# current value, and have to be replaced.)
#
# The alternative is to add collateral circuitry to shutdown the
# LM317HVT regulator under fault/exception situations-which is
# difficult to do since it does not have control shutdown pins; at
# least other than grounding the Adj pin, which reduces the regulator
# output voltage to 1.25 V with a 1.5 A current limit, minimum, (which
# is an inadequate solution for the collateral components.) The 120 V
# RMS input could be interrupted via a solid state relay, perhaps by
# current sensing the transformer current, or disconnecting the output
# of the regulator via a power FET with low on resistance. Almost any
# means would require R/S latching during fault/exception handling,
# and would add circuit complexity, which defeats the advantages of
# using "jiffy" regulators. Fusing for fault/exceptions seems an
# adequate and inexpensive alternative.
#
# Voltage rating of components; maximum output voltage from
# transformer = (1.05 * 20.7) * sqrt (2) = 30.74 V:
#
#     LM317HVT 60 V absolute maximum voltage, 60 / 30.74 = 1.95
#
#     4700 uF electrolytic filter capacitor(s) 63 V maximum voltage,
#     63 / 30.74 = 2.05
#
#     10 uF tantalum filter capacitor(s) 35 V maximum voltage, 35 /
#     16.35 = 2.14
#
#     Diode rectifier bridge, 100 V maximum reverse voltage, 100 /
#     30.74 = 3.25
#
# Or about a factor of 2 = 100% safety margin on component voltages.
#
# Current rating of components:
#
#     Transformer 1.388 A RMS per secondary, 1.4118 A for 5 seconds
#     maximum fuse opening time, a minimum of 4 hours at 0.6 A, 1.388
#     / 0.7021 = 1.98, a maximum of 1 hour at 0.6 * 1.35 = 0.81 A,
#     1.388 / 0.9498 = 1.46
#
#     Diode rectifier bridge, 35 A maximum forward current, 35 /
#     0.5609 = 62.4
#
#     Diode rectifier bridge, 35 A maximum forward current, in rush
#     current ~ ((1.05 * 20.7 * sqrt (2)) - (2 * 1.1)) / ((20.7 - 18)
#     / 1.388) = 14.7 A, 35 / 14.7 = 2.4
#
#     LM317HVT 1.5 A to 3.7 A minimum output current before thermal
#     shutdown or current limit, 1.5 A for 5 seconds fuse opening,
#     minimum of 4 hours at 0.6 A, 1.5 / 0.6 = 2.5, a maximum of 1
#     hour at 0.6 * 1.35 = 0.81 A, 1.5 / 0.81 = 1.85
#
#     4700 uF electrolytic filter capacitor 3.0 A ripple current,
#     2.3123 A for 5 seconds fuse opening, minimum of 4 hours at
#     1.1965 A, 3.0 / 1.1965 = 2.5, a maximum of 1 hour at 0.6 * 1.35
#     = 0.81 A, 3.0 / 1.5940 = 1.88
#
#     10 uF tantalum filter capacitor(s) N/A ripple current
#
# Or about a factor of 2 = 100% safety margin on component currents,
# fused for 5 second open on manufacturer's minimum operating maximum
# current, and a factor of 2, (e.g., sqrt (2 * current),) in power
# dissipation for a maximum of 1 hour.