maxima-circuits

Maxima functions for circuit calculations

See also the accompanying Hackaday article.

Usage

Load the functions using

load("circuits.mac")$

You may also place this line in your startup script (typically ~/.maxima/maxima-init.mac or ~/.maxima/wxmaxima-init.mac)

par(...)

Calculate reciprical of sum of reciprocals of supplied arguments, aka the harmonic sum. This calculates equivalents for parallel resistor or inductor values, or for series capacitors.

Examples:

(%i2) par(100, 75);
(%o2)                          42.85714285714285

(%i3) par(r1, r2, r3);
                                      1
(%o3)                            ------------
                                 1    1    1
                                 -- + -- + --
                                 r3   r2   r1

vdiv(r_top, r_bot)

Calculate output of a voltage divider as a ratio of the input voltage.

Examples:

(%i4) 5*vdiv(75, 150);
(%o4)                          3.333333333333333

(%i5) vdiv(r1, r2) = 3.3/5;
                           r2
(%o5)                    ------- = 0.6599999999999999
                         r2 + r1

pref(x, E)

Find preferred values from EIA standard "E series." The following values for E are allowed:

E3      : 50% tolerance
E6      : 20% tolerance
E12     : 10% tolerance
E24     : 5% tolerance
E48     : 2% tolerance
E96     : 1% tolerance
E192    : <1% tolerance
E48_24  : combined vales from E48 and E24 series
E96_24  : combined vales from E96 and E24 series
E192_24 : combined vales from E192 and E24 series

Examples:

(%i6) pref(50, E3);
(%o6)                                47.0
(%i7) pref(50, E24);
(%o7)                                51.0
(%i8) pref(50, E96);
(%o8)                                49.9

VrtodB(Vr)

Convert voltage ratio to dB

Example:

(%i2) VrtodB(sqrt(2));
(%o2)                          3.010299956639812

dBtoVr(dB)

Convert dB to voltage ratio

Example:

(%i6) dBtoVr(6);
(%o6)                          1.995262314968879

Comprehensive Example

Let's design a 6dB pi attenuator for a 50-ohm system. The circuit looks like this:

First, we write an expression for the output impedance:

(%i14) Zout: par(r2, r1+par(r2, 50));
                                       1
(%o14)                        -------------------
                              1          1
                              -- + --------------
                              r2       1
                                   --------- + r1
                                   1
                                   -- + 0.02
                                   r2

next, an expression for the input to output voltage ratio (Vout/Vin):

(%i15) Vout: vdiv(r1, par(r2, 50));
                                      1
(%o15)                   ----------------------------
                              1            1
                         (--------- + r1) (-- + 0.02)
                          1                r2
                          -- + 0.02
                          r2

Using these expressions, we create a system of two equations and solve for the resistor values:

(%i3) soln: solve([Zout = 50, Vout = dBtoVr(-6)], [r1, r2]);
(%o3) [[r1 = 37.35187703354015, r2 = 150.4760237537246], [r1 = 0, r2 = 0]]

The first solution is the one we're interested in. Now, we choose resistors from the E12 (10%) series:

(%i4) vals: pref(first(soln), E12);
(%o4)                       [r1 = 39.0, r2 = 150.0]

and evaluate the result:

(%i5) ev([Zout, VrtodB(Vout)], vals);
(%o5)              [50.66225165562914, - 6.192603348517975]

Let's see how much better 1% values are:

(%i6) vals: pref(first(soln), E96);
(%o6)                       [r1 = 37.4, r2 = 150.0]

How does this look?

(%i7) ev([Zout, VrtodB(Vout)], vals);
(%o7)              [49.95553579368609, - 6.009010999434952]

Close enough.