Re: Calculating RC networks the easy way.



On Sun, 08 Jun 2008 05:48:30 GMT, "Alex" <apogosso@xxxxxxxxxxx> wrote:


"The Phantom" <phantom@xxxxxxx> wrote in message
news:mr1m44tbq9m832tnvf4l80k41jj54u17gs@xxxxxxxxxx
Patrick Turner said:

"Reactive network math is anything but easily understandable."

then Doug Bannard said:

"Reactive network math is really not at all difficult."

to which Patrick replied:

"Oh yes it IS.

I don't know ANYONE with any clue how to calculate networks.

They all get stumped on 'j' and square root of -1, and other utterly
incomprehensible."

Let's see if we can help.

First, a little bit of history. If Patrick thinks it's difficult now, he
should see how it was done in the early days of electricity.

It was discovered early that the 3 classical 2-terminal circuit elements,
resistors, capacitors and inductors, caused a relationship to exist
between
the current through them and voltage across them:

Component e/i relationship

R e = i*R

de
C i = C*--
dt

di
L e = L*--
dt

where lower case e and i represent e(t) and I(t), time varying functions
of
voltage and current.

You can see right away that calculus is going to be involved here, and in
fact the way AC circuits were solved in the beginning was to set up the
differential equations of the circuit and solve them.

But, happily, it turns out that for steady state analysis there is a
shortcut. Around the 1890's, a man named Charles Proteus Steinmetz
immigrated to the United States and went to work for General Electric. He
realized that if you have a network composed of RLC elements which are
linear (this lets out inductors with magnetic cores unless the flux
density
in the core is kept well below saturation, and capacitors with certain
ceramic dielectrics, for example) and non-time-varying, the circuit
differential equations will have constant coefficients. Therefore the
solution will always be sums of exponentials with complex arguments.

This means that what would have been a thorny problem in differential
equations becomes a much easier problem in simple algebra, but with
complex
numbers.

Complex numbers are numbers of the form a + j*b, where a and b are
ordinary
numbers and j is the square root of -1. (Most everybody except Electrical
Engineers uses i to represent SQRT(-1), but i was already used for current
by EE's, so they use j for SQRT(-1).) When complex numbers were first
discovered (invented?), they seemed mysterious, and the a and b parts of
the complex number were given the unfortunate names of "real part" and
"imaginary part". There is nothing "imaginary" about SQRT(-1); it just
adds some additional complication to the arithmetic.

Complex numbers can be expressed in what is called rectangular form, a +
j*b, or in polar form r<theta (normally in a good text, the actual Greek
letter theta would be used, rather than spelling it out as I did here).
The letter r is the magnitude of the complex number, and is equal to
SQRT(a^2 + b^2), and theta is the angle of the number.

Here's the good part. The methods used to solve DC circuits, which only
involve resistors, can also solve AC circuits if you just use complex
numbers. Most readers will know how to calculate the equivalent
resistance
of several resistors in parallel or in series. And many will know how to
solve a circuit with a combination of series and parallel resistors.

The concept of resistance can be extended to AC circuits. Where DC
circuits have only resistors, AC circuits have resistors, capacitors and
inductors. The property of resistance which characterizes resistors is
called "reactance" when capacitors and inductors are considered. It is
the
property of "opposing" the current in the resistor, capacitor or inductor.
Sometimes it is easier in the arithmetic to deal with the reciprocal of
that property, and it is then called "conductance" (because rather than
opposing current, the reciprocal version "allows" it more as the
conductance value increases, which corresponds to a decrease in
resistance)
in connection with resistors, and "susceptance" in connection with
capacitors and inductors.

There are conventional symbols used for the resistance of resistors and
the
reactance of capacitors and inductors, and the conductance and susceptance
of those components:

Component Symbol

Resistor R (resistance) G (conductance)

Capacitor X (reactance) B (susceptance)

Inductor X (reactance) B (susceptance)

Sometimes Xc is used to denote the reactance of a capacitor, where it is
wanted to distinguish it from an inductor's reactance (and similarly Xl
for
an inductor).

When a circuit has resistors together with capacitors (and/or inductors),
the property of the combination of those components in opposing current is
called "impedance", and its reciprocal version is called "admittance".
For
doing arithmetic, the impedance is composed of the sum of the resistance
and the reactance multiplied by j, and similarly the admittance:

Property Formulation

impedance = resistance + j*reactance
Z = R + jX

admittance = conductance + j*susceptance
Y = G + jB


Just as the conductance of a resistor is the reciprocal of the resistance,
the susceptance of a capacitor or inductor is the reciprocal of the
reactance. The admittance is also the reciprocal of the impedance, but
this is NOT calculated by simply taking the reciprocals of the individual
parts (R and X) of the impedance; it has to be done according to the rules
of complex arithmetic.

The reactance of capacitors and inductors varies with frequency (f or F is
used as a symbol for frequency), and the lower case Greek letter omega is
used to represent the quantity w = 2*pi*f (I'll use lower case W, which
looks a lot like lower case omega), which is used to calculate reactance.
There are simple formulas for the reactance and susceptance of capacitors
and inductors (notice that 1/j = -j):

Component Reactance Susceptance

1 -j
C ----- or ----- j*w*c
j*w*C w*C

1 -j
L j*w*L ----- or -----
j*w*L w*L


Now we get to the meat of it. You don't have to understand how the rules
of complex algebra work, or why they are used in order to do calculations.
The key to it lies in the power of modern electronics. You get yourself a
calculator that can do complex arithmetic. I highly recommend either a
Hewlett-Packard or Texas Instruments calculator; some of the Casio, Sharp,
etc., low cost calculators can do it, but not as conveniently as the
somewhat more expensive HP's and TI's. The HP50G and the TI89 are good
candidates and can be had on eBay for about US$100, more or less.

The only new things you will have to remember are the formulas just above
for reactance and susceptance. You should already know how to combine
series and parallel resistors.

For AC circuits, you use the same methods that you use for DC circuits;
you
just have to do it with complex arithmetic. With a suitable calculator,
you don't have to understand how complex arithmetic works; the calculator
does it all for you.

You can select the calculator's mode so that the complex numbers are
displayed in rectangular form or in polar form.

I'm most familiar with the HP50 calculator, so I'll give some examples
using it. The HP50 is a stack oriented machine. It can hold several
numbers in a stack shown in the display, and can then operate on those
numbers with various mathematical functions, including the four simple
arithmetic operations.

The HP50 in rectangular mode displays a complex number 3 + j*4 like this:
(3.,4.)

The complex number is in parentheses, with the real part first and the
imaginary part second; the j is assumed to precede the second part, but is
not shown.

Example 1:

Calculate the impedance of a 1000 ohm resistor in parallel with a .01 uF
capacitor, at a frequency of 5000 Hz.

The well known product over the sum formula can be used. First type 1000
(the resistor value) and then press the enter key to put the value 1000 on
the stack. Now put zero on the stack to represent the real part of the
capacitor's reactance. Calculate 2*pi*5000*.01E-6 and press the 1/x key
on
the calculator and then the +/- key; you now have -1/(w*c); I get a
numerical value of -3183.1.

At this point you have zero on the stack and -1/(w*C) on the stack below
it. Use the calculator's R->C function to combine the two into a complex
number which looks like (0.,-3183.1); this is the reactance of the
capacitor. It would be a good idea to create a variable and save this
number in it for further use.

Remember that we put 1000 on the stack first, so right now we have 1000 on
the stack and the reactance of the capacitor on the stack. We're going to
use the product over the sum formula, so press the calculator's X key to
multiply the two; I get (0.,-3183098.86).

Leave that number on the stack, and re-type 1000 for the value of the
resistor. Recall the reactance of the capacitor from the variable where
you stored it. Now press the + key to get the sum of the resistance and
the reactance; I get (1000.,-3183.1).

At this point, we have the product on the stack and the sum on the stack
below the product. Now press the / key to calculate the product over the
sum; I get (901.17,-285.939). This means that the real part of the
impedance is 901.17 ohms and the imaginary part is -285.939 ohms, or
901.17
- j*285.939.

If you switch the calculator to polar mode and degrees mode, you will see
(954.028,<-17.44), which means the magnitude of the impedance is 954.028
ohms at an angle of -17.44 degrees. Switch back to rectangular mode.

Example 2:
Calculate the impedance of a 1000 ohm resistor in series with a .01 uF
capacitor at a frequency of 5000 Hz.

Leave the previous value of the parallel combination on the stack.

Type 1000 to put 1000 on the stack below the parallel value. Recall the
value of the capacitor reactance from the variable where it was save in
Example 1. Press the + key to get the impedance of the series
combination;
I get (1000.,-3183.1). Change the calculator mode to polar and see
(3336.48,<-72.56); the magnitude of the impedance is 3336.48 ohms at an
angle of -72.56 degrees.

The well known rule that for resistors (impedances) in series, resistances
add, has a counterpart involving conductances (admittances).

The product over the sum formula for the parallel equivalent of two
resistors (or impedances) is well known. Another method for any number of
parallel resistors is to convert the resistances (impedances) to
conductances (admittances), add the conductances and then convert back to
resistance.

For resistors (impedances) in parallel, their conductances (admittances)
add. Just remember that the conductance (admittance) is the reciprocal of
the resistance (impedance). The product over the sum formula can be
derived from this rule.

Example 3:

The impedance of the parallel combination of a 1000 ohm resistor and a .01
uF capacitor is still on the stack along with the impedance of the series
combination below it. Let's use the admittance method to find the
impedance of the parallel combination of the two impedances on the stack.
Press the 1/x key; use the swap function to exchange the two stack entries
;
press the 1/x key; press the + key; press the 1/x key.

We have now taken the reciprocal of the sum of the reciprocals of the two
impedances that were left on the stack from Example 1 and Example 2. This
is the impedance of a series combination of 1000 ohms and a .01 uF
capacitor at 5000 Hz, that combination in parallel with a 1000 ohm
resistor
and another .01 uF capacitor. I get a value of (704.09,-387.7) in
rectangular mode, or (803.78,<-28.84).

So we see that if we use a calculator that can do complex arithmetic, we
don't have to know how complex arithmetic works, and we can solve RC
networks by using the same rules we used for DC circuits. The only extra
thing we need to know is the formula for the reactance of a capacitor and
an inductor. This is easy to remember.

The voltage divider formula also works for RLC networks; just use complex
arithmetic on the calculator. All the other DC formulas work, too. The
Thevenin equivalent for an RLC network can be found in the same way as for
a DC circuit, etc., etc.

Everything is simple in theory, but complex in practice.
If you are connecting , say, several series RC networks in parallel, each
time you convert from impedamces to admittances, you need to multiply your
equation by a complex "mirror" value of the denominator

No, you don't need to do this. All you need to do is press the 1/x key which
calculates the reciprocal. The calculator takes care of the details of the
complex arithmetic. It is not complex (no pun intended) in practice. That's
the whole point of this thread. With the appropriate calculator, complex
arithmetic is no more difficult than ordinary arithmetic.

(I do not know the term in English, it has the same Re part and opposite Im part).

The term is "conjugate".

It is a
nightmare, and it is very easy to make a mistake.

It isn't a nightmare if you use this calculator. It is no easier to make a
mistake than when carrying out the same calculations for a DC circuit where the
only components are resistors.


It would be much easier if there existed a version of Excel spreadsheet
which handles complex values.

I disagree; it would not be much easier. The calculator handles complex values
just the way it should and is easier to use than Excel. It can't be any easier
to handle calculations involving complex numbers than with this calculator. The
only thing that is any more difficult than DC calculations is that you may have
to type in two parts (real and imaginary) for the value of a component, but once
the values are in the calculator, the arithmetic procedures are identical to
what you would do with a DC circuit.


Does such Excel exist?

Regards,
Alex


.



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