Original poster: "Jared Dwarshuis" <jdwarshuis@xxxxxxxxx>
Impedance, Mechanical, Electrical
I wrote this to help the novice coiler
understand a rather difficult topic. There is
nothing familiar or intuitive about the
mathematical approach to understanding
impedance. Brains are not "wired from the
factory" for grasping second order differential
equations and imaginary numbers. Such stuff
takes practice, and faith! Fortunately there are tangible,
" hands on" physical objects that can be related
to. This will help us to resolve the mystery.
So why should I care about this topic, where is the impetus ?
Always a valid question!
Hmmm, well??.
A Tesla coil should have a power factor correction for the inductive "load"
A Tesla coil might need radio frequency suppression, a "choke"
Optimal transfer of energy to the load will
require careful impedance matching.
You need to be armed with some theoretical
weaponry, or none of this will ever make any sense.
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When we have a mass tied to the end of a spring
we have an object that behaves mathematically
very much like a series inductor and capacitor.
Capacitors are like springs. Theoretically you
could store charge forever just like you could
keep a spring under tension forever. Both cases concern potential energy.
Inductors are very much like a mass. Lenz law
and Newton second law of motion are essentially
equivalent. Both cases concern forms of kinetic energy. (energy of motion!)
Since the lumped model of LCR and dampened
mass/spring share the exact same energy
equations, they also share the same equations
for impedance. The amplitude equations which
describe how much mass and spring bob up and
down along the X axis are very old, I don't know
who first wrote them. Here I will demonstrate
that the familiar impedance formula can actually
be derived from mass spring amplitude equations.
M dv/dt + CV +KX = F e(iwt) (where C is Kg/s)
L di/dt + RI + Q/C = V e(iwt) (where this C is capacitance)
Let:
X(t) = A e i(wt-phi)
Q(t) = Ae i(wt-phi)
Then:
M d2/dt2 A e i(wt-phi) + C d/dt A e i(wt-phi) + K A e i(wt-phi) = F e iwt
Where: M = Kg C = Kg/s K = Kg/ss F = Newton
L d2/dt2 A e i(wt-phi) + R d/dt A e i(wt-phi) + 1/C A e i(wt-phi) = V e iwt
Where: L = Kg (m/coul)sqrd R = Kg/s (m/coul)sqrd
1/C = Kg/ss (m/coul)sqrd V = N (m/coul)
Equating the real and imaginary parts of the
previous equations, squaring both sides, using
the trigonometric identity: (sine phi) sqrd + (cos phi)sqrd = 1
We get:
X = F / ((K- M(w)sqrd)sqrd + Csqrd(w)sqrd))^1/2
Q = V / ((1/C- L(w)sqrd)sqrd + Rsqrd(w)sqrd))^1/2
(w) can be pulled from the denominator.
Then:
X(w) = Velocity = F/ [ ( K/(w) M(w) )sqrd + Csqrd ]^1/2
Q(w) = Current = V/ [ ((w)L 1/(w)C )sqrd + Rsqrd ]^1/2
We recognize impedance in the denominator for both systems
On the surface it may seem a bit odd that X
corresponds to Q, but in reality both are just
describing displacements from an equilibrium
value. In the case of a dampened mass/spring,
the equilibrium is simply where the device comes to rest.
So far a lot of math. But what does it all mean??
Lets look at the easiest part first, namely
resistance. (named C in the mechanical world.)
C in the mechanical world has units of kg/s.
When we push a box across a uniform floor at a
constant velocity we can describe this with a
simple force equation. Force = velocity x C The
units are: Kg m/ ss = m/s x kg/s (C describes how slippery the floor is)
R in the electrical world is analogous. Voltage
takes the roll of force. Current becomes
velocity (moving charge) and R takes the roll of C.
*C and R are both linear in these models, not
always true in either the electric or mechanical worlds.*
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Now an informal look at reactance!
Let us imagine a smooth block attached to a
spring which is at the other end anchored to a
wall such that our smooth block can slide on a slightly rough floor.
We can always wiggle and jiggle the contraption
to find exactly when to push, so as to build up pushes for maximum jiggle.
Why is that?
Well suppose the block is really massive but
attached to a bitty joke of a spring.
Question?
Since it is attached to a bitty spring is it ok to give it a good kick?
Answer!
No! You will break your foot, nearly all of the
force is going into moving the large mass
Ok! Now push the contraption slowly. Did you
will find that nearly all the force went into the collapsing the spring?
This is the essence of reactance!
When we found our maximum jiggle, (likely
without realizing it) we had found the perfect
compromise point between pushing slowly and
giving it a good kick. And of course since we
are talking about different amounts of time
between pushes, we are actually talking about
frequency. (And so it is!, reactance is frequency dependant.)
We have not exhausted the topic, there is also
the matter of phase angle. Also there are
several other forms of impedance that we have
not even talked about. But enough for now.
End.
Jared Dwarshuis w/07