Impedance, Mechanical, Electrical

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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