Reactive impedance question

Discussion in 'Electronic Basics' started by Tim Williams, Oct 3, 2005.

1. Tim WilliamsGuest

I need to find the impedance (real, reactive) of the following network:
L +--/\/\/\--+
Vin o----333---+ R +---o GND
+----||----+
C

(R || C) in series with L. At resonance.

In particular I need to know the impedance presented to Vin as a function of
L, C and R, so I can determine the load on an inverter driving this circuit
with a square wave, and the impedance transformation.

So far I've been meyered in phasors, and in attempting to connect series
current with parallel voltage in terms of vectors I get a divide-by-vector
ugliness that obviously can't work...

Tim

2. John PopelishGuest

But a square wave is not a frequency, but a superposition of a great
many frequencies. Are you going to solve for the response of the
whole batch and combine those results to get a total current? Seems
like a differential equation approach might be more useful (or a spice
simulation doing that analysis).

3. Tim WilliamsGuest

True, but the next harmonic is the 3rd, easily blocked by inductance above
resonance.

Tim

4. John PopelishGuest

Okay, if you are only interested in the fundamental, lets see what we
know.

The voltage across the capacitor is identical to the voltage across
the resistor: Vr=Vc

The current through the capacitor is at right angles to the current
through the resistor, and the vector sum of those two must equal the
inductor current, since it is in series.
By Pythagorous: Il^2=Ic^2+Ir^2

The vector sum of either the resistor voltage or the identical
capacitor voltage and the inductor voltage is equal to the source
voltage. Vs=Vr+Vl or Vs=Vc+Vl

The inductor current and capacitor currents are at right angles to
their voltages (in opposite directions).

Since you are assuming resonance, the magnitude of the ratio of
inductor voltage divided by inductor current equals the magnitude of
the capacitor voltage divided by the capacitor current:
|Vl/Il|=|Vc/Ic|
or |j*(2*pi*f*L)|=|-j/(2*pi*f*C)|
or 2*pi*f*L=1/(2*pi*f*C

Unfortunately, I am too out of practice to combine these facts in a
phasor diagram so all I can suggest is a pair of loop equations solved
for the two complex currents, Ia and Ib.

Vs=Ia*[(j*2*pi*L)+R] -Ib*R
0=Ib*[R-j/(2*pi*f*C)] -Ia*R

and at resonance, I assume 2*pi*f*L=1/(2*pi*f*C)

Ztotal = Vs/Ia

But then, Ztotal = j*2*pi*f*L + 1/[(1/R) + (2*pi*f*C)/(-j)]

However, I would just plug the component values into LTspice and see
what current the simulation passes per volt of source.

5. Fred BartoliGuest

Why bother with phasors?

Simply fromù inspection:

R * 1/(j C w)
Z = j L w + ---------------
R + 1/(j C w)

which gives

Z = j L w + R / (1 + j R C w)

Where's the pb?

6. John - KD5YIGuest

Z(S) = LS + R/(RCS + 1)

7. Tim WilliamsGuest

"Fred Bartoli"
Uh, the lack of Pythagorean stuff, unless you can prove it.

Tim

8. John PopelishGuest

Pythagoras (the right angles) is hidden in those j's.

This is just Z1 in series with the parallel combination of Z2 and Z3.
Z1=jwL
Z2=R
Z3=1/jwC or -j/wC

9. Tim WilliamsGuest

Ah, well then. Imaginary numbers are fine for imagining things, but I need
Yes, yes it is.

Tim

10. Fred BartoliGuest

Tim,
Sorry if this sounds a bit rude (I don't intend to be) but I guess it's time
for you to read a good text on the basics.
This is basic stuff and if you don't understand this you're overcomplicating
your life with the "Pythagorean stuff" and probably calling for trouble with

11. You can eliminate the j's and get an expression for the magnitude of
the impedance with an applied sine wave(f is the sine wave frequency):

(4*f^2*L^2*Pi^2+R^2-8*C*f^2*L*Pi^2*R^2+16*C^2*f^4*L^2*Pi^4*R^2)
SQRT(-------------------------------------------------------------)
( 1+4*C^2*f^2*Pi^2*R^2 )

Looking at this ugly expression, you can see why using the complex
expression directly is easier and less mistake prone. (These days
there are calculators and computer programs that can do complex
arithmetic.)

Look up the magnitude of the harmonics in a square wave (the
magnitude of the fundamental is *not* just the amplitude of the square
wave; it's 4/Pi times the square wave amplitude) and compute the
current for several harmonics, with the first being your square wave
(resonance?) frequency, the next 3 times the frequency of the first
and 1/3 the magnitude of the first, etc., remembering that a square
wave has only odd harmonics. Add these currents root-mean-square
wise; that is, take the square root of the sum of the squares of the
individual harmonic currents, and that will be your RMS current with
square wave drive.

My quick calculation indicates that because of the series L
attenuating the harmonics, the fundamental current dominates, and only
a few harmonics are needed to get a very good result.

12. Tim WilliamsGuest

"Fred Bartoli"
In fact I haven't taken a single electronics class (that comes later, if the
prof says I even need to take it LOL). Over the years I've escaped phasors,
nasty reactives and phase shifts by winging it, and so far it's worked quite
well. No I don't build anything finicky in those regards without testing
and adjusting to account for my SWAGs. ;-)

I've had plenty of math classes (and finished near or at the top of all of
them), but we never did imaginary numbers outside of "sqrt(-1) = i and you
can't do anything about it except maybe a conjugate pair".
Since I picked up and read a textbook in depth two days ago I've gone from
blank stares to understanding what a phasor is doing, but that doesn't help
with a complex circuit like this. The only examples I've seen are
completely series or completely parallel circuits, not a combination like
this.

Hell, even the professionals on SEB can't seem to get a straight
answer...(except Phantom, I'm looking it through, thanks).
Nah. I've already got a working (if unprotected and open-loop) model that
would run probably a kW, I haven't taken it that high yet but I haven't
hooked up water cooling for the work coil, either. The only obstacle is
matching the load to the inverter using this series resonant matching
circuit, which although easy to treat qualitatively, is proving a bear
quantitatively. At present the only SWAG I have is to figure impedance
equal to Lmatch, which works for Vtank = 0V but falls apart, especially as
resonant voltage rises above the inverter's output (Q multiplication, which
if not for limiting circuitry and 400V-rated caps would push 10kW through
1/4" copper tubing until it melts at some 6,000VAC.)

Tim

13. Tim Williams wrote in message ...
Solve for admittance of the parallel RC section
Yrc = sqr(G^2 + Bc^2) where G = 1/R and Bc = 1/Xc

The reciprocal of the admittance is the impedance magnitude
of the parallel RC section
Zrc = 1/Yrc

The phase angle of the RC section is -
arctan Bc/G As an impedance , the phase angle is negative

In polar form, the RC section impedance is -
1/Yrc ohms at negative arctan(Bc/G) degrees

Convert the polar form of the RC section to rectangular form
so it is easier to add to the inductor. This is an
equivalent series circuit for the parallel RC section.
Rs = cos theta x Zrc
Xcs = sin theta x Zrc
In rectangular form, the eqivalent series RC impedance is
Rs - JXcs

Now add the series inductor to the equivalent series RC section
using rectangular forms for both sections.
(Rcoil + jXl) + (Rs -JXcs)
Sum both R's to get the resistive component and sum
both X's to get the reactive component

I added Rcoil to the inductor because it is there.

This is not difficult but it is tedious and easy to make a mistake.
Being a square wave, you would have to do this calculation at several
frequencies, and then sum the currents sqr(I1^2 + I2^2 +I3^2 etc).
Personally I'd find a way to cheat!

14. Tim WilliamsGuest

I've been looking at this today. Some algebra finds:
sqrt[1 - 2(C/L) + (2piFC)^2]
Z = 2piFRL ------------------------------
sqrt[1 + (2piFRC)^2]

I may be wrong on something in that long expression, but at any rate, in
your original equation, Z goes to zero as R goes to zero, no matter what the
frequency is (R^2 is a common term in the numerator radical). I know
matter-of-factly this circuit will show an R-limited peak in current (i.e.,
Z local minima) at resonance, whereas the above equation ultimately has F to
the first power (showing no resonance).

And you're certain that the above equation is the real derivation of the
imaginary equation, which is also correct? This is rather curious...

Tim

15. This would be correct if you hadn't taken the operator in front of the first R^2 in the
numerator to be a * instead of a +.
Look more closely; the 4 terms in the numerator are:

4*f^2*L^2*Pi^2

+R^2

-8*C*f^2*L*Pi^2*R^2

+16*C^2*f^4*L^2*Pi^4*R^2
It is correct.

16. Tim WilliamsGuest

Huh? Ah, so it is! Thanks for that...(I'm used to equations with spaces

Tim

17. And I usually put spaces around the operators,
but I wanted to get it all on one line. 18. Fred BartoliGuest

You know how to serie or parallel resistances. Just do the same with complex
impedances and build the overall impedance one step after another. That's as
simple as that.

Ah, but I missed the "at resonance" part.

So, when you have the impedance written down

Z = j L w + R / (1 + j R C w)

you eliminate the second term denominator imaginary part by multiplying it
by

(1 - j R C w)/(1 - j R C w)

Then compute the frequency that reduces the transformed Z imaginary part to
zero.

Inject this frequency into Z and you're done.

What does that last part mean?

19. Tim WilliamsGuest

"Fred Bartoli"
Ah, very good.
I mean if the tank capacitor could withstand open circuit voltage, the Q of
20 to 30 would take the 320Vp-p (hm, that's only 140V or so RMS) and
multiply it to around 4kV when it reaches equilibrium. At this point the
tank would be dissipating 10kW in its own loss.

Anyways, yesterday I hit something of a breakthrough and related the
currents, voltages and impedances at the proper angles to each other. I
managed to work it through and like magic, the sky opened forth and gave me
this: (For convienience I'm showing Z^2; Z is sqrt(blah).

Oh, and I realized that if I want to look at Z for F != Fo, I need
to include the tank coil as well. Hence the parallel circuit is now
L2 || C || R, with L1 in series from Z. Here L1, L2 and C represent the
/reactance/ of these components (e.g., "C" = 1/2piFC).

R^2*L2^2*C
Z^2 = ---------------------------------------- + L1^2
L2^2*C + R^2*L2^2*C + R^2*C - 2*R^2*L2

Also I calculated the voltage "Vx" (tank voltage) with respect to Vz (input
voltage) and R, C, L1 and L2.
Vx*R*L2*C
Vx = ---------------------------------------------------------------
sqrt[R^2*L1^2*C^2 + L1^2*C^2 + R^2*L1^2 + R^2*C^2 - 2*R^2*L1*C]
Thus real power dissipated in the effective loss resistance is
P = Vx^2 / R
Which I'm not going to write here.

Tim

20. Tim WilliamsGuest

(Oops it looks like I wrote Vx, not Vz in the numerator! )

I re-evaluated this (with ever-increasing tiredness) and now I get:
(Writing in terms of X(reactive) this time for clarity; space with no
operator means multiplication.)
Vz R XL2 Xc
Vx = --------------------------------------------------------------
sqrt(R^2 XL2^2 Xc^2 + XL1^2 R^2 Xc^2 + XL1^2 R^2 XL2^2 + ..
.. + XL1^2 XL2^2 Xc^2 + XL1^2 XL2^2 Xc^2 - 2 XL1^2 R^2 XL2 Xc)

(Yeah, so it splits the line, it's clear. <g>)

Thing is, I plug this into my graphing program, and as R > infinity
(practical values over 50 ohms, with L1 around 50uH, L2 around 10uH,
C around 20uF and F in kHz), the resonant peak only approaches Vz! Now
what's wrong!?

Tim  