Re: Tanks, impedance, the j operator and excel


"AnimalMagic" <AnimalMagic@xxxxxxxxxxxxxxxxxx> wrote in message
news:tb0n34pj1ihc6jjbamm3ccihus7pm7kltv@xxxxxxxxxx
On Sun, 25 May 2008 20:58:28 -0500, "Bob" <nobody@xxxxxxxx> wrote:

Hi gang,
This one is directed at you math whizzes..... I have a formula for tank
circuit impedance that I am attempting to analyze/plot in MS Excel. The
tank
consists of a resistor in series with the coil, a cap in parallel with the
coil resistor combination and a another resistor in parallel with the cap
and coil/resistor.
The formula consists of two quotients; One has a numerator that consists
of
several terms multiplied, added, squared, etc and the denominator is
similar. The second quotient is preceded by the "j" operator. (That is the
entire quotient appears to be "j"ed rather than any single term or
factor).
This quotient has a numerator and denominator similar to the first).
I assume that the first quotient is the real part of the impedance, while
the second one is the imaginary part, yielding the rectangular Z =R +/-j
form. Does this sound correct?
Assuming the math is done correctly, I believe I can use the IMABS
function
on the result to get the magnitude of the impedance. Yes?
In Excel, I have entered the first quotient as ordinary numbers, i.e.
"=A1*(C2+B3)..../D4*6.28*C3......"
I'm getting numbers that seem to make sense.
The second quotient has me scratching my head a bit. I have tried using
the
same process on this part, and then using "=IMABS(first quotient
results,+second quotient results)" to get the magnitude of the impedance
but
I think this may be incorrect.
Any thoughts? If someone wants it, I can post my Excel file to them.....

Thanks

Bob WB0POQ



You should post the spread*** workbook up in
alt.binaries.schematics.electronic

Then, we can all have a look at the real thing.

I am pretty good with excel, and this sounds pretty simple.

If you just post the entire formula for each element of your equation,
it can be worked out as well.

A diagram of the tank circuit will work as well.

Ok....here is the formula:

Z = Ry(RxRy+Rx2+w2L2) wRy2(L-CRx2-w2L2C)
---------------------------------------------------
+j --------------------------------------------------
(Rx+Ry-w2LCRy)2+(wL+wCRxRy)2 (Rx+Ry-w2LCRy)2+(wL+wCRxRy)2


Rx is the resistor in series with the coil (represents coil losses)
L is the coil
C is the capacitor in parallel with the coil/resistor network
Ry is the resistor in parallel with the capacitor coil/resistor combination
w is 2*PI*F

Sample values:
Rx = 0.2836 Ohms
L = 0.868uH
C = 335.2pF
Ry = 11 k Ohms
Fr = 9.336MHz



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