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Re: How do you measure couplin
Original poster: "Antonio Carlos M. de Queiroz" <acmq-at-compuland-dot-com.br>
Tesla list wrote:
> Original poster: Peter Lawrence <Peter.Lawrence-at-Sun.COM>
> is there an exact mathematical formula for the voltage and
current in
> the primary verses time in a TC (a hypothetical one with no losses) (ie a
> losely coupled dual resonator).
>
> I'm guessing its proportinal to either sin(C*t)+sin(D*t) or
sin(C*t)*sin(D*t)
> for suitable constants C,D that depend on Fres and K...
I prefer to make this analysis by synthesis:
Start with the desired mode, a:b, and two of the elements and
calculate the required element values making L1*C1=L2*C2.
The coupling coefficient is k12=(b^2-a^2)/(b^2+a^2).
This coil will oscillate at the frequencies b*w and a*w, where
w=(1/(a*b))sqrt((a^2+b^2)/(2*L2*C2)) rad/s.
The primary voltage is:
v1(t)=v1max*(0.5*cos(a*w*t)+0.5*cos(b*w*t))
To find the primary current, just calculate the current in the
primary capacitor with this voltage waveform (i1(t)=-C1*dv1/dt):
I1(t)=v1max*C1*(0.5*a*w*sin(a*w*t)+0.5*b*w*sin(b*w*t))
It's equally easy to find the secondary waveforms. Consider that
v2max=v1max*sqrt(C1/C2), by energy conservation, and that the
waveform vc2 has the same shape of vc1 (with maximum vc2max), but
starts at zero instead of the maximum value.
v2(t)=v2max*(0.5*cos(a*w*t)-0.5*cos(b*w*t)).
i2(t)=v2max*C2*(0.5*a*w*sin(a*w*t)-0.5*b*w*sin(b*w*t))
Ex: C1=10nF, L1=100 uH, L2=100 mH, C2=10 pF, mode 9:10, v1max=10 kV.
w=(1/90)*sqrt(181/(2*1e-12))=105.7 krad/s (16823 Hz)
A=a*w=951.31e3 (151.5 kHz)
B=b*w=1057.0e3 (168.2 kHz)
k12=19/181=0.1050
v1(t)=5000*(cos(A*t)+cos(B*t)); Maximum=10 kV
i1(t)=5.2851(9*sin(A*t)+10*sin(B*t)); Maximum=~100 A
v2(t)=158114*(cos(A*t)-cos(B*t)); Maximum=316 kV
i2(t)=167.13e-3*(9*sin(A*t)-10*sin(B*t)); Maximum=~3.16 A
Looks correct. The output voltage has a negative maximum for a
noninverting transformer.
I had not observed previously that it was so simple to obtain the
secondary waveforms. Interesting.
Antonio Carlos M. de Queiroz