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Single-reactance compensation circuit

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promach

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For the following single-reactance compensation circuit, how to derive equation 6.2 for w' ?

lZvxwfA.png
 

It seems you forgot the attachment.
 

I have solved the original question above (just substitute eq. 6.2 into eq. 6.1 will do). What do you mean by attachment ?

Now, I am stuck at deriving equation 6.6

Besides, what is the significance of maximizing susceptance (dB/dw = 0) ?

5ovK4of.png
 
promach said:
What do you mean by attachment ?
Click to expand...
For some reason I was not able to display the text you attached to the post. Now it's OK.
However the susceptance by definition is the imaginary part of the admitance, so starting from (6.3), analyzing each of the three terms separately we have:

Im(jwCp) = wCp
Im(1/(jwLp) = -1/(wLp)
Im(G)=Im[1/(R+jw'Ls)]=Im[ (R-jw'Ls)/(R^2+(w'*Ls)^2) ]=-w'Ls/[R^2+(w'*Ls)^2]

That is Bnet= Im(Ynet) = wCp - 1/(wLp) -w'Ls/[R^2+(w'*Ls)^2]

now simply take the derivative with respect to w and equate the result to 0 under the condition w=wo.
 

but how to get in particular the last term, 2*Ls / R2 ?
 

Take the derivative of -w'Ls/[R^2+(w'*Ls)^2] with respect to w, remembering that w'=w*[1-(w/wo)^2]
then put w=wo, that means all the terms [1-(w/wo)^2] will go to 0
 

remembering that w'=w*[1-(w/wo)^2]
then put w=wo, that means all the terms [1-(w/wo)^2] will go to 0
Click to expand...

but if I apply what you said about the last term, then the last term of eq.(6.6) should be ZERO instead of 2*Ls / R2 ?
 

No, you have first to take the derivative and then apply w=wo to the resut of the derivative calculation.
 

Yes, just apply the derivative properties.

Single reactance.jpg
 

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    promach

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There is a small mistake in your own expression of w'

wo is located at the nominator, not the denominator
 

Sorry, you are right. The derivative dw'/dw is then

1+(wo/w)^2

when w=wo

dw'/dw = 2

that is the same as before but with reversal in sign (in my previous post I didn't check the sign)
 

in post #3 , what is the significance of maximizing susceptance (dB/dw = 0) ?

and how is it related to quality factor of both the shunt and series circuits in figures 6.1a and 6.1b ?

Note: the picture screenshot is taken from Broadband RF and Microwave Amplifiers

WbMRr1j.png
 

No, I am trying to learn more about Class-E power amplifier before building it in LTSpice.
 

For the parallel - series resonant network the merit factor is given by Q=B/G (susceptance/conductance), while for the series - parallel resonant Q = X/R (reactance/resistance).
We know that the bandwidth is inversely proportional to Q, then to maximize the BW we have to lower the Q. Since the conductance or the resistance are fixed by the load, we can act only on susceptance or the reactance.
We can limit to the first topology. In this case at resonance (wo) B=0. In order to have a wide band we have to keep a low B around the resonance point.
We can do this forcing dB/dw=0 at w=wo
Since we have coupled parallel resonant followed by series resonant element with the same wo, the susceptance, starting from wo will be negative, while going back from infinite will be positive (as already said will be 0 for w=wo).
Then in-between positive and negative values (that is around wo) there will be an inflection point defined by dB/dw=0 at w=wo
The same if we consider the other topology with impedances (but the sign of X, with respect to the frequency, will be reversed compared with the sign of B).
 

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    promach

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Wait, why For the parallel - series resonant network the merit factor is given by Q=B/G (susceptance/conductance), while for the series - parallel resonant Q = X/R (reactance/resistance). ?

The attached document also does not explain this.
 

Attachments

  • RF Circuit Design - [Ch2-1] Resonator and Impedance Matching.pdf
    3.4 MB · Views: 117

We know Z=R+jX while Y=G+jB, furthermore Y = 1/Z this means:

Y = 1/(R+jX) = (R-jX)/(R^2+X^2) so that G = R/(R^2+X^2) and B = -X/(R^2+X^2)

Q = |X|/R = |B|/G

you can choose to use impedance or admittance depending from the simplicity of the calculation but, of course, the final result is always the same.

By the way, in my previous posts I've omitted the modulus (| |) in the Q definition since usually is implicit.
 

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    promach

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Since we have coupled parallel resonant followed by series resonant element with the same wo, the susceptance, starting from wo will be negative, while going back from infinite will be positive (as already said will be 0 for w=wo).
Then in-between positive and negative values (that is around wo) there will be an inflection point defined by dB/dw=0 at w=wo
Click to expand...

Why starting from negative ?
 

Simply calculate the limits of eq. 6.3 and eq. 6.10 for w-->0 and w--> infinite
 

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    promach

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