2
\$\begingroup\$

The diagram below is a schematic representation of a microwave lumped element notch filter. The goal is to analyse the parallel LRC circuit and derive an expression for \$S_{21}(\omega)\$, where port 1 is at the source and port 2 is measured across the load impedance \$Z_0\$, which could the input to an amplifier.

schematic

simulate this circuit – Schematic created using CircuitLab

My attempt at simplifying the circuit is shown below, where the resonator impedance has been condensed into \$Z_{res}\$ for clarity, where

$$Z_{res}= \left(\frac{1}{R_r}+j\omega C_r + \frac{1}{j\omega L_r}\right)^{-1} = \frac{R_r}{1+j\omega R_rC_rx},\,\\\text{where } x=\frac{\omega^2-\omega^2_0}{\omega\omega_0} \,\text{and } \omega_0^2 = \frac{1}{L_r C_r}$$

schematic

simulate this circuit

My question is, is the transformation from a to b legal? And if so, how are Cc and Zres related to Cc' and Zres'? Also if anyone has any advice on alternative methods to simplify this circuit, I would appreciate any feedback.

Thank you for your time.

\$\endgroup\$
4
  • \$\begingroup\$ it looks more like a band pass filter rather than a notch filter. \$\endgroup\$
    – Andy aka
    Commented Jan 21, 2014 at 22:56
  • \$\begingroup\$ Hi Andy - read your comment yesterday and have been trying to formulate an argument as to why it is a notch filter, other than simulation in ADS indicates it is. My intuition says that it should act as a bandpass, as the impedance on resonance should be large (assuming large R), so no current will flow resulting in the voltage being dropped across the load. Away from resonance the impedance is small so the load is shorted out. Hmm. \$\endgroup\$
    – bazza1988
    Commented Jan 23, 2014 at 9:13
  • \$\begingroup\$ @Andyaka, OK think I've figured it out. While the impedance of the LRC circuit is infinite at \$\omega_0^2=1/L_rC_r\$, there exists a condition where the impedance of the series combination of the coupling capacitor and resonator is zero, namely \$\omega_r^2=1/L_r(C_r+C_c)\$ . Under this condition, it acts as a notch filter. However, the question still remains of how to properly analyse it... \$\endgroup\$
    – bazza1988
    Commented Jan 23, 2014 at 16:16
  • \$\begingroup\$ Any ideas on how to analyse this circuit? \$\endgroup\$
    – bazza1988
    Commented Jun 24, 2014 at 11:08

1 Answer 1

1
\$\begingroup\$

By analyzing the filter, I feel this is not a notch filter. I will explain why it is not a notch filter. If any one has a different opinion please inform me.

Condition 1: Assume our input is dc. i e Vg=Vdc(means no frequency components.Hence ω=0).So Zlr = jωLr = 0 and Zcc = 1/jωCc= infinity. This means there will be no current flow through Cc and RLC parallel circuit, in this case. Hence whatever be the input, it will fully appear at output. Hence ouput Vo = Vmax.

Condition 2: Assume our input frequency component is infinite; ie (ω=infinity). In this case, both Zcc = Zcr = 0. This will make output grounded through Capacitors Cc and Cr. So here Vo = 0

Condition 3: Now assume frequency component is in between 0 and infinity.So the impedance of RLC circuit and capacitor Cc will be finite and hence surely this line holds a current correspond to this impedance. Hence current through Zo will be in between Vmax and 0 in this case.

According to this analysis, if we draw a graph between volatage and frequency, we can see that it will result to a low pass filter response. (Since output is maximum at ω=0 and minimum at ω=infinity ).

I will try to provide the analysis of this circuit later. Thanks

EDIT :

*Analysis**

Impedance of parallel circuit
     Zrlc = 1/[1/Z1 + 1/Z2 + 1/Z3] = Z1.Z2.Z3/(Z1.Z2+Z2.Z3+Z3.Z1)
     Substituting impdance values
     Zrlc = (R2).(1/sC2).(sL2)/[(R2/sC2)+(R2.sL2)+(sL2/sC2)] 
     Zrlc = s(R2.L2)/[R2-(R2.L2.C2)+sL2]
Impedance of circuit line that consist of C1 and RLC circuit
     Z = Zc1 + Zrlc
     Z = (1/sC1) + (s(R2.L2)/[R2-(R2.L2.C2)+sL2])
     From the equation, Z will be clearly a complex quantity. Let assume
     Z = (X - jω.Y)

The equation shows that, Z will become zero, when ω = ω0 (so that X = ω0.Y). So condition 3 can again divide into two section.

Condition 3.a: When 0 < ω < ω0(low frequency), capacitive impdance will be dominant and inductive impdance can be consider to be zero. This makes the RLC ciruit short circuited(since inductor impedance is zero). So the impedance exist is that of capacitor Cc. So the circuit will be look like this.

schematic

simulate this circuit – Schematic created using CircuitLab

So as ω increases, impedance of capacitor become low. This make an increase in current through capacitor. This leads to reduction of current flow through load. Hence Vo = Io x Zo become low. This will happen in the range 0 < ω < ω0

Condition 3.b When ω > ω0(High frequency), inductive load become dominant and capacitive impedance become very low.Hence we can assume capacitive loads as zero impdance. This makes the output is shunted to ground through capacitor Cc and capacitor Cr. So output remains zero from ω > ω0.

Final assumption: Now we have

      Vo = { Vmax ;  ω = 0
           { below Vmax and reduces as  ω increases ;  0< ω< ω0
           { 0 ;  ω = ω0
           { 0 ;  ω >  ω0

This is surely response of a 'low pass filter' (ideal case)

\$\endgroup\$
4
  • \$\begingroup\$ Hi @rajeevktomy, thank you for taking the time to respond. While your conditions 1 and 2 are correct, condition 3 skips over the behaviour around \$\omega_0\$, which isn't as simple as you've described. As I mentioned above, below \$\omega_0\$ there will exist a condition where the inductive reactance is cancelled by the capacitive reactance of the coupling capacitor and the input impedance goes to zero, hence shorting out \$Z_0\$. If you do get chance to simulate this circuit you should find condition 1 and 2 as you've stated, but around \$\omega_0\$, it will act like a notch filter. Cheers. \$\endgroup\$
    – bazza1988
    Commented Jan 29, 2014 at 8:41
  • \$\begingroup\$ @bazza1988 i have edited my answer. Please go through it and let me know your feedback \$\endgroup\$ Commented Jan 30, 2014 at 3:15
  • \$\begingroup\$ Again thank you for taking the time to discuss this. I can't see anything wrong with your analysis, but you seemed to have missed out the condition that I stated was the most important, namely when \$\omega \approx \omega_0\$. It is only around the resonant frequency of LRC+Coupling capacitor that it becomes a notch filter. For other frequencies you analysis holds. Cheers. \$\endgroup\$
    – bazza1988
    Commented Jan 30, 2014 at 19:39
  • \$\begingroup\$ @bazza1988 :cheers :) \$\endgroup\$ Commented Jan 31, 2014 at 1:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.