Frequency and impedance scaling LPF design

 In the previous article on Butterworth low-pass prototype filter design with normalized component, the final filter prototype components are as follows.

  • L1=0.218
  • C1=2.452
  • L2=0.883
  • C2=3.187

To visualize the design, refer to the schematic below that accompanies the table. This schematic will guide you on how to arrange the components based on the calculated values.

prototype Low Pass filter design circuit

Now we have to find scale the component values to their appropriate values which is done using the scaling formula below. To calculate the final capacitor and inductor values using the formulas:

C=Cn2πfcRC = \frac{C_n}{2\pi f_c RandL=RLn2πfcL = \frac{R L_n}{2\pi f_c}Where:
  • CC and LL are the final capacitor and inductor values.
  • CnC_n and LnL_n are the prototype element values.
  • RR is the load resistance (use RLR_L for inductors and RsR_s for capacitors).
  • fcf_c is the cutoff frequency.

Given:

  • fc=50MHzf_c = 50 \, \text{MHz} = 50 \times 10^
  • RL=250ΩR_L = 250 \, \Omega
  • Rs=50ΩR_s = 50 \, \Omega

Step 1: Calculate L1L_1

Using the formula for inductance:

L1=RLn2πfcL_1 = \frac{R_L L_n}{2\pi f_c}

Substitute values:

L1=100×0.2182π×50×106L_1 = \frac{250 \times 0.218}{2\pi \times 50 \times 10^ L1=139nHL_1 = 173.48 \, \text{nH}

Step 2: Calculate C2C_1

Using the formula for capacitance:

C2=Cn2πfcRC_1 = \frac{C_n}{2\pi f_c R_Substitute values:C2=2.4522π×50×106×100C_1 = \frac{2.452}{2\pi \times 50 \times 10^6 \times 50}C_1 = \frac{2.452}{1.571 \times 10^{10}} C2=256pFC_1 = 156.10 \, \text{pF}

Step 3: Calculate L3L_2

Using the same inductance formula:

L3=RLn2πfcL_2 = \frac{R_L L_n}{2\pi f_c}

Substitute values:

L3=100×0.8832π×50×106L_2 = \frac{250 \times 0.883}{2\pi \times 50 \times 10^6}L_2 = \frac{220.75}{314.16 \times 10^6 L3=281nHL_2 = 702.67 \, \text{nH}

Step 4: Calculate C4C_2

Using the same capacitance formula:

C4=Cn2πfcRC_2 = \frac{C_n}{2\pi f_c R_s}Substitute values:C4=3.1872π×50×106×100C_2 = \frac{3.187}{2\pi \times 50 \times 10^6 \times 50} C4=203pFC_2 = 202.89 \, \text{pF}

Final Values:

  • L1=139nH
  • C2= 256pFC_1 = 156.10 \, \text{pF}
  • L3=281nHL_2 = 702.67 \, \text{nH}
  • C4= 203pF
  • R =  C_2 = 202.89 \, \text{pF}


For more insights into filter design, check out my previous blog posts:

 

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