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.

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:

$$C = \frac{C_n}{2\pi f_c R$$and$$L = \frac{R L_n}{2\pi f_c}$$Where:

• $C$ and $L$ are the final capacitor and inductor values.
• $C_n$ and $L_n$ are the prototype element values.
• $R$ is the load resistance (use $R_L$ for inductors and $R_s$ for capacitors).
• $f_c$ is the cutoff frequency.

Given:

• $f_c = 50 \, \text{MHz} = 50 \times 10^$
• $R_L = 250 \, \Omega$
• $R_s = 50 \, \Omega$

Step 1: Calculate $L_1$

Using the formula for inductance:

$$L_1 = \frac{R_L L_n}{2\pi f_c}$$

Substitute values:

$$L_1 = \frac{250 \times 0.218}{2\pi \times 50 \times 10^$$ ​ $$L_1 = 173.48 \, \text{nH}$$

Step 2: Calculate $C_1$

Using the formula for capacitance:

$$C_1 = \frac{C_n}{2\pi f_c R_$$Substitute values:$$C_1 = \frac{2.452}{2\pi \times 50 \times 10^6 \times 50}$$$$C_1 = \frac{2.452}{1.571 \times 10^{10}}$$​ $$C_1 = 156.10 \, \text{pF}$$

Step 3: Calculate $L_2$

Using the same inductance formula:

$$L_2 = \frac{R_L L_n}{2\pi f_c}$$

Substitute values:

$$L_2 = \frac{250 \times 0.883}{2\pi \times 50 \times 10^6}$$$$L_2 = \frac{220.75}{314.16 \times 10^6$$​ $$L_2 = 702.67 \, \text{nH}$$

Step 4: Calculate $C_2$

Using the same capacitance formula:

$$C_2 = \frac{C_n}{2\pi f_c R_s}$$Substitute values:$$C_2 = \frac{3.187}{2\pi \times 50 \times 10^6 \times 50}$$ ​ $$C_2 = 202.89 \, \text{pF}$$

Final Values:

• $L_1=139\text{nH}$
• $C_1 = 156.10 \, \text{pF}$
• $L_2 = 702.67 \, \text{nH}$
• $C_4=\; 203\text{pF}$
• $C_2 = 202.89 \, \text{pF}$

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

• Second Order LC Low Pass Filter
• 1st Order Passive Filter Calculator
• How to Design State Variable Band Pass
• How to Design Twin-T Bandpass Filter
• LC Filter for Wireless Circuit Designs