In the previous article on Butterworth low-pass prototype filter design with normalized components, the final filter prototype components were determined as follows:
- $L_{n1} = 0.218$
- $C_{n1} = 2.452$
- $L_{n2} = 0.883$
- $C_{n2} = 3.187$
To visualize the design, refer to the schematic below. This diagram guides the arrangement of components based on the calculated scaling values.
Now, we must scale these prototype values to their final application values using the following scaling formulas:
$$C = \frac{C_n}{2\pi f_c R_s}$$$$L = \frac{R_L L_n}{2\pi f_c}$$Where:
- $C$ and $L$ are the final capacitor and inductor values.
- $C_n$ and $L_n$ are the prototype element values.
- $R_L$ is the load resistance (used for inductors) and $R_s$ is the source resistance (used for capacitors).
- $f_c$ is the cutoff frequency.
Given:
- $f_c = 50\text{ MHz} = 50 \times 10^6\text{ Hz}$
- $R_L = 250\text{ }\Omega$
- $R_s = 50\text{ }\Omega$
Step 1: Calculate $L_1$
Using the inductance formula with $R_L = 250\text{ }\Omega$ and $L_{n1} = 0.218$:
$$L_1 = \frac{250 \times 0.218}{2\pi \times 50 \times 10^6} \approx 173.48\text{ nH}$$Step 2: Calculate $C_1$
Using the capacitance formula with $R_s = 50\text{ }\Omega$ and $C_{n1} = 2.452$:
$$C_1 = \frac{2.452}{2\pi \times 50 \times 10^6 \times 50} \approx 156.10\text{ pF}$$Step 3: Calculate $L_2$
Using the inductance formula with $R_L = 250\text{ }\Omega$ and $L_{n2} = 0.883$:
$$L_2 = \frac{250 \times 0.883}{2\pi \times 50 \times 10^6} \approx 702.67\text{ nH}$$Step 4: Calculate $C_2$
Using the capacitance formula with $R_s = 50\text{ }\Omega$ and $C_{n2} = 3.187$:
$$C_2 = \frac{3.187}{2\pi \times 50 \times 10^6 \times 50} \approx 202.89\text{ pF}$$Final Scaled Values:
- $L_1$: $173.48\text{ nH}$
- $C_1$: $156.10\text{ pF}$
- $L_2$: $702.67\text{ nH}$
- $C_2$: $202.89\text{ pF}$
For more insights into filter design, check out these related posts:
