The Problem Statement: Designing a 100 kHz LC Low-Pass Filter

Exam Problem: Design a second-order passive LC low-pass filter with a cutoff frequency ($f_c$) of exactly 100 kHz. The filter must drive a $50\ \Omega$ resistive load (standard RF/coaxial impedance). Assume an impedance-matched design where the characteristic impedance ($Z_0$) of the filter equals the load resistance ($R_L = 50\ \Omega$). Calculate the required inductance ($L$) in microhenries ($\mu H$) and capacitance ($C$) in nanofarads ($nF$).

Step-by-Step Algebraic Solution & Method Selection

Before touching a calculator, we must establish which theorem and method apply. A simple $f_c = \frac{1}{2\pi\sqrt{LC}}$ formula gives us one equation with two unknowns ($L$ and $C$). To solve this, we use the characteristic impedance method for second-order L-section filters. By setting the filter's characteristic impedance $Z_0 = \sqrt{L/C}$ equal to the load resistance, we establish our second independent equation, ensuring the filter is properly damped and matched to the $50\ \Omega$ system without severe passband peaking.

Here is the exact algebraic sequence to isolate both variables.

  1. Define the knowns and convert to base SI units:
    $f_c = 100\text{ kHz} = 100,000\text{ Hz}$
    $R_L = Z_0 = 50\ \Omega$
  2. Calculate the angular cutoff frequency ($\omega_c$):
    $\omega_c = 2\pi f_c$
    $\omega_c = 2\pi(100,000) = 200,000\pi\text{ rad/s}$
  3. Express $L$ in terms of $C$ using the impedance equation:
    $Z_0 = \sqrt{\frac{L}{C}}$
    $50 = \sqrt{\frac{L}{C}}$
    Square both sides: $2500 = \frac{L}{C}$
    $L = 2500C$
  4. Substitute $L$ into the angular frequency equation:
    The natural frequency of an LC circuit is $\omega_c = \frac{1}{\sqrt{LC}}$.
    Substitute $L = 2500C$:
    $200,000\pi = \frac{1}{\sqrt{(2500C)(C)}}$
    $200,000\pi = \frac{1}{\sqrt{2500C^2}}$
    $200,000\pi = \frac{1}{50C}$
  5. Solve for Capacitance ($C$):
    Multiply both sides by $C$ and divide by $200,000\pi$:
    $C = \frac{1}{50 \times 200,000\pi}$
    $C = \frac{1}{10,000,000\pi} = \frac{1}{\pi \times 10^7}\text{ Farads}$
    $C \approx 3.183 \times 10^{-8}\text{ F}$
    Convert to nanofarads: $C = 31.83\text{ nF}$
  6. Solve for Inductance ($L$):
    Use the relation $L = 2500C$:
    $L = 2500 \times \left(\frac{1}{\pi \times 10^7}\right)$
    $L = \frac{2500}{\pi \times 10^7} = \frac{2.5 \times 10^3}{\pi \times 10^7} = \frac{2.5}{\pi} \times 10^{-4}\text{ Henrys}$
    $L \approx 0.79577 \times 10^{-4}\text{ H} = 7.9577 \times 10^{-5}\text{ H}$
    Convert to microhenries: $L = 79.58\ \mu\text{H}$

Sanity Check & Independent Verification

The Trap: The most common mistake on this specific exam problem is dropping the $2\pi$ when converting $f_c$ to $\omega_c$, or blindly applying the first-order RC filter formula ($f_c = \frac{1}{2\pi RC}$) by substituting $L$ for $R$. Another frequent error is mishandling the metric prefixes at the end, reporting $31.83\ \mu\text{F}$ instead of $nF$.

Order of Magnitude & Unit Check:
Let's verify the units under the square root: $\sqrt{H \cdot F} = \sqrt{\frac{V \cdot s}{A} \cdot \frac{A \cdot s}{V}} = \sqrt{s^2} = s$. The inverse of seconds is Hertz. The units are correct.
Physically, a 100 kHz cutoff sits at the top of the audio band and the bottom of the RF band. Inductors in the tens of $\mu H$ and capacitors in the tens of $nF$ are standard, readily available physical values for this frequency range (easily wound on a small ferrite toroid and sourced as a standard multilayer ceramic chip).

Independent Verification (Reverse Substitution):
Plug our final exact fractional answers back into the original frequency equation to prove they yield exactly 100 kHz.

$f_c = \frac{1}{2\pi\sqrt{LC}}$
$f_c = \frac{1}{2\pi\sqrt{\left(\frac{2.5}{\pi} \times 10^{-4}\right) \left(\frac{1}{\pi} \times 10^{-7}\right)}}$
$f_c = \frac{1}{2\pi\sqrt{\frac{2.5}{\pi^2} \times 10^{-11}}}$
Wait, let's use the simpler $50C$ substitution from Step 4: $\sqrt{LC} = 50C$.
$f_c = \frac{1}{2\pi(50C)} = \frac{1}{100\pi C}$
Substitute $C = \frac{1}{\pi \times 10^7}$:
$f_c = \frac{1}{100\pi \left(\frac{1}{\pi \times 10^7}\right)} = \frac{1}{\frac{100\pi}{\pi \times 10^7}} = \frac{1}{\frac{100}{10^7}} = \frac{10^7}{100} = 10^5\text{ Hz} = 100\text{ kHz}$.

The math holds perfectly. For real-world verification before soldering, simulate the netlist using the Analog Devices Filter Wizard or LTspice to verify the Bode plot shows a -3dB point exactly at 100 kHz with a maximally flat passband.

Frequently Asked Questions (FAQ)

Why use an LC filter instead of an RC filter for a 100 kHz cutoff?

An RC low-pass filter wastes power as heat across the resistor and only provides a -20 dB/decade roll-off. At 100 kHz, if you need to pass a high-frequency carrier while blocking switching noise, an RC filter would severely attenuate your passband signal. A passive LC filter provides a much steeper -40 dB/decade roll-off (second-order). Furthermore, the inductor's DC resistance (DCR) is typically a fraction of an ohm, meaning the LC filter passes DC and low-frequency AC with near-zero power loss, making it ideal for power supply ripple filtering and RF impedance matching.

How does load impedance affect the passive LC low-pass filter response?

The load impedance ($R_L$) directly dictates the filter's damping factor (Q factor). In our example, we matched $Z_0$ to $50\ \Omega$ for a clean response. If you connect this exact $79.58\ \mu\text{H}$ / $31.83\text{ nF}$ filter to a high-impedance load (like a $10\text{ k}\\Omega$ op-amp input), the filter will become severely underdamped. You will see a massive resonance peak (ringing) right at 100 kHz, amplifying noise instead of filtering it. Conversely, a very low impedance load will overdamp the circuit, degrading it into a sloppy first-order response. Always design $L$ and $C$ ratios around your specific load.

What type of physical capacitor is best for a 100 kHz LC filter?

For a 100 kHz signal path, you must use C0G (NP0) dielectric ceramic capacitors. Standard X7R or Y5V capacitors exhibit severe capacitance loss under DC bias voltage (sometimes dropping 50% of their rated value) and possess high microphonic/piezoelectric effects. If you use an X7R capacitor in a 100 kHz LC filter, the physical vibration from the AC ripple will modulate the capacitance, introducing nonlinear harmonic distortion and shifting your carefully calculated 100 kHz cutoff frequency dynamically. For the inductor, a shielded ferrite drum core or a powdered iron toroid prevents stray magnetic coupling to nearby sensitive traces.