# LCC T-Network Impedance Matching Design Given: - Source resistance (\(R_S\)) = **15 Ω** - Load resistance (\(R_L\)) = **60 Ω** - Quality factor (\(Q\)) = **12** - Frequency (\(f\)) = **125 MHz** We'll solve each partstep by step** --- ## (a) Find the Inductance ### **Step 1: Calculate Reactance** For a T-network with **LCC** configuration (L in series, C in shunt, then C in series): \[ Q = \sqrt{\frac{R_L}{R_S} - 1} \] But we're given Q, so let's verify if it matches: \[ Q = \sqrt{\frac{60}{15} - 1} = \sqrt{4 - 1} = \sqrt{3} \approx 1.732 \] But since Q is specified as 12, we'll use the problem's Q. Series reactance (inductor): \[ X_L = Q \cdot R_S = 12 \times 15 = 180~\Omega \] ### **Step 2: Calculate Inductance** \[ X_L = 2\pi f L \] \[ L = \frac{X_L}{2\pi f} = \frac{180}{2\pi \times 125 \times 10^6} \] \[ L \approx \frac{180}{785398163.4} \] \[ L \approx 229 \text{ nH} \] --- ## (b) Find the Two Capacitances ### **Step 3: Calculate Required Capacitances** #### **First Capacitance (\(C_1\)) (Shunt Capacitance)** \[ X_{C1} = \frac{R_L}{Q} = \frac{60}{12} = 5~\Omega \] \[ C_1 = \frac{1}{2\pi f X_{C1}} = \frac{1}{2\pi \times 125 \times 10^6 \times 5} \] \[ C_1 = \frac{1}{3.927 \times 10^9} \] \[ C_1 \approx 254 \text{ pF} \] #### **Second Capacitance (\(C_2\)) (Series Capacitance)** \[ X_{C2} = \frac{X_L \cdot X_{C1}}{X_L + X_{C1}} \] But in a T network, series capacitor reactance is given by: \[ X_{C2} = Q R_L = 12 \times 60 = 720~\Omega \] \[ C_2 = \frac{1}{2\pi f X_{C2}} = \frac{1}{2\pi \times 125 \times 10^6 \times 720} \] \[ C_2 = \frac{1}{5.654 \times 10^{10}} \] \[ C_2 \approx 17.7 \text{ pF} \] --- ## (c) Draw the Circuit ### **LCC T-Network Circuit Diagram** ``` Source (15Ω) ----L----+---- C2 ----+---- Load (60Ω) | | C1 | | | GND GND ``` - **L**: Series inductor (\(\approx 229~\text{nH}\)) - **C1**: Shunt capacitor (\(\approx 254~\text{pF}\)) - **C2**: Series capacitor (\(\approx 17.7~\text{pF}\)) --- ## **Summary Table** | Component | Value | |-----------|------------------| | L | 229 nH | | C1 | 254 pF | | C2 | 17.7 pF | --- ### **If you need a neat hand-drawn circuit, let me know!**

# LCC T-Network Impedance Matching Design Given: - Source resistance (\(R_S\)) = **15 Ω** - Load resistance (\(R_L\)) = **60 Ω** - Quality factor (\(Q\)) = **12** - Frequency (\(f\)) = **125 MHz** ## (a) Find the Inductance ### **Step 1: Calculate Inductor Reactance** Using the Quality factor (\(Q\)): \[ X_L = Q \cdot R_S = 12 \times 15 = 180~\Omega \] ### **Step 2: Calculate Inductance** Using the reactance formula: \[ X_L = 2\pi f L \] Rearranging gives: \[ L = \frac{X_L}{2\pi f} = \frac{180}{2\pi \times 125 \times 10^6} \] Calculating: \[ L \approx \frac{180}{785398163.4} \approx 229 \text{ nH} \] --- ## (b) Find the Two Capacitances ### **Step 3: Calculate Shunt Capacitance (\(C_1\))** Using the formula for \(C_1\): \[ X_{C1} = \frac{R_L}{Q} = \frac{60}{12} = 5~\Omega \] Then, we find \(C_1\): \[ C_1 = \frac{1}{2\pi f X_{C1}} = \frac{1}{2\pi \times 125 \times 10^6 \times 5} \] Calculating: \[ C_1 \approx \frac{1}{3.927 \times 10^9} \approx 254 \text{ pF} \] ### **Step 4: Calculate Series Capacitance (\(C_2\))** Using: \[ X_{C2} = \frac{X_L}{Q} = \frac{180}{12} = 15~\Omega \] Then, we find \(C_2\): \[ C_2 = \frac{1}{2\pi f X_{C2}} = \frac{1}{2\pi \times 125 \times 10^6 \times 15} \] Calculating: \[ C_2 \approx \frac{1}{4.712 \times 10^{10}} \approx 21.2 \text{ pF} \] --- ## (c) Draw the Circuit ### **LCC T-Network Circuit Diagram** ``` Source (15Ω) ----L----+---- C2 ----+---- Load (60Ω) | | C1 | | | GND GND ``` - **L**: Series inductor (\(\approx 229~\text{nH}\)) - **C1**: Shunt capacitor (\(\approx 254~\text{pF}\)) - **C2**: Series capacitor (\(\approx 21.2~\text{pF}\)) --- ## **Summary Table** | Component | Value | |-----------|------------------| | L | 229 nH | | C1 | 254 pF | | C2 | 21.2 pF | --- This completes the design for the LCC T-network impedance matching circuit.