Sure! Let's break down the problem step by step: --- # Part I Solution ## **Given:** - **Op-amp circuit in Fig. 6** - **Required output:** \( v_o = v_A + 2v_B + 3v_C \) - **To find:** a) Values of \(R_A\) and \(R_C\) b) Currents \(i_A\), \(i_B\), \(i_C\) for given voltages c) Confirm with SPICE (simulation step) --- ## **Step a: Find \(R_A\) and \(R_C\)** ### **Circuit Analysis** This is an **inverting summing amplifier**. The general output equation for an inverting op-amp with three inputs is: \[ v_o = -R_f \left( \frac{v_A}{R_A} + \frac{v_B}{R_B} + \frac{v_C}{R_C} \right) \] Where: - \( R_f \) = feedback resistor (here, 100 kΩ) - \( R_A, R_B, R_C \) = input resistors for \( v_A, v_B, v_C \) #### **But We Want:** \[ v_o = v_A + 2v_B + 3v_C \] Notice the desired output is **non-inverting** (positive gains), but the circuit is inverting. So, to get the positive output, the right side should be negative of what the circuit gives: \[ -v_o = -v_A - 2v_B - 3v_C \] \[ v_o = v_A + 2v_B + 3v_C \] So, set the inverting op-amp to produce \( -(v_A + 2v_B + 3v_C) \), and then we can invert it if needed (or just ensure the gain pattern matches). #### **Setting up ratios:** From the generic inverting summing amp formula: \[ v_o = -R_f \left( \frac{v_A}{R_A} + \frac{v_B}{R_B} + \frac{v_C}{R_C} \right) \] If you want: \[ v_o = -(v_A + 2v_B + 3v_C) \] Then compare coefficients: \[ \frac{R_f}{R_A} = 1, \quad \frac{R_f}{R_B} = 2, \quad \frac{R_f}{R_C} = 3 \] Given \( R_f = 100k\Omega \) and \( R_B = 50k\Omega \) (as shown in the figure): \[ \frac{100k}{R_A} = 1 \implies R_A = 100k\Omega \] \[ \frac{100k}{15k} = 2 \implies R_B = 50k\Omega \text{ (But actual R_B is 15kΩ, so let's use that)} \] \[ \frac{100k}{R_C} = 3 \implies R_C = \frac{100k}{3} \approx 33.33k\Omega \] But the circuit in the image shows \( R_B = 15k\Omega \), so let's use this value: \[ \frac{100k}{R_B} = \frac{100k}{15k} \approx 6.67 \] But the desired output is not matching because the resistor values in the image do not match the ratios for the requested output. ### **So, set:** - \( R_A = 100k\Omega \) - \( R_B = 50k\Omega \) - \( R_C = 33.33k\Omega \) (If you must use the values from the diagram, the output equation will be different.) --- ## **Step b: Find \(i_A, i_B, i_C\) for given voltages** Given: - \( v_A = .7V, v_B = .4V, v_C = 1.1V \) - Using the calculated resistors above: Currents into resistors: \[ i_A = \frac{v_A}{R_A} \] \[ i_B = \frac{v_B}{R_B} \] \[ i_C = \frac{v_C}{R_C} \] Plug in the numbers: \[ i_A = \frac{.7}{100,000} = 7 \times 10^{-6} \text{ A} = 7 \mu A \] \[ i_B = \frac{.4}{50,000} = 8 \times 10^{-6} \text{ A} = 8 \mu A \] \[ i_C = \frac{1.1}{33,333} \approx 33 \times 10^{-6} \text{ A} = 33 \mu A \] --- ## **Step c: SPICE Simulation** You would need to implement the circuit in a SPICE simulator (like LTspice or Multisim) using the resistor values above and check if the output equation \( v_o = v_A + 2v_B + 3v_C \) holds for your test voltages. --- # **Summary Table** | | Value | |-----|--------------------| | \( R_A \) | \( 100k\Omega \) | | \( R_B \) | \( 50k\Omega \) | | \( R_C \) | \( 33.3k\Omega \) | | \( i_A \) | \( 7\mu A \) | | \( i_B \) | \( 8\mu A \) | | \( i_C \) | \( 33\mu A \) | --- ## **If you have to use the resistor values from the diagram, let me know, as the output will not match the requested gain coefficients exactly!** --- Let me know if you want to see the SPICE netlist or how to simulate this in software!