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4.2 An op amp with an open-loop gain of 6 × 10° and Vec = 10 V has an output voltage of 3 V. If the voltage at the inverting input is –1 µV, what is the magnitude of the noninverting-input voltage?

4.2 An op amp with an open-loop gain of 6 × 10° and Vec = 10 V has an output voltage of 3 V. If the voltage at the inverting input is –1 µV, what is the magnitude of the noninverting-input voltage?

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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Please help with 4.2 and 4.7b.

### Problem 4.2 An operational amplifier (op amp) with an open-loop gain of \( 6 \times 10^5 \) and \( V_{cc} = 10 \) V has an output voltage of 3 V. If the voltage at the inverting input is \(-1 \, \mu \text{V}\), what is the magnitude of the noninverting-input voltage? #### Solution: Given: - Open-loop gain of the op amp, \( A = 6 \times 10^5 \) - Output voltage, \( V_{\text{out}} = 3 \, \text{V} \) - Inverting input voltage, \( V_{-} = -1 \, \mu \text{V} = -1 \times 10^{-6} \, \text{V} \) We need to find the noninverting input voltage \( V_{+} \). The relationship between the input and output of an op amp in open-loop configuration can be expressed as: \[ V_{\text{out}} = A \times (V_{+} - V_{-}) \] Solving for \( V_{+} - V_{-} \): \[ V_{+} - V_{-} = \frac{V_{\text{out}}}{A} \] \[ V_{+} - (-1 \times 10^{-6}) = \frac{3}{6 \times 10^5} \] \[ V_{+} + 1 \times 10^{-6} = \frac{3}{6 \times 10^5} \] \[ V_{+} = \frac{3}{6 \times 10^5} - 1 \times 10^{-6} \] \[ V_{+} = 0.5 \times 10^{-5} - 1 \times 10^{-6} \] \[ V_{+} = 5 \times 10^{-6} - 1 \times 10^{-6} \] \[ V_{+} = 4 \times 10^{-6} \, \text{V} \] Therefore, the magnitude of the noninverting-input voltage is \( 4 \, \mu \text{V} \).
Transcribed Image Text:### Problem 4.2 An operational amplifier (op amp) with an open-loop gain of \( 6 \times 10^5 \) and \( V_{cc} = 10 \) V has an output voltage of 3 V. If the voltage at the inverting input is \(-1 \, \mu \text{V}\), what is the magnitude of the noninverting-input voltage? #### Solution: Given: - Open-loop gain of the op amp, \( A = 6 \times 10^5 \) - Output voltage, \( V_{\text{out}} = 3 \, \text{V} \) - Inverting input voltage, \( V_{-} = -1 \, \mu \text{V} = -1 \times 10^{-6} \, \text{V} \) We need to find the noninverting input voltage \( V_{+} \). The relationship between the input and output of an op amp in open-loop configuration can be expressed as: \[ V_{\text{out}} = A \times (V_{+} - V_{-}) \] Solving for \( V_{+} - V_{-} \): \[ V_{+} - V_{-} = \frac{V_{\text{out}}}{A} \] \[ V_{+} - (-1 \times 10^{-6}) = \frac{3}{6 \times 10^5} \] \[ V_{+} + 1 \times 10^{-6} = \frac{3}{6 \times 10^5} \] \[ V_{+} = \frac{3}{6 \times 10^5} - 1 \times 10^{-6} \] \[ V_{+} = 0.5 \times 10^{-5} - 1 \times 10^{-6} \] \[ V_{+} = 5 \times 10^{-6} - 1 \times 10^{-6} \] \[ V_{+} = 4 \times 10^{-6} \, \text{V} \] Therefore, the magnitude of the noninverting-input voltage is \( 4 \, \mu \text{V} \).
### 4.7 For the circuit in Fig. P4.7: (a) Use the op-amp equivalent-circuit model to develop an expression for \( G = \frac{v_o}{v_s} \). (b) Simplify the expression by applying the ideal op-amp model parameters, namely \( A \to \infty \), \( R_i \to \infty \), and \( R_o \to 0 \). #### Figure P4.7: Circuit for Problem 4.7. **Description of the Circuit:** - **Voltage Source (\(v_s\))**: Provides the input signal to the circuit. - **Operational Amplifier (Op-Amp)**: Represented with the standard triangle symbol pointing to the right, with the inverting input (-), non-inverting input (+), and output. - **Non-inverting Input**: Connected to the ground. - **Inverting Input**: Connected to the voltage source (\(v_s\)). - **Output Voltage (\(v_o\))**: The voltage across the load resistor (\(R_L\)). - **Load Resistor (\(R_L\))**: Connected between the output of the op-amp and the ground. The circuit schematic can be interpreted as a typical configuration for analyzing the gain (\(G\)) of an operational amplifier inverting amplifier setup. The analysis involves exploring both the equivalent circuit model of the op-amp and simplifying it under the ideal op-amp assumptions.
Transcribed Image Text:### 4.7 For the circuit in Fig. P4.7: (a) Use the op-amp equivalent-circuit model to develop an expression for \( G = \frac{v_o}{v_s} \). (b) Simplify the expression by applying the ideal op-amp model parameters, namely \( A \to \infty \), \( R_i \to \infty \), and \( R_o \to 0 \). #### Figure P4.7: Circuit for Problem 4.7. **Description of the Circuit:** - **Voltage Source (\(v_s\))**: Provides the input signal to the circuit. - **Operational Amplifier (Op-Amp)**: Represented with the standard triangle symbol pointing to the right, with the inverting input (-), non-inverting input (+), and output. - **Non-inverting Input**: Connected to the ground. - **Inverting Input**: Connected to the voltage source (\(v_s\)). - **Output Voltage (\(v_o\))**: The voltage across the load resistor (\(R_L\)). - **Load Resistor (\(R_L\))**: Connected between the output of the op-amp and the ground. The circuit schematic can be interpreted as a typical configuration for analyzing the gain (\(G\)) of an operational amplifier inverting amplifier setup. The analysis involves exploring both the equivalent circuit model of the op-amp and simplifying it under the ideal op-amp assumptions.
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