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8 V 7V 2 +1 + 20 ΚΩ Mai Μ 20 kΩ 12 40 ΚΩ 13 4U ΚΩ Μ Vn |Vp + Vo 50 ΚΩ

8 V 7V 2 +1 + 20 ΚΩ Mai Μ 20 kΩ 12 40 ΚΩ 13 4U ΚΩ Μ Vn |Vp + Vo 50 ΚΩ

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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For all op amps, use ±Vcc = ±18V. 

Find V0 in terms V1 and V2. 

The diagram is a circuit schematic featuring operational amplifiers, resistors, and labeled voltages and currents. Below is a detailed description: 1. **Voltage Sources:** - \( V_1 = 6.98 \, \text{V} \) - \( V_2 = 7 \, \text{V} \) 2. **Operational Amplifiers:** - Two op-amps are depicted on the left side of the schematic, each connected to a voltage source. - The outputs of these op-amps are connected to resistors. 3. **Resistors:** - Several resistors, characterized by their resistance values, are present in the circuit: - Two resistors (\(20 \, \text{k}\Omega\) each) are connected in series from the operational amplifier outputs. - A \(40 \, \text{k}\Omega\) resistor is connected in parallel to \(V_p\). - The \(40 \, \text{k}\Omega\) resistor is grounded. - The output of the top op-amp is connected to a \(40 \, \text{k}\Omega\) resistor on the right. - Another \(50 \, \text{k}\Omega\) resistor connects the output (\(V_o\)) to ground. 4. **Currents and Voltages:** - The diagram indicates several currents: - \( i_1 \) flows to the inverting input (\(V_n\)) of the op-amp. - \( i_2 \) flows from the non-inverting input (\(V_p\)) of the op-amp. - \( i_3 \) through the top \(40 \, \text{k}\Omega\) resistor. - \( i_4 \) flows through the \(50 \, \text{k}\Omega\) resistor to ground. - Voltage points are labeled as \(V_n\), \(V_p\), and \(V_o\) (output of the op-amp on the right). This circuit can be analyzed for its behavior using operational amplifier rules, specifically with regard to the virtual short concept where the voltages \(V_n\) and \(V_p\) can be considered equal if the op-amp is in the linear region. Additionally, current flow and voltage drops across resistors obey Ohm’s Law, allowing for computation of unknown quantities
Transcribed Image Text:The diagram is a circuit schematic featuring operational amplifiers, resistors, and labeled voltages and currents. Below is a detailed description: 1. **Voltage Sources:** - \( V_1 = 6.98 \, \text{V} \) - \( V_2 = 7 \, \text{V} \) 2. **Operational Amplifiers:** - Two op-amps are depicted on the left side of the schematic, each connected to a voltage source. - The outputs of these op-amps are connected to resistors. 3. **Resistors:** - Several resistors, characterized by their resistance values, are present in the circuit: - Two resistors (\(20 \, \text{k}\Omega\) each) are connected in series from the operational amplifier outputs. - A \(40 \, \text{k}\Omega\) resistor is connected in parallel to \(V_p\). - The \(40 \, \text{k}\Omega\) resistor is grounded. - The output of the top op-amp is connected to a \(40 \, \text{k}\Omega\) resistor on the right. - Another \(50 \, \text{k}\Omega\) resistor connects the output (\(V_o\)) to ground. 4. **Currents and Voltages:** - The diagram indicates several currents: - \( i_1 \) flows to the inverting input (\(V_n\)) of the op-amp. - \( i_2 \) flows from the non-inverting input (\(V_p\)) of the op-amp. - \( i_3 \) through the top \(40 \, \text{k}\Omega\) resistor. - \( i_4 \) flows through the \(50 \, \text{k}\Omega\) resistor to ground. - Voltage points are labeled as \(V_n\), \(V_p\), and \(V_o\) (output of the op-amp on the right). This circuit can be analyzed for its behavior using operational amplifier rules, specifically with regard to the virtual short concept where the voltages \(V_n\) and \(V_p\) can be considered equal if the op-amp is in the linear region. Additionally, current flow and voltage drops across resistors obey Ohm’s Law, allowing for computation of unknown quantities
Expert Solution
Step 1: Summarize the details given:

Given Data:

An instrumentation amplifier circuit with,

  • Input voltages,
    • V subscript 1 equals 6.98 space text V end text
    • V subscript 2 equals 7 space text V end text
  • Saturation voltage V subscript c c end subscript equals plus-or-minus 18 space text V end text

To Find:

The output voltage V subscript 0 in terms V subscript 1 space & space V subscript 2.


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