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A motor is controlled by a H-bridge which can connect + or - 10V to the motor. Thus the voltage across the motor can be either +/- 5V or -+ 5V. An instrumentation amplifier, A1, is used to measure the motor current. If the motor draws up to ±1A and the motor current must be measured to within 10mA, what is the CMRR required for A1? Design Al using a 3-opamp instrumentation amplifier, considering only errors due to CMRR, indicating all resistor values and tolerances. The instrumentation amplifier is powered by +15V Key question: What is the CMRR required? +5V

A motor is controlled by a H-bridge which can connect + or - 10V to the motor. Thus the voltage across the motor can be either +/- 5V or -+ 5V. An instrumentation amplifier, A1, is used to measure the motor current. If the motor draws up to ±1A and the motor current must be measured to within 10mA, what is the CMRR required for A1? Design Al using a 3-opamp instrumentation amplifier, considering only errors due to CMRR, indicating all resistor values and tolerances. The instrumentation amplifier is powered by +15V Key question: What is the CMRR required? +5V

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
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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CMRR Calculation We want the common mode gain (worst case) x-5V to give an output Voltage which is no bigger than the differential gain x 500µV 5V Common Mode gain s 500u V. Differential gain +250µV Minimum 5V- Detectible Мaximum A1 A1 250μ ν Signal Voltage, Vmin Error Voltage, Verr -5V- The meaning of the specification is that Vmin >Verr CMRR Calculation 2 So for the minimum detectible signal to be larger than the common-mode error we have that Differential gain 500uV 2 Common - Mode gain · 5V Differential gain Common - Mode gain 5V Hence = CMRR > 500µV 5V So CMRR > 10 = 80dB %3D 500μν
Transcribed Image Text:CMRR Calculation We want the common mode gain (worst case) x-5V to give an output Voltage which is no bigger than the differential gain x 500µV 5V Common Mode gain s 500u V. Differential gain +250µV Minimum 5V- Detectible Мaximum A1 A1 250μ ν Signal Voltage, Vmin Error Voltage, Verr -5V- The meaning of the specification is that Vmin >Verr CMRR Calculation 2 So for the minimum detectible signal to be larger than the common-mode error we have that Differential gain 500uV 2 Common - Mode gain · 5V Differential gain Common - Mode gain 5V Hence = CMRR > 500µV 5V So CMRR > 10 = 80dB %3D 500μν
+5V A1 O/P 0.052 Motor -5V A motor is controlled by a H-bridge which can connect + or – 10V to the motor. Thus the voltage across the motor can be either +/- 5V or -/+ 5V. An instrumentation amplifier, Al1, is used to measure the motor current. If the motor draws up to +1A and the motor current must be measured to within 10mA, what is the CMRR required for A1? Design Al using a 3-opamp instrumentation amplifier, considering only errors due to CMRR, indicating all resistor values and tolerances. The instrumentation amplifier is powered by +15V Key question: What is the CMRR required? +5V On E 500μν +5V -5V → O/P A1 0.052 Motor Off On -5V Common mode voltage = (-5V + -4.9995V) / 2 -5V Differential voltage = 500µV for minimum detectible signal
Transcribed Image Text:+5V A1 O/P 0.052 Motor -5V A motor is controlled by a H-bridge which can connect + or – 10V to the motor. Thus the voltage across the motor can be either +/- 5V or -/+ 5V. An instrumentation amplifier, Al1, is used to measure the motor current. If the motor draws up to +1A and the motor current must be measured to within 10mA, what is the CMRR required for A1? Design Al using a 3-opamp instrumentation amplifier, considering only errors due to CMRR, indicating all resistor values and tolerances. The instrumentation amplifier is powered by +15V Key question: What is the CMRR required? +5V On E 500μν +5V -5V → O/P A1 0.052 Motor Off On -5V Common mode voltage = (-5V + -4.9995V) / 2 -5V Differential voltage = 500µV for minimum detectible signal
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