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Published byKristian Phillips Modified over 8 years ago
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V DC I V ref(erence) = 0V = ground R (Ω) + - VΩ com A DVM Vout= Vs / A + Vs - A Voltage Divider Variable Resistor V th + - Thevenin Equivalent th R Resistance- Controlled Voltage V
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V DC + - I V ref(erence) = 0V = ground R (Ω) R1 + - VΩ com A DVM R2 V2= VsR2 R1+R2 + Vs - Voltage Divider Variable Resistor I=Vth/Rth Norton Equivalent th R R+ΔR Sensor Bridge
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V in V out =AV in V V out =-AV in V V out =V in Non-Inverting Amplifier Inverting Amplifier Buffer (Unity Gain) V p V n Vcc -Vcc Voltage Controlled Switch V1 Vo=AV1+BV2 V2 Non-Inverting Summer Vo=-AV1-BV2 V2 V1 Inverting Summer V p V n Voltage Controlled Switch Vo=AV1-BV2 V2 V1 Subtracting Amp
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Non-Inverting Amplifier Vo= Vs (R1+R2)/R2 Inverting Amplifier Vo = Vs (-Rf/Rs) Buffer (Unity Gain) Vo = Vs Inverting Summer Non-Inverting Summer 0 0 (Off) (On) (Off) Differential Amp
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Example: Resistive Sensor System All rights reserved. Do not copy or distribute. © 2013 National Technology and Science Press Circuits, Second Edition by Fawwaz T. Ulaby and Michel M. Maharbiz, © NTS Press, Used with Permission by the Publisher
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Example: Averaging Circuit All rights reserved. Do not copy or distribute. © 2013 National Technology and Science Press Problem 4.27 Circuits, Second Edition by Fawwaz T. Ulaby and Michel M. Maharbiz, © NTS Press, Used with Permission by the Publisher
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Example: Linear Voltage Combination All rights reserved. Do not copy or distribute. © 2013 National Technology and Science Press Circuits, Second Edition by Fawwaz T. Ulaby and Michel M. Maharbiz, © NTS Press, Used with Permission by the Publisher Vo2 = 3V1 + 4V2 -5V3 – 8V4
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