lsli03

lsli03 - 03-1 03-1 Operational Amplifiers These are common...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 03-1 03-1 Operational Amplifiers These are common components in conditioning circuits. There are two inputs, v + and v i , two power supplies, + V s and- V s , and an output, v o . +- v o v +- v V S +- V S v o = min { + V s , max {- V s , ( v +- v- ) A }} , were A is the op-amp gain. Ignoring saturation, v o = A ( v +- v- ). Ideal Op-Amp Properties Infinite input impedance. Infinite gain. ( A = ∞ ) Zero output impedance. 03-1 EE 4770 Lecture Transparency. Formatted 9:45, 22 January 1999 from lsli03. 03-1 03-2 03-2 +- v o v +- v V S +- V S Where to Find Ideal Op-Amps An electronics textbook. However, in certain circuits a real op-amp performs almost the same as an ideal op-amp would. Simplifying Assumptions Current into inputs is zero. When used in a negative feedback configuration, v + = v- . Op-Amp Circuits to be Covered Non-inverting amplifier. Inverting amplifier. Summing amplifier. 03-2 EE 4770 Lecture Transparency. Formatted 9:45, 22 January 1999 from lsli03. 03-2 03-3 03-3 Non-Inverting Amplifier R A B R +- v v i o v o = R A + R B R A v i . 03-3 EE 4770 Lecture Transparency. Formatted 9:45, 22 January 1999 from lsli03. 03-3 03-4 03-4 Versatility of Inverting Amplifier Use of Non-Inverting Amplifier in Conditioning Circuits v o = R A + R B R A v i . Traditional Use, Voltage Amplifier Input is v i , output is v o . H c ( v i ) = R A + R B R A v i R A B R +- v v i o 03-4 EE 4770 Lecture Transparency. Formatted 9:45, 22 January 1999 from lsli03. 03-4 03-5 03-5 Non-Inverting Amplifier Example Problem Design a system with output v o = H ( x ) , where process variable x is water level, x ∈ [0 m , 1 m] , and H ( x ) = 10 x V m . Note: most example problems will not be as complete as the archetypical problem covered earlier. Solution: Use same float-and-cable system as in previous example problem. Use 100 kΩ three-terminal variable resistor with 1 V voltage source across fixed terminals: H t ( x ) = 1 x V m . Problem will be solved two ways: First way, we know what kind of conditioning circuit is needed. Second way, we have to determine algebraicly the type of conditioning circuit needed. 03-5 EE 4770 Lecture Transparency. Formatted 9:45, 22 January 1999 from lsli03. 03-5 03-6 03-6 First Way: Use Non-Inverting Amplifier Obviously, all that is needed is an amplifier with a gain of 10. A non- inverting amplifier will do....
View Full Document

Page1 / 21

lsli03 - 03-1 03-1 Operational Amplifiers These are common...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online