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U of M CHEM 4101 - Electronic Components in Analytical Instrumentation

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9/21/11 1 Lecture 8 – Sep 23 Chem 4101 – Fall 2011 1!Electronic Components in Analytical Instrumentation 1- Where are the electronic components? 2- What are the electronic components? 3- Fundamental laws of electrical circuits (DC currents) -Sections 1C, 2A Suggested exercises: 1-6, 2-1, 2-2, 2-10(a), 2-11 Lecture 8 – Sep 23 Chem 4101 – Fall 2011 Instrument components 2!A B C D e f g h Lecture 8 – Sep 23 Chem 4101 – Fall 2011 Domains in Analytical Information Figure 1-2, Section 1C -29/21/11 2 Lecture 8 – Sep 23 Chem 4101 – Fall 2011 Electronic Components Identify the following components: Power and signal wires Resistors Capacitors Potentiometers Transformer Fuse Semi-conductors - Transistor - Light emitting diode (LED) - Operational Amplifier (Op Amp) Lecture 8 – Sep 23 Chem 4101 – Fall 2011 Most Transducers Generate “Analog” Signals Analog signals are CONTINUOUS in both AMPLITUDE and TIME Voltage Current Lecture 8 – Sep 23 Chem 4101 – Fall 2011 Digital Signals V- V+ They have two discrete amplitude levels. Typical digital signals switch between low (0 volts) and high (5 volts). High Signal Low signal Figure 2-2 (a), Section 2A9/21/11 3 Lecture 8 – Sep 23 Chem 4101 – Fall 2011 Ohm’s Law V = I × R V: volts = Joule/Coul (work/unit charge) I: Amperes = Coul/s (charge flow) R: Ohms = Joule s/ Coul2 Resistors: 1. Fixed value (color coded) 2. Variable (Potentiometer, Pot) R V I Example R = 2000 ohms, I = 1.5 milliamps V = IR = 0.0015*2000 = 3.0 V Lecture 8 – Sep 23 Chem 4101 – Fall 2011 Kirchoff’s Current Law R1 R2 E I I2 I1 The sum of currents at any point (node) in a circuit must be zero A At point A: I + (-I1) + (- I2) =0 Therefore- I = I1 + I2 Convention Current into node is + Current out of node is - Lecture 8 – Sep 23 Chem 4101 – Fall 2011 Kirchoff’s Voltage Law V1,I1 R1 V2,I2, R2 V I The sum of voltages around a closed path (loop) must be zero B Two Loops: ABEFA +V + (-V1)=0 V=V1=I1R1 ABCDEFA V+(-V2)=0 V=V2 = I2R2 A F E D C I2R2 =V = I1R1 I1 = V/R1 I2 = V/R2 Note: voltage drop across resistor in direction of current arrow is negative + - + - Start here9/21/11 4 Lecture 8 – Sep 23 Chem 4101 – Fall 2011 Parallel connections - same voltage- V1,I1 R1 V2,I2, R2 V I B A F E D C I2R2 =V = I1R1 Therefore I2 = I1R1/R2 V1 = V2 Lecture 8 – Sep 23 Chem 4101 – Fall 2011 Series connections - same current - R1,I,V1 R2,I,V2 Voltage Law (ABCD): V + (-V1) +(- V2)=0 A B V = V1 + V2! = I R1 + I R2! = I (R1 + R2)! = I (Req);! Req = R1 + R2!Req V, I V I + - I + - C D I Equivalent circuit Lecture 8 – Sep 23 Chem 4101 – Fall 2011 Voltage Divider Figure 2-2 (a), Section 2A9/21/11 5 Lecture 8 – Sep 23 Chem 4101 – Fall 2011 Current (I) to Voltage (V) converter circuit Note OP AMP Vout = Ipmt* R Photomultiplier Tube (PMT) Photon


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U of M CHEM 4101 - Electronic Components in Analytical Instrumentation

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