EECS 42, Spring 2005 Week 5b 1Week 5bTopic 1: Decibels – Logarithmic Measure for Power, Voltage, Current, Gainand LossTopic 2: Operational AmplifiersEECS 42, Spring 2005 Week 5b 2Decibels – A Logarithmic MeasureCurious units called “decibels” are used by EEsto measure electric power, voltage, current, the gain or loss of amplifiers, and the insertion loss of filters.The decibel (dB) always refers to the ratio of the value of a quantity to a reference amount of that quantity.The word decibel is a reference to powers of ten and to Alexander Graham Bell.EECS 42, Spring 2005 Week 5b 3Logarithmic Measure for PowerTo express a power, P, in terms of decibels, one starts by choosing a reference power, Preference, and writingPower P in decibels = 10log10(P/Preference)Exercise: Express a power of 50 mW in decibels referred to 1 watt. Solution:P (dB) =10log10(50 x 10-3/1) = - 13 dBW. (The symbol dBWmeans “decibels referred to one watt”.)EECS 42, Spring 2005 Week 5b 4Aside About Resonant CircuitsWhen dealing with resonant circuits it is convenient to refer to the frequency difference between points at which the power from the circuit is half that at the peak of resonance. Such frequencies are known as “half-power frequencies”, and the power output there referred to the peak power (at the resonant frequency) is10log10(Phalf-power/Presonance) = 10log10(1/2) = -3 dB.EECS 42, Spring 2005 Week 5b 5Logarithmic Measures for Voltage or CurrentFrom the expression for power ratios in decibels, we can readily derive the corresponding expressions for voltage or current ratios.Suppose that the voltage V (or current I) appears across (or flows in) a resistor whose resistance is R. The corresponding power dissipated, P, is V2/R (or I2R). We can similarly relate the reference voltage or current to the reference power, asPreference= (Vreference)2/R or Preference= (Ireference)2R.Hence,Voltage, V in decibels = 20log10(V/Vreference)Current, I, in decibels = 20log10(I/Ireference)EECS 42, Spring 2005 Week 5b 6Note that the voltage and current expressions are just like the power expression except that they have 20 as the multiplier instead of 10 because power is proportional to the square of the voltage or current.Example: How many decibels larger is the voltage of a 9-volt transistor battery than that of a 1.5-volt AA battery? Let Vreference= 1.5. The ratio in decibels is20 log10(9/1.5) = 20 log10(6) = 16 dB.EECS 42, Spring 2005 Week 5b 7Gain or Loss Expressed in DecibelsThe gain produced by an amplifier or the loss of a filter is often specified in decibels.The input voltage (current, or power) is taken as the reference value of voltage (current, or power) in the decibel defining expression:Voltage gain in dB = 20 log10(Voutput/Vinput)Current gain in dB = 20log10(Ioutput/IinputPower gain in dB = 10log10(Poutput/Pinput)Example: The voltage gain of an amplifier whose input is 0.2 mV and whose output is 0.5 V is 20log10(0.5/0.2x10-3) = 68 dB.EECS 42, Spring 2005 Week 5b 8Change of Voltage or Current withA Change of FrequencyOne may wish to specify the change of a quantity such as the output voltage of a filter when the frequency changes by a factor of 2 (an octave) or 10 (a decade).For example, a single-stage RC low-pass filter has at frequencies above ω = 1/RC an output that changes at the rate -20dB per decade.EECS 42, Spring 2005 Week 5b 9Midterm 1Midterm 1 – Thursday February 24 in class.Covers through text Sec. 4.3, topics of HW 4. GSIs will review material in discussion sections prior to the exam. No books at the exam, no cell phones, you may bring one 8-1/2 by 11 sheet of notes (both sides of page OK), you may bring a calculator, and you don’t need a blue book.EECS 42, Spring 2005 Week 5b 10Homework and LabsThere will be no homework assignment for the next week, in view of the midterm.There will be a lab next week on RC Filters and the use of LabVIEW, a computer system for taking experimental data in a lab.EECS 42, Spring 2005 Week 5b 11OUTLINE• The operational amplifier (“op amp”)• Feedback• Comparator circuits• Ideal op amp• Unity-gain voltage follower circuitReadingBegin Ch. 14EECS 42, Spring 2005 Week 5b 12The Operational Amplifier• The operational amplifier (“op amp”) is a basic building block used in analog circuits.– Its behavior is modeled using a dependent source.– When combined with resistors, capacitors, and inductors, it can perform various useful functions:• amplification/scaling of an input signal• sign changing (inversion) of an input signal• addition of multiple input signals• subtraction of one input signal from another• integration (over time) of an input signal• differentiation (with respect to time) of an input signal• analog filtering• nonlinear functions like exponential, log, sqrt, etcEECS 42, Spring 2005 Week 5b 13Op Amp Circuit Symbol and Terminals+–V +V –non-inverting inputinverting inputpositive power supplynegative power supplyoutputThe output voltage can range from V –to V +(“rails”) The positive and negative power supply voltages do not have to be equal in magnitude.EECS 42, Spring 2005 Week 5b 14Op Amp Terminal Voltages and Currents+–V +V –• All voltages are referenced to a common node.• Current reference directions are intothe op amp.–Vcc++Vcc–+vn–+vp–common node(external to the op amp)ipinio+vo–ic+ic-EECS 42, Spring 2005 Week 5b 15Op Amp Voltage Transfer Characteristic• In the linear region, vo= A (vp– vn) = A vidwhere A is the open-loop gain• Typically, Vcc≤ 20 V and A > 104Æ linear range: -2 mV ≤ vid= (vp– vn) ≤ 2 mVThus, for an op amp to operate in the linear region, vp≅ vn(i.e., there is a “virtual short” between the input terminals.)vovid= vp–vnThe op amp is adifferentiating amplifier:“linear”“positive saturation”“negative saturation”Vcc-VccRegions of operation:slope = A >>1EECS 42, Spring 2005 Week 5b 16Achieving a “Virtual Short”• Recall the voltage transfer characteristic of an op amp:Q: How does a circuit maintain a virtual short at the input of an op amp, to ensure operation in the linear region?A: By using negative feedback. A signal is fed back from the output to the invertinginput terminal, effecting a stable circuit connection. Operation in the linear region enforces the virtual short circuit.vovp–vnVcc-Vccslope = A >>1~1 mV~10 Vvovp–vnVcc-Vccslope = A >>1Plotted using different scales for voand vp–vn~10
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