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Shantilal Shah Government Engineering College Bhavnagar Electrical Engg. Department Branch :- ELECTRICAL Semester :Batch:- B1 rd 3 SUBJECT: ANALOG ELECTRONICS SUBJECTED BY: PROF. A. M.UPADHYAY Group members NAME 1) Gamit sujal J. ENROLL.NO (130430109018) 2) Chaudhari piyush U. (130430109008) What is an Op-Amp? – The Surface An Operational Amplifier (Op-Amp) is an integrated circuit that uses external voltage to amplify the input through a very high gain. We recognize an Op-Amp as a massproduced component found in countless electronics. What an Op-Amp looks like to a lay-person What an Op-Amp looks like to an engineer History of the Op-Amp – The Dawn The Vacuum Tube Age The First Op-Amp: (1930 – 1940) Designed by Karl Swartzel for the Bell Labs M9 gun director Uses 3 vacuum tubes, only one input, and ± 350 V to attain a gain of 90 dB Loebe Julie then develops an Op-Amp with two inputs: Inverting and Non-inverting Mathematics of the Op-Amp The gain of the Op-Amp itself is calculated as: G = Vout/(V+ – V-) The maximum output is the power supply voltage When used in a circuit, the gain of the circuit (as opposed to the op-amp component) is: Av = Vout/Vin Op-Amp Characteristics • Open-loop gain G is typically over 9000 • But closed-loop gain is much smaller • Rin is very large (MΩ or larger) • Rout is small (75Ω or smaller) • Effective output impedance in closed loop is very small Symbols for Ideal and Real Op Amps OpAmp uA741 LM111 LM324 Terminals on an Op Amp Positive power supply (Positive rail) Non-inverting Input terminal Output terminal Inverting input terminal Negative power supply (Negative rail) Open Circuit Output Voltage vo = A vd Ideal Op Amp vo = ∞ (vd) Open Circuit Output Voltage Real Op Amp Voltage Range Positive Saturation A vd > V+ Linear Region V- < A vd < V+ Negative Saturation A vd < V- Output Voltage vo ~ V+ vo = A vd vo ~ V- The voltage produced by the dependent voltage source inside the op amp is limited by the voltage applied to the positive and negative rails. Voltage Transfer Characteristic Range where we operate the op amp as an amplifier. vd Ideal Op Amp v2 i2 = 0 v1 = v2 vd = 0 V i1 = 0 v1 Because Ri is equal to ∞W, the voltage across Ri is 0V. Ideal Operational Amplifier The “ideal” op amp is a special case of the ideal differential amplifier with infinite gain, infinite Rid and zero Ro . v v o id A and lim vid 0 A If A is infinite, vid is zero for any finite output voltage. Infinite input resistance Rid forces input currents i+ and i- to be zero. The ideal op amp operates with the following assumptions: It has infinite common-mode rejection, power supply rejection, open-loop bandwidth, output voltage range, output current capability and slew rate It also has zero output resistance, input-bias currents, inputoffset current, and input-offset voltage. Almost Ideal Op Amp Ri = ∞ W Therefore, i1 = i2 = 0A Ro = 0 W Usually, vd = 0V so v1 = v2 The op amp forces the voltage at the inverting input terminal to be equal to the voltage at the noninverting input terminal if there is some component connecting the output terminal to the inverting input terminal. Rarely is the op amp limited to V- < vo < V+. The output voltage is allowed to be as positive or as negative as needed to force vd = 0V. Inverting Amplifier: Input and Output Resistances Rout is found by applying a test current (or voltage) source to the amplifier output and determining the voltage (or current) after turning off all independent sources. Hence, vs =0 vx i R i R 2 2 11 But i1=i2 vx i ( R R ) 1 2 1 vs R R since v 0 in i 1 s Since v- = 0, i1=0. Therefore vx = 0 irrespective of the value of ix . Rout 0 Inverting Amplifier:Voltage Gain The negative voltage gain vs isR i R vo 0 1 2 2 But is= i2 and v- = 0 (since vid= v+ - v-= 0) R vs vo is 2 and Av R vs R 1 1 implies that there is a 1800 phase shift between both dc and sinusoidal input and output signals. The gain magnitude can be greater than 1 if R2 > R1 The gain magnitude can be less than 1 if R1 > R2 The inverting input of the op amp is at ground potential (although it is not connected directly to ground) and is said to be at virtual ground. The Non-inverting Amplifier: Configuration • The input signal is applied to the non-inverting input terminal. • A portion of the output signal is fed back to the negative input terminal. • Analysis is done by relating the voltage at v1 to input voltage vs and output voltage vo . Non-inverting Amplifier: Voltage Gain, Input Resistance and Output Resistance R vs v v 1 and v vo id 1 1 R R 1 2 vs v But vid =0 1 R R vo vs 1 2 R 1 R v o R1 R2 Av 1 2 R vs R 1 1 vs R Since i+=0 in i Rout is found by applying a test current source to the amplifier output after setting vs = 0. It is identical to the output resistance of the inverting amplifier i.e. Rout = 0. Since i-=0 Non-inverting Amplifier: Example Problem: Determine the output voltage and current for the given non-inverting amplifier. Given Data: R1= 3kW, R2 = 43kW, vs= +0.1 V Assumptions: Ideal op amp Analysis: R 43kW Av 1 2 1 15.3 R 3kW 1 vo Avvs (15.3)(0.1V)1.53V Since i-=0, vo 1.53V io 33.3A R R 43kW 3kW 2 1 Types of Gain if is i1 io i i2 Gain Error Gain Error is given by GE = (ideal gain) - (actual gain) For the non-inverting amplifier, GE 1 A 1 1 A (1 A ) Gain error is also expressed as a fractional or percentage error.1 A 1 1 FGE 1 A 1 1 A A PGE 1 100% A Types of Closed Loop Gain Gain Variable Name Equation Units Voltage Gain Current Gain AV AI vo/vs io/is None or V/V None or A/A Transresistance Gain Transconductance Gain AR AG vo/is io/vs V/A or W A/V or W1 Finite Common-Mode Rejection Ratio (CMRR) A(or Adm) = differential-mode gain Acm = common-mode gain vid = differential-mode input voltage vic = common-mode input voltage v v v v id v v id 1 ic 2 2 ic 2 A real amplifier responds to signal common to both inputs, called the common-mode input voltage (vic). In general, v v vo A (v v ) Acm 1 2 dm 1 2 2 vo A (v ) Acm(v ) ic dm id An ideal amplifier has Acm = 0, but for a real amplifier, Acm v v ic A v ic vo A v dm id dm id CMRR A dm A CMRR dm Acm and CMRR(dB) 20log (CMRR) 10 Op-Amp Summing Amplifier Op-Amp Differential Amplifier If R1 = R2 and Rf = Rg: Op Amps Applications Audio amplifiers Speakers and microphone circuits in cell phones, computers, mpg players, boom boxes, etc. Instrumentation amplifiers Biomedical systems including heart monitors and oxygen sensors. Power amplifiers Analog computers Combination of integrators, differentiators, summing amplifiers, and multipliers Applications of Op-Amps Simple EKG circuit Uses differential amplifier to cancel common mode signal and amplify differential mode signal Realistic EKG circuit Uses two non-inverting amplifiers to first amplify voltage from each lead, followed by differential amplifier Forms an “instrumentation amplifier” Applications of Op-Amps Filters Low pass filter Types: •Low pass filter •High pass filter •Band pass filter •Cascading (2 or more filters connected together) R2 R1 Low pass filter transfer function + Low pass filter Cutoff frequency C + Vcc + - Vcc + V0 __ Summary The output of an ideal op amp is a voltage from a dependent voltage source that attempts to force the voltage at the inverting input terminal to equal the voltage at the non-inverting input terminal. Almost ideal op amp: Output voltage limited to the range between V+ and V-. Ideal op amp is assumed to have Ri = ∞ W and Ro = 0 W. Almost ideal op amp: vd = 0 V and the current flowing into the output terminal of the op amp is as much as required to force v1 = v2 when V+< vo< V-. Operation of an op amp was used in the analysis of voltage comparator and inverting amplifier circuits. Effect of Ri < ∞ W and Ro > 0 Wwas shown. THANK YOU