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ANALOG ELECTRONICS 3rd ELECTRICAL DEPARTMENT Prajapati Omprakash 130960109028 ARUN MUCHHALA ENGINEERING COLLEGEDHARI ANALOG ELECTRONICS [2014-2015] Chapter Goals • Understand the “magic” of negative feedback and the characteristics of ideal op amps. • Understand the conditions for non-ideal op amp behavior so they can be avoided in circuit design. • Demonstrate circuit analysis techniques for ideal op amps. • Characterize inverting, non-inverting, summing and instrumentation amplifiers, voltage follower and first order filters. • Learn the factors involved in circuit design using op amps. • Find the gain characteristics of cascaded amplifiers. • Special Applications: The inverted ladder DAC and successive approximation ADC ANALOG ELECTRONICS Differential Amplifier Model: Basic Represented by: A = open-circuit voltage gain vid = (v+-v-) = differential input signal voltage Rid = amplifier input resistance Ro = amplifier output resistance The signal developed at the amplifier output is in phase with the voltage applied at the + input (non-inverting) terminal and 180° out of phase with that applied at the - input (inverting) terminal. ANALOG ELECTRONICS LM741 Operational Amplifier: Circuit Architecture Current Mirrors ANALOG ELECTRONICS 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, openloop bandwidth, output voltage range, output current capability and slew rate – It also has zero output resistance, input-bias currents, input-offset current, and input-offset voltage. ANALOG ELECTRONICS The Inverting Amplifier: Configuration • The positive input is grounded. • A “feedback network” composed of resistors R1 and R2 is connected between the inverting input, signal source and amplifier output node, respectively. ANALOG ELECTRONICS Inverting Amplifier:Voltage Gain • The negative voltage gain 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 vs isR i R vo 0 • The inverting input of the op 1 2 2 amp is at ground potential But is= i2 and v- = 0 (since vid= v+ - v-= 0) (although it is not connected directly to ground) and is said to R vs vo is and Av 2 be at virtual ground. R vs R 1 1 ANALOG ELECTRONICS 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 ANALOG ELECTRONICS Inverting Amplifier: Example • • • • Problem: Design an inverting amplifier Given Data: Av= 20 dB, Rin = 20kW, Assumptions: Ideal op amp Analysis: Input resistance is controlled by R1 and voltage gain is set by R2 / R1. AvdB 20log Av , Av 1040dB/20dB 100 and Av = -100 10 A minus sign is added since the amplifier is inverting. R R 20kW 1 in R Av 2 R 100R 2MW 2 1 R 1 ANALOG ELECTRONICS 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 . ANALOG ELECTRONICS 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 ANALOG ELECTRONICS Non-inverting Amplifier: Example • Problem: Determine the output voltage and current for the given noninverting 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 Av vs (15.3)(0.1V)1.53V Since i-=0, vo 1.53V io 33.3A R R 43kW 3kW 2 1 ANALOG ELECTRONICS Finite Open-loop Gain and Gain Error vo Av A(vs v ) A(vs vo ) 1 id vo A Av v s 1 A A is called loop gain. R 1 v v v o 1 R R o 1 2 R 1 is called the R R feedback factor. 1 2 For A >>1, R 1 Av 1 2 R 1 This is the “ideal” voltage gain of the amplifier. If A is not >>1, there will be “Gain Error”. ANALOG ELECTRONICS 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 ANALOG ELECTRONICS Gain Error: Example • Problem: Find ideal and actual gain and gain error in percent • Given data: Closed-loop gain of 100,000, open-loop gain of 1,000,000. • Approach: The amplifier is designed to give ideal gain and deviations from the ideal case have to be determined. Hence, 1 . 10 5 Note: R1 and R2 aren’t designed to compensate for the finite open-loop gain of the amplifier. 6 A 10 Av 9.09x104 • Analysis: 1 A 106 1 105 105 9.09x10 4 PGE 100% 9.09% 5 10 ANALOG ELECTRONICS Output Voltage and Current Limits Practical op amps have limited output voltage and current ranges. Voltage: Usually limited to a few volts less than power supply span. Current: Limited by additional circuits (to limit power dissipation or protect against accidental short circuits). The current limit is frequently specified in terms of the minimum load resistance that the amplifier can drive with a given output voltage swing. Eg: i 5V 10mA o 500W v vo v io i i o o L F R R R R L 2 1 EQ R R (R R ) EQ L 1 2 For the inverting amplifier, ANALOG ELECTRONICS R R R EQ L 2 Example PSpice Simulations of Non-inverting Amplifier Circuits ANALOG ELECTRONICS ANALOG ELECTRONICS ANALOG ELECTRONICS ANALOG ELECTRONICS ANALOG ELECTRONICS The Unity-gain Amplifier or “Buffer” • This is a special case of the non-inverting amplifier, which is also called a voltage follower, with infinite R1 and zero R2. Hence Av = 1. • It provides an excellent impedance-level transformation while maintaining the signal voltage level. • The “ideal” buffer does not require any input current and can drive any desired load resistance without loss of signal voltage. • Such a buffer is used in many sensor and data acquisition system applications. ANALOG ELECTRONICS The Summing Amplifier Since the negative amplifier input is at virtual ground, v v 1 i i 2 i vo 1 R 2 R R 1 2 3 3 Since i-=0, i3= i1 + i2, R R vo 3 v 3 v R 1 R 2 1 2 • Scale factors for the 2 inputs can be independently adjusted by the proper choice of R2 and R1 . • Any number of inputs can be connected to a summing junction through extra resistors. • This circuit can be used as a simple digital-to-analog converter. This will be illustrated in more detail, later. ANALOG ELECTRONICS The Difference Amplifier R Since v-= v+ vo 2 (v v ) R 1 2 1 For R2= R1 vo (v1 v2) • This circuit is also called a differential amplifier, since it amplifies the difference between the input signals. v o v- i R v- i R • Rin2 is series combination of R1 2 2 1 2 and R2 because i+ is zero. R R R R 2 v- 2 v • For v2=0, Rin1= R1, as the circuit v- 2 ( v v- ) 1 R 1 R R 1 reduces to an inverting amplifier. 1 1 1 • For general case, i1 is a function R of both v1 and v2. 2 v Also, v R R 2 1 2 ANALOG ELECTRONICS Difference Amplifier: Example • • • • Problem: Determine vo Given Data: R1= 10kW, R2 =100kW, v1=5 V, v2=3 V Assumptions: Ideal op amp. Hence, v-= v+ and i-= i+= 0. Analysis: Using dc values, R 100kW A 2 10 dm R 10kW 1 Vo A V V 10(5 3) dm 1 2 Vo 20.0 V Here Adm is called the“differential mode voltage gain” of the difference amplifier. ANALOG ELECTRONICS Finite Common-Mode Rejection Ratio (CMRR) 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 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 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 ANALOG ELECTRONICS Finite Common-Mode Rejection Ratio: Example • Problem: Find output voltage error introduced by finite CMRR. • Given Data: Adm= 2500, CMRR = 80 dB, v1 = 5.001 V, v2 = 4.999 V • Assumptions: Op amp is ideal, except for CMRR. Here, a CMRR in dB of 80 dB corresponds to a CMRR of 104. • Analysis: v 5.001V 4.999V id v 5.001V 4.999V 5.000V ic 2 v 5.000 ic V 6.25V vo A v 25000.002 dm id CMRR 104 In the "ideal" case, vo A v 5.00 V dm id 6.255.00 % output error 100% 25% 5.00 The output error introduced by finite CMRR is 25% of the expected ideal output. ANALOG ELECTRONICS uA741 CMRR Test: Differential Gain ANALOG ELECTRONICS Differential Gain Adm ANALOG = 5 V/5ELECTRONICS mV = 1000 uA741 CMRR Test: Common Mode Gain ANALOG ELECTRONICS Common Mode Gain Acm = 160 mV/5 V = .032 ANALOG ELECTRONICS CMRR Calculation for uA741 Adm 1000 CMRR 3.125x10 4 Acm .032 CMRR(dB) 20log 10 CMRR 89.9 dB ANALOG ELECTRONICS