opamp1 - ELEC 101 1 Op Amp Circuits 1 2009/10 Spring...

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Unformatted text preview: ELEC 101 1 Op Amp Circuits 1 2009/10 Spring Voltage Amplifier also called Voltage-Controlled VoltageAmplifier or V-to-V converter a Circuit Model of voltage amplifier Vi RO Ri VO AVVi Vi : input voltage VO : output voltage Ri : input resistance RO : output resistance AV : voltage gain Example: HiFi pre-amp ELEC 101 Op Amp Circuits 2 2009/10 Spring b Circuit Model of source – voltage amplifier – load RS 0 RO Vi source AVVi RL With amplifier With RO = 0 R L R O For ideal voltage amplifier, VO A V Vi ( Ri >> RS ) A V VS ( RO << RL ) VO A V VS Vi VS RL : load resistance VS : source voltage Ri = RO = 0 Ri = R i R S load RS : source resistance c Proof VO Ri VS d RO Vi Ri AVVi VO ELEC 101 Example Op Amp Circuits 3 2009/10 Spring Find voltage gain AV of the voltage amplifier. 100C 0C 0V 1V Temperature sensor VS = 0V at 0C +1mV/C RS = 300 + V - i Voltage amplifier Ri = 2k RO = 400 + VO Meter - display RL = 10k Need to have VO = 1V when VS = 100mV (at 100C) ELEC 101 a Op Amp Circuits 4 Circuit Model 300 2k 400 Vi AVVi VS sensor b 2009/10 Spring amplifier VO 10k meter Use voltage divider 10k 10k 400 2k 10k A V VS 2k 300 10k 400 VO A V Vi 1V A V 100mV 2k 10k 2.3k 10.4k A V 12 ELEC 101 2 Op Amp Circuits 5 Current Amplifier also called Current-Controlled CurrentAmplifier or I-to-I converter a Circuit Model IO Ii Ri AIIi RO Example: HiFi power amp AI : current gain Ii : input current IO : output current Ri : input resistance RO : output resistance 2009/10 Spring ELEC 101 b Op Amp Circuits 6 2009/10 Spring Circuit Model of source – current amplifier – load Ii 0 d IO Proof With Ri RS IS source AIIi RO Current amplifier RL c IO AIIi AIIS IO AIIi RO RO = R L R O Ri = 0 ( Ri << RS ) Ro = ( RO >> RL ) Ri Ii IS With For ideal current amplifier, Ii R i R S load IS : source current Ri = 0 IO AIIS ELEC 101 3 Op Amp Circuits 7 2009/10 Spring Resistance Amplifier also called Current-Controlled Voltage-Amplifier or Transimpedance Amplifier or I-to-V converter a Ii RO Ri b VO ARIi For ideal resistance amplifier, Ri = 0 ( Ri << RS ) , Ro = 0 ( RO << RL ) Ii 0 Ri 0 RO ARIi VO VO A R I S Example: optical sense amplifier ELEC 101 4 Op Amp Circuits 8 2009/10 Spring Conductance Amplifier also called Voltage-Controlled Current-Amplifier or Transconductance Amplifier or V-to-I converter a IO Vi AGVi Ri RO For ideal conductance amplifier, b Ri = ( Ri >> RS ) , RO = ( RO >> RL ) IO Vi Ri AGVi RO I O A G VS Example: CMOS amplifier ELEC 101 Op Amp Circuits 9 2009/10 Spring Ideal Amplifier Vi RO Ri Vi 0 Ri AGVi 0 Ri 0 RO ARIi I O A G Vi A G VS RO IO AIIi RO Ri Ii IO 0 VO A V Vi A V VS AVVi Ii VO IO AIIi AIIS VO VO A R I i ARIS Voltage amplifier (V-AMP), also called Voltage-Controlled VoltageAmplifier or V-to-V converter Conductance amplifier (G-AMP), also called Voltage-Controlled CurrentAmplifier or V-to-I converter Current amplifier (I-AMP), also called Current-Controlled CurrentAmplifier or I-to-I converter Resistance amplifier (R-AMP) , also called Current-Controlled VoltageAmplifier or I-to-V converter ELEC 101 5 Op Amp Circuits 10 Operational Amplifier (op amp) - most popular integrated circuit (IC) - first made in 1960, companies : Fairchild, LM, A .. - basic important component for analog electronics and computing Operational : can perform mathematical operations such as d log exp etc dx It is the basic component of analog computer and can amplify continuous signal. 2009/10 Spring ELEC 101 a Op Amp Circuits 11 Symbol Vi 2009/10 Spring Example V VO Find VO. 1V 1V A 200 k VO V VO A Vi A ( V V ) 200k (1V 1V ) 0V b Basic equations A 200 k VO = A Vi Vi = V - V A is called open-loop voltage gain Op amp is a differential input, single-ended output device. Hence op amp is also a difference (differential) amplifier. Example Find VO. 1V VO 0V VO A Vi A (V V ) 200k (1V 0V) 0.2V ELEC 101 c Op Amp Circuits 12 2009/10 Spring Circuit Model V1 : Positive supply voltage V : Noninverting Input Ii RO Ri AVi Vi V : Inverting Input VO VO : Output a Ground V2 : Negative supply voltage 6 741 Op Amp most typical used op amp Ri 2M, RO 75, A 200,000 15V power supply 741 op amp requires supply voltage ( 8 to 22V) to operate. It is usually powered by 15V. Maximum VO V1 and minimum VO V2 . Vi + 741 op amp V1 = 15V VO V2 = 15V ELEC 101 b Op Amp Circuits 13 Operating regions of 741 op amp maxVO 15V maxVi 75V A 200k 2009/10 Spring Example Find VO. 741 op amp A 200 k Negative Saturation Region VO = 15V VO Positive Saturation Region VO = 15V 15V 999mV 1000mV VO 15V VO A Vi A ( V V ) 15V 75V 75V 15V Vi 200k (999mV 1000mV ) 200V VO 15V Linear (or amplifier) region VO = AVi Linear region of op amp is quite narrow (range of Vi 150 V). Op amp is impractical to use. op amp saturates VO 15V ELEC 101 7 Op Amp Circuits 14 Ideal Op Amp Assumptions b 15V 2009/10 Spring Vi 0V Vi 0V or V = V called virtual short assumption Vi Ii V RO VO Ri AVi Short Proof IO Vi VO 15V 0V A V 15V c Ideal op amp assumptions are commonly used to simplify op amp circuit analysis. Ideal op amp has several assumptions. a R i , Ii 0A I i 0A called virtual open assumption Short Proof Ii Vi 0V 0A Ri R O 0 , A , Bandwidth Note that IO is not necessary equal to 0A. ELEC 101 Op Amp Circuits 15 Example Find IL. Assume ideal op amp. VS = 1V VO i 400 I a 200 Vi V V 0V c 2009/10 Spring VO I 200 i 400 5mA 600 3V V 3V IL O 1.5mA 2k 2k IL load RL 2k VO is controlled by VS and is independent of the load RL Vi 0V 0A V V VS 1V 1V V I 5mA 200 200 b Ii 0A i I 5mA Ii 0A 6.5mA VO 3V 1V VS = 1V 0A 5mA 400 5mA 200 6.5mA I L 1.5mA load RL 2k ELEC 101 Op Amp Circuits 16 Example Find VO. Assume ideal op amp. 2009/10 Spring Example Find VO. Assume ideal op amp. I VO VS R VS V R R R switch V VO R Ii 0A apply KCL to V , I Ii 0A Ii VS VO apply KCL to V , V V 0V 0A VS V V VO R R VO VS When switch is closed, circuit is an inverter. VS V V VO R R VO VS V VS V R I i 0A Ii VO VS switch R R R R R R R VS 0A VS 0A VO VS R VS When switch is open, circuit is a follower. ELEC 101 8 Op Amp Circuits 17 Inverting Amplifier Short proof A fundamental op amp feedback circuit. R1 VS apply KCL to V R2 V 2009/10 Spring VO V VS V V VO R1 R2 V V 0V V is also called virtual ground. VO R2 AV VS R1 AV is called closed loop voltage gain, R2 is a feedback resistor. AV is precise, stable and easy to control. The voltage gain AV is always negative, it can be some values between 0 and . VS 0V 0V VO R1 R2 VO R 2 VS R1 ELEC 101 Op Amp Circuits 18 Example Find VO. Assume ideal op amp. 2009/10 Spring Example Find VO(t). Assume ideal op amp. 6k 6k 1k VO VS 1k VS(t) VO VS 1V 1V 3k R2 R1 6k / /3k 2 V 1k 2V 0V 2V 3k VO(t) 4V VO(t) 0V 4V VO ( t ) 2 VS ( t ) ELEC 101 9 Op Amp Circuits 19 Summing Amplifier 2009/10 Spring Proof (a) apply KCL Extension of inverting amplifier. Can be extended to N inputs. R2 VS1 V VS2 V V VO R S1 R S2 R2 V V 0V RS2 VO VS2 RS1 VS1 R R VO 2 VS1 2 VS2 R S1 R S2 This is also called as an “Adder” which adds several input signals with different weighting factors. VO R2 R VS1 2 VS2 R S1 R S2 (b) Or using superposition. ELEC 101 Example Op Amp Circuits 20 Find VO. Assume ideal op amp. 2k 2009/10 Spring Example Find VO(t). Assume ideal op amp. 2k 6k VS2 =2V VO 3k VS1 =1V 2V 0V 2V 6k VS2(t) VO(t) 3k VS1 =1V R2 R VS1 2 VS2 R S1 R S2 VO ( t ) 6k 6k 1V 2V 3k 2 k 2V 6V 8V R2 R VS1 2 VS2 R S1 R S2 VO ( t ) 6k 6k VS1 ( t ) VS2 3k 2 k 2V 3VS2 ( t ) VO(t) 4V 2V 8V ELEC 101 Op Amp Circuits 21 Extended to N inputs, with all resistors have the same value. 2009/10 Spring With different resistor values. RSn R R VS2 V R VS1 R2 VSn R VSn VO V VO (VS1 VS2 VSn ) This is also called as an “Adder” which adds several input signals together. VS2 RS2 V RS1 VS1 VO V R R R VO 2 VS1 2 VS2 2 VSn R S2 R Sn R S1 This is also called as an “Weighted Summer” in which the output is a weighted sum of the input signals. ELEC 101 Op Amp Circuits 22 With all RS’s are variable resistors, the gain of each individual inputs can be adjusted. VSn RSn VS2 R2 RS2 V RS1 VS1 VO V R2 R2 R2 VO VS1 VS2 VSn R S1 R S2 R Sn Applications: Karaoke, audio mixer, equalizer, etc. 2009/10 Spring ELEC 101 10 Op Amp Circuits 23 2009/10 Spring Non-inverting Amplifier Short proof Another fundamental op amp feedback circuit. R1 R2 V VS VO V VO R2 AV 1 VS R1 The voltage gain AV is always positive, it can be some values between 1 and . apply KCL to V 0V V V VO R1 R2 V V VS VS VS VO R1 R2 R2 VS R1 VO R 1 2 VS R1 VO VS ELEC 101 Op Amp Circuits 24 Example Find VO. Assume ideal op amp. 2k 2009/10 Spring Example Find VO(t). Assume ideal op amp. 4k 2k VO VS=2V VO 4 k R 1 2 1 3 2 k R1 VS 4k VO(t) VS(t) 2V 0V 2V VO ( t ) R 4k 1 2 1 3 VS ( t ) R1 2k VO 3 VS 3 2V 6V VO(t) 6V 0V 6V ELEC 101 Op Amp Circuits 25 2009/10 Spring Voltage Follower or Buffer Amplifier 11 When we set R1 = , we have a AV VO R R 1 2 1 2 1 VS R1 R2 R1= V VS V R 2 can be 0 R2 VO V VS VO V V VS V Since the voltage gain is 1, the amplifier is called Voltage Follower. The value of the output voltage always follows the value of the input voltage. VO ELEC 101 Op Amp Circuits 26 b VS For an ideal op amp, VO R i , R O 0 Voltage follower has very high input resistance and very low output resistance. The amplifier can be used to connect a high input resistance source to a low resistance load (act as a buffer amplifier). 2009/10 Spring ELEC 101 Op Amp Circuits 27 Example A sensor has an internal resistance of 19k and outputs a sensed voltage of 5mV. If the sensor is connected to a voltmeter with input resistance of 1k, what is the measured voltage? sensor resistance = 19k sensor voltage Vsensed = 5mV sensor Vmeasured 5mV 2009/10 Spring When a buffer amplifier is added between the senor and the voltmeter, what is the measured voltage? I voltmeter + Vmeasured input resistance of meter = 1k 1k 0.25mV 19k 1k Vsensed The measured voltage is much smaller than the sensed voltage. Vsensed 19k VS VO + 1k Vmeasured Since I = 0, VS = Vsensed, VO = VS (buffer amplifier). With RO = 0, Vmeasured = VO, Therefore, Vmeasured = Vsensed. The measured voltage is exactly the same as the sensed voltage from the sensor. ELEC 101 12 Op Amp Circuits 28 Difference Amplifier Proof The output voltage of the Difference Amplifier, also called as Differential Amplifier, is proportional to the difference of signal V2V1. R3 2009/10 Spring apply KCL to V_ V1 V V VO R3 R4 V V V2 R4 V1 R3 VO V2 R4 VO R4 V2 V1 R3 VO V R4 R3 R4 R4 (V1 V ) R3 R4 R ) V1 4 R3 R3 R4 R R VO V2 (1 4 ) V1 4 R3 R4 R3 R3 V (1 R4 (V2 V1 ) R3 ELEC 101 Example Op Amp Circuits 29 Find VO. 2k 10k 1V VO 1.01V 2k VO 10k R4 10k (V2 V1 ) (1.01V 1V) 0.05V R3 2k 2009/10 Spring ELEC 101 Op Amp Circuits 30 2009/10 Spring This Difference Amplifier has two disadvantages: (1) Need to change two resistors (R4 or R3) to change the gain of the amplifier. I V2V1 R3 R4 R3 I VO VO R4 V2 V1 R3 R4 virtual short (2) Input resistance is not very high (=2R3). Hence it draw current from inputs. Higher the gain means to have a lower R3. This leads to a lower input resistance. Recall that an idea voltage amplifier should have an infinite large input resistance. ELEC 101 13 Op Amp Circuits 31 Instrumentation Amplifier A special purpose differential amplifier. V2 R Advantages: The input resistance is very large ( ~ ). Hence draws NO current from source. Only need to change one resistor ( R ) to change the gain of the Instrumentation Amplifier. R1 I 2009/10 Spring VO R1 V1 Disadvantage: The output voltage is a differential voltage. V2 V1 (R 2R 1 ) R R V2 V1 (1 2 1 ) R VO I (R 2R 1 ) ELEC 101 Op Amp Circuits 32 VO V1 V2 2V1 V2 R 1 V1 V2 R2 RG Another Instrumentation Amplifier, single-ended output. V1 V1 V2 I2 R2 VO V1 R2 R1 RG I1 V2 R1 R2 2009/10 Spring VO V1 V2 RG V2 R2 R R 2R VO 1 V1 V2 1 1 1 R R RG 2 2 R R VO 1 2 2 2 V1 V2 R RG 1 Advantages: V2 I1R1 I2R1 V1 The input resistance is very large ( ~ ). Hence draws NO current from source. V2 V1 V2 V V1 V1 V2 R 1 O R 1 V1 V2 R RG RG 2 R2 Only need to change RG to adjust the gain. V2 VO V1 V1 V2 V1 V2 R 1 V1 V2 R2 RG The output voltage VO is reference to Ground. ELEC 101 14 a Op Amp Circuits 33 2009/10 Spring Other Applications of Op Amp Circuits This circuit is a good current source. Current Source If R1 = 100k , RF = 1M , R2 =10k , V1 = 2V, then iL = 2mA. R1 RF iF V1 i1 i1 V1 iF R1 V2 i F R F V1 RF R1 i L i F i 2 i1 i 2 V2 Z iL V1 V2 R1 R 2 V1 V1R F V1 R (1 F ) R 1 R 1R 2 R 1 R2 iL is independent of Z. iL depends on V1 and the resistors. And i1 = 20A and draws almost no current from V1 . R2 i2 ELEC 101 b Op Amp Circuits 34 Negative Impedance Converter R1 i V i3 R i3 V R1 i 2 i1 V1 R1 V(1 R 2 / R1 ) V R 1 V 2 R R1 R R in V V V R R 1 i i3 V R 2 1 R2 R1 R If we set R1 = R2, R in R . This op amp circuit generates a negative resistance. R2 i2 i1 i1 2009/10 Spring R V 1 2 R 1 R may in general be replaced by an arbitrary impedance Z. It generates a negative impedance, Z. ELEC 101 c Op Amp Circuits 35 2009/10 Spring Voltage-to-Current Converter (V-to-I converter) r r R VS VS R R iL VS R iL Z R Z Z iL VS R This is a current source (voltage-to-current converter). Current IL is directly proportional to VS (IL = VS/R) and is independent of the value of the load impedance. R R Z iL ELEC 101 d Op Amp Circuits 36 Analog-to-Digital Converter and Digital-to-Analog Converters (A/D, D/A Converters) Digital Signal Processing (DSP) A digital system with analog I/O Example: Digitizing an analog signal to with 3-bit resolution 2009/10 Spring ELEC 101 Op Amp Circuits 37 2009/10 Spring Analog-to-Digital Converter (A/D Converter) Vref v Op amp functions as a Comparator i +V R R CMP B R CMP C Binary encoder CMP Vref A vi CMP VO 2 Binary output b1b0 When vi > Vref, VO = +V (logic “1”). When vi < Vref, VO = 0 (logic “0”). R Binary encoder Analog signal A B C b1 b0 vi < ¼Vref 0 0 0 0 0 ¼Vref < vi < ½Vref 0 0 1 0 1 ½Vref < vi < ¾Vref 0 1 1 1 0 ¾Vref < vi 1 1 1 1 1 Note: A 2-bit A/D converter needs 3 op amps. An n-bit A/D converter needs 2n1 op amps. This is called a parallel or flash A/D converter. ELEC 101 Op Amp Circuits 38 2009/10 Spring Binary encoder Analog signal A B C b1 b0 vi < ¼Vref 0 0 0 0 0 ¼Vref < vi < ½Vref 0 0 1 0 1 ½Vref < vi < ¾Vref 0 1 1 1 0 ¾Vref < vi 1 1 1 1 1 BC b1 BC b0 00 01 11 10 A The binary encoder can be implemented with a simple logic function. 00 01 11 10 A 0 0 0 1 X 0 0 1 0 X 1 X X 1 X 1 X X 1 X b1 BC b0 BC AC ELEC 101 Op Amp Circuits 39 2009/10 Spring Digital-to-Analog Converter (D/A Converter) V LSB 16R V 8R V 4R V 2R R r r CD Player VO MSB If all switches are closed, VO R( V V V V 15 V V V V )( ) V 2R 4R 8R 16R 2 4 8 16 16 If all switches are open, VO R( 0 0 0 0 ) ( 0 0 0 0 ) 0 V 2R 4R 8R 16R 2 4 8 16 16 V 0 V 0 10 0 0 V V If the 4-bit digital data is 1010, VO R( )( ) V 2R 4R 8R 16R 2 4 8 16 16 This is a 4-bit Digital-to-Analog Converter which provides 16-level output voltages from 0V, (1/16)V, (2/16)V, …, up to (15/16)V. ELEC 101 e Op Amp Circuits 40 Synthesizer V4 V3 V2 V1 VO R( 2009/10 Spring R4 R3 R r r R2 R1 VO R R R R V1 V2 V3 V4 V3 V2 ) V1 R R R V4 R R1 R 2 R 3 R 4 1 2 4 3 A synthesizer is an electronic instrument that can produce various sounds by generating and combining different frequencies. It can produce thousands of different sounds and sound combinations by controlling different resistor ratios R/R1, R/R2, etc. ELEC 101 15 Op Amp Circuits 41 Real Op Amp calculation When op amp is real, ideal op amp assumptions (Ri = , RO = 0 , A = , bandwidth = ) cannot be used. The real op amp model is used. 15V Vi I V i RO Ri AVi VO IO V 15V That’s mean a voltage-control voltage-source (dependent source) with finite input and output resistance is used in the calculation. 2009/10 Spring ELEC 101 Op Amp Circuits 42 Example Find VO/VS . For the op amp, b 2009/10 Spring V VS IR1 VS R i , R O 0, A 50 VO VS R1 R1 R 2 but VO AVi A(V V ) AV R1 800 R 2 10k VS VO VO V V VS O S R1 A R1 R 2 R1 800 1 VO R1 R 2 800 10k 9.84 1 800 1 R1 VS 50 800 10k A R1 R 2 1 R2 10k a R1 800 VS I V Vi VO c When A = 1000, VO/VS = 12.33 AVi V When A = 100, VO/VS = 11.01 d When op amp is ideal A = VO R 10k 2 12.5 VS R1 800 ELEC 101 Op Amp Circuits 43 2009/10 Spring Find VO for an ideal voltage follower. Example 19k sensor voltage = 5mV 1k voltage follower voltmeter sensor 5mV 19k Vi 5mV 5mV AVi 1k + VO 5mV Recall in page 27, without the voltage follower, VO 5mV 1k 0.25mV 19k 1k ELEC 101 Op Amp Circuits 44 Example For a non-ideal op amp, Ri =10k, RO = 100 , A = 104. Find VO. Ii Vi 5mV 10k 100 IO 104Vi 1k + VO 19k 5mV VO 10 4 Vi VO VO 19k 10k 100 1k But Vi I i 10k 5mV VO 10k 19k 10k 10 5mV VO 29 105 5mV VO VO V 5mV VO 29 O 19k 10k 100 1k 5mV VO 107 5mV VO 290VO 29VO 2009/10 Spring 107 5mV 5mV VO 7 10 290 29 1 4.99984mV 5mV Even the op amp is not ideal, the output voltage is closed to the one with an ideal op amp. Ii 5mV 4.99984mV 5.517 pA 19k 10k R iF 5mV 5mV 906.3M Ii 5.517pA Even with Ri =10k, the effective input resistance of the op amp circuit has a very large input resistance. This feedback characteristics will be discussed in advanced electronic circuit course. ...
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This note was uploaded on 01/28/2011 for the course ELEC 101 taught by Professor Chan during the Fall '09 term at HKUST.

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