* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Chapter 9
Chirp spectrum wikipedia , lookup
Sound reinforcement system wikipedia , lookup
Ringing artifacts wikipedia , lookup
Loudspeaker wikipedia , lookup
Utility frequency wikipedia , lookup
Flip-flop (electronics) wikipedia , lookup
Ground loop (electricity) wikipedia , lookup
Signal-flow graph wikipedia , lookup
Switched-mode power supply wikipedia , lookup
Stage monitor system wikipedia , lookup
Audio power wikipedia , lookup
Control system wikipedia , lookup
Resistive opto-isolator wikipedia , lookup
Scattering parameters wikipedia , lookup
Dynamic range compression wikipedia , lookup
Two-port network wikipedia , lookup
Zobel network wikipedia , lookup
Public address system wikipedia , lookup
Oscilloscope history wikipedia , lookup
Schmitt trigger wikipedia , lookup
Rectiverter wikipedia , lookup
Opto-isolator wikipedia , lookup
Regenerative circuit wikipedia , lookup
Electronics Principles & Applications Fifth Edition Charles A. Schuler Chapter 9 Operational Amplifiers ©1999 Glencoe/McGraw-Hill INTRODUCTION • The Differential Amplifier • The Operational Amplifier • Determining Gain • Frequency Effects • Applications • Comparators A differential amplifier driven at one input +VCC Noninverted output Inverted output C B E C E -VEE B Both outputs are active because Q1 drives Q2. Q1 serves as an emitter-follower amplifier in this mode to drive Q2. +VCC C B Q1 E Q2 serves as a common-base amplifier in this mode. It’s driven at its emitter. C E Q 2 -VEE B A differential amplifier driven at both inputs Common mode input signal +VCC Reduced output Reduced output C B E C E -VEE B A differential amplifier driven at both inputs Differential mode input signal +VCC Increased output Increased output C B E C E -VEE B Differential amplifier dc analysis IRE = IE = VEE - VBE RE IRE 2 9 V - 0.7 V = 2.13 mA = 3.9 kW VCC VRL = IC x RL = 1.06 mA x 4.7 kW +9 V = 4.98 V = 1.06 mA 4.7 kW RL RL IC = IE = 1.06 mA C B 10 kW RB E 3.9 kW VEE VCE = VCC - VRL - VE 4.7 kW = 9 - 4.98 -(-0.7) C E RE -9 V = 4.72 V B RB 10 kW Differential amplifier dc analysis continued Assume b = 200 IB = IC b VB = VRB = IB x RB 1.06 mA = 200 = 5.3 mA VCC 4.7 kW 10 kW RB +9 V = 53 mV RL RL 4.7 kW C B = 5.3 mA x 10 kW E 3.9 kW VEE C E RE -9 V B RB 10 kW Differential amplifier ac analysis 50 mV 50 mV (50 mV is conservative) = 47 W = rE = IE 1.06 mA RL VCC +9 V AV(CM) = RL 2 x RE AV(DIF) = 2 x rE 4.7 kW 4.7 kW RL RL 4.7 kW = 4.7 kW 2 x 3.9 kW = = 50 2 x 47 W C C = 0.6 B 10 kW RB E 3.9 kW VEE E RE -9 V B RB 10 kW Differential amplifier ac analysis continued CMRR = 20 x log AV(DIF) AV(CM) VCC 4.7 kW 10 kW RB 50 = 38.4 dB 0.6 +9 V RL RL 4.7 kW C B = 20 x log E 3.9 kW VEE C E RE -9 V B RB 10 kW AV(CM) = A current source can replace RE to decrease the common mode gain. RL VCC 2 x RE Replaces this with a very high resistance value. 4.7 kW C B 10 kW RL RL 4.7 kW RB E * C E B RB 10 kW 2 mA *NOTE: Arrow shows conventional current flow. A practical current source IC 9 V - 5.1 V IZ = = 10 mA 390 W IE = 390 W 5.1 V - 0.7 V 2.2 kW IC = IE = 2 mA 5.1 V 2.2 kW -9 V = 2 mA Differential amplifier quiz When a diff amp is driven at one input, the number of active outputs is _____. two When a diff amp is driven at both inputs, there is high gain for a _____ signal. differential When a diff amp is driven at both inputs, there is low gain for a ______ signal. common-mode The differential gain can be found by dividing the collector load by ________. 2rE The common-mode gain can be found by dividing the collector load by ________. 2RE Op amps have two inputs Inverting input Non-inverting input Output Op-amp Characteristics • • • • High CMRR High input impedance High gain Low output impedance • • • • Available as ICs Inexpensive Reliable Widely applied With both inputs grounded through equal resistors, VOUT should be zero volts. +VCC VOUT -VEE Imperfections can make VOUT non-zero. The offset null terminals can be used to zero VOUT. Dt DV 741 DV Slew rate = Dt 0.5 V ms The output of an op amp cannot change instantaneously. VP f > fMAX fMAX = Slew-rate distortion Slew Rate 2p x VP Operational amplifier quiz The input stage of an op amp is a __________ amplifier. differential Op amps have two inputs: one is inverting and the other is ________. noninverting An op amp’s CMRR is a measure of its ability to reject a ________ signal. common-mode The offset null terminals can be used to zero an op amp’s __________. output The ability of an op amp output to change rapidly is given by its _________. slew rate Op-amp follower AV(OL) = the open loop voltage gain AV(CL) = the closed loop voltage gain This is a closed-loop circuit with a voltage gain of 1. RL It has a high input impedance and a low output impedance. Op-amp follower AV(OL) = 200,000 AV(CL) = 1 The differential input approaches zero due to the high open-loop gain. Using this model, VOUT = VIN. VDIF = 0 VIN VOUT RL Op-amp follower AV(OL) = 200,000 B=1 A AB +1 VIN VOUT The feedback ratio = 1 200,000 @1 AV(CL) = (200,000)(1) + 1 VIN VOUT RL The closed-loop gain is increased by decreasing the feedback with a voltage divider. R1 200,000 AV(CL) = = 11 (200,000)(0.091) + 1 RF R1 B= 100 kW RF + R1 10 kW = VIN VOUT RL 10 kW 100 kW + 10 kW = 0.091 It’s possible to develop a different model for the closed loop gain by assuming VDIF = 0. R1 VIN = VOUT x R1 + RF RF Divide both sides by VOUT and invert: 100 kW R1 10 kW VOUT VDIF = 0 VIN VOUT VIN RF =1+ R1 RL AV(CL) = 11 In this amplifier, the assumption VDIF = 0 leads to the conclusion that the inverting op amp terminal is also at ground potential. This is called a virtual ground. Virtual ground RF We can ignore the op amp’s input current since it is so small. Thus: IR1 = IRF 10 kW R1 By Ohm’s Law: 1 kW VIN VDIF = 0 VIN R1 VOUT RL VOUT VIN = = -VOUT RF -RF R1 = -10 The minus sign designates an inverting amplifier. Due to the virtual ground, the input impedance of the inverting amplifier is equal to R1. Virtual ground RF 10 kW R1 1 kW VDIF = 0 VIN Although op amp input currents are small, in some applications, offset error is minimized by providing equal paths for the input currents. R2 = R1 || RF = 910 W This resistor reduces offset error. A typical op amp has internal frequency compensation. R Output C Break frequency: 1 fB = 2pRC Bode plot of a typical op amp Break frequency 120 100 80 60 Gain in dB 40 20 0 1 10 100 1k 10 k 100 k 1M Frequency in Hz Op amps are usually operated with negative feedback (closed loop). This increases their useful frequency range. AV(CL) = RF VIN RF =1+ R1 100 kW =1+ = 101 1 kW 100 kW R1 1 kW VOUT dB Gain = 20 x log 101 = 40 dB VIN VOUT RL Using the Bode plot to find closed-loop bandwidth: 120 100 Gain in dB 80 Break frequency 60 AV(CL) 40 20 0 1 10 100 1k 10 k 100 k 1M Frequency in Hz A 741 op amp slews at 0.5 V ms 70 V A 318 op amp slews at ms There are two frequency limitations: Slew rate determines the large-signal bandwidth. Internal compensation sets the small-signal bandwidth. The Bode plot for a fast op amp shows increased small-signal bandwidth. 120 100 80 Gain in dB 60 40 fUNITY 20 0 1 10 100 1k 10 k 100 k 1M 10M Frequency in Hz fUNITY can be used to find the small-signal bandwidth. AV(CL) = RF VIN RF =1+ R1 100 kW =1+ = 101 1 kW 100 kW R1 1 kW VIN VOUT VOUT 318 Op amp RL fUNITY fB = AV(CL) 10 MHz = = 99 kHz 101 Op amp feedback quiz The open loop gain of an op amp is reduced with __________ feedback negative The ratio RF/R1 determines the gain of the ___________ amplifier. inverting 1 + RF/R1 determines the gain of the ___________ amplifier. noninverting Negative feedback makes the - input of the inverting circuit a ________ ground. virtual Negative feedback _________ small signal bandwidth. increases R Amplitude response of RC lag circuit Vout C 1 fb = 2pRC fb 0 dB -20 dB Vout -40 dB -60 dB 10fb 100fb 1000fb f R Phase response of RC lag circuit = tan -1 Vout -45o -90o C -XC R 0.1fb 0o Vout fb 10fb f Interelectrode capacitance and Miller effect The gain from base to collector makes CBC effectively larger C BE in the input circuit. CBC R CMiller = AVCBC CInput = CMiller + CBE CMiller CBE 1 fb = 2pRCInput Bode plot of an amplifier with two break frequencies. 50 dB 40 dB 20 dB/decade 30 dB 20 dB 40 dB/decade 10 dB 0 dB 10 Hz 100 Hz 1 kHz fb1 10 kHz 100 kHz fb2 Multiple lag circuits: Vout R1 C1 R2 C2 R3 0o Vout Phase reversal -180o Negative feedback becomes positive C3 f Op amp compensation • Interelectrode capacitances create several break points. • Negative feedback becomes positive at some frequency due to cumulative phase lags. • If the gain is > 0 dB at that frequency, the amplifier is unstable. • Frequency compensation reduces the gain to 0 dB or less. Op amp compensation quiz Beyond fb, an RC lag circuit’s output drops at a rate of __________ per decade. 20 dB The maximum phase lag for one RC network is __________. 90o An interelectrode capacitance can be effectively much larger due to _______ effect. Miller Op amp multiple lags cause negative feedback to be ______ at some frequency. positive If an op amp has gain at the frequency where feedback is positive, it will be ______. unstable Inverted sum of three sinusoidal signals RF 10 kW 5 kW 5 kHz 3.3 kW 3 kHz 1 kW 1 kHz Summing Amplifier Amplifier scaling: 1 kHz signal gain is -10 3 kHz signal gain is -3 5 kHz signal gain is -2 Difference of two sinusoidal signals (V1 = V2) RF 1 kW Subtracting Amplifier (A demonstration of common-mode rejection) 1 kW 1 kW 1 kW V1 V2 VOUT = V2 - V1 Active low-pass filter VIN -3 dB Gain Frequency fC VOUT Active high-pass filter VIN -3 dB Gain fC Frequency VOUT VIN VOUT Active band-pass filter -3 dB Gain Frequency Bandwidth VOUT VIN Active band-stop filter -3 dB Gain Frequency Stopband Integrator C R VIN V Slope = s VOUT 1 Slope = -VIN x RC Comparator with a 1 Volt reference +VSAT 1V 0V -VSAT VOUT VIN 1V Comparator with a noisy input signal +VSAT 1V 0V -VSAT VOUT VIN 1V Schmitt trigger with a noisy input signal +VSAT UTP LTP Trip points: R1 VSAT x R1 + RF -VSAT VIN R1 VOUT RF Hysteresis = UTP - LTP +5 V Window comparator 4.7 kW VUL 3V R1 311 VOUT R2 4.7 kW VIN 311 VLL 1 V VOUT is LOW (0 V) when VIN is between 1 V and 3 V. Window comparator VUL 3V +5 V 311 VOUT VIN 311 VLL 1V Many comparator ICs require pull-up resistors in applications of this type. +5 V Window comparator 4.7 kW VUL 3V R1 311 VOUT R2 4.7 kW VIN 311 VLL 1V VOUT is TTL logic compatible. Op amp applications quiz A summing amp with different gains for the inputs uses _________. scaling Frequency selective circuits using op amps are called _________ filters. active An op amp integrator uses a _________ as the feedback element. capacitor A Schmitt trigger is a comparator with __________ feedback. positive A window comparator output is active when the input is ______ the reference points. between REVIEW • The Differential Amplifier • The Operational Amplifier • Determining Gain • Frequency Effects • Applications • Comparators