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**Principles & Applications Operational Amplifiers**

Electronics Principles & Applications Sixth Edition Charles A. Schuler Chapter 9 Operational Amplifiers (student version) © Glencoe/McGraw-Hill

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**INTRODUCTION The Differential Amplifier The Operational Amplifier**

Determining Gain Frequency Effects Applications Comparators

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Dear Student: This presentation is arranged in segments. Each segment is preceded by a Concept Preview slide and is followed by a Concept Review slide. When you reach a Concept Review slide, you can return to the beginning of that segment by clicking on the Repeat Segment button. This will allow you to view that segment again, if you want to.

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**Concept Preview Differential amplifiers always have two inputs.**

Differential amplifiers can have one or two outputs. Driving one input provides a difference signal. Both outputs will be active and will be out of phase with each other. Driving both inputs with the same signal results in reduced output. Driving both inputs with a difference signal results in increased output.

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**C C B B E E A differential amplifier driven at one input +VCC -VEE**

Inverted output Noninverted output +VCC C C B B E E -VEE

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**Both outputs are active because Q1 drives Q2.**

Q1 serves as an emitter-follower amplifier in this mode to drive Q2. Q2 serves as a common-base amplifier in this mode. It’s driven at its emitter. +VCC C C B B E E Q1 Q2 -VEE

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**C C B B E E A differential amplifier driven at both inputs +VCC -VEE**

Common mode input signal Reduced output Reduced output +VCC C C B B E E -VEE

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**C C B B E E A differential amplifier driven at both inputs +VCC -VEE**

Differential mode input signal Increased output Increased output +VCC C C B B E E -VEE

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**Concept Review Differential amplifiers always have two inputs.**

Differential amplifiers can have one or two outputs. Driving one input provides a difference signal. Both outputs will be active and will be out of phase with each other. Driving both inputs with the same signal results in reduced output. Driving both inputs with a difference signal results in increased output. Repeat Segment

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Concept Preview The current in the emitter resistor divides equally between the two transistors in a differential amp. The differential gain is determined by the collector load and the ac emitter resistance. The common mode gain is determined by the collector load and the emitter resistor. The ratio of the differential gain to the common mode gain is called the CMRR. The CMRR is greatly improved by using a current source in the emitter circuit.

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**Differential Amplifier dc Analysis**

IRE = VEE - VBE RE 9 V V 3.9 kW = = 2.13 mA VRL = IC x RL = 1.06 mA x 4.7 kW IE = IRE 2 = 1.06 mA VCC +9 V = 4.98 V VCE = VCC - VRL - VE 4.7 kW RL RL 4.7 kW = (-0.7) IC = IE = 1.06 mA C C = 4.72 V B B E E 10 kW RB RB 10 kW 3.9 kW RE VEE -9 V

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**Differential Amplifier dc Analysis (continued)**

Assume b = 200 VB = VRB = IB x RB = 5.3 mA x 10 kW IB = IC b 1.06 mA 200 = VCC +9 V = 53 mV = 5.3 mA 4.7 kW RL RL 4.7 kW C C B B E E 10 kW RB RB 10 kW 3.9 kW RE VEE -9 V

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**Differential Amplifier ac Analysis**

50 mV IE = 1.06 mA = 47 W (50 mV is conservative) VCC +9 V AV(CM) = RL 2 x RE AV(DIF) = RL 2 x rE 4.7 kW 2 x 3.9 kW = 4.7 kW RL RL 4.7 kW = 50 4.7 kW 2 x 47 W = C C = 0.6 B B E E 10 kW RB RB 10 kW 3.9 kW RE VEE -9 V

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**Differential Amplifier ac Analysis (continued)**

CMRR = 20 x log AV(DIF) AV(CM) = 20 x log 50 0.6 = 38.4 dB VCC +9 V 4.7 kW RL RL 4.7 kW C C B B E E 10 kW RB RB 10 kW 3.9 kW RE VEE -9 V

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*** C C B B E E A current source can replace RE to decrease**

the common mode gain. AV(CM) = RL 2 x RE VCC Replaces this with a very high resistance value. 4.7 kW RL RL 4.7 kW C C B B E E 10 kW RB RB 10 kW * 2 mA NOTE: Arrow shows conventional current flow.

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**A Practical Current Source**

IZ = 9 V V 390 W = 10 mA IC IE = = 2 mA 5.1 V V 2.2 kW 390 W IC = IE = 2 mA 5.1 V 2.2 kW -9 V

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**The common-mode signal cannot be seen in the output.**

A Demonstration of Common-mode Rejection The common-mode signal cannot be seen in the output. The amplitude of the common-mode signal is almost 30 times the amplitude of the differential signal. 6.3 V 60 Hz 212 mV 1 kHz

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**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

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Concept Review The current in the emitter resistor divides equally between the two transistors in a differential amp. The differential gain is determined by the collector load and the ac emitter resistance. The common mode gain is determined by the collector load and the emitter resistor. The ratio of the differential gain to the common mode gain is called the CMRR. The CMRR is greatly improved by using a current source in the emitter circuit. Repeat Segment

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Concept Preview Operational amplifiers have one output and two inputs: inverting and non-inverting. Some op amps have offset null terminals which can be used to zero the dc output. The output of an op can change no faster than its slew rate. Slew rate is specified in volts per microsecond. The slew rate and the amplitude of the output signal determine the power bandwidth of an op amp.

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Op amps have two inputs Inverting input Output Non-inverting input

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**Op-amp Characteristics**

High CMRR High input impedance High gain Low output impedance Available as ICs Inexpensive Reliable Widely applied

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**With both inputs grounded through equal **

resistors, VOUT should be zero volts. +VCC VOUT Imperfections can make VOUT non-zero. The offset null terminals can be used to zero VOUT. -VEE

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Dt DV DV Dt Slew rate = 741 0.5 V ms The output of an op amp cannot change instantaneously.

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Slew-rate distortion VP f > fMAX fMAX = Slew Rate 2p x VP fMAX is known as the power bandwidth.

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**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

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Concept Review Operational amplifiers have one output and two inputs: inverting and non-inverting. Some op amps have offset null terminals which can be used to zero the dc output. The output of an op can change no faster than its slew rate. Slew rate is specified in volts per microsecond. The slew rate and the amplitude of the output signal determine the power bandwidth of an op amp. Repeat Segment

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**Concept Preview An op amp follower has a closed loop gain of 1.**

The input and output signals are in-phase in a follower amplifier. The closed loop gain can be increased by decreasing the feedback ratio. The input and output signals are out of phase in an inverting amplifier. The – terminal of an inverting amplifier acts as a virtual ground. The input impedance of an inverting amplifier is equal to the input resistor.

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**It has a high input impedance and a low output impedance.**

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. It has a high input impedance and a low output impedance. RL

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**The differential input**

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 VOUT RL VIN

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**Op-amp Follower AV(OL) = 200,000 B = 1 AB +1 A VIN VOUT**

The feedback ratio = 1 200,000 (200,000)(1) + 1 @ 1 AV(CL) = VOUT RL VIN

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**The closed-loop gain is increased by decreasing **

the feedback with a voltage divider. 200,000 (200,000)(0.091) + 1 = 11 AV(CL) = RF B = R1 RF + R1 100 kW R1 10 kW 10 kW 100 kW + 10 kW = VOUT RL VIN = 0.091

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**It’s possible to develop a different model for the closed loop gain**

by assuming VDIF = 0. VIN = VOUT x R1 R1 + RF RF Divide both sides by VOUT and invert: 100 kW R1 10 kW = VOUT VIN 1 + RF R1 VDIF = 0 VOUT RL VIN AV(CL) = 11

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**-VOUT -RF 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. We can ignore the op amp’s input current since it is so small. Thus: Virtual ground RF IR1 = IRF 10 kW 1 kW By Ohm’s Law: R1 VIN R1 = -VOUT RF VDIF = 0 VIN VOUT RL VOUT VIN = -RF R1 = -10 The minus sign designates an inverting amplifier.

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**Due to the virtual ground, the input impedance **

of the inverting amplifier is equal to R1. Virtual ground RF Although op amp input currents are small, in some applications, offset error is minimized by providing equal paths for the input currents. 10 kW R1 1 kW VDIF = 0 VIN R2 = R1 || RF = 910 W This resistor reduces offset error.

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**Concept Review An op amp follower has a closed loop gain of 1.**

The input and output signals are in-phase in a follower amplifier. The closed loop gain can be increased by decreasing the feedback ratio. The input and output signals are out-of-phase in an inverting amplifier. The – terminal of an inverting amplifier acts as a virtual ground. The input impedance of an inverting amplifier is equal to the input resistor. Repeat Segment

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**Concept Preview Most op amps have built-in frequency compensation.**

The internal frequency compensation produces a break frequency of 10 Hz or so. The closed loop small signal bandwidth is greater than the break frequency. A Bode plot can be used to determine the small signal bandwidth of a closed loop amplifier. The gain-bandwidth product can also be used to determine the closed loop small signal bandwidth.

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**A typical op amp has internal frequency compensation.**

Output C Break frequency: fB = 2pRC 1

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**Bode Plot of a Typical Op Amp**

Break frequency 120 100 80 60 Gain in dB 40 20 1 10 100 1k 10 k 100 k 1M Frequency in Hz

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**Op amps are usually operated with negative feedback **

(closed loop). This increases their useful frequency range. = VOUT VIN 1 + RF R1 AV(CL) = RF = 1 + 100 kW 1 kW = 101 100 kW R1 1 kW dB Gain = 20 x log 101 = 40 dB VOUT RL VIN

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**Using the Bode plot to find closed-loop bandwidth:**

120 100 Gain in dB 80 Break frequency 60 AV(CL) 40 20 1 10 100 1k 10 k 100 k 1M Frequency in Hz

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**There are two frequency limitations: **

0.5 V ms 70 V ms A 741 op amp slews at A 318 op amp slews at There are two frequency limitations: Slew rate determines the large-signal bandwidth. Internal compensation sets the small-signal bandwidth.

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**The Bode plot for a fast op amp shows increased bandwidth.**

120 100 fUNITY is also called the gain-bandwidth product. 80 Gain in dB 60 40 fUNITY 20 1 10 100 1k 10 k 100 k 1M 10M Frequency in Hz

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**fUNITY can be used to find the small-signal bandwidth.**

= VOUT VIN 1 + RF R1 AV(CL) = RF = 1 + 100 kW 1 kW = 101 100 kW R1 1 kW fB = fUNITY AV(CL) VOUT RL VIN 318 Op amp 10 MHz 101 = 99 kHz =

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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

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**Concept Review Most op amps have built-in frequency compensation.**

The internal frequency compensation produces a break frequency of 10 Hz or so. The closed loop small signal bandwidth is greater than the break frequency. A Bode plot can be used to determine the small signal bandwidth of a closed loop amplifier. The gain-bandwidth product can also be used to determine the closed loop small signal bandwidth. Repeat Segment

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Concept Preview The amplitude response of an RC lag network is –20 dB per decade beyond the break frequency. The phase response of an RC lag network is –45 degrees at the break frequency. The Miller effect makes some interelectrode capacitances appear to be larger. Multiple lag networks inside an op amp make negative feedback become positive at some frequency. Frequency compensation insures that the gain is less than 0 dB at that frequency.

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**Amplitude Response of RC Lag Circuit**

Vout C fb = 2pRC 1 fb 10fb 100fb 1000fb f 0 dB -20 dB Vout -40 dB -60 dB

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**Phase Response of RC Lag Circuit**

Vout C R -XC = tan-1 0.1fb fb 10fb 0o f Vout -45o -90o

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**Interelectrode Capacitance and Miller Effect**

The gain from base to collector makes CBC effectively larger in the input circuit. CBC CBE R CMiller CBE CMiller = AVCBC fb = 2pRCInput 1 CInput = CMiller + CBE

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**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 10 kHz 100 kHz fb1 fb2

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**Multiple Lag Circuits:**

Vout R1 R2 R3 C1 C2 C3 0o f Vout Phase reversal -180o Negative feedback becomes positive!

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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.

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**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

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Concept Review The amplitude response of an RC lag network is –20 dB per decade beyond the break frequency. The phase response of an RC lag network is –45 degrees at the break frequency. The Miller effect makes some interelectrode capacitances appear to be larger. Multiple lag networks inside an op amp make negative feedback become positive at some frequency. Frequency compensation insures that the gain is less than 0 dB at that frequency. Repeat Segment

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**Concept Preview Op amps can be used to sum (add) two or more signals.**

Scaling in a summing amp provides different gain for each signal. Op amps can be used to subtract two signals. Cascade RC filters have relatively poor performance. Active filters combine op amps with RC networks. Feedback in an op amp active filter sharpens the knee of the frequency response curve.

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**RF Summing Amplifier Amplifier scaling: 1 kHz signal gain is -10**

Inverted sum of three sinusoidal signals RF 5 kHz 5 kW 10 kW Summing Amplifier 3.3 kW Amplifier scaling: 1 kHz signal gain is -10 3 kHz signal gain is -3 5 kHz signal gain is -2 3 kHz 1 kW 1 kHz

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**common-mode rejection)**

Difference of two sinusoidal signals (V1 = V2) RF 1 kW Subtracting Amplifier (A demonstration of common-mode rejection) 1 kW VOUT = V2 - V1 1 kW 1 kW V1 V2

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**A cascade RC low-pass filter**

(A poor performer since later sections load the earlier ones.) An active low-pass filter (The op amps provide isolation and better performance.)

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Active filter -20 Cascade RC -40 Amplitude in dB -60 10 100 Frequency in Hz

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**Active low-pass filter**

with feedback VOUT C1 C2 VIN feedback At relatively low frequencies, Vout and Vin are about the same. Thus, the signal voltage across C1 is nearly zero. C1 has little effect at these frequencies.

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**Active low-pass filter**

with feedback VOUT C1 C2 VIN As fIN increases and C2 loads the input, Vout drops. This increases the signal voltage across C1. This sharpens the knee. -3 dB Feedback can make a filter’s performance even better! Gain Frequency fC

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**The slope eventually reaches 24 dB/octave or 80 db/decade **

Note the flat pass band and the sharp knee. Active filter using feedback (two stages) -20 -40 Amplitude in dB The slope eventually reaches 24 dB/octave or 80 db/decade for all the filters (4 RC sections). -60 10 100 Frequency in Hz

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**Concept Review Op amps can be used to sum (add) two or more signals.**

Scaling in a summing amp provides different gain for each signal. Op amps can be used to subtract two signals. Cascade RC filters have relatively poor performance. Active filters combine op amps with RC networks. Feedback in an op amp active filter sharpens the knee of the frequency response curve. Repeat Segment

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Concept Preview Other active filters include high-pass, band-pass and band-stop. An active rectifier will work with millivolt level signals. The output slope of an op amp integrator is equal to the dc input voltage times the reciprocal of the time constant. Comparators can be used to change analog waveforms to digital waveforms. A Schmitt trigger uses positive feedback to produce hysteresis and noise immunity.

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**Active high-pass filter**

VOUT VIN feedback -3 dB Gain fC Frequency

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**Active band-pass filter**

VIN VOUT Active band-pass filter (multiple feedback) -3 dB Gain Frequency Bandwidth

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**Active band-stop filter**

VOUT VIN Active band-stop filter (multiple feedback) -3 dB Gain Frequency Stopband

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0 V 56.6 mV Active rectifier 40 mV 0 V mV

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Integrator C V Slope = s R VOUT VIN Slope = -VIN x 1 RC

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**Comparator with a 1 Volt Reference**

+VSAT 1 V 0 V -VSAT VOUT VIN 1 V

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**Comparator with a Noisy Input Signal**

+VSAT 1 V 0 V -VSAT VOUT VIN 1 V

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**Schmitt Trigger with a Noisy Input Signal**

+VSAT UTP LTP -VSAT Trip points: R1 + RF R1 VSAT x VOUT VIN RF R1 Hysteresis = UTP - LTP

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**VOUT is LOW (0 V) when VIN is between 1 V and 3 V.**

Window Comparator 4.7 kW R1 311 VOUT VUL 3 V R2 4.7 kW VIN 311 VLL VOUT is LOW (0 V) when VIN is between 1 V and 3 V. 1 V

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**require pull-up resistors in applications of this type.**

+5 V Window Comparator 311 VOUT VUL 3 V VIN 311 Many comparator ICs require pull-up resistors in applications of this type. VLL 1 V

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**VOUT is TTL logic compatible.**

Window Comparator 4.7 kW R1 311 VOUT VUL 3 V R2 4.7 kW VIN 311 VLL VOUT is TTL logic compatible. 1 V

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**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

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Concept Review Other active filters include high-pass, band-pass and band-stop. An active rectifier will work with millivolt level signals. The output slope of an op amp integrator is equal to the dc input voltage times the reciprocal of the time constant. Comparators can be used to change analog waveforms to digital waveforms. A Schmitt trigger uses positive feedback to produce hysteresis and noise immunity. Repeat Segment

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**REVIEW The Differential Amplifier The Operational Amplifier**

Determining Gain Frequency Effects Applications Comparators

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© 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-1 Electronics Principles & Applications Eighth Edition Chapter 6 Introduction.

© 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 6-1 Electronics Principles & Applications Eighth Edition Chapter 6 Introduction.

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