11 Chapter Operational Amplifiers
Microelectronic Circuit Design Richard C. Jaeger Travis N. Blalock
Chap 11-1
Chapter Goals
Understand behavior and characteristics of ideal differential and op amps. Demonstrate circuit analysis techniques for ideal op amps. Characterize inverting, non-inverting, summing and instrumentation amplifiers, voltage follower and integrator. Learns factors involved in circuit design using op amps. Provide and introduction to active filters
Explore applications of op amps in nonlinear circuits, such as precision rectifiers. Provide examples of multivibrator circuits employing positive feedback. Demonstrate use of ac analysis capability of SPICE.
Chap 11-2
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
Signal developed at amplifier output is in phase with the voltage applied at + input (non-inverting) terminal and 180 out of phase with that applied at - input (inverting) terminal.
Chap 11-3
Differential Amplifier Model: With Source and Load
RL = load resistance RS = Thevenin equivalent resistance of signal source vs = Thevenin equivalent voltage of signal source R R L and vid = vs R idR vo* = Av + id Ro + R S id L R R id L Av = = vs Rid + RS Ro + RL vo Op amp circuits are mostly dc-coupled amplifiers. Signals vo and vs may have a dc component representing a dc shift of the input away from Q-point. Op-amp amplifies both dc and ac components.
Chap 11-4
Differential Amplifier Model: With Source and Load (Example)
Problem: Calculate voltage gain Given Data: A=100, Rid =100k, Ro = 100, RS =10k, RL =1000 Analysis: R R vo id L Av = = vs Rid + RS Ro + RL 100k 1000 = 100 = 82.6 = 38.3dB 10k +100k 100 +1000 Ideal amplifiers output depends only on input voltage difference and not on source and load resistances.This can be achieved by using fully mismatched resistance condition (Rid >> RS or infinite Rid and Ro << RL or zero Ro ). vo Av = =A vo = Av or id v id A = open-loop gain (maximum voltage gain available from the device)
Chap 11-5
Ideal Operational Amplifier
Ideal op amp is a special case of ideal differential amplifier with infinite gain, infinite Rid and zero Ro . v v = o and id A 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.
Ideal op amp has following assumptions:
Infinite common-mode rejection, power supply rejection, open-loop bandwidth, output voltage range, output current capability and slew rate Zero output resistance, input-bias currents and offset current, input-offset voltage.
Chap 11-6
Inverting Amplifier: Configuration
Positive input is grounded. Feedback network, resistors R1 and R2 connected between inverting input and signal source and amplifier output node respectively.
Chap 11-7
Inverting Amplifier:Voltage Gain
Negative voltage gain implies 1800 phase shift between dc/sinusoidal input and output signals. Gain greater than 1 if R2 > R1 Gain less than 1 if R1 > R2 vs isR i R vo = 0 1 22 But is=i2 and v-=0 (since vid=v+-v-=0) R vs vo is = = 2 and Av = R vs R 1 1 Inverting input of op amp is at ground potential (not connected directly to ground) and is said to be at virtual ground.
Chap 11-8
Inverting Amplifier: Input and Output Resistances
Rout is found by applying a test current (or voltage) source to amplifier output and determining the voltage(or current) and turning off all independent sources. Hence, vs = 0 vx = i R + i R 2 2 11 But i1=i2 v x = i ( R + R ) 12 1 R = =R in i 1 s vs Since v- = 0, i1=0 and vx = 0 irrespective of the value of ix .
Rout = 0
Chap 11-9
Inverting Amplifier: Example
Problem:Design an inverting amplifier Given Data: Av=20 dB, Rin =20k, Assumptions: Ideal op amp Analysis: Input resistance is controlled by R1 and voltage gain is set by R2 / R1. Av =1040dB / 20dB = 100 and A =-100
v
A minus sign is added since the amplifier is inverting. R = R = 20k 1 in R Av = 2 R = 100R = 2M 2 1 R 1
Chap 11-10
Non-inverting Amplifier: Configuration
Input signal is applied to the non-inverting input terminal. Portion of the output signal is fed back to the negative input terminal. Analysis is done by relating voltage at v1 to input voltage vs Chap 11-11 and output voltage vo .
Non-inverting Amplifier:Voltage Gain, Input Resistance and Output Resistance
Since i-=0 But vid =0 R 1 v = vo 1 R +R 12 vs = v 1
vs v = v id 1
and
Since i+=0 Rout is found by applying a test current source to amplifier output and setting vs = 0 and is identical to the output resistance of inverting amplifier i.e. Rout =0 Chap 11-12
R +R vo = vs 1 2 R 1 R v o R1 + R2 Av = = = 1+ 2 R vs R 1 1 vs R = = in i +
Unity-gain Buffer
A special case of non-inverting amplifier, also called voltage follower with infinite R1 and zero R2. Hence Av =1. Provides excellent impedance-level transformation while maintaining signal voltage level. Ideal voltage buffer does not require any input current and can drive any desired load resistance without loss of signal voltage. Unity-gain buffer is used in may sensor and data acquisition systems.
Chap 11-13
Summing Amplifier
Since i-=0, i3= i1 + i2, R R 3v 3v vo = R1R 2 1 2 Scale factors for the 2 inputs can be independently adjusted by proper choice of R2 and R1. Since negative amplifier input is at virtual ground, v v 1 i = 2 i = vo i= 1R 2R R 1 23 3 Any number of inputs can be connected to summing junction through extra resistors. This is an example of a simple digital-to-analog converter.
Chap 11-14
Difference Amplifier
R Since v-= v+ vo = 2 (v1 v2 ) R 1 For R2= R1 vo = (v1 v2 ) Also called a differential subtractor, amplifies difference between input signals. v o = v- i R = v - i R Rin2 is series combination of R1 22 12 and R2 because i+ is zero. R +R R R 2 v- 2 v For v =0, R = R , as the circuit = v- 2 ( v v- ) = 1 2 in1 1 R1 R R1 reduces to an inverting amplifier. 1 1 1 R For general case, i1 is a function 2v Also, v + = of both v1 and v2. R +R 2 12
Chap 11-15
Difference Amplifier: Example
Problem:Determine Vo, V+, V-, Io, I1, I2, I3 . Given Data: R1= 10k, R2 =100k, V1=5 V, V2=3 V Assumptions: Ideal op amp. Hence, V-= V+ and I-= I+= 0. Analysis:Using dc values, R 2 V = 3V 100k V+ = V- = = 2.73V 2 R +R 10k +100k 12 V V- 5V - 2.73V I =I = 1 = = 227A 12 R 10k 1 Vo = V I R I R = 5V (227A)(110k) = 20.0V 1 11 2 2 Io = I = 227A 2
Chap 11-16
Instrumentation Amplifier
R vo = 4 (va v ) b R 3 va iR i(2R ) iR = v 2 1 2b v v i= 1 2 2R 1 R R vo = 4 1+ 2 (v v ) R R 1 2 3 1 Combines 2 non-inverting amplifiers with the difference amplifier to provide higher gain and lower input resistance. Input resistance is infinite because input current to both op amps is zero, output resistance is set to zero by difference amplifier.
Chap 11-17
Active Low-pass Filter
A generalized inverting amplifiers gain is Z (s) Vo (s) Av (s) = = 2 Vs (s) Z (s) 1 In a single-pole low-pass filter, 1 R 2 sC = R2 Z ( s) = R Z ( s) = 1 1 2 1 sCR +1 R+ 2 2 sC R R 1 1 2 Av (s) = = 2 R (1+ sCR ) R (1+ s ) 1 2 1 H 1 = 2f = H H RC 2
Chap 11-18
Active Low-pass Filter (contd.)
Single pole at H At frequencies below H amplifier is an inverting amplifier with gain set by ratio of resistors R2 and R1. At frequencies above H, the amplifier response rolls of at 20dB/decade. Cutoff frequency and gain can be independently set.
Chap 11-19
Active Low-pass Filter: Example
Problem:Design an active low-pass filter Given Data: Av=40 dB, Rin =5 k, fH =2 kHz Assumptions: Ideal op amp, specified gain represents low-frequency gain. Av =1040dB / 20dB = 100 Analysis: Input resistance is controlled by R1 and voltage gain is set by R2 / R1. R R = R = 5k Av = 2 R = 100R = 500k 1 in 2 1 R and 1 1 1 C= = = 156pF 2f R 2 (200kHz)(510k) H2 Closest capacitor value of 160 pF lowers cutoff frequency to 1.95 kHz. Another choice of 150 pF raises cutoff frequency to 2.08 kHZ.
Chap 11-20
Integrator
Since ic= is 1 dvo = vsd RC t vo(t)= 1 vs ()d +vo(0) RC 0 vo(0)=Vc (0) Feedback resistor R2 in the inverting amplifier is replaced by capacitor C. The circuit uses frequency-dependent feedback. dvo vs ic = C is = dt R Voltage at the circuits output at time t is given by the initial capacitor voltage integral of the input signal from start of integration interval, here, t=0. Integration of an input step signal results in a ramp at the output.
Chap 11-21
Differentiator
v i = o R R Since iR= is vo = RC Input resistor R1 in the inverting amplifier is replaced by capacitor C. Derivative operation emphasizes highfrequency components of input signal, hence is less often used than the integrator. dvs dt is = C dvs dt
Output is scaled version of derivative of input voltage.
Chap 11-22
Cascaded Amplifiers
Connecting several amplifiers in cascade can meet design specifications not met by a single amplifier (output of one stage connected to input of next). Each amplifer is built by using an op amp with parameters A, Rid, Ro,called open loop parameters that describe the op amp with no external elements. Av, Rin, Rout are closed loop parameters that describe both the closed-loop op amp with the feedback network as well as the overall composite (cascaded) amplifier.
Chap 11-23
Two-port Model for 3-stage Cascade Amplifier
Each amplifier in the 3-stage cascaded amplifier replaced is by its 2-port model. R R inB inC vo = A vs A A vB R vA R vC + +R +R inB inC outA outB vo Since Rout=0 Av = =A A A vA vB vC vs Rin= RinA and Rout= RoutC =0
Chap 11-24
12 CC Vo(s) 12 A ( s) = = LP Vs(s) 2 G + G G G s +s 1 2 + 1 2 C CC 1 12 In standard form, s2 A ( s) = LP Op amp is voltage follower with unity s2 + s o + o2 Q gain over a wide range of frequencies. C RR 1 o = 12 Uses positive feedback through C1 at Q= 1 RR CC C R1 + R2 1212 frequencies above dc to realize complex 2
poles without inductors. Feedback network provides dc path for amplifiers input bias currents.
Active Filters: 2nd order Lowpass Filter function is: G G The transfer
Often, circuits are designed with C1 = C2 = C.
Chap 12 25
Active Filters: Low-pass (Frequency Response)
For Q=0.71,magnitude response is maximally flat (Butterworth Filter: Maximum bandwidth without peaking) For Q>0.71, response shows undesired peaking. For Q<0.71: Filters bandwidth capability is wasted. At <<o, filter has unity gain. At >>o,response exhibits twopole roll-off at 40dB/decade. At =o, gain of filter =Q. Sensitivity, S represents fractional change in parameter, P due to a given fractional change in value of Z. Sensitivity of with respect to R and C is: S = S =1 RC 2
Chap 12 26
Active Filters: Low-pass (Example)
Problem: Design second-order low-pass filter with maximally flat response. Given data: fH = 5 kHZ Analysis:C1 = 2C2 = 2C and R1 = R2 = R.
R=
1 2oC
Q= 1 2
1/oC is the reactance of C at o, R is 30% smaller than this value. Thus impedance level of filter is set by C. If impedance level is too low, op amp will not be able to supply current required to drive feedback network. At 5 kHz, for a 0.01 F capacitor, 1 1 = = 3180 oC 104 (10 8 ) 3180 R= = 2250 2 Final values: = R1 = R2 = 2.26k, C1 = 0.02 F, C2 = 0.01 F
Chap 12 27
Active Filters: High-pass with Gain
The transfer function is: s2 A ( s) = HP s2 + s o + o2 Q Q=
o = 1 RC
R C +C RC 1 1 2 + (1 K ) 2 2 R CC RC 2 12 11
1
Voltage follower in low-pass filter replaced by non-inverting amplifier with gain K, which gives an added degree of freedom in design. dc paths for both op amp input bias currents through R2 and feedback resistors.
For R1 = R2 = R and C1 = C2 = C, o = 1 Q= 1 RC 3 K For K=3, Q is infinite, poles are on j axis causing sinusoidal oscillations. K>3 causes instability due to right-half plane poles. 1 K 3
Chap 12 28
Active Filters: High-pass with Gain (Frequency Response)
For Q=0.71,magnitude response is maximally flat (Butterworth Filter response). Amplifier gain is constant at >o, the lower cutoff frequency of the filter.
Chap 12 29
Active Filters: Band-pass
Uses inverting op amp and its full loop gain (ideally infinite). V ( s) sC V (s) = o 21 R 2 G V = sC + C + G V ( s) sC Vo(s) 1 th th 1 2 th 1 R RC V ( s) so 3 22 A ( s) = o = BP R + R R C s2 + s o + 2 V ( s) 1 3 11 th o Q
1 o = R R CC th 2 1 2
R CC 12 2 Q= R C1 + C2 th R 2 Q= R th
2 RC 2
For C1 = C2 = C,
1 o = CRR th 2
BW =
Chap 12 30
Active Filters: Band-pass
(Frequency Response)
Response peaks at o and gain at center frequency is 2Q2. At <<o or >>o, filter response corresponds to single-pole high-pass or low-pass filter changing at a rate of 20dB/decade.
Chap 12 31
Active Filters: Tow-Thomas Biquad
1 1 1 - V ( s) V ( s) = Vs ( s) V ( s) General biquadratic transfer function to bp lp sR C bp sR C sRC represent low-pass, high-pass, band-pass, 1 2 1 all-pass and notch filters: V ( s) = V ( s) lp 2 +a s+a sRC bp as 0 so T ( s) = 2 1 A ( s) = K bp s2 + s o + o2 s2 + s o + o2 Q Q R 1 1 R Q = 2 BW = K= o = In Tow-Thomas biquad, first op amp is a R R RC RC multi-input integrator and third op amp is 1 2 o 2 simply an inverter. A ( s) = K lp s2 + s o + o2 Q
Thus, center frequency, Q and gain can each be adjusted independently.
Chap 12 - 32
Active Filters: Complete Tow-Thomas Biquad
The Tow-Thomas Biquad can achieve all filter functions with
addition of extra passive components as shown. C R 2 1 + s 1 3 + 1 s C C R RR RR C 2 Vo(s) 5 1 4 Av (s) = = Vs(s) 2 + s R 1 + 1 s R RC R 2C 2 2
Chap 12 - 33
Active Filters: Tow-Thomas Biquad (Example)
Problem: Design band-pass filter using Tow-Thomas circuit Given data: fo = 5 kHZ, BW = 200 Hz, midband gain =20 f Unknowns: R, R1, R2, R3, C Q = o = 10 BW Analysis:
Input resistance to the filter is set by R1.At the center frequency, X = 1 = R = 2R C oC 1 Also, first op amp must supply ac signal current to parallel combination of R, R2, C, second op amp must drive parallel combination of R3, C third must drive R3 in parallel with R. If we choose C = 2700 pF, R R 1 R = 10R = 294k R = 2 = = 14.7k R= = 29.4k 2 1 20 2 4000C R3 can be chosen arbitrarily as long as it doesnt load down second and third op amps. R3 =49.9 k
Chap 12 - 34
Active Filters: Tow-Thomas Biquad (Circuit and Simulation Results)
QuickTime and a TIFF (Uncompressed) decompressor are needed to see this picture.
Chap 12 - 35
Magnitude Scaling
Magnitude of filter impedances may all be increased or decreased by a magnitude scaling factor KM, without changing o or Q of the filter. To scale the magnitude of the impedance of the filter elements:
R' = K
M
R
C' =
C K M
1 Z '= =K Z MC C C ' 1 =o RR CC 1212
Applying magnitude scaling to low-pass filter:
o '=
K R K M 1 M C 1 K M C 2 K M
1 R 2K CC 1 2 K MM
=
Q' =
R K R C RR M 1 M 2 1 2 =Q 1 = K R +K R R +R C12 M1 M2 2 K
Chap 12 - 36
Frequency Scaling
Cutoff or center frequencies of filter may be scaled by a frequency scaling factor KF, without changing Q of the filter if each capacitor value is divided by KF and resistor values are left unchanged.
R' = R
C' =
C
K
F
Applying frequency scaling to low-pass filter:
o ' =
K 1 F = = K o F CC RR CC 1212 RR 1 2 1 2K K FF
Q' =
C 1 C K R R2 RR 1= 1 2 =Q F 1 R +R R +R C C12 212 2 K F
Chap 12 - 37
Nonlinear Circuit Applications: Precision Half-Wave Rectifier
For vS >0, vO = vS, i>0, diode is forwardbiased and feedback loop is closed. Rectification is perfect even for small input voltages.. For vS <0,diode is cutoff, i=0, vO=0. Primary sources of error are gain error and offset error due to nonideal op amp. For negative input voltages, output voltage v1 is saturated at negative limit. Large negative voltages across input can destroy unprotected op amps. Response time of circuit is slowed down due to slow recovery of internal circuits from saturation.
vO is rectified replica of vS without loss of voltage drop as in diode rectifier circuit.
Chap 12 - 38
Nonlinear Circuit Applications: NonSaturating Precision Half-Wave Rectifier
For vS >0, v1 is negative (one diode-drop below zero), D2 is forward biased, current in R2 is zero, vO = 0, D1 is reverse biased. Feedback loop is closed through D2.
Chap 12 - 39
Nonlinear Circuit Applications: AC Voltmeter
For a sinusoidal input of amplitude VM and frequency o, output is a rectified sine wave given by its Fourier series. If cutoff frequency of low-pass filter c << o, output consists primarily of dc voltage component. R R V M vo = 4 2 R R 3 1
Half-wave rectifier is combined with lowpass filter to form basic ac voltmeter.
Voltmeter range can be adjusted through the 4 resistors.
Chap 12 - 40
Circuits with Positive Feedback: Comparator
For inputs>VREF,output saturates at VCC. For inputs<VREF,output saturates at -VEE. Amplifiers built for use as comparators can handle saturation at the voltage extremes without incurring excessive internal time delays. For noisy inputs, multiple transitions may occur as input signal crosses reference level.
Chap 12 - 41
Schmitt Trigger
Schmitt trigger uses positive feedback and is bistable. For positive output voltages,VREF=VCC. For positive output voltages,VREF=-VEE. Reference level changes when output changes state. Voltage transfer characteristic exhibits hysteresis and doesnt respond to noise voltage magnitude smaller than the difference between the 2 threshold levels set by the reference voltage
Vn < V (V ) = (V +V ) CC EE EE CC
Chap 12 - 42
Astable Multivibrator
Uses positive and negative feedback to generate rectangular output. Output voltage switches periodically between VCC and -VEE.
For symmetrical power supplies, output of circuit is square wave with period T 1+ T = T + T = 2RC ln 12 1
Chap 12 - 43
Astable Multivibrator (contd.)
Astable multivibrator can be used to generate square, triangular and sine wave outputs as shown at frequencies up to few MHz. Frequency is varied by changing R3 or C3, C3 is changed in decade steps, R3 may be varied continuously using potentiometer.
Chap 12 - 44
Monostable Multivibrator or One Shot
Operates with one stable state, generates single pulse of known duration on application of trigger signal. D1 couples trigger signal into circuit, clamping diode D2 limits negative voltage excursion on capacitor C.
Chap 12 - 45
Monostable Multivibrator (contd.)
Output of circuit consists of positive pulse with fixed duration T given by V 1+ D V CC T = RC ln 1 For well-defined pulse, circuit should not be triggered till voltages on all nodes return to their quiescent steady-state values. Recovery time (return of circuit to state before trigger pulse was V applied) is given by: CC 1+ V EE Tr = RC ln V D 1 V EE
Chap 12 - 46
End of Chapter 11