Instrumentation Amplifiers Filters Integrators Differentiators Frequency-Gain Relation Non-Linear Op-Amp Applications DC Imperfections
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1 Lecture Op-Amp Building Blocks and Applications Instrumentation Amplifiers Filters Integrators Differentiators Frequency-Gain elation Non-Linear Op-Amp Applications DC Imperfections ELG439
2 Check List for Selecting Op-Amps Power Supply Gain Bandwidth Cost Voltage Offset Stability Output Current Noise Special Functions ELG439
3 Types of Op-Amps Precision Amplifiers: Low noise Low Offset Voltage Low Input Bias Current. Low Power/Low Bandwidth Video Amplifiers Audio Amplifiers High Voltage/High Current Differential/Instrumentation Amplifiers ELG439 3
4 Instrumentation Amplifier (IA) Applications Used in Environments with High Noise. Offers high Input Impedance. Low Bias Currents. Gain Determined by One esistor. Commercial IA is based on Standard 3-OP Amps. equires Several Matched esistors. ELG439 4
5 Instrumentation Amplifier Combines non-inverting amplifiers with the difference amplifier to provide higher gain and higher input resistance. ) b v (v a 3 4 v o b v i ) i( i a v v v i ) v (v 3 4 v o Ideal input resistance is infinite because input current to both op amps is zero. The CM is determined only by Op-Amp 3. NOTE ELG439 5
6 Instrumentation Amplifier Input (a) output (b) Stages of IA Figure 8.4, 8.5 A V v v out v F ELG439 6
7 Instrumentation Amplifier: Example Problem: Determine V o Given Data: = 5 kw, = 50 kw, 3 = 5 kw, 4 = 30 kw V =.5 V, V =.5 V Assumptions: Ideal op amp. Hence, v - = v + and i - = i + = 0. Analysis: Using dc values, A 4 30kW dm 5kW 50kW 5kW 3 V o A (V V )(.5.5)5.50V dm ELG439 7
8 Op-amp Circuits Employing Complex Impedances V V V V out S out S ( ( j) j) Z Z F S Z Z F S Figure 8.0 ELG439 8
9 Active Low-Pass Filter Figure 8., 8.3 Normalized esponse of Active Low-Pass Filter ELG439 9
10 The Active Low-Pass Filter Use a phasor approach to gain analysis of this inverting amplifier. Let s = j. A v v o ( j) v ( j) Z ( j ) Z ( j ) Z ( j) jc jc jc A v ( jc ) Z j e j ( j c ) c f c C f c C f c is called the high frequency cutoff of the low-pass filter. ELG439 0
11 Active Low-Pass Filter A v e j ( j ) c c e j e jtan (/ c ) At frequencies below f c (f H in the figure), the amplifier is an inverting amplifier with gain set by the ratio of resistors and. At frequencies above f c, the amplifier response rolls off at -0dB/decade. Notice that cutoff frequency and gain can be independently set. c e j[ tan (/ c )] magnitude ELG439 phase
12 Active Low-Pass Filter: Example Problem: Design an active low-pass filter Given Data: A v = 40 db, in = 5 kw, f H = khz Assumptions: Ideal op amp, specified gain represents the desired lowfrequency gain. Analysis: A 40dB/0dB v 0 00 Input resistance is controlled by and voltage gain is set by /. The cutoff frequency is then set by C. 5kW in and A v kW The closest standard capacitor value of 60 pf lowers cutoff frequency to.99 khz. C f (khz)(500kw) 59pF H ELG439
13 Active High-Pass Filter Figure 8.4, 8.5 Normalized esponse of Active High-Pass Filter ELG439 3
14 Active Band-Pass Filter Figure 8.6, 8.7 Normalized Amplitude esponse of Active Band-Pass Filter ELG439 4
15 Cascaded Amplifiers Connecting several amplifiers in cascade (output of one stage connected to the input of the next) can meet design specifications not met by a single amplifier. Each amplifer stage is built using an op amp with parameters A, id, o, called open loop parameters, that describe the op amp with no external elements. A v, in, out are closed loop parameters that can be used to describe each closed-loop op amp stage with its feedback network, as well as the overall composite (cascaded) amplifier. ELG439 5
16 Two-Port Model for a 3-Stage Cascade Amplifier Each amplifier in the 3-stage cascaded amplifier is replaced by its -port model. v o A v inb va s A inc A vb vc outa inb outb inc v Since A A A A v o out = 0 v va vb vc s in = ina and out = outc = 0 ELG439 6
17 Op-Amp Integrator t v out( t) vs ( t) dt C S F Figure 8.30 ELG439 7
18 Inverting Integrator Now replace resistors a and f by complex components Z a and Z f, respectively, therefore Z f Supposing Vo Vin Z a (i) The feedback component is a capacitor C, i.e., Z f jc Vin ~ Za + Zf Vo (ii) The input component is a resistor, Z a = Therefore, the closed-loop gain (V o /V in ) become: vo ( t) vi ( t) dt C where v ( t) i V e i jt Vin ~ + C Vo What happens if Z a = /jc whereas, Z f =? Inverting differentiator ELG439 8
19 Op-Amp Integrator Example: (a) (b) Determine the rate of change of the output voltage. Draw the output waveform. Solution: (a) ate of change of the output voltage V t o Vi C 50 mv/ s 5 V (0 kw)(0.0f) (b) In 00 s, the voltage decrease V o ( 50 mv/ s)(00μs) 5V ELG439 9
20 Op-Amp Differentiator v out ( t) F C S dvs ( t) dt Figure to t t Vi C + Vo 0 to t t ELG439 0
21 Frequency-Gain elation Ideally, signals are amplified from DC to the highest AC frequency Practically, bandwidth is limited 74 family op-amp have an limit bandwidth of few KHz. Unity Gain frequency f : the gain at unity Cutoff frequency f c : the gain drop by 3dB from dc gain G d GB Product : f = G d f c (Voltage Gain) Gd 0.707Gd 0 fc 0log(0.707)=3dB (frequency) f ELG439
22 Building Filters: Gain and Bandwidth Gain Bandwidth Product = Gain x Bandwidth ELG439
23 ELG439 3
24 Gain-Bandwidth Product Example: Determine the cutoff frequency of an op-amp having a unit gain frequency f = 0 MHz and voltage differential gain G d = 0V/mV Sol: Since f = 0 MHz By using GB production equation (Voltage Gain) Gd 0.707Gd? Hz f = G d f c f c = f / G d = 0 MHz / 0 V/mV = / = 500 Hz 0 fc (frequency) 0MHz f ELG439 4
25 Ideal Versus Practical Op-Amp Ideal Actual Open Loop gain A 0 5 Bandwidth BW 0-00Hz Input Impedance Z in >MW Output Impedance Z out 0 W 0-00 W Output Voltage V out Depends only on V d = (V + V ) Differential mode signal Depends slightly on average input V c = (V + +V )/ Common-Mode signal CM 0-00dB ELG439 5
26 Analysis Ideal Op-Amp Main Properties: () The voltage between V + and V is zero V + = V () The current into both V + and V termainals is zero For ideal Op-Amp circuit: () Write the kirchhoff node equation at the noninverting terminal V + () Write the kirchhoff node eqaution at the inverting terminal V (3) Set V + = V and solve for the desired closed-loop gain ELG439 6
27 vi v+ v- + vo vi v+ v- + vo a f a f Noninverting amplifier f v o ( ) v a i Noninverting input with voltage divider f v o ( )( ) v a i vi v+ v- + vo vi v+ v- + vo f Voltage follower v o v i f Less than unity gain vo v i ELG439 7
28 Non-Linear Op-Amp Applications Applications using saturation Comparators Comparator with hysteresis (Schmitt Trigger) Oscillators. Applications using active feedback components Log, antilog, squaring etc. amplifiers Precision rectifier ELG439 8
29 Comparators A comparator is a device that compares the magnitude of two inputs and gives an output that indicates which of the two is larger! V OUT Ideal response V OUT = A 0 V IN Practical response (clipped) V IN If A 0 is large, practical response can be approximated as : V IN > 0 V + > V - V OUT = +V SAT V IN < 0 V + < V - V OUT = -V SAT ELG439 9
30 Hysteresis A comparator with hysteresis has a safety margin. One of two thresholds is used depending on the current output state. V Upper threshold time Lower threshold ELG439 30
31 Schmitt Trigger The Schmitt trigger is an op-amp comparator circuit featuring hysteresis. The inverting variety is the most commonly used. Switching occurs when: VIN V V VOUT But, V OUT V V THESH SAT V SAT ELG439 3
32 Schmitt Trigger Op-amp Circuit The open-loop comparator from the previous two slides is very susceptible to noise on the input Noise may cause it to jump erratically from + rail to rail voltages The Schmitt Trigger circuit (see at the left) solves this problem by using positive feedback It is a comparator circuit in which the reference voltage is derived from a divided fraction of the output voltage, and fed back as positive feedback. The output is forced to either V POS or V NEG when the input exceeds the magnitude of the reference voltage The circuit will remember its state even if the input comes back to zero (has memory) The transfer characteristic of the Schmitt Trigger is shown at the left Note that the circuit functions as an inverter with hysteresis Switches from + to rail when v IN > V POS (/( + )) Switches from to + rail when v IN < V NEG (/( + )) ELG439 3
33 Asymmetrical Thresholds We do not need always want the threshold levels to be symmetrical around 0 V. More general configuration features an arbitrary reference level. Using Kirchoff s current law: VOUT V VEF V 0 VOUT VEF V V V V VOUT VEF ELG439 33
34 ealising V EF V THESH V SAT V EF But, Providing r r and V EF V S r r r ELG439 34
35 DC Imperfections Three DC Imperfections of eal Op-Amps Input Bias Current; Input Offset Current; and Input Offset Voltage (output voltage may not be zero for zero input voltage) Bias Current: All op-amps draw a small constant DC bias currents at their inputs. Typical value for a 74 is around 00 na. This is only notable when very high impedance sources are used. In such cases, an alternative op-amp with lower bias current should be used. Bias Current I B = I B++I B_ Offset Current I off = I B+ I B ELG439 35
36 Offset Voltage When both input voltages are equal, the output should be zero. Actually it probably won t be due to an offset voltage between the inputs. Typically, this is around mv. Offset voltage is automatically compensated by a negative feedback network. It can be a problem for precision comparator applications. Both the offset voltage and bias current are DC. A.C. operation is not affected by them (they just add an offset) Negative feedback reduces the effect of both. Steps can be taken to reduce them (further reading) ELG439 36
37 Example: Find the worst-case DC Output Voltage of an Inverting Amplifier assuming v in = 0. The maximum bias current of the Op-Amp is 00 na. The maximum offset current is 40 na, and the maximum offset voltage is mv. ELG439 37
38 First, Consider the Offset Voltage V 0,voff = + V off V 0,off = V off V 0,off = and + mv ELG439 38
39 Second, Bias Current Sources V 0,bias = I I I = 0 I = I B V 0,bias = I B ELG439 39
40 Third, Offset Current Source ELG439 V 0,ioff = I off = - and mv V 0 = V 0,voff + V o,bias + V o,ioff V 0 = + 0 +=34mV V 0 = + 0 =-4mV 40
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