OPERATIONAL AMPLIFIER PREPARED BY, PROF. CHIRAG H. RAVAL ASSISTANT PROFESSOR NIRMA UNIVRSITY


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1 OPERATIONAL AMPLIFIER PREPARED BY, PROF. CHIRAG H. RAVAL ASSISTANT PROFESSOR NIRMA UNIVRSITY
2 INTRODUCTION OpAmp means Operational Amplifier. Operational stands for mathematical operation like addition, subtraction, Integration etc. The general purpose OpAmp is IC741. It is a 8 pin DIP (Dual in line packaging).
3 INTRODUCTION Generally +15V is supplied at pin no. 7 & 15V is supplied at pin no. 4. Generally inside the opamp (IC741), there are 22 transistors, 11 resistors, 1 capacitor, and 1 diode.
4 Properties: INTRODUCTION
5
6
7 OPAMP PARAMETERS Common Mode Rejection Ratio Commonmode Input Voltage Input Offset Voltage Input Bias Current Input Impedance Input Offset Current Output Impedance Slew Rate
8 COMMONMODE REJECTION RATIO (CMRR) The ability of amplifier to reject the commonmode signals (unwanted signals) while amplifying the differential signal (desired signal). Ratio of openloop gain, A ol to commonmode gain, A cm. The higher the CMRR, the openloop gain is high and commonmode gain is low. CMRR is usually expressed in db & decreases with frequency.
9 INPUT OFFSET VOLTAGE Ideally, output of an opamp is 0 Volt if the input is 0 Volt. But practically, a small dc voltage will appear at the output when no input voltage is applied. Thus, differential dc voltage is required between the inputs to force the output to zero volts. This is called the Input Offset Voltage, V os. Range between 2 mv or less.
10 INPUT BIAS CURRENT The input bias current should be ideally zero. The dc current required by the inputs of the amplifier to properly operate the first stage. It is the average of both input currents.
11 INPUT IMPEDANCE It is the total resistance between the inverting and noninverting inputs. Differential input impedance : It is the total resistance between the inverting and noninverting inputs. Commonmode input impedance: It is the total resistance between each input and ground.
12 INPUT OFFSET CURRENT It is the difference of input bias currents of opamps.
13 SLEW RATE It is the maximum rate of change of the output voltage in response to a step input voltage.
14 SLEW RATE It s a measure of how fast the output can follow or track the input signal.
15 TOTAL OUTPUT OFFSET VOLTAGE For the circuit shown above, the output offset voltage V oo is caused by V io  can be positive or negative with respect to ground. Similarly, the V oib is caused by I B can also be positive or negative with respect to ground.
16 TOTAL OUTPUT OFFSET VOLTAGE If these two V oo and V oib of different polarity V oot will be very little. If these two V oo and V oib of same polarity V oot will be, V oot = V oo + V oib V oot = V oo + V oiio Since I io < I B, because the use of R OM, the current generated output offset voltage will be reduced.
17 TOTAL OUTPUT OFFSET VOLTAGE (EXAMPLE) V io = 7.5 mv maximum I io = 50 na maximum I B = 250 na maximum at T A = 25 o C V oot =? Here, the current generated output offset voltage is due to bias current I B in the inverting lead. So, V oot will be,
18 TOTAL OUTPUT OFFSET VOLTAGE (EXAMPLE) V io = 7.5 mv maximum I io = 50 na maximum I B = 250 na maximum at T A = 25 o C V oot =? Now, the current generated output offset voltage is due to input offset current I io in the inverting lead. So, V oot will be,
19 THERMAL DRIFT The values V io, I io, and I B are assumed to be constant but in practice these values vary with: 1. Change in temperature 2. Change in supply voltages : +V CC and V EE 3. Time The most serious variation is due to change in temperature. Thermal voltage drift: The average rate of change of input offset voltage per unit change in temperature is called thermal voltage drift and is denoted by ΔV io / ΔT. It is expressed in µv / o C. By same concept, ΔI io ΔT = Thermal drift in input offset current (pa / o C) ΔI B ΔT = Thermal drift in input bias current (pa / o C)
20 ERROR VOLTAGE The drifts in V io and I io can cause the change in V oot. So, consider the below figure which uses the compensating network and resistor R OM.
21 ERROR VOLTAGE Here it is assumed that the effects of V io and I io are reduced to zero at room temperature. In order to find the effects of V io and I io on the opamp performance, it is needed to find ΔV oot / ΔT (average change in total output offset voltage per unit change in temperature). It is assumed that the two drift effects are cumulative or additive. So, the maximum possible change in ΔV oot resulting from change in temperature by multiplying both the sides by ΔT. So, ΔV oot = error voltge E v.
22 ERROR VOLTAGE An error can be either positive or negative. So, the expression for the error voltage for inverting amplifier is, Generally we want to have E v = 0 which is possible if the drifts are zero. But practically no any drifts are zero. So, practical amplifiers can have finite value of error voltage E v. The solution is high performance precision Opamp such as LH0044A which has extremely low thermal voltage and thermal current drifts. It has ΔV io = 0.1 µv / o C and ΔI io = 5 pa / o C ΔT ΔT These drifts can cause serious problems in dc as well ac amplifiers both. In any given amplifier, the saturation voltages are assumed to be < the supply voltages + V CC and V EE.
23 ERROR VOLTAGE (EXAMPLE) Given: ΔVio = 30 µv / o C maximum and ΔIio = 300 pa / o C maximum. ΔT ΔT V s = + 15 V, R 1 = 1 kω, R F = 100 kω and R L = 10 kω. Find E v =? and V O =? at 35 o C. If V in = 1 mv dc and 10 mv dc. (Assume) an amplifier is nulled at 25 o C.
24 ERROR VOLTAGE (EXAMPLE) Solution: First find the change in temperature. So, ΔT = 35 o C 25 o C = 10 o C. Now for V in = 1 mv, the V o will be, This mv dc error can cause the V O to change between mv to mv.
25 ERROR VOLTAGE (EXAMPLE) Now for V in = 10 mv, the V o will be, The percent error is lager when 1 mv signal has to be amplified. i.e mv mv = [ mv / (actual voltage) 100 mv] * 100 = 61.2 %. The percent error for 10 mv signal is mv mv = [ mv / (actual voltage) 1000 mv] * 100 = 6.12 %. Thus for a given opamp, % error will increase as the amplitude of input signal will decrease.
26 OPEN LOOP VOLTAGE GAIN AS FUNCTION OF FREQUENCY Since, j = 1 / j and X C = 1 / 2πfC So, the open loop voltage gain is, High frequency model of an opamp with single break frequency
27 OPEN LOOP VOLTAGE GAIN AS FUNCTION OF FREQUENCY Let, f o = 1 / 2πR O C; then, Where, A OL (f) = open loop voltage gain as a function of frequency A = gain of opamp at 0 Hz (dc) f = operating frequency (Hz) = break frequency of the opamp f O The f o is dependent on C and R O (output resistor). So, for a given opamp, the f o is fixed. Now the open loop gain in magnitude and phase angle (expressed in polar forms),
28 OPEN LOOP VOLTAGE GAIN AS FUNCTION OF FREQUENCY The expression of gain magnitude in decibels, we get, Now for 741C, for f o = 5Hz the gain A = 2,00,000. So, by putting these values, So, by putting the values of f = 0 Hz, 5 Hz, 50 Hz, 500 Hz, 5 khz, 50 khz, 100 khz & 1 MHz; closed loop gain will be obtained in decibels.
29 OPEN LOOP VOLTAGE GAIN AS FUNCTION OF FREQUENCY Similarly for phase angle, For similar values of frequencies, the phase angles will be obtained. For f = 0 Hz, 5 Hz, 50 Hz, 500 Hz, 5 khz, 50 khz, 100 khz & 1 MHz;
30 OPEN LOOP VOLTAGE GAIN AS FUNCTION OF FREQUENCY For open loop gain: Operating frequency (f) A OL (f) 0 Hz Hz Hz Hz 66 5 khz khz khz 20 1 MHz 0
31 OPEN LOOP VOLTAGE GAIN AS FUNCTION OF FREQUENCY For phase angle: Operating frequency (f) A OL (f) 0 Hz Hz Hz Hz khz khz khz MHz 90 0
32 OPEN LOOP VOLTAGE GAIN AS FUNCTION OF FREQUENCY Open loop voltage versus frequency graph:
33 OPEN LOOP VOLTAGE GAIN AS FUNCTION OF FREQUENCY Phase angle (in degrees) versus frequency:
34 OPEN LOOP VOLTAGE GAIN AS FUNCTION OF FREQUENCY Conclusion: A OL (f) db is approximately constant from 0 Hz to break frequency 5 Hz. When f = f 0 : A OL (f) db is db down from its value at 0 Hz. So, break frequency is also sometimes called as 3db frequency or corner frequency. A OL (f) db decreases by 20 db as tenfold (one decade) increase in frequency. So, it says as gain rolls of at 20 db / decade. Gain rolls off at 6 db/octave : octave represents twofold increase in frequency. It can be seen from 50 khz to 100 khz, gain decreases from 26 db to 20 db. A OL (f) db = 0 ; that frequency is called unity gain bandwidth / gain bandwidth product / closed loop bandwidth / small signal bandwidth / unity gain crossover frequency.
35 OPEN LOOP VOLTAGE GAIN AS FUNCTION OF FREQUENCY Gain: Here, magnitude of gain and phase angle of gain is a function of frequency. By examine the equations, at any frequency less than f o, the gain is approximately constant and equal to A. (for f = 0 Hz, A OL (f) = A) For frequencies > f o ; the gain A OL (f) decreases because the denominator term increases. Phase shift: For phase shift equation: At 0 Hz, the phase shift between input an output is zero. For any frequency < f o ; the absolute value of phase shift is less than 45 o. As frequency increases, the phase angle increase towards or absolute 90 0.
36 SLEW RATE Maximum rate of change of output voltage with respect to time. Specified in V / μs. Ideally we want infinite slew rate i.e opamps output voltage simultaneously changes with input voltage. Slew rates ranging from 0.1 V / μs to 1000 V / μs, the opamps are available. National semiconductor LH0063C has a slew rate of 6000 V / μs. Generally the slew rates are specified for unity gain and is measured by applying step dc input voltage. Data sheets can have output voltage swing as a function of frequency or voltage follower large signal pulse response to show slew rate. Slew rate improves as closed loop gain and dc supply voltages are increased. Slew rate can be as function of temperature; slew rate decreases as temperature increases.
37 SLEW RATE Output Voltage Swing as a Function of Frequency voltage follower large signal pulse response
38 Causes of slew rate: SLEW RATE UGB product : small signal high frequency limitation of an opamp Slew rate: large signal phenomenon; large signal high frequency signal. It can cause by current limiting saturation of internal stages of an opamp when high frequency large amplitude signal is applied. The internal capacitor prevents the output voltage to respond immediately to fast changing input because the capacitor has finite charging and discharging time. The rate at which the voltage across capacitor rises is,
39 SLEW RATE Slew rate equation: Assume that the input is large amplitude with high frequency sine wave. Then, The rate of change of output: The maximum rate of change of output occurs when cos ωt = 1:
40 SLEW RATE But ω = 2* π*f. so, slew rate will be obtained as, Where, SR = Slew Rate f = input frequency (Hz) V P = peak value of output sine wave (volts) The slew rate determines the maximum frequency (f max ) of operation for a desired output voltage swing. With lower frequency the SR determines the maximum undistorted output voltage swing. So, the value of right hand side of above equation < Slew Rate the output waveform will always be undistorted. If the V p or frequency of input signal is increased to exceed the slew rate of an opamp, the output will be distorted.
41 SLEW RATE EFFECTS (EXAMPLE) The slew rate has important effects on both, open loop and close loop configurations. Below figure shows the open loop configuration using 741C Opamp. Here the output voltage goes from+14 V to 14 V as the input sine wave crosses zero volts. The time taken for the output to go from +14 V to 14 V can be determined by slew rate of an opamp listed in 741C data sheet.
42 SLEW RATE EFFECTS (EXAMPLE) The 741C has a typical slew rate of 0.5 V / µs. Therefore the time taken is, 28 V 0.5 V/µs = 56 µs This 56 μs must be the minimum time between the two zero crossings. Hence the maximum input frequency at which the output will be distorted is, µs 1 = = khz = 8.93 khz 112 µs So, at f max the output will be triangular instead of square wave.
43 SLEW RATE EFFECTS (EXAMPLE) Now consider the effect of slew rate on inverting amplifier with feedback. The below figure has as gain of 50 and operate at a gain of 50 up to 20 khz. So, the maximum output voltage at 20 khz is,
44 SLEW RATE EFFECTS (EXAMPLE) Hence for the output to be sine wave the maximum input signal (V in max ) can be, Output / input = gain So, Output / gain = input
45 AC AMPLIFIERS AC Inverting Amplifier AC Non inverting Amplifier
46 AC AMPLIFIERS The opamp can amplify ac as well as dc signals. If it is used to amplify ac signals it can be called as ac amplifier. If designer needs the ac response characteristics of opamp or need low frequency and high frequency limits or if the ac input is riding on some dc level an ac amplifier with coupling capacitor is used. By using the voltage divider rule, V A can be
47 So by rearranging the equation, we get AC AMPLIFIERS The output and gain of this inverting amplifier,
48 AC AMPLIFIERS If R in = 50 Ω, C i = 0.1 µf, R 1 = 100 Ω R F = 1 kω, R L = 10 kω Supply voltages = + 15 V Find bandwidth of the amplifier. Solution: The inverting amplifier with feedback, the R if = R 1 = 100 Ω. The source resistance is the output resistance of first stage i.e. R in = R O = 50 Ω. So,
49 AC AMPLIFIERS Since closed loop gain A F is R F / R 1 = 10. The bandwidth is [ higher frequency cut of (f H ) lower frequency cut off (f L )].
50 THE PEAKING AMPLIFIER Peaking Amplifier Schematic Diagram Peaking Amplifier Frequency Response
51 THE PEAKING AMPLIFIER The frequency response that peaks at certain frequency i.e. peaking response. This can be obtained by parallel LC network as shown in previous figure. Opamp uses this parallel LC network in feedback path. The resonant or peak frequency at which peak occurs, Here, Q coil = figure of merit of coil. The LC network impedance is very large at resonant frequency. So gain of the amplifier is also maximum at resonance and given by, R P = equivalent parallel resistance of the tank circuit = Q coil2 * R
52 THE PEAKING AMPLIFIER The frequencies below peak frequency and frequencies above peak frequency, the impedance of the parallel LC network is less than R P. So, the gain is also less than (R F R P / R 1 ) at any other frequency than f P. The bandwidth can easily be computed by using the below equation. Here, f P = Peaking frequency Q P = figure of merit of the parallel resonant circuit (R F R P / X L )
53 THE PEAKING AMPLIFIER (EXAMPLE) Given: The peaking amplifier with a gain of 10 at peak frequency of 16 khz. Determine all the values. Solution: Also 10 * 16 khz = 160 khz which is in the band of 741C. Its unity gain bandwidth is 1 MHz. The capacitor should be less than 1 μf to avoid leakage problems. The inductance can also be less than 1 H to avoid large size.
54 THE PEAKING AMPLIFIER (EXAMPLE) Now by using the f P equation, Let L = 10 mh and R = internal resistance of an inductor = 30 Ω And
55 THE PEAKING AMPLIFIER (EXAMPLE) Let R 1 = 100 Ω, then R F can be found as follow. Hence the components values are,
56 INTEGRATOR (EXAMPLE) With basic integrator circuit, R 1 C F = 1 second. Input wave is given below. Find the output voltage and sketch it. The output is given by following equation. So, 0 1 t=4 ( 2 ) V O =  2 dt 2 dt 2 dt 2 dt 2 dt = V O = ( ) = 8 V The output is a ramp function with a slop of 2 V / s
57 DIFFERENTIATOR (EXAMPLE) Given : Design the differentiator to differentiate the input signal that varies from 10 Hz to 1 khz. If the input is of sine wave 1 V peak at 1000 Hz is applied to differentiator of previous part. Draw its output waveforms. Solution: The first step is to find f a, the frequency at which gain is 0 db. Let C 1 be the 0.1 µf Take R F = 1.5 kω The second step is to find f b, the gain limiting frequency.
58 DIFFERENTIATOR (EXAMPLE) The third step is to find C F, the gain limiting frequency. Let R 1 = 82 Ω. So, R 1 C 1 = R F C F Let C F = µf R OM = R F = 1.5 kω. The complete circuit is shown.
59 DIFFERENTIATOR (EXAMPLE) With V P = 1 V and f = 1000 Hz. the input voltage is given as, Hence the output voltage is, Input voltage Output voltage
60
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