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Lecture No. 15
Ideal Operational Amplifiers An ideal op-amp is a device which acts as an ideal voltage controlled voltage source. The device will have the following characteristics:
1. No current flows into the input terminals of the device. This is equivalent to having an infinite input resistance Ri=∞. In practical terms this implies that the amplifier device will make no power demands on the input signal source. 2. Have a zero output resistance (Ro=0). This implies that the output voltage is independent of the load connected to the output. In addition the ideal op-amp model will have infinite open loop gain (A →∞). The ideal op-amp model is shown schematically on Figure.
The input marked “ + ” is called the noninverting input . . . The input marked “ - ” is called the inverting input . . .
The output voltage of operation amplifier is: Vo = Ao (V+ - V-)
The gain of operation amplifier is:
Vo A o V - V -
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Ideal Operational Amplifier Operation With Ao = ∞ , we can conceive of three rules of operation:
1. If V+ > V- then Vo increases . . . 2. If V+ < V- then Vo decreases . . . 3. If V+ = V- then Vo does not change . . .
In a real op amp, Vo cannot exceed the dc power supply voltages. In normal use as an amplifier, an operational amplifier circuit employs negative feedback - a fraction of the output voltage is applied to the inverting input.
Op Amp Operation with Negative Feedback Consider the effect of negative feedback: 1. If V+ > V- then Vo increases . . . Because a fraction of Vo is applied to the inverting input, V- increases . . . The “gap” between V+ and V- is reduced and will eventually become zero. Thus, Vo takes on the value that causes V+ - V- = 0!!!
2. If V+ < V- then Vo decreases . . . Because a fraction of Vo is applied to the inverting input, V- decreases . . . The “gap” between V+ and V- is reduced and will eventually become zero. Thus, Vo takes on the value that causes V+ - V- = 0!!!
In either case, the output voltage takes on whatever value that causes V+ - V- = 0!!!
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Operational Amplifier Circuits
1- The Inverting Amplifier Let’s put our ideal op amp concepts to work in this basic circuit:
Voltage Gain Because the ideal op amp has Ri = ∞ , the current into the inputs will be zero. This means i1 = i2 , i.e., resistors R1 and R2 form a voltage divider. Therefore, we can use superposition to find the voltage V- .
Now, because there is negative feedback, vo takes on whatever value that causes v+ - v- = 0 , and v+ = 0 !!!
Input Resistance This means resistance “seen” by the signal source vi, not the input resistance of the op amp, which is infinite. Because V- = 0, the voltage across R1 is vi. Thus:
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Output Resistance This is the Thevenin resistance which would be “seen” by a load looking back into the circuit. Our op amp is ideal; its Thevenin output resistance is zero:
2. The Current-to-Voltage Converter The circuit diagram of a current-to-voltage converter is shown in Figure below. The circuit is a special case of an inverting amplifier where the input resistor is replaced with a short circuit.
- Because the V− terminal is a virtual ground, the input resistance is zero. The output resistance is also zero. - Because i1 +iF = 0 and vo = iF RF , it follows that the trans- resistance (trans-impedance) gain is given by:
vo RF i1
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- Because RS connects from a virtual ground to ground, the current through RS is zero. Then;
i1 = iS and vo = −RF iS
Thus the output voltage is independent of RS.
3-The Non-inverting Amplifier If we switch the vi and ground connections on the inverting amplifier, we obtain the noninverting amplifier:
Voltage Gain This time our rules of operation and a voltage divider equation lead to:
Input and Output Resistance The source is connected directly to the ideal op amp, so:
A load “sees” the same ideal Thevenin resistance as in the inverting case:
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Lecture No. 16
4-The Voltage Follower
Voltage Gain This one is easy:
i.e., the output voltage follows the input voltage. Input and Output Resistance By inspection, we should see that these values are the same as for the noninverting amplifier . . .
In fact, the follower is just a special case of the noninverting amplifier, with R1 = ∞ and R2 = 0!!!
5-The Inverting Summer This is a variation of the inverting amplifier:
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Voltage Gain We could use the superposition approach as we did for the standard inverter, but with three sources the equations become unnecessarily complicated . . . so let’s try this instead . . . Recall . . . vO takes on the value that causes v- = v+ = 0 . . . So the voltage across RA is vA and the voltage across RB is vB :
Because the current into the op amp is zero:
Finally, the voltage rise to vO equals the drop across RF:
6- Differential Amplifier The op amp is a differential amplifier to begin with, so of course we can build one of these!!!
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Voltage Gain Again, vO takes on the value required to make v+ = v- . Thus:
We can now find the current i1 , which must equal the current i2 :
Knowing i2 , we can calculate the voltage across R2 . . .
Then we sum voltage rises to the output terminal:
Working with just the v2 terms from equation, yield;
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And, finally, returning the resulting term to equation, yield;
7- Integrator
From our rules and previous experience we know that v- = 0 and iR = iC , so . . .
Combining the two previous equations, and recognizing that vO = - vC :
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Normally vC (0) = 0 (but not always). Thus the output is the integral of vi , inverted, and scaled by 1/RC.
8- The Differentiator
From the i-v relationship of a capacitor:
Recognizing that vO = -vR :
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Lecture No. 17
9- The Comparator • Open loop! • Only two output states... vref +vsat if vin > vref -vsat if vin < vref • Problems with noise! Vo
vin
vin
vref
t
Vo +vsat
t
-vsat
10- The Schmitt trigger • Positive feedback (not open loop)... • Hysteresis.... There are now two distinct threshold voltages that determine the state of the output...one threshold to go high if it's low another threshold to go low if its high... • Good for rejecting noise.
Voltage at non-inverting input By superposition:
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To find the trip-points let Vin = Vu Or VL and solve...
Subtract the trip-points to get the hysteresis...
To design: 1) Determine R1 and R2 based on trip-points:
2) Determine Vref (R3 and R4) from R1 and R2:
vin
Vu VL
t
Vo +vsat
t
-v sat
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Op Amp Circuits - Designing with Real Op Amps
1-Resistor Values Our ideal op amp can supply unlimited current; real ones can’t . . .
Now, we can note the following: 1. To limit iF + iL to a reasonable value, we adopt the “rule of thumb” that resistances should be greater than approx. 100 . 2. Larger resistances render circuits more susceptible to noise and more susceptible to environmental factors. To limit these problems we adopt the “rule of thumb” that resistances should be less than approximately 1 M.
2- Source Resistance - In some designs RS will affect desired gain. - If we wish to ignore source resistance effects, resistances must be much larger than RS (if possible).
Slew Rate of Op Amp Circuits The slew rate (SR) is defined as the maximum rate of change of the output of an op amp circuit. The SR in general describes the degradation effect on the high frequency response of the active amplifier (one with an op amp) near or at the rated maximum output voltage swing. This effect is generally due to the compensating capacitor and not to the transistor circuits internal to the op amp. In short, the SR effect is due to the maximum supplied current available for charging up the compensating capacitor. We know that the current required to charge a capacitor is
The Slew Rate is found from
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Example (1): an Op Amp has a maximum output current of 1 mA. If the compensating capacitor is 1000 pF, find the Slew rate. Solution:
Example(2): Suppose that the input signal to a unity gain amplifier configuration is a 20kHz sine wave and slew rate = 0.5V/μs. What is the largest possible amplitude of the input signal to avoid distortion due to slewing? Solution
If the input voltage is vi(t) =Msin2πft and the magnitude of the gain at that frequency is unity,
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Lecture No. 18
Examples Sheet: Operational Amplifiers
Q1. An operational amplifier has an open loop gain of 160dB. If the saturation voltage of the op-amp is ±12V, what is the differential input voltage range for operation in the linear region?
Answer: Operation in the linear region is the output from the op-amp satisfies the differential gain equation:
Where, Ao = 160dB = 100,000,000, and vo = vsat = ± 12V.
Hence, the input voltage range that gives linear operation is not very large!
Q2. Plot the transfer characteristic of an op-amp with the following characteristics: Ao = 1,000,000, vsat = ±10V.
v 10 Answer: v o 10V d 6 Ao 10 A rough sketch is all that is required, with important characteristics highlighted:
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Q3. Calculate the closed loop gain for the inverting amplifier shown below. State any assumptions made in your analysis. What is the input resistance of the circuit?
Answer: Assumption: The op-amp is ideal, Inverting amplifier closed loop gain equation:
Input resistance is given by: Rin = R1 = 30k
Q4. Calculate the closed loop gain for the non-inverting amplifier shown below. State any assumptions made in your analysis. What is the input resistance of the circuit?
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Answer: Assumption: The op-amp is ideal, Non-inverting amplifier closed loop gain equation:
Input resistance is infinite since the input is connected directly to the non-inverting input of the Op Amp.
Q5. Design an amplifier circuit with a closed loop gain of 40dB and an input impedance of 10k.
Answer: Some thinking required here! You are asked to design an amplifier with a closed loop gain of 100. The non-inverting amplifier is not suitable to use here because the non-inverting amplifier has infinite input impedance, we are required to design an amplifier with finite input impedance. Hence, we will have to use two inverting amplifiers with a total closed loop gain of 100 – the signal inversion will cancel over the two stages.
Avt = Av1 Av2
R1 = 10kW since this forms the input impedance of the circuit.
Lets choose R2 = 100k, R 100 103 Then A 2 10 v1 3 R1 10 10
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Av2 =10 We choose similar values for stage 2, give a total gain of 100. R2 = 100k and R1 = 10k
Q6. For the inverting amplifier shown below, plot its transfer characteristics over the range -10V Answer: The saturation voltage of the op-amp is given by Vsat = ±15V, therefore the output from the op-amp cannot exceed this value. We are required to plot the transfer characteristic (output voltage against input voltage) over the input voltage range –10V When vo = Vsat Hence, if the output is 15V, the input is –4.54V, and if the output is –15V, the input 4.54V. For input signals greater than these the output will saturate and vo=±15V. In between, the relationship will be linear with slope given by the closed loop gain (negative). Linear electronic 74 Q7. a) Drive the closed loop gain of the op-amp circuit illustrated below. b) What is the gain when R1= R2=10kW, R3=5kW, and R4=15kW? Answer: a) In order to determine the gain of this circuit (in terms of vo and vs) we need to simplify the input network by calculating the Thevenin equivalent, i.e. The Thevenin voltage is simply given by the voltage across R2 (R1 and R2 form a voltage divider network), i.e. Linear electronic 75 the resistance as seen at the output, Rth, is given by: b) For the values given the gain is: R R 15 103 10 103 Av 4 2 0.75 6 R1R2 R1R3 R2 R3 (10 10 10 5 10 5 )10 Linear electronic 76 Lecture No. 19 Q8. Derive an expression for the closed loop gain of the circuit below. Answer: Assume that the currents will flow as illustrated and the op-amp is ideal, and a virtual earth exists, v- = 0V and i- = 0A. Kirchhoff’s Current Law at node A, We know the inverting input is held at ground, we know that vs is seen across R1 Kirchhoff’s current law at node B And We know that vB is seen across R3 hence Using the equations for i2 and i3 Linear electronic 77 Using Kirchhoff’s voltage law at the output Q9. a) Design a non-inverting amplifier to have a closed loop gain of 10. The op-amp has an output saturation voltage of 10V, what is its maximum signal amplitude that can be applied to the input without saturation occurring? b) If the output of the op-amp is connected to a load of 5k, what will be the output current from the op-amp when the maximum voltage is observed at the output of the op. Amp? Note: State any assumptions made in your analysis. Q10. For the noninverting amplifier illustrated below, what is? a) The closed loop gain b) The closed loop gain assuming R2>>R1 Note: State any assumptions made in your analysis. Linear electronic 78 Q11. Consider the circuit below consisting of a bridge circuit and amplifier. Determine the output voltage, vo, in terms of the source voltage, vs. State any assumptions made in your analysis. Q12. Design a circuit to sum three signals, v1, v2 and v3, in the following manner: vo = - (5 v1 +10 v2 + 20 v3 ) Answer: we choose a summing amplifier with three inputs. The output voltage is given by: Linear electronic 79 It must be choosing resistor values so that: Let Rf = 100k R1= Rf / 5 = 20 k, R2= Rf / 10 = 10 k and R3= Rf / 20 = 5 k Q13. Design a circuit to sum three signals, v1, v2 and v3, in the following manner: vo = - (5 v1 +10 v2 + 20 v3 ) Answer: the answer is similar to the previous question except you must invert the output. Q14. Design a circuit to sum three signals, v1, v2 and v3, in the following manner: vo = 3 v1 +2 v2 - 4 v3 Answer: the answer is similar to the previous question except you must invert inputs v1 and v2 prior to the inverting summing amplifier. Q15. Design a circuit to calculate the difference between two signals, v1 and v2, where vo =10 (v1 -v2) Answer: We must be used the difference amplifier as shown below. Choose resistor values such that: Let R1 = 10k R2 = R1 10=100 k Linear electronic 80 Q16. In Figure below, Consider an op-amp connected to a power supply of ±15V used without feedback. The op-amp saturates at 80% of the power supply. Plot the transfer characteristics and the output waveform if the input signal is vs=5sin (wt). Answer: vsat is given by 80% of ±15V, hence vsat= ±12V. The output will switch at an input of 3V, since the 3V reference is connected to the inverting input as shown in figure below. vo vo 12V t 3V vs -12V 5V vs -5V t Linear electronic 81 Lecture No. 20 Oscillators Oscillation: an effect that repeatedly and regularly fluctuates about the mean value Oscillator: circuit that produces oscillation Characteristics: wave-shape, frequency, amplitude, distortion, stability Application of Oscillators 1 Oscillators are used to generate signals, e.g. – Used as a local oscillator to transform the RF signals to IF signals in a receiver; – Used to generate RF carrier in a transmitter – Used to generate clocks in digital systems; – Used as sweep circuits in TV sets and CRO. Oscillators divided to two main types: 1. Linear Oscillators: generate sinusoidal waves utilizing the resonance phenomena. 1 Wien Bridge Oscillators 2 RC Phase-Shift Oscillators 3 LC Oscillators 2. Nonlinear Oscillators (function generators): generate the square, triangle, and pulse waveform like Multivibrators. Linear electronic 82 1- Linear Oscillators + V Vs Amplifier (A) Vo + Positive Vf Frequency-Selective Feedback Network () Feedback For sinusoidal input is connected “Linear” because the output is approximately sinusoidal A linear oscillator contains: - A frequency selection feedback network - An amplifier to maintain the loop gain at unity + V Vs A(f) Vo + Vf SelectiveNetwork (f) V AV A(V V ) o s f Vf Vo V A o Vs 1 A If Vs = 0, the only way that Vo can be nonzero is that loop gain A=1 which implies that Linear electronic 83 | A | 1 Barkhausen Criterion A 0 Wien Bridge Oscillator Rf R1 + C R Vo R Z1 C Z2 Z1 R1 C1 Z2 Vi C2 R2 Vo Linear electronic 84 1 1 X C 2 and XC1 C2 C1 Z R jX 1 1 C1 1 1 1 jR X Z 2 C2 2 R2 jXC2 R2 jXC2 Therefore, the feedback factor, V Z (jR X / R jX ) o 2 2 C2 2 C2 V Z Z (R jX ) (jR X / R jX ) i 1 2 1 C1 2 C2 2 C2 jR X 2 C2 (R jX )(R jX ) jR X 1 C1 2 C2 2 C2 can be rewritten as: R X 2 C2 R X R X R X j(R R X X ) 1 C2 2 C1 2 C2 1 2 C1 C2 For Barkhausen Criterion, imaginary part = 0, i.e., R1R2 XC1XC2 0 1 1 or R1R2 C1 C2 1/ R1R2C1C2 Linear electronic 85 Supposing, R1=R2=R and XC1= XC2=XC, RX C 3RX j(R2 X2 ) C C 1 RC 1 3 Rf Av 1 Av 3 1 R1 Therefore, R f 2 R1 Linear electronic 86 Lecture No. 21 RC Phase-Shift Oscillator Rf R1 C C C + R R R Using an inverting amplifier The additional 180o phase shift is provided by an RC phase- shift network - Applying KVL to the phase-shift network, we have C C C V1 Vo R R R 1 2 3 I I I Linear electronic 87 V1 I1(R jXC ) I2R 0 I R I (2R jX ) I R 1 2 C 3 0 I2R I3(2R jXC ) Solve for I3, we get RjXC R V1 R 2RjXC 0 I 0 R 0 3 RjXC R 0 R 2RjX R C 0 R 2RjXC 2 V1R I3 2 2 2 (R jXC )[(2R jXC ) R ] R (2R jXC ) The output voltage, 3 V1R Vo I3R 2 2 2 (R jXC )[(2R jXC ) R ] R (2R jXC ) Hence the transfer function of the phase-shift network is given by: 3 Vo R 3 2 3 2 V1 (R 5RXC ) j(XC 6R XC ) Linear electronic 88 For 180o phase shift, the imaginary part = 0, i.e., 3 2 XC 6R XC 0 or XC 0 (Rejected) 2 2 XC 6R 1 6RC and, 1 29 Note: The –ve sign mean the phase inversion from the voltage Linear electronic 89 Lecture No. 22 LC Oscillators Av Ro ~ + 2 Z1 Z2 1 Z3 Zp 1 The frequency selection network (Z1, Z2 and Z3) provides a phase shift of 180o 2 The amplifier provides an addition shift of 180o Two well-known Oscillators: 1 Colpitts Oscillator 2 Harley Oscillator Av Ro ~ + Vf Z1 Z2 Vo Z3 Zp Linear electronic 90 Z1 Vf Vo Vo Z1 Z3 Z Z //(Z Z ) p 2 1 3 Z2(Z1 Z3 ) Z1 Z2 Z3 For the equivalent circuit from the output Ro Io + + Zp Vo AvVi A V V V A Z v i o or o v p Ro Zp Zp Vi Ro Zp Therefore, the amplifier gain is obtained, V A Z (Z Z ) A o v 2 1 3 V R (Z Z Z ) Z (Z Z ) i o 1 2 3 2 1 3 The loop gain, A Z Z A v 1 2 Ro (Z1 Z2 Z3 ) Z2(Z1 Z3 ) Linear electronic 91 If the impedance are all pure reactances, i.e., Z jX , Z jX and Z jX 1 1 2 2 3 3 The loop gain becomes, AvX1X2 A jRo (X1 X2 X3 ) X2(X1 X3 ) The imaginary part = 0 only when X1+ X2+ X3=0 - It indicates that at least one reactance must be –ve (capacitor). - X1 and X2 must be of same type and X3 must be of opposite type With imaginary part = 0, A X A X A v 1 v 1 X1 X3 X2 For Unit Gain & 180o Phase-shift, X2 A 1 Av X1 C1 R L1 R C L C2 L2 Harley Oscillator Colpitts Oscillator Linear electronic 92 Colpitts Oscillator C1 R L C2 L I1 node 1 I2 V 3 + I C2 V R C1 4 m I g V In the equivalent circuit, it is assumed that: 1 Linear small signal model of transistor is used 2 The transistor capacitances are neglected 3 Input resistance of the transistor is large enough At node 1, V V i (jL) 1 1 Linear electronic 93 where, i jC V 1 2 V V (1 2LC ) 1 2 Apply KCL at node 1, we have V1 jC2V gmV jC1V1 0 R 2 1 jC2V gmV V (1 LC2 ) jC1 0 R For Oscillator V must not be zero, therefore it enforces, 1 2LC g 2 j(C C ) 3LC C 0 m R R 1 2 1 2 Imaginary part = 0, we have 1 o LC T And C C C 1 2 T C C 1 2 Real part = 0, yields C g 2 m RC 1 Linear electronic 94 Lecture No. 23 Nonlinear Oscillators generate square, triangle, and pulse waveforms. Generation of Square and triangle waveforms using Multivibrator The nonlinear oscillator devices make use of special class of circuit known as Multivibrator. The Multivibrator is divided into tree types: - Bistable Multivibrator - Astable Multivibrator - Mono-stable Multivibrator I- Operation of the Bistable Multivibrator R1 R2 v+ + vo v1 R1 The voltage divider (R1, R2) will feed a fraction of R1 R2 the output signal back to the positive input terminal of OP-Amp. If A is grater than unity, the feedback signal will greater the original increment in V+. This regenerative process continues until Op-Amp saturation (Vsat+).When this happens, the V+ = Vsat+ . Also, the Op-Amp would have ended up saturated in Linear electronic 95 negative direction with (Vsat-) and V+ = Vsat- . The transfer characteristic of bistable Multivibrator When v1 increase from 0V, we can see from the circuit that nothing happens until v1 reaches a value equal to V+ (i.e Vth = Vsat+). Then vo goes to Vsat- because the difference between the V+ and V- terminals becomes negative. Now consider v1 is decrease and V+ = Vsat-, we see that the circuit remains in the negative saturation atate until v1 goes negative to the point that it equals –Vth = Vsat-. Then OP Amp goes to the positive saturation (vo =Vsat+)because the difference between the V+ and V- terminals becomes positive. vo vo +Vcc +Vcc Vth -Vth v1 v1 -Vcc -Vcc vo +Vcc -Vth Vth v1 -Vcc Linear electronic 96 A Square-wave Oscillator by using Astable Multivibrator A square waveform can be generated by arranging for a bistable multivibrator to switch states periodically. This can be done by connecting bistable with an RC circuit in a feedback loop. R vc C vo vf + R R 1 2 Let vo = Vsat+, the capacitor will charge toward this level through resistor R. Thus the voltage across C, which is applied to the negative input terminal of Op-Amp, will rise exponentially toward Vsat+ with time constant = C R. In this case V+ = Vth. When the capacitor voltage (Vc) reaches the positive threshold (Vth) the Op-Amp will switch to other stable state on which vo = Vsat-. The capacitor will then start discharging and its voltage will decrease exponential toward Vsat-. When the capacitor voltage equal to the negative threshold (-Vth), the Op-Amp will switch to other stable state on which vo = Vsat+ - During the charging interval T1: the voltage of negative terminal (capacitor voltage) at any time is: V Vc Vsat (Vsat Vsat )exp( t / ) At Vc=Vsat+ V 1 sat V T ln sat 1 1 Linear electronic 97 During the discharging interval T2: the voltage of negative terminal (capacitor voltage) at any time is: V Vc Vsat (Vsat Vsat )exp( t / ) Put, Vc=Vsat- V 1 sat V T ln sat 2 1 If Vsat+ = -Vsat-, Then: 1 T 2 ln 1 L+ = Vsat+ and L- =Vsat- Linear electronic 98 Lecture No. 24 Triangle Waveform Generator The exponential waveforms generated in the astable circuit can be changed to triangular by replacing the RC circuit with an integrator. The integrator causes linear charging and discharging of capacitor, thus providing a triangular waveform. v Let, L+ = Vsat+ and L- =Vsat- Let the output of the bistable circuit be at L+, a current equal to L+/R will flow into the resistor R and through capacitor C causing the output of the integrator to linearly decrease with slop of –L+/CR. This will continue until the integrator output reaches the lower threshold VTL of the bistable circuit, at which point the bistable circuit will switch states, its output becoming negative and equal to L-. Linear electronic 99 At this the current through R and C will reverse direction. The integrator will strart to increase linearly with a positive slop equal to |L-| / RC. This will continue until the integrator output voltage reaches the positive threshold of the bistable circuit Vth. At this point, the bistable circuit switches, its output becomes (L+). - During T1 interval : VTH VTL L T1 CR VTH VTL T1 CR L - During T2 interval: VTH VTL L T2 CR VTH VTL T2 CR L Linear electronic 100 Example: Consider the triangular waveform generator circuit with Op Amp have saturation voltage 10V . If a capacitor C=0.01F and resistor R1 = 10k are used. Find the values of R and R2 such that the frequency of oscillation is 1kHz and the triangular waveform has a 10V peak to peak amplitude Sol Because The waveform with 10V peak to peak , then one peak is equal to 5V, i.e VTH = 5V and VTL= -5V VTH = Vsat+ = VTH / Vsat+ = 5/10 = 0.5 But R1 10 0.5 R2= 10k R1 R2 10 R2 T1 =T2 because Vsat+ = Vsat- T = 1/f = 1/1kHz = 1msec and T1= 0.5msec VTH VTL T1 CR L 5 ( 5 ) 0.5 10 3 0.01 10 6 R 10 0.5 10 3 R 5 10 4 0.01 10 6 R= 50k