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Operational Amplifiers

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1 Operational Amplifiers
EE 174 Spring 2016 Operational Amplifiers

2 Contents Introduction Brief of History Fundamentals of Op-Amps
Basic operation Gain Offset Applications

3 Introduction Operational Amplifier (Op-Amp) name comes from the fact that it was originally used to perform mathematical operations. Op-Amp is an active circuit element which is basic component used to build analog circuits. Op-Amp is a low cost integrating circuit consisting of transistors, resistors, diodes and capacitors. Op-Amp amplify an input signal produces an output voltage equal to the difference between the two input terminals multiplied by the gain A. Op-Amps are two-port networks in which the output voltage or current is directly proportional to either input voltage or current. Four different kind of ampliers exits: Voltage amplifier: Av = Vo / Vi Current amplifier: Ai = Io / Ii Transconductance amplifier: Gm = Io / Vi Transresistance amplifier: Rm = Vo / Ii Op-Amps are commonly used for both linear and nonlinear applications: Inverting/Non-inverting Amplifiers, Variable Gains Amplifiers, Summers, Integrators/Differentiators, Filters (High, Low, Band Pass and Notch Filters), Schmitt trigger, Comparators, A/D converters.

4 Brief History of Op-Amp
Monolithic IC Op-Amp First created in 1963 μA702 by Fairchild Semiconductor μA741 created in 1968, became widely used due to its ease of use 8 pin, dual in-line package (DIP) Further advancements include use of field effects transistors (FET), greater precision, faster response, and smaller packaging Vacuum Tube Op-Amps (1930’s-1940’s) Dual-supply voltage of +300/-300 V Output swing +/- 50 volts Open-loop voltage gain of 15,000 to 20,000, Slew rate of +/- 12 volts/µsecond Maximum output current of 1 mA George Philbrick Solid State Discrete Op-Amps (1960’s) Dual-supply voltage of +15/-15 V Output swing +/- 11 volts Open-loop voltage gain of 40,000, Slew rate of +/- 1.5 volts/µsecond Maximum output current of 2.2 mA

5 Internal Op-Amp Circuits

6 Op-Amp Ideal, Equivalent Circuit, Characteristics and Features
Op-Amp Equivalent Circuit Ideal Op-Amp Op-Amp Symbol Op-Amp Characteristics 741 Op-Amp Features +V2: Non-inverting input -V1: Inverting input +Vs: Positive source PS -Vs: Negative source PS Vout: Output voltage ON: Offset Null NC: Not Connected

7 The Ideal Op-Amp Assumptions
vd = v+ - v- vo = Avd = A(v+ – v-) Note: v+ = v2, v- = v1 1) The input impedance Ri is infinite - i.e. no current flows into either input. 2) The output impedance Ro is zero - i.e. the op-amp can drive any load impedance to any voltage. 3) The open-loop gain (A) is infinite. 4) The bandwidth is infinite. 5) The output voltage is zero when the input voltage difference is zero. 6) The slew rate is infinite

8 Op-Amp two basic operations
The Non-inverting Op Amp The Inverting Op Amp

9 Op-Amp Gain Open loop gain:   This form of gain is measured when no feedback is applied to the op amp. Closed loop gain:   This form of gain is measured when the feedback loop is closed and the overall gain of the circuit is much reduced. It has two forms, signal gain and noise gain. The expression for the gain of a closed-loop amplifier involves the open-loop gain. If G is the actual gain, NG is the noise gain, and AVOL is the open-loop gain of the amplifier, then: Since the open-loop gain is very high, the closed-loop gain of the circuit is simply the noise gain. Loop gain: The difference between the open-loop gain and the closed-loop gain is known as the loop gain. This is useful information because it gives you the amount of negative feedback that can apply to the amplifier system.

10 Signal Gain and Noise Gain
Signal gain: This is the gain applied to the input signal, with the feedback loop connected. It can be inverting or non-inverting. It can even be less than unity for the inverting case. Signal gain is the gain that we are primarily interested in when designing circuits. Noise gain: This is the gain applied to a noise source in series with an op amp input. It is also the gain applied to an offset voltage. Noise gain is equal to the signal gain of a non-inverting amp. It is the same for either an inverting or non-inverting stage. It is the noise gain that is used to determine stability. Summary of Signal Gain and Noise Gain

11 Op-Amp Gain and Bandwidth
The Voltage Gain (A) of the operational amplifier can be found using the following formula: and in Decibels or (dB) is given as: Gain versus Bandwidth Applying feedback will reduce the gain but increase the bandwidth.

12 Gain Bandwidth Product (GBP)
Gain-bandwidth (GBW) product is defined as the open-loop gain multiplied by the bandwidth. The GBW product can be used to calculate the closed-loop gain bandwidth. GBW = A x BW where A is in ratio (not in dB) GBW/closed-loop gain = closed-loop BW Closed-loop BW = GBW / A Example 1: If the GBW is 10 unity gain and the closed-loop gain is 10. Closed-loop BW = 10 MHz/10 = 1 MHz Example 2: GBW 1 unity gain, closed loop gain 70 dB. Closed-loop BW? Gain: 20 log A = 70 dB  log A = 70/20 = 3.5 A = = 3162 Closed-loop BW = 1 MHz/3162 = 316 Hz

13 Gain Bandwidth Product Examples
Given the open loop gain graph for an op-amp as shown. Find the closed loop BW for gain of 88 dB. Find the closed loop gain in dB for a BW of 200 Hz Find the closed loop gain in dB for a BW of 250K Hz. Solutions: a) 88 dB = 20 log A  A = = 25,119 Closed loop BW = 1 M / 25,119 ≈ 40 Hz b) Gain A = 1M / 200 = 5,000 or dB c) Gain A = 1M / 250K = 4 or 12 dB 12dB 250K

14 Bode Plot Example Given a transfer function:
H(s) = 100 𝑠+30 = ( 𝑠 ) = ( 𝑠 ) = ( 𝑠 ) Constant gain = 3.3 Gain in dB = 20 log (3.3) ≈ 10 dB Pole at 30 rad/sec Phase = 45o at 30 rad/sec

15 Op-Amp Saturation The op amp has three distinct regions of operation: Linear region (−Vsat < Vo < Vsat), positive (Vo > Vsat) and negative saturation Vo < −Vsat . The output voltage of a practical amplifier cannot exceed certain threshold value is called saturation. A voltage amplifier behaves linearly, i.e., Vo/Vi = Av = constant as long as the output voltage remains below the “saturation” voltage, −Vsat < Vo < Vsat Note that the saturation voltage, in general, is not symmetric. For an amplifier with a given gain, Av, the above range of Vo translate into a certain range for Vi − Vsat < Vo < Vsat − Vsat < Av Vi < Vsat − Vsat / Av < Vi < Vsat / Av Any amplifier will enter its saturation region if Vi is raised above certain limit. The figure shows how the amplifier output clips when amplifier is not in the linear region. Example: For Av = 105, Vsat,1 = L- = -12V, Vsat,2 = L+ = +15V, find range of input Vi to prevent saturation. Solution: -12V / 105 < Vi < 15V / 105 or -0.12mV < Vi < 0.15mV

16 Op-Amp slew rate basics
Square Wave: The slew rate (SR)of an op amp is the rate of change in the output voltage caused by a step change on the input. It is measured as a voltage change in a given time - typically V / µs. The slew rate may be very important in many applications. For the given graph, it takes 2 µs to swing 10 V so SR = 5V/ µs. Sine Waves: The maximum rate of change for a sine wave occurs at the zero crossing and may derived as follows: Given vo = VP sin(2π f t) Slew Rate (SR) = 𝑑vo 𝑑𝑡 = 2π f VP cos(2π f t) SR = 2π fmax VP Example: Given an application where an op amp is required to amplify a signal with a peak amplitude of 5 volts at a frequency of 25kHz. What is the slew rate requirement? Solution: SR = 2 π x x 5 = 785,000V/s or 0.785V/µs or higher would be required V/µs max = 2π f VP t = 0

17 Op-Amp slew rate Example
Given: Slew rate S0 = 1 V/μs; input voltage vin (t) = sin(2π × 105t); closed loop gain AV =10. Sketch the theoretically correct output and the actual output of the amplifier in the same graph. What is the maximum frequency that will not violating the given slew rate? Solutions: 1) With the closed-loop voltage gain of 10  vo (t) = 10 sin(2π × 105t) = Aω = A x 2π x f = 10 × 2π × 105 = 6.28 V/μs max 2) fmax = SR / (A x 2π) = 1 x 106 / (10 x 2π) = 15.9 kHz

18 Op-Amp Offset Parameters
Input Offset Voltage: Input offset voltage is defined as the voltage that must be applied between the two input terminals of an OPAMP to null or zero the output.  fig. 2, shows that two dc voltages are applied to input terminals to make the output zero. Vio = Vdc1 – Vdc2 Vdc1 and Vdc2 are dc voltages and RS represents the source resistance. Vio is the difference of Vdc1 and Vdc2. It may be positive or negative. For a 741C OPAMP the maximum value of Vio is 6mV. It means a voltage ± 6 mV is required to one of the input to reduce the output offset voltage to zero. The smaller the input offset voltage the better the differential amplifier, because its transistors are more closely matched. Input Offset Current: The input offset current Iio is the difference between the currents into inverting and non-inverting terminals of a balanced amplifier. Iio = |   IB1 – IB2 | The Iio for the 741C is 200nA maximum. As the matching between two input terminals is improved, the difference between IB1 and IB2 becomes smaller, i.e. the Iio value decreases further. For a precision OPAMP 741C, Iio is 6 nA fig. 2

19 Op-Amp Offset Parameters
Input Bias Current: The input bias current IB is the average of the current entering the input terminals of a balanced amplifier i.e. IB = (IB1 + IB2 ) / 2 For 741C IB(max) = 700 nA and for precision 741C IB = ± 7 nA Offset Voltage Adjustment Range: 741 OPAMP have offset voltage null capability. Pins 1 and 5 are marked offset null for this purpose. It can be done by connecting 10 K ohm pot between 1 and 5 as shown in fig. 3. By varying the potentiometer, output offset voltage (with inputs grounded) can be reduced to zero volts. Thus the offset voltage adjustment range is the range through which the input offset voltage can be adjusted by varying 10 K pot. For the 741C the offset voltage adjustment range is ± 15 mV. fig. 3

20 Op-Amp Offset Example Example: Determine the output voltage in each of the following cases for the open loop differential amplifier of fig. 4: vin 1 = 5 m V dc, vin 2 = -7 µVdc vin 1 = 10 mV rms, vin 2= 20 mV rms Specifications of the OPAMP are given below:  A = 200,000, Ri = 2 M Ω , R O = 75Ω, + VCC = + 15 V, - VEE = - 15 V, and output voltage swing = ± 14V. Solution: (a). The output voltage of an OPAMP is given by Remember that vo = 2.4 V dc with the assumption that the dc output voltage is zero when the input signals are zero. (b). The output voltage equation is valid for both ac and dc input signals. The output voltage is given by Thus the theoretical value of output voltage vo = V rms. However, the OPAMP saturates at ± 14 V. Therefore, the actual output waveform will be clipped as shown fig. 5. This non-sinusoidal waveform is unacceptable in amplifier applications.

21 Op-Amp Applications Differential Amplifier If R1 = R2 and R3 = R4

22 Op-Amp Applications Find i, if and vo. i = 0 A
For the ideal op amp shown below, what should be the value of resistor Rf to obtain a gain of 5?

23 Op-Amp Applications Calculate all voltage drops and currents in this circuit, complete with arrows for current direction and polarity markings for voltage polarity. Then, calculate the overall voltage gain of this amplifier circuit (AV), both as a ratio and as a figure in units of decibels (dB): Solutions: Gain = AV = = dB 2) Given gain RF/RS = 100, GBW product = A0 ω0 = K = 4π × Determine the overall 3-dB bandwidth of the cascade amplifier below: Solutions: The 3-dB bandwidth for each amplifier is: Thus, the bandwidth of the cascade amplifier is 4π × 104, the cascade amplifier gain is A1A1= 100 × 100 = 104. To achieve the same gain with a single-stage amplifier having the same K, the bandwidth would be

24 Op-Amp Applications Problem: Suppose that we need an amplifier with input resistance of 500 kΩ or greater and a voltage gain of -10. The feedback resistors are to be implemented in integrated form and have values of 10 kΩ or less to conserve chip area. Choose a suitable circuit configuration and specify the resistance values. Solution: Gain of -10  Use inverting Op-Amp where gain A = - R2/R1 = -10 To attain desired input resistance R1 = 500 kΩ  R2 = 5MΩ. These values exceed the maximum values allowed. To attain large input resistance with moderate resistances for an inverting amplifier, we cascade a voltage follower with an inverter. To achieve the desired gain = - 10. Choose R1 = 1 kΩ  R2 = 10 kΩ. R2

25 The Op-Amp Comparators
The comparator is an electronic decision making circuit that makes use of an operational amplifiers very high gain in its open-loop state, that is, there is no feedback resistor. The Op-amp comparator compares one analogue voltage level with another analogue voltage level, or some preset reference voltage, VREF and produces an output signal based on this voltage comparison. In other words, the op-amp voltage comparator compares the magnitudes of two voltage inputs and determines which is the largest of the two. Then an op-amp comparator can be configured to operate in what is called an inverting or a non-inverting configuration.

26 The Op-Amp Comparators

27 The Schmitt Trigger The Schmitt trigger is a comparator application which switches the output negative when the input passes upward through a positive reference voltage. It then uses positive feedback of a negative voltage to prevent switching back to the other state until the input passes through a lower threshold voltage, thus stabilizing the switching against rapid triggering by noise as it passes the trigger point. That is, it provides feedback which is not reversed in phase, but in this case the signal that is being fed back is a negative signal and keeps the output driven to the negative supply voltage until the input drops below the lower design threshold.

28 The Schmitt Trigger A variation of the previous circuit is the single power supply op-amp Schmitt trigger circuit. This circuit does not require positive and negative voltage for the op-amp to operate. The negative voltage of the op-amp is connected directly to the ground (0V) of the circuit. The calculation of the two threshold points is different. To calculate the High Threshold Level, you solve the following formula: RTOT = (R1 x RFB) / (R1 + RFB) VTHRESHOLD_HIGH = V * R2 / (R2 + RTOT) For the Low Threshold Level, you solve the following formula: RTOT = (R2 x RFB) / (R2 + RFB) VTHRESHOLD_LOW = V x RTOT / (R1 + RTOT) with Schmitt Trigger without Schmitt Trigger 3.4 V 1.7 V 5.0 V 0.0 V Let V = 5V, R1 = R2 = RFB = 10 kΩ. RTOT = 10 kΩ // 10 kΩ = 5 kΩ VTH = 5 V x (10 kΩ / 15 kΩ) = 3.4 V VTL = 5 V x (5 kΩ / 15 kΩ) = 1.7 V

29

30 References: Fundamentals of Electrical Engineering, Giorgio Rizzoni, McGraw-Hill Higher Education


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