Question 1 Questions An operational amplifier is a particular type of differential amplifier. Most op-amps receive two input voltage signals and output one voltage signal: power 1 2 - power Here is a single op-amp, shown under two different conditions (different input voltages). Determine the voltage gain of this op-amp, given the conditions shown: 12 V 12 V 12 V 1 = 1.00 V = 1.5 V 2 = 1.00003 V -12 V 12 V 12 V 12 V 1 = 1.00 V = 6.8 V 2 = 1.00004 V -12 V Also, write a mathematical formula solving for differential voltage gain (A V ) in terms of an op-amp s input and output voltages. file 00848 1
Question 2 A helpful model for understanding opamp function is one where the output of an opamp is thought of as being the wiper of a potentiometer, the wiper position automatically adjusted according to the difference in voltage measured between the two inputs: Positive power supply "rail" V () V- Voltmeter (-) Negative power supply "rail" To elaborate further, imagine an extremely sensitive, analog, zero-center voltmeter inside the opamp, where the moving-coil mechanism of the voltmeter mechanically drives the potentiometer wiper. The wiper s position would then be proportional to both the magnitude and polarity of the difference in voltage between the two input terminals. Realistically, building such a voltmeter/potentiometer mechanism with the same sensitivity and dynamic performance as a solid-state opamp circuit would be impossible, but the point here is to model the opamp in terms of components that we are already very familiar with, not to suggest an alternative construction for real opamps. Describe how this model helps to explain the output voltage limits of an opamp, and also where the opamp sources or sinks load current from. file 02290 Question 3 What does it mean if an operational amplifier has the ability to swing its output rail to rail? Why is this an important feature to us? file 00844 2
Question 4 Determine the output voltage polarity of this op-amp (with reference to ground), given the following input conditions: V V?????? V V?????? V V?????? file 03761 Question 5 What does the phrase open-loop voltage gain mean with reference to an operational amplifier? For a typical opamp, this gain figure is extremely high. Why is it important that the open-loop voltage gain be high when using an opamp as a comparator? file 00873 3
Question 6 The voltage gain of a single-ended amplifier is defined as the ratio of output voltage to input voltage: Amplifier A V A V = Often voltage gain is defined more specifically as the ratio of output voltage change to input voltage change. This is generally known as the AC voltage gain of an amplifier: A V (AC) = In either case, though, gain is a ratio of a single output voltage to a single input voltage. How then do we generally define the voltage gain of a differential amplifier, where there are two inputs, not just one? Differential amplifier (-) () A V file 02287 Question 7 Write the transfer function (input/output equation) for an operational amplifier with an open-loop voltage gain of 100,000. In other words, write an equation describing the output voltage of this op-amp ( ) for any combination of input voltages (() and () ): () (-) file 00925 4
Question 8 How much voltage would have to be dialed up at the potentiometer in order to stabilize the output at exactly 0 volts, assuming the opamp has no input offset voltage? 12 V 12 V??? -12 V 5 V -12 V V - Voltmeter file 00924 Question 9 Complete the table of voltages for this opamp voltage follower circuit: 15 V -15 V 0 volts 0 volts 5 volts 10 volts 15 volts 20 volts -5 volts -10 volts -15 volts -20 volts file 02289 5
Question 10 This operational amplifier circuit is often referred to as a voltage buffer, because it has unity gain (0 db) and therefore simply reproduces, or buffers, the input voltage: Voltage buffer circuit = What possible use is a circuit such as this, which offers no voltage gain or any other form of signal modification? Wouldn t a straight piece of wire do the same thing? Explain your answers. A simpler voltage buffer? = file 03801 Question 11 For all practical purposes, how much voltage exists between the inverting and noninverting input terminals of an op-amp in a functioning negative-feedback circuit? file 00930 6
Question 12 Determine the trip voltage of this comparator circuit: the value of input voltage at which the opamp s output changes state from fully positive to fully negative or visa-versa: 12 V 1 kω 3.3 kω Now, what do you suppose would happen if the output were fed back to the noninverting input through a resistor? You answer merely has to be qualitative, not quantitative: 12 V 1 kω 3.3 kω R feedback For your information, this circuit configuration is often referred to as a Schmitt trigger. file 02294 7
Question 13 Assume that the comparator in this circuit is capable of swinging its output fully from rail to rail. Calculate the upper and lower threshold voltages, given the resistor values shown: 12 V 10 kω -12 V 5 kω V UT = V LT = file 01169 Question 14 Assume that the comparator in this circuit is only capable of swinging its output to within 1 volt of its power supply rail voltages. Calculate the upper and lower threshold voltages, given the resistor values shown: 10 V 1 kω -10 V 3.3 kω V UT = V LT = file 02662 8
Question 15 Write the transfer function (input/output equation) for an operational amplifier with an open-loop voltage gain of 100,000, and the inverting input connected to a voltage divider on its output terminal (so the inverting input receives exactly one-half the output voltage). In other words, write an equation describing the output voltage of this op-amp ( ) for any given input voltage at the noninverting input (() ): () (-) R R 1 2 Then, once you have an equation written, solve for the output voltage if the noninverting input voltage is -2.4 volts. file 00928 Question 16 Calculate the overall voltage gain of this amplifier circuit (A V ), both as a ratio and as a figure in units of decibels (db). Also, write a general equation for calculating the voltage gain of such an amplifier, given the resistor values of R 1 and R 2 : 27 kω R 1 27 kω R 2 file 02457 9
Question 17 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 (A V ), both as a ratio and as a figure in units of decibels (db): 22 kω 47 kω =??? R 2 R 1 = 3.2 volts file 02459 Question 18 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 (A V ), both as a ratio and as a figure in units of decibels (db): =??? R a 10 kω = -2.35 V R b 2.7 kω file 02460 10
Question 19 The equation for voltage gain (A V ) in a typical noninverting, single-ended opamp circuit is as follows: A V = R 1 R 2 1 Where, R 1 is the feedback resistor (connecting the output to the inverting input) R 2 is the other resistor (connecting the inverting input to ground) Suppose we wished to change the voltage gain in the following circuit from 5 to 6.8, but only had the freedom to alter the resistance of R 2 : R 2 1k175 R 1 4k7 Algebraically manipulate the gain equation to solve for R 2, then determine the necessary value of R 2 in this circuit to give it a voltage gain of 6.8. file 02707 Question 20 Calculate the necessary resistor value (R 1 ) in this circuit to give it a voltage gain of 30: 39 kω R 1 file 02725 11
Question 21 Calculate the voltage gain for each stage of this amplifier circuit (both as a ratio and in units of decibels), then calculate the overall voltage gain: Stage 1 Stage 2 1 kω 3.3 kω 470 Ω 2.7 kω file 02727 Question 22 Calculate the necessary resistor value (R 1 ) in this circuit to give it a voltage gain of 15: R 1 22 kω file 02729 Question 23 Calculate the necessary resistor value (R 1 ) in this circuit to give it a voltage gain of 7.5: 8.3 kω R 1 file 02730 12
Question 24 Operational amplifier circuits employing negative feedback are sometimes referred to as electronic levers, because their voltage gains may be understood through the mechanical analogy of a lever. Explain this analogy in your own words, identifying how the lengths and fulcrum location of a lever relate to the component values of an op-amp circuit: Fulcrum Lever Fulcrum Lever file 00933 Question 25 Compare and contrast inverting versus noninverting amplifier circuits constructed using operational amplifiers: Inverting amplifier circuit Noninverting amplifier circuit How do these two general forms of opamp circuit compare, especially in regard to input impedance and the range of voltage gain adjustment? file 02469 13
Answer 1 A V = 530,000 Answers A V = (2 1 ) Follow-up question: convert this voltage gain figure (as a ratio) into a voltage gain figure in decibels. Answer 2 The output voltage of an opamp cannot exceed either power supply rail voltage, and it is these rail connections that either source or sink load current. Follow-up question: does this model realistically depict the input characteristics (especially input impedance) of an opamp? Why or why not? Answer 3 Being able to swing the output voltage rail to rail means that the full range of an op-amp s output voltage extends to within millivolts of either power supply rail (V and ). Challenge question: identify at least one op-amp model that has this ability, and at least one that does not. Bring the datasheets for these op-amp models with you for reference during discussion time. 14
Answer 4 In these illustrations, I have likened the op-amp s action to that of a single-pole, double-throw switch, showing the connection made between power supply terminals and the output terminal. V V (-) (-) V V () (-) V V (-) (-) Answer 5 Open-loop voltage gain simply refers to the differential voltage gain of the amplifier, without any connections feeding back the amplifier s output signal to one or more of its inputs. A high gain figure means that a very small differential voltage is able to drive the amplifier into saturation. Answer 6 Voltage gain for a differential amplifier is defined as the ratio of output voltage to the difference in voltage between the two inputs. 15
Answer 7 = 100,000(() () ) Answer 8 5 volts Answer 9 0 volts 0 volts 5 volts 5 volts 10 volts 10 volts 15 volts 15 volts 20 volts 15 volts -5 volts -5 volts -10 volts -10 volts -15 volts -15 volts -20 volts -15 volts Follow-up question: the output voltage values given in this table are ideal. A real opamp would probably not be able to achieve even what is shown here, due to idiosyncrasies of these amplifier circuits. Explain what would probably be different in a real opamp circuit from what is shown here. Answer 10 While this circuit offers no voltage gain, it does offer current gain and impedance transformation. Much like the common-collector (or common-drain) single transistor amplifier circuits which also had voltage gains of (near) unity, opamp buffer circuits are useful whenever one must drive a relatively heavy (low impedance) load with a signal coming from a weak (high impedance) source. Answer 11 Zero volts Answer 12 With no feedback resistor, the trip voltage would be 9.21 volts. With the feedback resistor in place, the trip voltage would change depending on the state of the opamp s output! Follow-up question: describe what effect this changing trip voltage value will have on the operation of this comparator circuit. Answer 13 V UT = 8 volts V LT = -8 volts Challenge question: how would you recommend we change the circuit to give threshold voltages of 6 volts and -6 volts, respectively? Answer 14 V UT = 2.093 volts V LT = -2.093 volts 16
Answer 15 = 100,000(() 1 2 ) (I ve left it up to you to perform the algebraic simplification here!) For an input voltage of -2.4 volts, the output voltage will be -4.7999 volts. Follow-up question: what do you notice about the output voltage in this circuit? What value is it very close to being, in relation to the input voltage? Does this pattern hold true for other input voltages as well? Answer 16 A V = 2 = 6.02 db A V = R 1 R 2 1 (expressed as a ratio, not db) Follow-up question: explain how you could modify this particular circuit to have a voltage gain (ratio) of 3 instead of 2. Answer 17 I R2 = I R1 = 68.09 µa V R2 = 1.498 V V R1 = 3.2 V 22 kω 47 kω = 4.698 V Current arrows drawn in the direction of conventional flow notation. = 3.2 volts A V = 1.468 = 3.335 db 17
Answer 18 Current arrows drawn in the direction of conventional flow = -11.054 V V Ra = 8.704 V 10 kω I Ra = I Rb = 870.4 µa = -2.35 V 2.7 kω V Rb = 2.35 V A V = 4.704 = 13.449 db Follow-up question: how much input impedance does the -2.35 volt source see as it drives this amplifier circuit? Answer 19 R 2 = R 1 A V 1 For the circuit shown, R 2 would have to be set equal to 810.3 Ω. Answer 20 R 1 = 1.345 kω Answer 21 Stage 1: A V = 4.3 = 12.669 db Stage 2: A V = 6.745 = 16.579 db Overall: A V = 29.002 = 29.249 db Answer 22 R 1 = 1.467 kω Answer 23 R 1 = 62.25 kω 18
Answer 24 The analogy of a lever works well to explain how the output voltage of an op-amp circuit relates to the input voltage, in terms of both magnitude and polarity. Resistor values correspond to moment arm lengths, while direction of lever motion (up versus down) corresponds to polarity. The position of the fulcrum represents the location of ground potential in the feedback network. Answer 25 The noninverting configuration exhibits a far greater input impedance than the inverting amplifier, but has a more limited range of voltage gain: always greater than or equal to unity. 19