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| - | ====== Real Analog: Chapter 5 ====== | + | ===== Real Analog: Chapter 5 ====== |
| - | ====== 5. Introduction and Chapter Objectives | + | [[{}/ |
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| + | * Lecture Material: | ||
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| + | * Chapter 5 Video: | ||
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| + | <-- | ||
| + | ===== 5. Introduction and Chapter Objectives ===== | ||
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| The above constraint is based on pure inequalities – in general, the output voltage range will be somewhat less than the range specified by V- and V+. The margin between the output and the supply voltages will vary depending on the specific op-amp. Any attempt to drive the output voltage beyond the range specified by the supply voltages will cause the output to // | The above constraint is based on pure inequalities – in general, the output voltage range will be somewhat less than the range specified by V- and V+. The margin between the output and the supply voltages will vary depending on the specific op-amp. Any attempt to drive the output voltage beyond the range specified by the supply voltages will cause the output to // | ||
| - | |||
| - | Similarly it makes sense that the power supply voltages will constrain the range of allowable input voltages, as provided below: | ||
| **Input Voltage Constraint**: | **Input Voltage Constraint**: | ||
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| * The voltages at the input terminals are the same: $v_n = v_p$ | * The voltages at the input terminals are the same: $v_n = v_p$ | ||
| * The output voltage is constrained to be between the positive and negative power supply voltages: $V^- < v_{OUT} < V^+$ | * The output voltage is constrained to be between the positive and negative power supply voltages: $V^- < v_{OUT} < V^+$ | ||
| - | * The input voltages are constrained to be between | + | * Nothing is known about the current out of the op-amp, $i_{OUT}$ |
| * All voltages on the above diagram are relative to the same reference. | * All voltages on the above diagram are relative to the same reference. | ||
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| $$V_{OUT} = K \left(v_p - v_n \right) = K \cdot \Delta v_{in} | $$V_{OUT} = K \left(v_p - v_n \right) = K \cdot \Delta v_{in} | ||
| - | Where in $\Delta v_{in}$ is the difference between the voltages at the input terminals and //K// is a very large number. (Values of //K// for typical commercially available operational amplifiers can be on the order of $10^6$$ or higher.) Since the output voltage is constrained to be less than the supply voltages, | + | Where in $\Delta v_{in}$ is the difference between the voltages at the input terminals and //K// is a very large number. (Values of //K// for typical commercially available operational amplifiers can be on the order of $10^6$ or higher.) Since the output voltage is constrained to be less than the supply voltages, |
| $$V^- < K \cdot \Delta v_{in} < V^+$$ | $$V^- < K \cdot \Delta v_{in} < V^+$$ | ||
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| ==== Exercises ==== | ==== Exercises ==== | ||
| - | - An operational amplifier has a gain K = 10,000. The voltage supplies are $V+ = 20V$ and $V-= -10V$. Determine the output voltage if the voltage difference between the input terminals $\left( v_p - v_n \right) is: | + | - An operational amplifier has a gain K = 10,000. The voltage supplies are $V+ = 20V$ and $V-= -10V$. Determine the output voltage if the voltage difference between the input terminals $\left( v_p - v_n \right)$ is: |
| - 1mV | - 1mV | ||
| - 2mV | - 2mV | ||
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| Using Ohm’s law to write these currents in terms of node voltages and taking advantage of the fact that the voltage at the inverting terminal of the op-amp is zero (because there is no voltage difference across the input terminals of the op-amp and we have chosen the non-inverting terminal voltage as our reference) results in: | Using Ohm’s law to write these currents in terms of node voltages and taking advantage of the fact that the voltage at the inverting terminal of the op-amp is zero (because there is no voltage difference across the input terminals of the op-amp and we have chosen the non-inverting terminal voltage as our reference) results in: | ||
| - | $$\frac{V_{in}-0}{R_{in} = \frac{0-V_{OUT}}{R_f}$$ | + | $$\frac{V_{in} - 0}{R_{in}} = \frac{0 - V_{OUT}}{R_f}$$ |
| - | Solving the above for $V_OUT$ results in: | + | Solving the above for $V_{OUT}$ results in: |
| $$V_{OUT} = - \left( \frac{R_f}{R_{in}} \right) V_{in}$$ | $$V_{OUT} = - \left( \frac{R_f}{R_{in}} \right) V_{in}$$ | ||
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| ==== Example 5.3 ==== | ==== Example 5.3 ==== | ||
| Determine $V_{OUT}$ as a function of $V_1$ and $V_2$ for the circuit shown below. | Determine $V_{OUT}$ as a function of $V_1$ and $V_2$ for the circuit shown below. | ||
| + | {{ : | ||
| + | |||
| + | Denoting the non-inverting terminal of the op-amp as node a and the inverting terminal as node b, and applying the ideal op-amp rules results in the figure below: | ||
| + | |||
| + | {{ : | ||
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| + | The voltage $V_a$ can be determined from a voltage divider relation (or by applying KCL at node a) as $V_a = \frac{V_2}{2}$. | ||
| + | |||
| + | Thus, the voltage at the inverting terminal is $V_B = V_a = \frac{V_2}{2}$. Applying KCL at node b results in: | ||
| + | |||
| + | $$\frac{V_1-V_b}{R} = \frac{V_b-V_{OUT}}{R} \Rightarrow \frac{V_1 - \frac{V_1}{2}}{R} = \frac{frac{V_2}{2} - V_{OUT}}{R}$$ | ||
| + | |||
| + | Simplification of the above results in $V_{OUT} = V_2 - V_1$. | ||
| + | |||
| + | Comments: | ||
| + | * The above circuit performs a subtraction operation. The voltage $V_1$ is subtracted from the voltage $V_2$. | ||
| + | * The inverting and non-inverting terminals of the op-amp are treated as separate nodes in this analysis, even though the op-amp constrains the voltages at these nodes to be the same. Thus, we apply KCL at __each__ input terminal of the op-amp. | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 5.4 ==== | ||
| + | Determine $V_{OUT}$ as a function of $V_1$ and $V_2$ for the circuit shown below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Choosing the non-inverting terminal voltage as our reference voltage and applying the ideal operational amplifier rules allows us to label the voltages and currents shown in red below: | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Applying KCL at the inverting terminal of the op-amp results in: | ||
| + | |||
| + | $$\frac{V_1-0}{R} + \frac{V_2 -0}{R} = \frac{0-V_{OUT}}{R}$$ | ||
| + | |||
| + | Or | ||
| + | |||
| + | $$V_{OUT} = - \left( V_1 + V_2 \right)$$ | ||
| + | |||
| + | Comments: | ||
| + | * The circuit inverts the sum of the inputs. One can use an inverting amplifier with a gain of one in conjunction with the above circuit to obtain a non-inverted sum of the inputs. | ||
| + | * An arbitrary number of inputs can be summed, by simply increasing the number of input signals and resistors applied at the inverting terminal of the op-amp, which is often referred to as the “summing node”. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 5.5 ==== | ||
| + | Determine $V_{OUT}$ as a function of $V_{IN}$ for the circuit shown below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Since there is no circuit element in the feedback loop, the inverting terminal voltage is identical to the output voltage, $V_{OUT}$. The ideal op-amp rules require that the inverting and non-inverting terminal voltages are the same, so: | ||
| + | |||
| + | $$V_{OUT} = V_{in}$$ | ||
| + | |||
| + | The circuit is called a //voltage follower//, since the output voltage simply “follows” the input voltage. This circuit, though it appears to do nothing, is actually extremely useful. Since the input voltage is applied directly to an op- amp input terminal, the input resistance to the circuit is infinite and no current is drawn from the source. Thus, __the source provides no power__ in order to generate the output voltage -- all power provided to the load comes from the op-amp power supplies. This can be extremely useful in isolating different portions of a circuit from one another. | ||
| + | |||
| + | Consider, as an example, the following case. We have a loading circuit with an equivalent resistance of $100\Omega$. We wish to apply 6V to the circuit, but only have access to a 12V source. It is decided that we will use a voltage divider containing two $100\Omega$ resistors in series to reduce the supply voltage to the desired 6V level as shown in the circuit to the left below. However, adding the loading circuit to the voltage divider changes the voltage provided to the load, as shown to the right below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Addition of a voltage follower to the circuit isolates the voltage divider from the load, as shown below. Power to the op-amp can be provided by connecting the 12V source to $V^+$ and grounding $V^-$, as shown, since the desired op-amp output is between 0V and 12V. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | ==== Section Summary ==== | ||
| + | * Analysis of linear operational amplifier circuits typically consists of the following components: | ||
| + | * Assume that the voltage difference across the input terminals is zero. | ||
| + | * Assume that the currents into the input terminals is zero. | ||
| + | * Apply KCL at op-amp input terminals. | ||
| + | * Apply KCL at other circuit nodes, if necessary. | ||
| + | * Check to ensure that output voltage remains within range specified by op-amp power supply voltages. | ||
| + | * Op-amp circuits which perform the following functions are presented in this section: | ||
| + | * Inverting voltage amplification | ||
| + | * Non-inverting voltage amplification | ||
| + | * Summation (addition) | ||
| + | * Differencing (subtraction) | ||
| + | * Buffering | ||
| + | |||
| + | The reader should be able to sketch circuits, which perform the functions above. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Exercises ==== | ||
| + | - Represent the circuit of Example 5.2 as a voltage controlled voltage source. | ||
| + | - Represent the circuit of Example 5.2 as a voltage controlled current source. | ||
| + | - Find $V_{out}$ for the circuit below | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | 4. Find V in the circuit below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ===== 5.5: Comparators ===== | ||
| + | Operational amplifiers are intended to be incorporated into circuits which //feeds back// the op-amp output to one or both of the input terminals. That is, the output voltage is connected in some way to the op-amp inputs. Typically, for stable operation, the output is fed back to the __inverting__ input terminal for stable operation (as in all of our circuit examples in section 5.4). If the output is not fed back to the input of the op-amp, the op-amp may not function as expected. | ||
| + | |||
| + | Comparators are operational amplifier –like devices which are intended to be operated without feedback from the output to the input. The circuit symbol for a comparator looks like an op-amp symbol, reflecting their similarities. Figure 5.15 provides a typical comparator symbol, with applicable voltages labeled. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Operation of the comparator is simple: if $v_p$ is greater than $v_n$, the output goes to the high supply voltage, $V^+$. If $v_p$ is less than $v_n$, the output goes to the low supply voltage, $V^-$. The comparator is essentially checking the sign between the voltage at the inverting and non-inverting inputs, and adjusting the output voltage accordingly. Mathematically, | ||
| + | |||
| + | $$V_{OUT} = \begin{cases} | ||
| + | V^+, v_p - v_n > 0\\ | ||
| + | V^-, v_p - v_n < 0 | ||
| + | \end{cases}$$ | ||
| + | |||
| + | Because of this behavior, comparators are sometimes used to implement logical functions. Commercial comparators are also commonly equipped with a //latch// input. Once the latch is enabled, the output responds to only the time the polarity between the input terminal voltages changes. | ||
| + | |||
| + | ==== Section Summary ==== | ||
| + | * Unlike operational amplifiers, comparators are intended to operate without feedback. | ||
| + | * A comparator essentially operates on the difference in the voltage across the input terminals and puts out either a high or low voltage, depending upon the sign of the difference. The behavior of a comparator is typically modeled as: | ||
| + | |||
| + | $$V_{OUT} = \begin{cases} | ||
| + | V^+, v_p - v_n > 0\\ | ||
| + | V^-, v_p - v_n < 0 | ||
| + | \end{cases}$$ | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Exercises ==== | ||
| + | - A comparator like that shown in Figure 5.14 has the sinusoidal signal below applied across the input terminals. (E.g. the plot below is $v_p-v_n$ vs. time.) Sketch the output voltage $v_{out}(t)$. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ===== 5.6: A Few Non-ideal Effects ===== | ||
| + | In section 5.2, we indicated that operational amplifiers are designed to have high input resistances, | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | In section 5.2, we also provided the assumptions applicable to ideal operational amplifier operation, along with their associated conclusions, | ||
| + | * The output voltage is bounded by the power supply voltages: $V^- < v_{OUT} < V^+$ | ||
| + | * $K \rightarrow \infty$. Thus, $\Delta v_{in} = 0$ and $v_p = v_n$ | ||
| + | * $R_{in} \rightarrow \infty$. Thus, $i_p = -i_n = 0$, and the operational amplifier draws no power at its input. | ||
| + | * $R_{OUT} = 0$. Thus, there is no limit on the output current (or power) which can be provided by the op-amp. | ||
| + | |||
| + | Practical operational amplifiers have finite gains (K for most amplifiers is in the range $10^5 – 10^7$), finite input resistances (typical values are on the order of a few mega-ohms to hundreds or thousands of mega-ohms) and non-zero output resistances (generally on the order of 10 to 100 ohms). In this section, we will very briefly discuss a few of the ramifications of these non-ideal parameters. | ||
| + | |||
| + | ==== Input Resistance Effects ==== | ||
| + | The high input resistance of the operational amplifier means that circuits connected to the op-amp input do not have to provide much power to the op-amp circuit. This is the op-amp property that is employed in buffer amplifiers and instrumentation amplifiers. Instrumentation systems, for example, have very limited power output capabilities; | ||
| + | |||
| + | It is apparent from Figure 5.17 that the output resistance of the system, if it is large enough, can have an effect on the voltage difference across the op-amp input terminals, since: | ||
| + | |||
| + | $$\Delta v_{in} = V_S \left( \frac{R_{in}}{R_{in} + R_S} \right)$$ | ||
| + | |||
| + | Since we generally want to amplify $V_S$ directly, any difference between $V_S$ and in $\Delta v_{in}$ will degrade our output voltage from its desired value. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | ==== Output Resistance Effects ==== | ||
| + | The op-amp output resistance essentially limits the amount of power the op-amp can provide at its output terminal. This can become a problem if we want to connect very low resistance loads to the output of an operational amplifier. For example, audio speakers commonly have an 8Ω resistance. Figure 5.18 shows an 8Ω speaker connected to the output of an operational amplifier which has an 80Ω output resistance. In this case, we expect the maximum output voltage to be: | ||
| + | |||
| + | $$v_{out} = K \cdot V_S \left( \frac{8\Omega}{8\Omega + 80\Omega} \right) = 0.09KV_s$$ | ||
| + | |||
| + | If the maximum output voltage of the op-amp is low, we may not have nearly enough power to operate the speaker. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | ==== Finite Gain Effects ==== | ||
| + | As an example of the effects of a finite voltage gain, let us assume that an operational amplifier has a gain of K = 10,000 and supply voltages $V^+ = 10V$ and $V^- = -10V$. From equation (5.2), the linear operating range of the operational amplifier is over the range of input voltage differences: | ||
| + | |||
| + | $$\frac{V^-}{K} \leq \Delta v_{in} \leq \frac{V^+}{K}$$ | ||
| + | |||
| + | The non-ideal operational amplifier of interest can then allow input terminal voltage differences of up to $-1mV \leq \Delta v_{in} \leq 1mV$. Although voltage differences of a millivolt will be considered to be essentially zero for any of the voltage levels we will deal with in this class, these voltages are definitely __not__ zero for some applications. | ||
| + | |||
| + | ==== Section Summary ==== | ||
| + | * The effect of a finite input resistance on an operational amplifier’s operation is that the current into the input terminals will not be identically zero. Thus, a real operational amplifier with finite input resistance will always draw some power from a circuit connected to it. Whether this has a significant effect on the overall circuit’s operation is generally a function of the output resistance of the circuit to which the amplifier is connected. | ||
| + | * The effect of a non-zero output resistance on an operational amplifier’s operation is that the power output of the amplifier is limited. Thus, a realistic operational amplifier will not be able to provide any arbitrary current to a load. Whether this has a significant effect on the overall circuit’s operation is primarily dependent upon the value of the load resistance. | ||
| + | * The effect of a finite op-amp gain is that the voltage difference across the input terminals may not be identically zero. | ||
| + | |||
| + | [[{}/ | ||
| + | [[{}/ | ||