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7.7: Extended Topic

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  • A Precision Log Amp


    The basic log amplifier discussed earlier suffers from two major problems: repeatability and temperature sensitivity. Let's take a closer look at Equation 7.12. First, if you work backward through the derivation, you will notice that the constant 0.0259 is actually K T/q using a temperature of 300 K. Thus, we may rewrite this equation as,

    $$ V_{out} =− \frac{K T}{q} ln \frac{V_{in}}{R_i I_s} \label{7.19}\tag{7.19} $$


    Note that the output voltage is directly proportional to the circuit temperature, T. Normally, this is not desired. The second item of interest is \(I_s\). This current can vary considerably between devices, and is also temperature sensitive, approximately doubling for each 10 C° rise. For these reasons, commercial log amps such as those mentioned earlier, are preferred. This does not mean that it is impossible to create stable log amps from the basic op amp building blocks. To the contrary, a closer look at practical circuit solutions will point out why specialised log ICs are successful. We shall treat the two problems separately.


    One way of removing the effect of \(I_s\) from our log amp is to subtract away an equal effect. The circuit of Figure 7.65 does exactly this. This circuit utilizes two log amps. Although each is drawn with an input resistor and input voltage source, removal of \(R_1\) and \(R_2\) would allow current sensing inputs. Each log amp's transistor is part of a matched pair. These two transistors are fabricated on the same silicon wafer and exhibit nearly identical characteristics (in our case, the most notable being \(I_s\)).

    Based on Equation 7.19 we find

    $$ V_A = \frac{−K T}{q} ln \frac{V_{in1}}{R_1 I_{s1}} \notag $$

    $$ V_B = \frac{−K T}{q} ln \frac{V_{in2}}{R_2 I_{s2}} \notag $$


    Where typically, \(R_1 = R_2\). These two signals are fed into a differential amplifier comprised of op amp 3, and resistors \(R_3\) through \(R_6\). It is normally set for a gain of unity. The output of the differential amplifier (point C) is

    $$ V_C = V_B − V_A \notag $$

    $$ V_C = \frac{−K T}{q} ln \frac{V_{in2}}{R_2 I_{s2}} – \left( − \frac{K T}{q} ln \frac{V_{in1}}{R_1 I_{s1}} \right) \notag $$

    $$ V_C = \frac{K T}{q} \left( ln \frac{V_{in1}}{R_1 I_{s1}} − ln \frac{V_{in2}}{R_2 I_{s2}} \right) \label{7.20}\tag{7.20} $$


    Using the basic identity that subtracting logs is the same as dividing their arguments, 7.20 becomes

    $$ V_C = \frac{K T}{q} ln \frac{\frac{V_{in1}}{R_1 I_{s1}}}{\frac{V_{in2}}{R_2 I_{s2}}} \label{7.21}\tag{7.21} $$


    Because \(R_1\) is normally set equal to \(R_2\), and \(I_{s1}\) and \(I_{s2}\) are identical due to the fact that \(Q_1\) and \(Q_2\) are matched devices, 7.21 simplifies to

    $$ V_C = \frac{K T}{q} ln \frac{V_{in1}}{V_{in2}}  \label{7.22}\tag{7.22} $$


    As you can see, the effect of \(I_s\) has been removed. \(V_C\) is a function of the ratio of the two inputs. Therefore, this circuit is called a log ratio amplifier. The only remaining effect is that of temperature variation. Op amp 4 is used to compensate for this. This stage is little more than a standard SP noninverting amplifier. What makes it unique is that \(R_8\) is a temperature-sensitive resistor. This component has a positive temperature coefficient of resistance, meaning that as temperature rises, so does its resistance. Because the gain of this stage is 1 + \((R_7/R_8)\), a temperature rise causes a decrease in gain. Combining this with 7.22 produces

    $$ V_{out} = \left( 1+ \frac{R_7}{R_8} \right) \frac{K T}{q} ln \frac{V_{in1}}{V_{in2}}  \label{7.23}\tag{7.23} $$


    If the temperature coefficient of \(R_8\) is chosen correctly, the first two temperature dependent terms of 7.23 will cancel, leaving a temperature stable circuit. This coefficient is approximately 1/300 K, or 0.33% per C°, in the vicinity of room temperature.


    Our log ratio circuit is still not complete. Although the major stability problems have been eliminated, other problems do exist. It is important to note that the transistor used in the feedback loop loads the op amp, just as the ordinary feedback resistor Rf would. The difference lies in the fact that the effective resistance, which the op amp sees is, the dynamic base-emitter resistance, \(r^{'}_e\). This resistance varies with the current passing through the transistor and has been found to equal 26m \(V/I_E\) at room temperature. For higher input currents this resistance can be very small and might lead to an overload condition. A current of 1 mA, for example, would produce an effective load of only 26 Ω. This problem can be alleviated by inserting a large value resistance in the feedback loop. This resistor, labeled \(R_E\), is shown in Figure 7.66. An appropriate value for \(R_E\) may be found by realizing that at saturation, virtually all of the output potential will be dropped across \(R_E\), save \(V_{BE}\). The current through \(R_E\) is the maximum expected input current plus load current. Using Ohm's Law,


    $$ R_E = \frac{V_{sat} − V_{BE}}{I_{max}}  \label{7.24}\tag{7.24} $$


    Figure 7.66 also shows a compensation capacitor, \(C_c\). This capacitor is used to roll off high frequency gain in order to suppress possible high frequency oscillations. An optimum value for \(C_c\) is not easy to determine, as the feedback element's resistance changes with the input level. It may be found empirically in the laboratory. A typical value would be in the vicinity of 100 pF.

    One possible application for the log ratio circuit is found in Figure 7.67. This system is used to measure the light transmission of a given material. Because variations in the light source will affect a direct measurement, the measurement is instead made relative to a known material. In this example, light is passed through a known medium (such as a vacuum) while it is simultaneously passed through the medium under test. On the far side of both materials are light-sensitive devices such as photodioides, phototransistors, or photomultiplier tubes. These devices will produce a current proportional to the amount of light striking them. In this system, these currents are fed into the log ratio circuit, which will then produce an output voltage proportional to the light transmission abilities of the new material. This setup eliminates the problem of light source fluctuation, because each input will see an equal percentage change in light intensity. Effectively, this is a common-mode signal that is suppressed by the differential amplifier section. This system also eliminates the difficulty of generating calibrated light intensity readings for comparisons. By its very nature, this system performs relative readings.






    In this chapter, you have examined a variety of non-linear op amp circuits. They are deemed non-linear because their input/output characteristic is not a straight line.

    Precision rectifiers use negative feedback to compensate for the forward drop of diodes. By doing so, very small signals may be successfully rectified. Both halfwave and full-wave versions are available. An extension of the precision rectifier is the precision peak detector. This circuit uses a capacitor to effectively lengthen the duration of peaks. It can also be used as an envelope detector.


    Other active diode circuits include clampers and limiters. Clampers adjust the DC level of an input signal so that the output is either always positive or always negative. The furthest peaks will just hit 0 V. Clampers may also be used with a DC offset for greater control of the output signal. Limiters force their output to be no greater than a preset value. It may be thought of as a programmable signal clipper.


    Function generation circuits are used to create piece-wise linear approximations of transfer functions. These can be used to compensate for the non-linearities of transducers and measurement devices. There are two basic approaches for realizing a design: (1) using Zener diodes, and (2) using a biased-signal diode network. The Zener scheme is a bit easier to set up, but does not offer the performance of the biased-diode form

    In order to improve the performance of comparators, positive feedback may be used. The resulting circuits are called Schmitt Triggers and have much greater noise immunity than simple open loop op amp comparators do. Specialized comparators are also available. These devices are generally faster than simple op amps, and offer more flexibility for logic family interfacing.


    Log and anti-log amplifiers utilize a transistor in their feedback loops. The result is a form of signal compression for the log amp, and an inverse signal expansion for the anti-log amp. Specialized log/anti-log ICs are available that take much of the tedium out of amplifier design and testing.

    Finally, a four-quadrant multiplier produces an output that is proportional to the product of its two inputs. These devices may be used for a number of applications, including balanced modulation, frequency doubling, and more.

    In short, although op amps are linear devices, they may be used with other components to form non-linear circuits.


    Review Questions


    1. What are the advantages of active rectifiers versus passive rectifiers?

    2. What are the disadvantages of active rectifiers versus passive rectifiers?

    3. What is a peak detector?

    4. What is a limiter?

    5. What is the function of a clamper?

    6. What are the differences between active and passive clampers?

    7. What is a transfer function generator circuit, and what is its use?

    8. Explain how a function circuit might be used to linearize an input transducer.

    9. What is a Schmitt trigger?

    10. What is the advantage of a Schmitt-type comparator versus an ordinary open-loop op amp comparator?

    11. What are the advantages of dedicated comparators such as the LM311 versus ordinary op amp comparators?

    12. How are log and anti-log amplifiers formed?

    13. What is the effect of passing a signal through a log or anti-log amplifier?

    14. What is a four-quadrant multiplier?

    15. How might a multiplier be used?

    16. What is the difference between a multiplier and a VCA?





    Analysis Problems


    1. Sketch the output of Figure 7.2 if the input is a 100 Hz 2 V peak triangle wave.

    2. Repeat Problem 1 for a square wave input.

    3. Sketch the output of Figure 7.8 if \(R_f\) = 20 kΩ and \(R_i\) = 5 kΩ. Assume that Vin is a 0.5 V peak triangle wave.

    4. Repeat Problem 3, but with the diode reversed.

    5. Sketch the output of Figure 7.9 if \(C\) = 50 nF and \(R\) = 5 MΩ. Assume that the input is a 1 V peak pulse with 10% duty cycle. The input frequency is 1 kHz.

    6. Repeat Problem 5 with the diode reversed

    7. Sketch the output of Figure 7.14 if \(R\) = 25 kΩ and Vin = 2 sin 2 π 500 t.

    8. Repeat Problem 7 with \(V_{in}\) equal to a 3 V peak square wave.

    9. Sketch the output of Figure 7.15 if the diodes are reversed. Assume that \(V_{in}\) is a 100 mV peak-to-peak sine wave at 220 Hz.

    10. In Figure 7.23, assume that \(R\) = 10 kΩ, \(R_l\) = 1 MΩ, \(C\) = 100 nF and \(V_{offset}\) = 2 V. Sketch the output for a 10 kHz 1 V peak sine wave input.

    11. Sketch the output of Figure 7.27 if \(R_i\) = 10 kΩ, \(R_f\) = 40 kΩ and the Zener potential is 3.9 V. The input signal is a 2 V peak sine wave.

    12. Sketch the transfer curve for the circuit of Problem 11.

    13. In Figure 7.31, sketch the transfer curve if \(R_i\) = 5 kΩ, \(R_f\) = 33 kΩ, \(R_a\) = 20 kΩ and \(V_{Zener}\) = 5.7 V.

    14. Sketch the output of the circuit in Problem 13 for an input signal equal to a 2 V peak triangle wave.

    15. Repeat Problem 14 with a square wave input.

    16. If \(R_i\) = 4 kΩ, \(R_a\) = 5 kΩ, \(R_f\) = 10 kΩ, and \(V_z\) = 2.2 V in Figure 7.33, determine the minimum and maximum input impedance.

    17. Draw the transfer curve for the circuit of Problem 16.

    18. Sketch the output voltage for the circuit of Problem 16 if the input signal is a 3 V peak triangle wave.

    19. Sketch the transfer curve for the circuit of Figure 7.34 if \(R_i\) = 1 kΩ, \(R_f\) = 10 kΩ, \(R_a\) = 20 kΩ, \(R_b\) = 18 kΩ, \(V_{Zener-a}\) = 3.9 V, and \(V_{Zener-b}\) = 5.7 V.

    20. Sketch the output of the circuit in Problem 19 if the input is a 1 V peak triangle wave.

    21. If \(R_1\) = 10 kΩ and \(R_2\) = 33 kΩ in Figure 7.47, determine the upper and lower thresholds if the power supplies are ±15 V.

    22. Determine the upper and lower thresholds for Figure 7.48 if \(R_1\) = 4.7 kΩ and \(R_2\) = 2.2 kΩ, with ±12 V power supplies.

    23. Sketch the output of Figure 7.53 if \(V_{in} = 2 sin 2 π 660 t\), \(V_{ref} = 1 V DC\), and \(V_{strobe} = 5 V DC.

    24. Repeat Problem 23 for \(V_{strobe}\) = 0 V DC.

    25. Determine the output of Figure 7.56 if \(V_{in} = 0.1 V\), \(R_i = 100 kΩ\), and \(I_s = 60 nA\).

    26. Determine the output of Figure 7.58 if \(V_{in} = 300 mV\), \(R_f = 20 kΩ\) and \(I_s = 40 nA\).

    27. Sketch the output signal of Figure 7.61 if \(V_{in} = 2 sin 2 π 1000 t\), \(K\) = 0.1, and \(V_{control} = 1 V\) DC, -2 V DC, and 5 V DC.

    28. Sketch the output of Figure 7.62 if \(V_{in} = 1 sin 2 π 500 t\), and \(K\) = 0.1.

    29. Sketch the output of Figure 7.63 if \(V_{in} = 5 sin 2 π 2000 t\), \(K\) = 0.1, \(V_x\) = 4 V DC, and \(R_1\) = \(R_2\) = 10 kΩ.



    Design Problems


    30. Determine values of \(R\) and \(C\) in Figure 7.9 so that stage 1 slewing is at least 1V/μs, along with a time constant of 1 ms. Assume that op amp 1 can produce at least 20 mA of current.

    31. In Figure 7.21, assume that \(C\) = 100 nF. Determine an appropriate value for Rl if the input signal is at least 2 kHz. Use a time constant factor of 100 for your calculations.

    32. Determine new values for \(R_a\) and \(R_f\) in Problem 13, so that the slopes are -5 and -3.

    33. Determine new resistor values for the circuit of Figure 7.33 such that the slopes are -10 and -20. The input impedance should be at least 3 kΩ.

    34. Determine the resistor values required in Figure 7.37 to produce slopes of -5, -8, and -12, if the input impedance must be at least 1 kΩ.

    35. Determine the resistor values required in Figure 7.42 for \(S_0\) = -1, \(S_1\) = \(S_{-1}\) = -5, \(S_2\) = \(S_{-2}\) = -12. Also \(V_1\) = |\(V_{-1}\)| = 3 V, and \(V_2\) = |\(V_{-2}\)| = 4 V. Use \(R_f\) = 20 kΩ.

    36. Sketch the transfer curve for Problem 35.

    37. Determine a value for \(R_i\) in Figure 7.56 so that a 3 V input will produce an output of 0.5 V. Assume \(I_s\) = 60 nA.

    38. Determine a value for \(R_f\) in Figure 7.58 such that a 0.5 V input will produce a 3 V output. Assume \(I_s\) = 40 nA.

    39. If the multiplier of Figure 7.63 can produce a maximum current of 10 mA, what should the minimum sizes of \(R_1\) and \(R_2\) be? (Assume ±15 V supplies).


    Challenge Problems


    40. Assuming that 5% resistors are used in Figure 7.14, determine the worstcase mismatch between the two halves of the rectified signal.

    41. Design a circuit that will light an LED if the input signal is beyond ±5 V peak. Make sure that you include some form of pulse-stretching element so that the LED remains visible for short duration peaks.

    42. A pressure transducer produces an output of 500 mV per atmosphere up to 5 atmospheres. From 5 to 10 atmospheres, the output is 450 mV per atmosphere. Above 10 atmospheres, the output falls off to 400 mV per atmosphere. Design a circuit to linearize this response using the Zener form

    43. Repeat the Problem above using the biased-diode form.

    44. Design a circuit to produce the transfer characteristic shown in Figure 7.68

    45. Sketch the output waveform of Figure 7.48 if \(R_1\) = 22 kΩ, \(R_2\) = 4.7 kΩ, and \(V_{in} = 2 + 8 sin 2 π 120 t\). Assume that the power supplies are ±15 V.


    46. Determine the output of Figure 7.60 if \(K = 0.1\), \(V_x = 1 sin 2 π 440 t\), and \(V_y = 3 sin 2 π 200000 t\).


    Computer Simulation Problems


    47. In preceding work it was noted that the use of an inappropriate device model can produce computer simulations that are way off mark. An easy way to see this is to simulate the circuit of Figure 7.2 using an accurate model (such as the 741 simulation presented in the chapter) and a simple model (such as the controlled-voltage source version presented in Chapter 2). Run two simulations of this circuit for each of these models. One simulation should use a lower input frequency, such as 1 kHz, and the second simulation should use a higher frequency where the differences in the models is very apparent, such as 500 kHz.


    48. Simulate the circuit designed in Problem 30 using a square wave input. Perform the simulation for several different input frequencies between 10 Hz and 10 kHz. Do the resulting waveforms exhibit the proper shapes?

    49. Perform simulations for the circuit of Figure 7.14. Use R = 12 kΩ. For the input waveform, use both sine and square waves, each being 1 V peak at 200 Hz.

    50. Component tolerances can directly affect the rectification accuracy of the full-wave circuit shown in Figure 7.14. This effect is easiest to see using a square wave input. If the positive and negative portions of the input signal see identical gains, the output of the circuit will be a DC level. Any inaccuracy or mismatch will produce a small square wave riding on this DC level. This effect can be simulated using the Monte Carlo analysis option. If Monte Carlo analysis is not available on your system, you can still see the effect by manually altering the resistor values within a preset tolerance band for each of a series of simulation runs.

    51. Use a simulator to verify the results of the limiter examined in Example 7.3.

    52. Use a simulator to verify the response of the circuit shown in Figure 7.37. For the stimulus, use a low frequency triangle wave of 4 V peak amplitude. Does the resulting waveform exhibit the same hard “corners” shown on the transfer curve of Figure 7.37?