Homework Given January 30, 2007

Due February 6, 2007

This assignment is due on February 6, 2007.  Chapter 1 exercises are on pages 28 through 36.   Chapter 2 exercises are on pages 87 through 105.

Live Links to the Exercises

*      Exercise 1.37

*      Exercise 1.51

*      Exercise 1.65 (FET transistor simplified circuits)

*      Exercise 1.99

*      Exercise 2.6

*      Exercise 2.7 (Norton equivalent circuit for output of op-amp)

*      Exercise 2.19

Exercise 1.37

Prove that a resistor is a linear circuit element…

Show that  using Kirchhoff’s laws and Ohm’s law.


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Exercise 1.51

Find the Thévenin equivalent of the circuit of Figure P1.2 and show the role of resistor .  Figure P1.2 is on page 28.


The open circuit voltage is determined by  and  as a voltage divider, and the source voltage of 12 V.  The short circuit current is determined by  and the source voltage.  The role of  is to load the source with a current but  does not affect the Thévenin equivalent at the terminals a and a’.


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Exercise 1.65 (FET transistor simplified circuits)

Find the Thévenin equivalents of each of the circuits (a), (b), and (c) of figure P1.65, using the test source method to find the Thévenin equivalent resistance.  For each of the three circuits, we will determine the open circuit voltage to obtain the Thevenin voltage, then, we modify the circuit to make the Thevenin voltage zero by setting  to zero and use the test source method to find the Thevenin resistance.  We set  to zero and apply a test voltage  to terminals  and  and solve the circuit for the current out of the test source to find the Thevenin resistance.

Part (a)

Here,  is a voltage drop across a resistance .   Note that the designation  is used twice in this circuit  We will modify the problem to avoid confusion by designating the first one .  There is a voltage divider that determines ,



Given , we used the controlled source and Ohm's law to find the Thevenin equivalent voltage as



We see from the circuit that the test source does not cause a voltage across  so the controlled current is zero amperes.  This the Thevenin equivalent resistance is ,


Part (b)

This circuit is simplified by making  and open circuit and adding a resistor  to what would be an emitter resistor in a bipolar transistor.  The voltage  that drives the controlled source is the difference between  and the voltage drop across , which we find by Ohm's law to be


This gives us an equation for ,


which we solve as




We again use Ohm's law to find the Thevenin equivalent voltage,



For  equal to zero, the output is the same as for part (a), so the Thevenin equivalent resistance is the same,


Part (c)

Here we have the same circuit as part (b), but the problem statement is to find the Thevenin equivalent for the terminals across .  Since the circuit is the same as for part (b), we can use the results from part (b) to find  and thus the controlled current, which we use with Ohm's Law to find the Thevenin controlled voltage,




We note that setting  to zero leaves the ungrounded terminal of  connected to an open circuit, but unlike part (b) the test source will have an effect on ,


The current out of the test source will be the draw through  by Ohm's law minus the current from the controlled source,




so the Thevenin equivalent resistance is



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Exercise 1.99

Find the relationships between the resistances and capacitances for a voltage divider as shown in Figure P1.99 on page 35, both of whose impedances are a resistor in parallel with a capacitor.


The impedance of a resistor and a capacitor in parallel is



The circuit in the figure is a voltage divider of two such impedances.  Any of several methods will show that the ratio of the resistances must be the inverse of the ratio of the capacitances.  I suggest writing the voltage divider equation for the two impedances and finding a relationship for the transfer function, e.g. the voltage divider output voltage divided by the input voltage in terms of the two impedances,










Any one of the last three forms given here suggest that the transfer function is independent of frequency when



or, the ratios of the resistances and capacitances are the inverse of each other,




The problem statement points out that the circuit is similar to that of an oscilloscope probe. The input, , is the voltage under test.  The top resistance, , is a high resistance such as  to offer minimum effect on the circuit under test.  The top capacitance  is the stray capacitance in the probe tip resistance, with a value such as .


The resistance  is the oscilloscope impedance, which is typically .  Note that resistances of  in the probe and  in the oscilloscope are typical of a 10X probe, and at DC and low frequencies, the oscilloscope will see 1/10 of the voltage under test.


The capacitor  must be 10 times the stray capacitance  for the oscilloscope to see 1/10 of the voltage under test at high frequency.  If the high frequency response isn't the same as the low frequency response, not only will the oscilloscope be out of calibration at high frequencies, the waveforms that it displays will be distorted.  Thus there is a trimmer capacitor that adjusts  in the jack of the oscilloscope probe where it plugs into the oscilloscope, and a typical trimmed value of this capacitance is .


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Exercise 2.6

In the simple op-amp model shown in Figure 2.2 on page 39, current may flow either direction through the output terminal.  In a real op-amp, from where do these currents come?


If you look at the general block diagram of an op-amp shown on page 37, which I also drew on the board in class on more than one occasion, the output stage of an op-amp is a follower amplifier.  The currents that it provides come from the power supplies.


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Exercise 2.7 (Norton equivalent circuit for output of op-amp)

Draw the dual of the op-amp of figure 2.2, with Norton equivalent output driven by the input current between the inverting and non-inverting input.


Ohm's law gives the input current as



and we have the Norton equivalent current of the output as



The Norton equivalent resistance is the same as the Thevenin equivalent resistance, .  We can use these equations together to eliminate  and find



to give us the Norton equivalent circuit for the output, in terms of the current between the terminals in the input.


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Exercise 2.19

A vacuum sensor provides a resistance that varies from  to  as the vacuum varies from atmospheric pressure to 10-3 torr.  Design a circuit that provides a voltage that uses this variable resistance to vary its output from 0 V to 10 V.  Voltages available are -12 V, 0 V (ground), and +12 V.  Use standard 5% resistor values.


The Preface of our text, on page vi, states that homework problems marked with a script "D" as does 2.19 have significant design content.  As noted at the top of the page where the homework problems begin, such as page 87, the problems are marked according to level of effort required:  no marking is easiest, an open circle requires more effort, a half-filled circle is harder, and a filled circle denotes the most difficult category.  Problem 2.19 is marked with an open circle.  This, problem 2.19 requires that you formulate a circuit – the part of the problem where you need to design – and you will need to refer to the book and give it a little thought – hence the open circle marking.


The first clue is that 2.19 falls in a block of problems involving non-inverting amplifiers as treated in section 2.4.1, pages 41 through 43.  Thus you can reasonably expect that there is a non-inverting amplifier in a simple circuit that will solve the problem.  So, part of your circuit will look like Figure 2.5 on page 41.


The rest of our problem states that we have a sensor that provides a resistance that varies from  to , and that an output is desired that varies from 0 Volts to 10 Volts as the resistance varies over that range.  The op-amp must drive a   load, which is no problem with any general purpose op-amp.  We have  and available for the sensor circuitry and to power the op-amp.


A last requirement is that we use 5% standard value resistors.  In the Index on page 1123, under "Resistor color code/standard values," Appendix D on pages 1097-1099 is listed.  A table on page 1099 lists standard resistor values for 5%, 10%, and 20% tolerances over the range 10 to 91.  Note that "22" means any of , , , ,  etc. are standard 5% values.  The text of Appendix D and table D.1 show that the color code of any one of these resistors will be red, red, a multiplier value, and gold to indicate 5% tolerance, so colored bands beginning at one end will have these colors.  The multiplier color will be black for , brown for , red for , etc.


We now have enough information to block out our design problem.  We need to place our variable resistor in a simple circuit with one or two other fixed standard value resistors.  Voltages available to apply to this circuit are , 0 Volts (ground), and .  We want a node on this circuit to provide 0 Volts when the resistor is  and a positive voltage when the resistor is , and this voltage is to be provided to the non-inverting input of the op-amp in the circuit on page 41.  Note that, for an ideal op-amp, there is no current being drawn from our sensor circuit, which simplifies the design and analysis of the sensor circuit.


We now have reduced our problem to answering three questions:

  1. What is the simplest possible circuit that includes our variable resistor and hooks to two of the voltages , ground or 0 Volts, and  that produces 0 Volts when our variable resistor is  and a positive voltage when our variable resistor is ?  Can we do this with only one other fixed resistor?  Is this resistor a standard 5% value -- listed in the 5% column of the table on page 1099, times a power of 10?
  2. The non-inverting amplifier will give us 0 Volts when the sensor resistor circuit gives us 0 Volts.  What is the gain required to give an output of 10 Volts from the non-inverting amplifier when the sensor resistor is ?
  3. The gain of a non-inverting amplifier is  as given in Figure 2.5 on page 41, the discussion and equation (2.6) on the top of page 42, Example 2.1 on page 42, and equation (2.12) on page 43.  What is a ratio of standard 5% values from the table on page 1099 that gives us the closest possible gain we need according to step 2?  We use those values, with an appropriate multiplier -- power of ten -- to keep the resistors in the range of  to , to complete our design.


The simplest possible circuit is a voltage divider.  A voltage divider that produces zero volts must have a positive and a negative voltage, so we have two resistors connected to the  and  voltage sources.  We need zero volts when the sensor resistor  has a value of , so the added resistor has a fixed value of .  We will be using a non-inverting op-amp configuration by direction, so the voltage divider must provide a positive voltage when  takes a value of , so  is connected to the  voltage source.


The output of the voltage divider is




so when  takes on the value of , we have  at the output of the voltage divider.  The non-inverting amplifier must provide a gain of 2.5 to provide an  of  from an input of  .  From Section 2.4.1 and Figure 2.5 on page 41, we see that the gain of the non-inverting op-amp circuit is , so we need a ratio of 1.5 to 1 for .  From Table D.2 on page 1099, we see that 10 and 15 are standard values for 5% resistors, so we pick  and  as convenient values.  The final circuit is shown below in Figure 1.   The equation for the output voltage as a function of the sensor resistance is




which is seen to meet the requirements of a voltage swing from 0 V to  as  increases from  to .  The final requirement, that the circuit drive a  load, is met by using an op-amp output as the circuit output.



Figure 1  Design for Problem 2.19


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