Homework Given February 15, 2007

*      Due February 23, 2007

*      Chapter 2 exercises are on pages 87 through 105.

Live Links to the Exercises

*      Exercise 2.20

*      Exercise 2.23

*      Exercise 2.26

*      Exercise 2.57

*      Exercise 2.69

*      Exercise 2.83

*      Exercise 2.97

Exercise 2.20

The non-inverting op-amp of figure P2.20 on page 88 shows a DC reference source  in series with the resistor , which has five times the resistance of the resistor .  We use Ohm’s law to find the current through , the reference source, and  as

.

 

Using the perfect op-amp principle of the virtual short, we know that this current has a voltage drop through  that is equal to the input voltage .  Setting these equal, we find that

 

 

which tells us that the effect of the DC reference source as placed in the circuit shown in Figure P2.20(a) is to add that voltage to the output.

 

We now examine the circuit of Figure P2.20(b).  Placing the source in the branch with  instead produces the same current  but the voltage at the non-inverting input is the drop across  plus the voltage .  From the virtual short principle, we have

.

 

Solving this equation for  we have

 

 

We see that the DC source has more effect on the output in the configuration shown in Figure P2.20(b).

 

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Exercise 2.23

We could use the results of problem 2.20(b) here but we will instead follow the step-by-step instructions in the problem statement, using the previous homework problem as background and illumination.

 

We have the non-inverting op-amp of Figure P2.23, in which the voltage divider of a non-inverting op-amp circuit is returned to the input voltage instead of ground.  Part (a) of the problem statement asks for us to find the voltage and current across  using the virtual short principle.

 

Using the virtual short principle, we write the voltages at the inverting and non-inverting inputs to the op-amp as equal,

 

from which we immediately see that the current through  is zero, and thus by Ohm’s law the voltage across  is zero.  The current through  is the same as the current through  and thus is also zero.  Given that the currents through  and  are zero, the voltages across them are, by Ohm’s law, zero, and we have

.

 

 

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Exercise 2.26

We have a circuit, shown in Figure P2.26, in which the output of a non-inverting amplifier is connected to the input through a resistor  and we are asked to find an equation for the current  through this resistor.  From the equation for the gain of a non-inverting amplifier, we have

.

 

The current  is, by Ohm’s law,

 

 

 

This current flow is back into the battery as a charging current.

 

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Exercise 2.57

This problem is from the block about Section 2.4.5, Op-Amp Voltage Follower, pages 48-50.  That section uses the perfect op-amp model.  The problem statement is that the op-amp gain K is 200,000, and that the op-amp drives a  load.  Here we use the finite gain equation

.

 

Note that this equation does not depend on the output load because we are neglecting the output impedance of the op-amp.  From this equation we have

so we have

 

 

 

 

The current into the load is

.

 

 

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Exercise 2.69

We refer to Figure 2.16 page 55, and use the equation for gain as given in Equation (2.49),

 

where  and  are the input voltages.  Thus the gain is

.

 

We are given these values:

 

 

 

where c is a constant that may vary from zero to one.  Thus the gain is

.

 

Thus, the maximum gain for c=0 and the minimum gain for c=1 are

 

 

which shows a range of gain of almost 10:1.

 

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Exercise 2.83

We have a hybrid circuit shown in Figure P2.83 page 96, with a two-input non-inverting amplifier added to the circuit of an inverting summing amplifier.  For a circuit with  inverting amplifier inputs and  amplifier inputs, we can analyze the output by superposition.

 

For the inverting summation amplifier, we have the circuit treated in detail in Section 2.4.8, Summation Amplifier, on pages 56-59, except that the amplifier has several resistors between the non-inverting input and ground.  Since we neglect the op-amp bias currents with the perfect op-amp model, the inverting summation amplifier is unaffected by the modifications, and

 

for each input .  For the non-inverting amplifier, we have a voltage divider on the input, so the voltage on the non-inverting input is

 

 

 

and a non-inverting amplifier with gain

 

 

so that the total gain is

.

 

 

If we have another op-amp with an inverting summation for the non-inverting inputs, we can have a circuit that is independent of  but if we feed it directly to the non-inverting input of our existing amplifier we still have the second gain factor, which is a function of .  One simple solution is to have two inverting summation amplifiers, with a difference amplifier as shown in Figure 2.13 page 51 combing their outputs.

 

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Exercise 2.97

This problem poses the interesting question of what happens to the generic summation amplifier of Figure 2.17 page 56 if the feedback resistor is replaced by a capacitor.  This problem is in a block that addresses issues in Section 2.4.10, Op-Amp Integrator on pages 60-64.  In this section we know that the virtual short makes the voltage at the node with the inverting input equal to virtually zero, so the current into that node from the inputs is the sum of the input voltages divided by the respective resistances, and the current through the capacitor must be equal to that current.  Since the voltage across a capacitor must be equal to the time integral of the current, the voltage must be

.

 

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