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 Ohmís 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).

 

Back to Top

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 Ohmís 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 Ohmís law, zero, and we have

.

 

 

Back to Top

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 Ohmís law,

 

 

 

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

 

Back to Top

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

.

 

 

Back to Top

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.

 

Back to Top

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.

 

Back to Top

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

.

 

Back to Top