CHAPTER 1



CHAPTER 11

Solutions for Exercises

E11.1 (a) A noninverting amplifier has positive gain. Thus

[pic]

(b) An inverting amplifier has negative gain. Thus

[pic]

E11.2

[pic]

[pic]

[pic]

[pic]

[pic]

E11.3 Recall that to maximize the power delivered to a load from a source with fixed internal resistance, we make the load resistance equal to the internal (or Thévenin) resistance. Thus we make [pic] Repeating the calculations of Exercise 11.2 with the new value of RL, we have

[pic]

[pic]

[pic]

E11.4

[pic]

By inspection, [pic]

[pic]

[pic]

E11.5 Switching the order of the amplifiers of Exercise 11.4 to 3-2-1, we have

[pic]

[pic]

[pic]

E11.6 [pic]

[pic]

[pic]

E11.7 The input resistance and output resistance are the same for all of the amplifier models. Only the circuit configuration and the gain parameter are different. Thus we have [pic]and we need to find the open-circuit voltage gain. The current amplifier with an open-circuit load is:

[pic]

[pic]

E11.8 For a transconductance-amplifier model, we need to find the short-circuit transconductance gain. The current-amplifier model with a short-circuit load is:

[pic]

[pic]

The impedances are the same for all of the amplifier models, so we have [pic]

E11.9 For a transresistance-amplifier model, we need to find the open-circuit transresistance gain. The transconductance-amplifier model with an open-circuit load is:

[pic]

[pic]

The impedances are the same for all of the amplifier models, so we have [pic]

E11.10 The amplifier has [pic]

(a) We have [pic] which is much less than [pic], and we also have [pic] which is much larger than [pic]. Therefore for this source and load, the amplifier is approximately an ideal voltage amplifier.

(b) We have [pic] which is much greater than [pic], and we also have [pic] which is much smaller than [pic]. Therefore for this source and load, the amplifier is approximately an ideal current amplifier.

(c) We have [pic] which is much less than [pic], and we also have [pic] which is much smaller than [pic]. Therefore for this source and load, the amplifier is approximately an ideal transconductance amplifier.

(d) We have [pic] which is much larger than [pic], and we also have [pic] which is much larger than [pic]. Therefore for this source and load, the amplifier is approximately an ideal transresistance amplifier.

(e) Because we have [pic], the amplifier does not approximate any type of ideal amplifier.

E11.11 We want the amplifier to respond to the short-circuit current of the source. Therefore, we need to have [pic]. Because the amplifier should deliver a voltage to the load that is independent of the load resistance, the output resistance [pic] should be very small compared to the smallest load resistance. These facts ([pic]very small and [pic] very small) indicate that we need a nearly ideal transresistance amplifier.

E11.12 The gain magnitude should be constant for all components of the input signal, and the phase should by proportional to the frequency of each component. The input signal has components with frequencies of 500 Hz, 1000 Hz and 1500 Hz, respectively. The gain is [pic] at a frequency of 1000 Hz. Therefore the gain should be [pic] at 500 Hz, and [pic] at 1500 Hz.

E11.13 We have

[pic]

[pic]

The corresponding phasors are [pic] and [pic]. Thus the complex gain is

[pic]

E11.14 [pic]

E11.15 Equation 11.13 states

Percentage tilt [pic]

Solving for fL and substituting values, we obtain

[pic]

as the upper limit for the lower half-power frequency.

E11.16 (a) [pic]

[pic]

[pic]

The desired term has an amplitude of V1 = 100 and a second-harmonic distortion term with an amplitude of V2 = 0.5. There are no higher order distortion terms so we have [pic] or 0.5%.

[pic]

(b) [pic]

[pic]

[pic]

The desired term has an amplitude of V1 = 500 and a second-harmonic distortion term with an amplitude of V2 = 12.5. There are no higher order distortion terms so we have [pic] or 2.5%.

[pic]

E11.17 With the input terminals tied together and a 1-V signal applied, the differential signal is zero and the common-mode signal is 1 V. The common-mode gain is[pic] which is equivalent to -20 dB. Then we have [pic]

E11.18 (a) [pic] [pic]

[pic]

Thus [pic].

(b) [pic] [pic]

[pic]

Thus [pic].

(c) [pic]

[pic]

[pic]

[pic]

E11.19 Except for numerical values this Exercise is the same as Example 11.13 in the book. With equal resistances at the input terminals, the bias currents make no contribution to the output voltage. The extreme contributions to the output due to the offset voltage are

[pic]

The extreme contributions to the output voltage due to the offset current are

[pic]

Thus, the extreme output voltages due to all sources are [pic] V.

E11.20 This Exercise is similar to Example 11.13 in the book with [pic]= 50 kΩ and [pic]= 0. With unequal resistances at the input terminals, the bias currents make a contribution to the output voltage given by

[pic]

The extreme contributions to the output due to the offset voltage are

[pic]

The extreme contributions to the output voltage due to the offset current are

[pic]

Thus, the extreme output voltages due to all sources are a minimum of 2.5 V and a maximum of 10.83 V.

Answers for Selected Problems

P11.4* [pic]

[pic][pic]

[pic]

[pic]

P11.5* [pic]

[pic]

P11.10* [pic](.

P11.15* [pic]

P11.20* [pic]

[pic]

[pic][pic]

P11.22* Five amplifiers must be cascaded to attain a voltage gain in excess of 1000.

P11.25* [pic]

P11.32* [pic]

[pic]

[pic]

[pic]

[pic]

P11.33* [pic]

[pic]

[pic]

P11.38* The voltage-amplifier model is:

[pic]

The transconductance-amplifier model is:

[pic]

P11.40* [pic]

[pic]

[pic]

P11.41* [pic]

[pic]

[pic]

P11.52* [pic]

P11.55* To sense the open-circuit voltage of a sensor, we need an amplifier with very high input resistance (compared to the Thévenin resistance of the sensor). To avoid loading effects by the variable load resistance, we need an amplifier with very low output resistance (compared to the smallest load resistance). Thus, we need a nearly ideal voltage amplifier with a gain of 1000.

P11.56* The input resistance is that of the ideal transresistance amplifier which is zero. The output resistance of the cascade is the output resistance of the ideal transconductance amplifier which is infinite. An amplifier having zero input resistance and infinite output resistance is an ideal current amplifier. Also, we have [pic].

P11.61* We need a nearly ideal transconductance amplifier.

[pic] [pic]

P11.67* The complex gain for the 1000-Hz component is

[pic]

The complex gain for the 2000-Hz component is

[pic]

P11.68* The signal to be amplified is the short-circuit current of an electrochemical cell (or battery). This signal is dc and therefore a dc-coupled amplifier is needed.

P11.70* [pic]

P11.75* The gain at 2000 Hz must be [pic]. The output signal is

[pic]

The plots are:

[pic]

P11.82* [pic]

[pic]

P11.83* (a)

[pic]

(b)

[pic]

(c)

[pic]

P11.86* [pic]

[pic]

[pic]

[pic]

P11.93* [pic]

P11.98* The extreme values of the output voltage are:

[pic]

If the resistors are exactly equal, then the output voltage is zero.

P11.99* The output voltage can range from -3.333 to +3.333 V.

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