Homework 8 solution

Concept (Review) Questions

4-1. How is the linear range of an op amp defined?(5 pts)

-As a linear input-output this equivalence allows us to apply the tools we developed to analyze a large array of op-amp circuits and to do so with relative ease.

4-2. What is the difference between the op-amp gain A and the circuit gain G?(5 pts)

-An op-amp gain (open-loop) is valid only when the op amp is not connected to an external circuit on the output side (open loop), if continues to hold approximately if the output circuit satisfies certain conditions. A circuit gain (closed loop) defines the gain of the entire circuit

4-3. An op amp is characterized by three important input-output attributes. What are they?(5 pts)

-negative saturation

-linear saturation

-positive saturation

4-4. Why is negative feedback used in op-amp circuits?(5 pts)

- negative feedback keeps the circuit in a self-stabilizing system. The stability gives the op-amp the capacity to work in its linear (active) mode, as opposed to merely being saturated fully "on" or "off" as it was when used as a comparator, with no feedback at all.

4-5. How large is the circuit gain G in the absence of feedback? How large is it with 100 percent feedback (equivalent to setting R1= 0 in the circuit)?(5 pts)

- the circuit gain G is 10^6.

4-6. What are the current and voltage constraints of the ideal op amp?(5 pts)

- Ip= In= 0 : input current constraint

-Vn=Vp: input voltage constraint

4-17. In a digital-to-analog converter, what dictates the maximum value that Rf can assume?(5 pts)

-The magnitude of G is selscted to suit the range of the output voltage. If the input is a 3-bit sequence whose range of decimal values extends from 0 to 7, one might design the circuit so that G=1, because in that case. the maximum output voltage is 7V, which is below Vcc for most of the op amps. For digital signals with longer sequences, G needs to be smaller than 1 in order to avoid saturating the op amp.

Exercises

4-1. In the circuit, insert a series resistance R(s) between V(s) and V(p) and then repeat the solution to obtain an expression for G. Evaluate G for R(s)= 10Ω and use the same values listed for the other quantities. What impact does the insertion of R(s) have on the magnitude of G?(5 pts)

- G= [A(Ri+Rs)(R1+R2)+R2R0]/ [AR2(R1+Rs)+R0(R2+R1+Rs)+R1R2+(Ri+Rs)(R1+R2)] = 4.999977

4-2. To evaluate the tradeoff between circuit gain G and the linear dynamic range of Vs, apply to find the magnitude of G and then determine the corresponding dynamic range of Vs for each of the following values of R2: 0 (no feedback), 800 Ω, 8.8 kΩ, 40 kΩ, 80 kΩ, and 1MΩ. Except for R2, all other quantities remain unchanged. (5 pts)

R2 / G / V(s) Range
0 / 10^6 / -10μV to +10μV
800Ω / 101 / -99 mV to +99 mV
8.8 kΩ / 10.1 / -0.99 V to +0.99 V
40 kΩ / 3 / -3.3 V to +3.3V
80 kΩ / 2 / -5V to +5V
1 MΩ / 1.08 / -9.26 V to +9.26

4-3. Consider the noninverting amplifier circuit under the conditions of the ideal op-amp model. Assume V(CC)= 10 V. Determine the value of G and the corresponding dynamic range of Vs for each of the following values of R1/R2: 1,1,9,99, 10^3, 10^6(5 pts)

R1/R2 / G / Vs Range
0 / 1 / -10V to +10V
1 / 2 / -5V to +5V
9 / 10 / -1V to +1V
99 / 100 / -0.1V to +0.1V
1000 / ~1000 / -10 mV to +10mV
10^6 / ~10^6 / -10μV to + 10μV

4-4. The input to an inverting-amplifier circuit consists of Vs=0.2 V and Rs=10Ω. If V(CC)=12V, what is the maximum value that Rf can assume before saturating the op amp?(5 pts)

- Gmax= -60, Rf= 600Ω

4-5. The circuit is to be used to perform the operation V0= 3V1+6V2. If R1=1.2 kΩ, Rs2= 2 kΩ, and Rf2= 4 kΩ, select values for R2 and Rf1 so as to realize the desired result.(5 pts)

-Rf1= 1.8 kΩ, R2= 600Ω

4-6. The difference-amplifier circuit is used to realize the operation V0= (6V2-2)V. Given that R3= 5 kΩ, R4=6 kΩ, and R2= 20 kΩ, specify values for V1 and R1.(5 pts)

-V1= 0.2 V, R1= 2kΩ

4-7. Express V0 in terms on V1, V2, and V3 for the circuit.(5 pts)

V0= 12V1 + 6V2 + 3V3

4-9. A 3-bit DAC uses an R-2R ladder design with R= 3kΩ and Rf= 24kΩ. If V(CC)= 10V, write an expression for Vout and evaluate it for [V1V2V3]=[111].(5 pts)

- Vout= -Rt/8R (4V1+2V2+V3) = -(4V1+2V2+V3). For [V1V2V3]=[111], Vout= -7V, whose magnitude is smaller thanVcc= 10V.

Problem 4-46(10 pts)

4-51(10 pts)