For the circuit of Fig. 6.62, calculate the differential input voltage and the input bias current if the op amp is a(n) ( a ) μ A741; ( b ) LF411; ( c ) AD549K.

Question

Want to see more full solutions like this?

Answer 1

Textbook Question

Chapter 6, Problem 50E

For the circuit of Fig. 6.62, calculate the differential input voltage and the input bias current if the op amp is a(n) (a) μA741; (b) LF411; (c) AD549K.

Chapter 6, Problem 50E, For the circuit of Fig. 6.62, calculate the differential input voltage and the input bias current if

(a)

Expert Solution

To determine

Find the differential input voltage and input bias current of op amp.

Answer to Problem 50E

The differential voltage vd of μA741 is 1.548×10−5 V and input bias current id is 7.743×10−12 A.

Explanation of Solution

Given Data:

Differential op amp is μA741.

Calculation:

The redrawn circuit of op amp is shown in Figure 1 as follows:

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 6, Problem 50E

Refer to the redrawn Figure 1:

The expression for the nodal analysis at the nodal voltage vd is:

−vd+450 mVR1+−vd+450 mV−voutRf+−vdRi=0 (1)

Here,

vd is the differential voltage at node terminal of input across op amp,

vin is the applied input across op amp,

R1 is the input resistance of op amp,

Rf is the feedback resistance of op amp and

vout is the output voltage of μA741 op amp.

The expression for nodal analysis at nodal voltage vout is:

vout−(−vd+450 mV)Rf+vout−AvdR0=0 (2)

Here,

A is the open loop gain of op amp and

R0 is the output resistance of op amp.

The expression for the bias input current of differential op amp is:

vd=(id×Ri) (3)

Here,

id is the bias input current of practical op amp.

Simplify the equation (1) for vd.

−vd+450 mVR1+−vd+450 mV−voutRf+−vdRi=0−vdR1−vdRf−vdRi−voutRf+450 mVRf=−450 mVR1−vd(1R1+1Rf+1Ri)−voutRf+450 mVRf=−450 mVR1

−vd(RiRf+R1Ri+R1RfR1RfRi)=−450 mVR1+voutRf−450 mVRf (4)

Refer to the TABLE 6.3 in the textbook.

Substitute 1.4 kΩ for Rf, 250 Ω for R1 and 2 MΩ for Ri in the equation (4).

−vd(2 MΩ×1.4 kΩ+250 Ω×2 MΩ+250 Ω×1.4 kΩ2 MΩ×1.4 kΩ×250 Ω)=−450 mV250 Ω+vout1.4 kΩ−450 mV1.4 kΩ−vd(2800 MΩ+500 MΩ+0.35 MΩ700000 MΩ)=−450×10−3V250 Ω+vout1.4×103 Ω−3.214×10−4−vd(4.714×10−3)=−1.8×10−3+7.143×10−4vout−3.214×10−4

Rearrange for vout.

−vd(4.714×10−3)=−1.8×10−3+7.143×10−4vout−3.214×10−4−vd(4.714×10−3)=7.143×10−4vout−2.121×10−37.143×10−4vout=2.121×10−3−vd(4.714×10−3)vout=2.121×10−3−vd(4.714×10−3)7.143×10−4

vout=2.969−6.599vd (5)

Substitute 2.969−6.599vd for vout from equation (5) in equation (2).

(2.969−6.599vd)−(−vd+450 mV)1.4 kΩ+(2.969−6.599vd)−A×vd75 Ω=0−6.599vd1.4 kΩ+vd1.4 kΩ−6.599vd75 Ω−A×vd75 Ω+2.9691.4×103 Ω−450×10−3 V1.4×103 Ω+2.96975 Ω=0vd(−6.5991.4 kΩ+11.4 kΩ−6.59975 Ω−A75 Ω)+2.121×10−3−3.214×10−4+0.0395=0

vd(−4.713×10−3+7.14×10−4−0.0879−A75 Ω)+0.0413=0 (6)

Substitute 2×105 V/V for A in the equation (6).

vd(−4.713×10−3+7.14×10−4−0.0879−2×105 V/V75 Ω)+0.0413=0vd(−2666.758)+0.0413=0vd=0.04132666.758 Vvd=1.548×10−5 V

Substitute 1.548×10−5 V for vd in equation (3).

1.548×10−5 V=(id×2 MΩ)1.548×10−5 V=(id×2×106Ω) {∵1 MΩ=1×106Ω}

Rearrange for id.

id=7.743×10−12 A

Conclusion:

Thus, the differential voltage vd of μA741 is 1.548×10−5 V and input bias current id is 7.743×10−12 A.

(b)

Expert Solution

To determine

Find the differential input voltage and input bias current of op amp.

Answer to Problem 50E

The differential voltage vd of LF411 is 1.485×10−5 V and input bias current id is 1.485×10−17 A.

Explanation of Solution

Given Data:

Differential op amp is LF411.

Calculation:

Refer to the TABLE 6.3 in the textbook.

Substitute 1.4 kΩ for Rf, 250 Ω for R1 and 1 TΩ for Ri in the equation (4).

−vd(1 TΩ×1.4 kΩ+250 Ω×1 TΩ+250 Ω×1.4 kΩ1 TΩ×1.4 kΩ×250 Ω)=−450 mV250 Ω+vout1.4 kΩ−450 mV1.4 kΩ−vd(1400 TΩ+250 TΩ+0.00035 TΩ350000 TΩ)=−450×10−3V250 Ω+vout1.4×103 Ω−3.214×10−4−vd(4.714×10−3)=−1.8×10−3+7.143×10−4vout−3.214×10−4

Rearrange for vout.

−vd(4.714×10−3)=7.143×10−4vout−2.121×10−37.143×10−4vout=2.121×10−3−vd(4.714×10−3)vout=2.121×10−3−vd(4.714×10−3)7.143×10−4

vout=2.969−6.599vd

Substitute 2.969−6.599vd for vout in equation (2).

(2.969−6.599vd)−(−vd+450 mV)1.4 kΩ+(2.969−6.599vd)−A×vd1 Ω=0−6.599vd1.4 kΩ+vd1.4 kΩ−6.599vd1 Ω−A×vd1 Ω+2.9691.4×103 Ω−450×10−3 V1.4×103 Ω+2.9691 Ω=0vd(−6.5991.4 kΩ+11.4 kΩ−6.5991 Ω−A1 Ω)+2.121×10−3−3.214×10−4+2.969=0

vd(−4.713×10−3+7.14×10−4−6.599−A)+2.970=0 (7)

Substitute 2×105 V/V for A in the equation (7).

vd(−4.713×10−3+7.14×10−4−6.599−2×105)+2.970=0vd(−200006.603)+0.0349=0vd=2.970200006.603 Vvd=1.485×10−5 V

Substitute 1.485×10−5 V for vd in equation (3).

1.485×10−5 V=(id×1 TΩ)1.485×10−5 V=(id×1×1012 Ω) {∵1 TΩ=1×1012 Ω}

Rearrange for id.

id=1.485×10−17 A

Conclusion:

Thus, the differential voltage vd of LF411 is 1.485×10−5 V and input bias current id is 1.485×10−17 A.

(c)

Expert Solution

To determine

Find the differential input voltage and input bias current of AD549K op amp.

Answer to Problem 50E

The differential voltage vd of AD549K is 1.99×10−7 V and input bias current id is 1.996×10−20 A.

Explanation of Solution

Given Data:

Differential op amp is AD549K.

Calculation:

Refer to the TABLE 6.3 in the textbook.

Substitute 1.4 kΩ for Rf, 250 Ω for R1 and 10 TΩ for Ri in the equation (4).

−vd(10 TΩ×1.4 kΩ+250 Ω×10 TΩ+250 Ω×1.4 kΩ10 TΩ×1.4 kΩ×250 Ω)=−450 mV250 Ω+vout1.4 kΩ−450 mV1.4 kΩ−vd(14000 TΩ+2500 TΩ+0.00035 TΩ3500000 TΩ)=−450×10−3V250 Ω+vout1.4×103 Ω−3.214×10−4−vd(4.714×10−3)=−1.8×10−3+7.143×10−4vout−3.214×10−4

Rearrange for vout.

−vd(4.714×10−3)=7.143×10−4vout−2.121×10−37.143×10−4vout=2.121×10−3−vd(4.714×10−3)vout=2.121×10−3−vd(4.714×10−3)7.143×10−4

vout=2.969−6.599vd

Substitute 2.969−6.599vd for vout in equation (2).

(2.969−6.599vd)−(−vd+450 mV)1.4 kΩ+(2.969−6.599vd V)−A×vd15 Ω=0−6.599vd1.4 kΩ+vd1.4 kΩ−6.599vd15 Ω−A×vd15 Ω+2.9691.4×103 Ω−450×10−3 V1.4×103 Ω+2.96915 Ω=0vd(−6.5991.4 kΩ+11.4 kΩ−6.59915 Ω−A15 Ω)+2.121×10−3−3.214×10−4+0.198=0

vd(−4.713×10−3+7.14×10−4−0.439−A)+0.1997=0 (8)

Substitute 106 V/V for A in equation (8).

vd(−4.713×10−3+7.14×10−4−0.439−106)+0.1997=0vd(−1000000.443)+0.1997=0vd=0.19971000000.443 Vvd=1.99×10−7 V

Substitute 1.99×10−7 V for vd in equation (3).

1.99×10−7 V=(id×10 TΩ)1.99×10−7 V=(id×10×1012 Ω) {∵1 TΩ=1×1012 Ω}

Rearrange for id,

id=1.996×10−20 A

Conclusion:

Thus, the differential voltage vd of AD549K is 1.99×10−7 V and input bias current id is 1.996×10−20 A.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

HW_#1 HW_01.pdf EE 213-01 Assignments P Pearson MyLab and Mastering uah.instructure.com P Course Home Watch out for units (i.e, kQ, mA, etc), show all units on answers and clearly mark all answers 1)(5 pts) Specify if the following elements are absorbing or delivering power, and determine the amount of power being absorbed or delivered. 10 V + a) 5A + 5 V - b) 2A + 10 V -2A 2)(5 pts) Two circuits, shown by boxes A and B are connected as shown below. Use the current reference direction provided and the voltage reference polarity shown to determine the power for the interconnection. Also state the direction of power flow for the connection: form A to B or B to A. A I + I a) I = -6 A V = -20 Volts b) 1 = 8 A V = 30 Volts c) = 4 A V = -80 Volts d) I = -5 A V = 40 Volts B 3) (5 pts) Use Ohm's Law, KCL and KVL to determine values for V1, V2, V₁, I₁, and I2. 200 Ohms A, + 12 + V1 75 Ohms 3 Amps + V2 300 Ohms 25 Ohms 4) (5 pts) For the circuit in Problem 2, determine the power (expressed as a…

Find the power delivered across the 10 ohm resistor

For the Circuit Below Find: A) io in terms of Rb and the numerical resistor values provided in the schematic. Reduce your solution to a minimal Equations B)the power delivered across Rb in terms of Rb and the numerical resistor values provided in the schematic. Reduce your solution to a minimal equation. C) the power delivered across Rc if Rb is 5 Q Ra $50 15Ω M 120 90V +1 Rb 150 150 m Rc

Answer 2

Textbook Question

Chapter 6, Problem 50E

For the circuit of Fig. 6.62, calculate the differential input voltage and the input bias current if the op amp is a(n) (a) μA741; (b) LF411; (c) AD549K.

Chapter 6, Problem 50E, For the circuit of Fig. 6.62, calculate the differential input voltage and the input bias current if

(a)

Expert Solution

To determine

Find the differential input voltage and input bias current of op amp.

Answer to Problem 50E

The differential voltage vd of μA741 is 1.548×10−5 V and input bias current id is 7.743×10−12 A.

Explanation of Solution

Given Data:

Differential op amp is μA741.

Calculation:

The redrawn circuit of op amp is shown in Figure 1 as follows:

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 6, Problem 50E

Refer to the redrawn Figure 1:

The expression for the nodal analysis at the nodal voltage vd is:

−vd+450 mVR1+−vd+450 mV−voutRf+−vdRi=0 (1)

Here,

vd is the differential voltage at node terminal of input across op amp,

vin is the applied input across op amp,

R1 is the input resistance of op amp,

Rf is the feedback resistance of op amp and

vout is the output voltage of μA741 op amp.

The expression for nodal analysis at nodal voltage vout is:

vout−(−vd+450 mV)Rf+vout−AvdR0=0 (2)

Here,

A is the open loop gain of op amp and

R0 is the output resistance of op amp.

The expression for the bias input current of differential op amp is:

vd=(id×Ri) (3)

Here,

id is the bias input current of practical op amp.

Simplify the equation (1) for vd.

−vd+450 mVR1+−vd+450 mV−voutRf+−vdRi=0−vdR1−vdRf−vdRi−voutRf+450 mVRf=−450 mVR1−vd(1R1+1Rf+1Ri)−voutRf+450 mVRf=−450 mVR1

−vd(RiRf+R1Ri+R1RfR1RfRi)=−450 mVR1+voutRf−450 mVRf (4)

Refer to the TABLE 6.3 in the textbook.

Substitute 1.4 kΩ for Rf, 250 Ω for R1 and 2 MΩ for Ri in the equation (4).

−vd(2 MΩ×1.4 kΩ+250 Ω×2 MΩ+250 Ω×1.4 kΩ2 MΩ×1.4 kΩ×250 Ω)=−450 mV250 Ω+vout1.4 kΩ−450 mV1.4 kΩ−vd(2800 MΩ+500 MΩ+0.35 MΩ700000 MΩ)=−450×10−3V250 Ω+vout1.4×103 Ω−3.214×10−4−vd(4.714×10−3)=−1.8×10−3+7.143×10−4vout−3.214×10−4

Rearrange for vout.

−vd(4.714×10−3)=−1.8×10−3+7.143×10−4vout−3.214×10−4−vd(4.714×10−3)=7.143×10−4vout−2.121×10−37.143×10−4vout=2.121×10−3−vd(4.714×10−3)vout=2.121×10−3−vd(4.714×10−3)7.143×10−4

vout=2.969−6.599vd (5)

Substitute 2.969−6.599vd for vout from equation (5) in equation (2).

(2.969−6.599vd)−(−vd+450 mV)1.4 kΩ+(2.969−6.599vd)−A×vd75 Ω=0−6.599vd1.4 kΩ+vd1.4 kΩ−6.599vd75 Ω−A×vd75 Ω+2.9691.4×103 Ω−450×10−3 V1.4×103 Ω+2.96975 Ω=0vd(−6.5991.4 kΩ+11.4 kΩ−6.59975 Ω−A75 Ω)+2.121×10−3−3.214×10−4+0.0395=0

vd(−4.713×10−3+7.14×10−4−0.0879−A75 Ω)+0.0413=0 (6)

Substitute 2×105 V/V for A in the equation (6).

vd(−4.713×10−3+7.14×10−4−0.0879−2×105 V/V75 Ω)+0.0413=0vd(−2666.758)+0.0413=0vd=0.04132666.758 Vvd=1.548×10−5 V

Substitute 1.548×10−5 V for vd in equation (3).

1.548×10−5 V=(id×2 MΩ)1.548×10−5 V=(id×2×106Ω) {∵1 MΩ=1×106Ω}

Rearrange for id.

id=7.743×10−12 A

Conclusion:

Thus, the differential voltage vd of μA741 is 1.548×10−5 V and input bias current id is 7.743×10−12 A.

(b)

Expert Solution

To determine

Find the differential input voltage and input bias current of op amp.

Answer to Problem 50E

The differential voltage vd of LF411 is 1.485×10−5 V and input bias current id is 1.485×10−17 A.

Explanation of Solution

Given Data:

Differential op amp is LF411.

Calculation:

Refer to the TABLE 6.3 in the textbook.

Substitute 1.4 kΩ for Rf, 250 Ω for R1 and 1 TΩ for Ri in the equation (4).

−vd(1 TΩ×1.4 kΩ+250 Ω×1 TΩ+250 Ω×1.4 kΩ1 TΩ×1.4 kΩ×250 Ω)=−450 mV250 Ω+vout1.4 kΩ−450 mV1.4 kΩ−vd(1400 TΩ+250 TΩ+0.00035 TΩ350000 TΩ)=−450×10−3V250 Ω+vout1.4×103 Ω−3.214×10−4−vd(4.714×10−3)=−1.8×10−3+7.143×10−4vout−3.214×10−4

Rearrange for vout.

−vd(4.714×10−3)=7.143×10−4vout−2.121×10−37.143×10−4vout=2.121×10−3−vd(4.714×10−3)vout=2.121×10−3−vd(4.714×10−3)7.143×10−4

vout=2.969−6.599vd

Substitute 2.969−6.599vd for vout in equation (2).

(2.969−6.599vd)−(−vd+450 mV)1.4 kΩ+(2.969−6.599vd)−A×vd1 Ω=0−6.599vd1.4 kΩ+vd1.4 kΩ−6.599vd1 Ω−A×vd1 Ω+2.9691.4×103 Ω−450×10−3 V1.4×103 Ω+2.9691 Ω=0vd(−6.5991.4 kΩ+11.4 kΩ−6.5991 Ω−A1 Ω)+2.121×10−3−3.214×10−4+2.969=0

vd(−4.713×10−3+7.14×10−4−6.599−A)+2.970=0 (7)

Substitute 2×105 V/V for A in the equation (7).

vd(−4.713×10−3+7.14×10−4−6.599−2×105)+2.970=0vd(−200006.603)+0.0349=0vd=2.970200006.603 Vvd=1.485×10−5 V

Substitute 1.485×10−5 V for vd in equation (3).

1.485×10−5 V=(id×1 TΩ)1.485×10−5 V=(id×1×1012 Ω) {∵1 TΩ=1×1012 Ω}

Rearrange for id.

id=1.485×10−17 A

Conclusion:

Thus, the differential voltage vd of LF411 is 1.485×10−5 V and input bias current id is 1.485×10−17 A.

(c)

Expert Solution

To determine

Find the differential input voltage and input bias current of AD549K op amp.

Answer to Problem 50E

The differential voltage vd of AD549K is 1.99×10−7 V and input bias current id is 1.996×10−20 A.

Explanation of Solution

Given Data:

Differential op amp is AD549K.

Calculation:

Refer to the TABLE 6.3 in the textbook.

Substitute 1.4 kΩ for Rf, 250 Ω for R1 and 10 TΩ for Ri in the equation (4).

−vd(10 TΩ×1.4 kΩ+250 Ω×10 TΩ+250 Ω×1.4 kΩ10 TΩ×1.4 kΩ×250 Ω)=−450 mV250 Ω+vout1.4 kΩ−450 mV1.4 kΩ−vd(14000 TΩ+2500 TΩ+0.00035 TΩ3500000 TΩ)=−450×10−3V250 Ω+vout1.4×103 Ω−3.214×10−4−vd(4.714×10−3)=−1.8×10−3+7.143×10−4vout−3.214×10−4

Rearrange for vout.

−vd(4.714×10−3)=7.143×10−4vout−2.121×10−37.143×10−4vout=2.121×10−3−vd(4.714×10−3)vout=2.121×10−3−vd(4.714×10−3)7.143×10−4

vout=2.969−6.599vd

Substitute 2.969−6.599vd for vout in equation (2).

(2.969−6.599vd)−(−vd+450 mV)1.4 kΩ+(2.969−6.599vd V)−A×vd15 Ω=0−6.599vd1.4 kΩ+vd1.4 kΩ−6.599vd15 Ω−A×vd15 Ω+2.9691.4×103 Ω−450×10−3 V1.4×103 Ω+2.96915 Ω=0vd(−6.5991.4 kΩ+11.4 kΩ−6.59915 Ω−A15 Ω)+2.121×10−3−3.214×10−4+0.198=0

vd(−4.713×10−3+7.14×10−4−0.439−A)+0.1997=0 (8)

Substitute 106 V/V for A in equation (8).

vd(−4.713×10−3+7.14×10−4−0.439−106)+0.1997=0vd(−1000000.443)+0.1997=0vd=0.19971000000.443 Vvd=1.99×10−7 V

Substitute 1.99×10−7 V for vd in equation (3).

1.99×10−7 V=(id×10 TΩ)1.99×10−7 V=(id×10×1012 Ω) {∵1 TΩ=1×1012 Ω}

Rearrange for id,

id=1.996×10−20 A

Conclusion:

Thus, the differential voltage vd of AD549K is 1.99×10−7 V and input bias current id is 1.996×10−20 A.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Textbook Question

Chapter 6, Problem 50E

For the circuit of Fig. 6.62, calculate the differential input voltage and the input bias current if the op amp is a(n) (a) μA741; (b) LF411; (c) AD549K.

Chapter 6, Problem 50E, For the circuit of Fig. 6.62, calculate the differential input voltage and the input bias current if

(a)

Expert Solution

To determine

Find the differential input voltage and input bias current of op amp.

Answer to Problem 50E

The differential voltage vd of μA741 is 1.548×10−5 V and input bias current id is 7.743×10−12 A.

Explanation of Solution

Given Data:

Differential op amp is μA741.

Calculation:

The redrawn circuit of op amp is shown in Figure 1 as follows:

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 6, Problem 50E

Refer to the redrawn Figure 1:

The expression for the nodal analysis at the nodal voltage vd is:

−vd+450 mVR1+−vd+450 mV−voutRf+−vdRi=0 (1)

Here,

vd is the differential voltage at node terminal of input across op amp,

vin is the applied input across op amp,

R1 is the input resistance of op amp,

Rf is the feedback resistance of op amp and

vout is the output voltage of μA741 op amp.

The expression for nodal analysis at nodal voltage vout is:

vout−(−vd+450 mV)Rf+vout−AvdR0=0 (2)

Here,

A is the open loop gain of op amp and

R0 is the output resistance of op amp.

The expression for the bias input current of differential op amp is:

vd=(id×Ri) (3)

Here,

id is the bias input current of practical op amp.

Simplify the equation (1) for vd.

−vd+450 mVR1+−vd+450 mV−voutRf+−vdRi=0−vdR1−vdRf−vdRi−voutRf+450 mVRf=−450 mVR1−vd(1R1+1Rf+1Ri)−voutRf+450 mVRf=−450 mVR1

−vd(RiRf+R1Ri+R1RfR1RfRi)=−450 mVR1+voutRf−450 mVRf (4)

Refer to the TABLE 6.3 in the textbook.

Substitute 1.4 kΩ for Rf, 250 Ω for R1 and 2 MΩ for Ri in the equation (4).

−vd(2 MΩ×1.4 kΩ+250 Ω×2 MΩ+250 Ω×1.4 kΩ2 MΩ×1.4 kΩ×250 Ω)=−450 mV250 Ω+vout1.4 kΩ−450 mV1.4 kΩ−vd(2800 MΩ+500 MΩ+0.35 MΩ700000 MΩ)=−450×10−3V250 Ω+vout1.4×103 Ω−3.214×10−4−vd(4.714×10−3)=−1.8×10−3+7.143×10−4vout−3.214×10−4

Rearrange for vout.

−vd(4.714×10−3)=−1.8×10−3+7.143×10−4vout−3.214×10−4−vd(4.714×10−3)=7.143×10−4vout−2.121×10−37.143×10−4vout=2.121×10−3−vd(4.714×10−3)vout=2.121×10−3−vd(4.714×10−3)7.143×10−4

vout=2.969−6.599vd (5)

Substitute 2.969−6.599vd for vout from equation (5) in equation (2).

(2.969−6.599vd)−(−vd+450 mV)1.4 kΩ+(2.969−6.599vd)−A×vd75 Ω=0−6.599vd1.4 kΩ+vd1.4 kΩ−6.599vd75 Ω−A×vd75 Ω+2.9691.4×103 Ω−450×10−3 V1.4×103 Ω+2.96975 Ω=0vd(−6.5991.4 kΩ+11.4 kΩ−6.59975 Ω−A75 Ω)+2.121×10−3−3.214×10−4+0.0395=0

vd(−4.713×10−3+7.14×10−4−0.0879−A75 Ω)+0.0413=0 (6)

Substitute 2×105 V/V for A in the equation (6).

vd(−4.713×10−3+7.14×10−4−0.0879−2×105 V/V75 Ω)+0.0413=0vd(−2666.758)+0.0413=0vd=0.04132666.758 Vvd=1.548×10−5 V

Substitute 1.548×10−5 V for vd in equation (3).

1.548×10−5 V=(id×2 MΩ)1.548×10−5 V=(id×2×106Ω) {∵1 MΩ=1×106Ω}

Rearrange for id.

id=7.743×10−12 A

Conclusion:

Thus, the differential voltage vd of μA741 is 1.548×10−5 V and input bias current id is 7.743×10−12 A.

(b)

Expert Solution

To determine

Find the differential input voltage and input bias current of op amp.

Answer to Problem 50E

The differential voltage vd of LF411 is 1.485×10−5 V and input bias current id is 1.485×10−17 A.

Explanation of Solution

Given Data:

Differential op amp is LF411.

Calculation:

Refer to the TABLE 6.3 in the textbook.

Substitute 1.4 kΩ for Rf, 250 Ω for R1 and 1 TΩ for Ri in the equation (4).

−vd(1 TΩ×1.4 kΩ+250 Ω×1 TΩ+250 Ω×1.4 kΩ1 TΩ×1.4 kΩ×250 Ω)=−450 mV250 Ω+vout1.4 kΩ−450 mV1.4 kΩ−vd(1400 TΩ+250 TΩ+0.00035 TΩ350000 TΩ)=−450×10−3V250 Ω+vout1.4×103 Ω−3.214×10−4−vd(4.714×10−3)=−1.8×10−3+7.143×10−4vout−3.214×10−4

Rearrange for vout.

−vd(4.714×10−3)=7.143×10−4vout−2.121×10−37.143×10−4vout=2.121×10−3−vd(4.714×10−3)vout=2.121×10−3−vd(4.714×10−3)7.143×10−4

vout=2.969−6.599vd

Substitute 2.969−6.599vd for vout in equation (2).

(2.969−6.599vd)−(−vd+450 mV)1.4 kΩ+(2.969−6.599vd)−A×vd1 Ω=0−6.599vd1.4 kΩ+vd1.4 kΩ−6.599vd1 Ω−A×vd1 Ω+2.9691.4×103 Ω−450×10−3 V1.4×103 Ω+2.9691 Ω=0vd(−6.5991.4 kΩ+11.4 kΩ−6.5991 Ω−A1 Ω)+2.121×10−3−3.214×10−4+2.969=0

vd(−4.713×10−3+7.14×10−4−6.599−A)+2.970=0 (7)

Substitute 2×105 V/V for A in the equation (7).

vd(−4.713×10−3+7.14×10−4−6.599−2×105)+2.970=0vd(−200006.603)+0.0349=0vd=2.970200006.603 Vvd=1.485×10−5 V

Substitute 1.485×10−5 V for vd in equation (3).

1.485×10−5 V=(id×1 TΩ)1.485×10−5 V=(id×1×1012 Ω) {∵1 TΩ=1×1012 Ω}

Rearrange for id.

id=1.485×10−17 A

Conclusion:

Thus, the differential voltage vd of LF411 is 1.485×10−5 V and input bias current id is 1.485×10−17 A.

(c)

Expert Solution

To determine

Find the differential input voltage and input bias current of AD549K op amp.

Answer to Problem 50E

The differential voltage vd of AD549K is 1.99×10−7 V and input bias current id is 1.996×10−20 A.

Explanation of Solution

Given Data:

Differential op amp is AD549K.

Calculation:

Refer to the TABLE 6.3 in the textbook.

Substitute 1.4 kΩ for Rf, 250 Ω for R1 and 10 TΩ for Ri in the equation (4).

−vd(10 TΩ×1.4 kΩ+250 Ω×10 TΩ+250 Ω×1.4 kΩ10 TΩ×1.4 kΩ×250 Ω)=−450 mV250 Ω+vout1.4 kΩ−450 mV1.4 kΩ−vd(14000 TΩ+2500 TΩ+0.00035 TΩ3500000 TΩ)=−450×10−3V250 Ω+vout1.4×103 Ω−3.214×10−4−vd(4.714×10−3)=−1.8×10−3+7.143×10−4vout−3.214×10−4

Rearrange for vout.

−vd(4.714×10−3)=7.143×10−4vout−2.121×10−37.143×10−4vout=2.121×10−3−vd(4.714×10−3)vout=2.121×10−3−vd(4.714×10−3)7.143×10−4

vout=2.969−6.599vd

Substitute 2.969−6.599vd for vout in equation (2).

(2.969−6.599vd)−(−vd+450 mV)1.4 kΩ+(2.969−6.599vd V)−A×vd15 Ω=0−6.599vd1.4 kΩ+vd1.4 kΩ−6.599vd15 Ω−A×vd15 Ω+2.9691.4×103 Ω−450×10−3 V1.4×103 Ω+2.96915 Ω=0vd(−6.5991.4 kΩ+11.4 kΩ−6.59915 Ω−A15 Ω)+2.121×10−3−3.214×10−4+0.198=0

vd(−4.713×10−3+7.14×10−4−0.439−A)+0.1997=0 (8)

Substitute 106 V/V for A in equation (8).

vd(−4.713×10−3+7.14×10−4−0.439−106)+0.1997=0vd(−1000000.443)+0.1997=0vd=0.19971000000.443 Vvd=1.99×10−7 V

Substitute 1.99×10−7 V for vd in equation (3).

1.99×10−7 V=(id×10 TΩ)1.99×10−7 V=(id×10×1012 Ω) {∵1 TΩ=1×1012 Ω}

Rearrange for id,

id=1.996×10−20 A

Conclusion:

Thus, the differential voltage vd of AD549K is 1.99×10−7 V and input bias current id is 1.996×10−20 A.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

Chapter 6, Problem 50E

For the circuit of Fig. 6.62, calculate the differential input voltage and the input bias current if the op amp is a(n) (a) μA741; (b) LF411; (c) AD549K.

Chapter 6, Problem 50E, For the circuit of Fig. 6.62, calculate the differential input voltage and the input bias current if

(a)

Expert Solution

To determine

Find the differential input voltage and input bias current of op amp.

Answer to Problem 50E

The differential voltage vd of μA741 is 1.548×10−5 V and input bias current id is 7.743×10−12 A.

Explanation of Solution

Given Data:

Differential op amp is μA741.

Calculation:

The redrawn circuit of op amp is shown in Figure 1 as follows:

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 6, Problem 50E

Refer to the redrawn Figure 1:

The expression for the nodal analysis at the nodal voltage vd is:

−vd+450 mVR1+−vd+450 mV−voutRf+−vdRi=0 (1)

Here,

vd is the differential voltage at node terminal of input across op amp,

vin is the applied input across op amp,

R1 is the input resistance of op amp,

Rf is the feedback resistance of op amp and

vout is the output voltage of μA741 op amp.

The expression for nodal analysis at nodal voltage vout is:

vout−(−vd+450 mV)Rf+vout−AvdR0=0 (2)

Here,

A is the open loop gain of op amp and

R0 is the output resistance of op amp.

The expression for the bias input current of differential op amp is:

vd=(id×Ri) (3)

Here,

id is the bias input current of practical op amp.

Simplify the equation (1) for vd.

−vd+450 mVR1+−vd+450 mV−voutRf+−vdRi=0−vdR1−vdRf−vdRi−voutRf+450 mVRf=−450 mVR1−vd(1R1+1Rf+1Ri)−voutRf+450 mVRf=−450 mVR1

−vd(RiRf+R1Ri+R1RfR1RfRi)=−450 mVR1+voutRf−450 mVRf (4)

Refer to the TABLE 6.3 in the textbook.

Substitute 1.4 kΩ for Rf, 250 Ω for R1 and 2 MΩ for Ri in the equation (4).

−vd(2 MΩ×1.4 kΩ+250 Ω×2 MΩ+250 Ω×1.4 kΩ2 MΩ×1.4 kΩ×250 Ω)=−450 mV250 Ω+vout1.4 kΩ−450 mV1.4 kΩ−vd(2800 MΩ+500 MΩ+0.35 MΩ700000 MΩ)=−450×10−3V250 Ω+vout1.4×103 Ω−3.214×10−4−vd(4.714×10−3)=−1.8×10−3+7.143×10−4vout−3.214×10−4

Rearrange for vout.

−vd(4.714×10−3)=−1.8×10−3+7.143×10−4vout−3.214×10−4−vd(4.714×10−3)=7.143×10−4vout−2.121×10−37.143×10−4vout=2.121×10−3−vd(4.714×10−3)vout=2.121×10−3−vd(4.714×10−3)7.143×10−4

vout=2.969−6.599vd (5)

Substitute 2.969−6.599vd for vout from equation (5) in equation (2).

(2.969−6.599vd)−(−vd+450 mV)1.4 kΩ+(2.969−6.599vd)−A×vd75 Ω=0−6.599vd1.4 kΩ+vd1.4 kΩ−6.599vd75 Ω−A×vd75 Ω+2.9691.4×103 Ω−450×10−3 V1.4×103 Ω+2.96975 Ω=0vd(−6.5991.4 kΩ+11.4 kΩ−6.59975 Ω−A75 Ω)+2.121×10−3−3.214×10−4+0.0395=0

vd(−4.713×10−3+7.14×10−4−0.0879−A75 Ω)+0.0413=0 (6)

Substitute 2×105 V/V for A in the equation (6).

vd(−4.713×10−3+7.14×10−4−0.0879−2×105 V/V75 Ω)+0.0413=0vd(−2666.758)+0.0413=0vd=0.04132666.758 Vvd=1.548×10−5 V

Substitute 1.548×10−5 V for vd in equation (3).

1.548×10−5 V=(id×2 MΩ)1.548×10−5 V=(id×2×106Ω) {∵1 MΩ=1×106Ω}

Rearrange for id.

id=7.743×10−12 A

Conclusion:

Thus, the differential voltage vd of μA741 is 1.548×10−5 V and input bias current id is 7.743×10−12 A.

(b)

Expert Solution

To determine

Find the differential input voltage and input bias current of op amp.

Answer to Problem 50E

The differential voltage vd of LF411 is 1.485×10−5 V and input bias current id is 1.485×10−17 A.

Explanation of Solution

Given Data:

Differential op amp is LF411.

Calculation:

Refer to the TABLE 6.3 in the textbook.

Substitute 1.4 kΩ for Rf, 250 Ω for R1 and 1 TΩ for Ri in the equation (4).

−vd(1 TΩ×1.4 kΩ+250 Ω×1 TΩ+250 Ω×1.4 kΩ1 TΩ×1.4 kΩ×250 Ω)=−450 mV250 Ω+vout1.4 kΩ−450 mV1.4 kΩ−vd(1400 TΩ+250 TΩ+0.00035 TΩ350000 TΩ)=−450×10−3V250 Ω+vout1.4×103 Ω−3.214×10−4−vd(4.714×10−3)=−1.8×10−3+7.143×10−4vout−3.214×10−4

Rearrange for vout.

−vd(4.714×10−3)=7.143×10−4vout−2.121×10−37.143×10−4vout=2.121×10−3−vd(4.714×10−3)vout=2.121×10−3−vd(4.714×10−3)7.143×10−4

vout=2.969−6.599vd

Substitute 2.969−6.599vd for vout in equation (2).

(2.969−6.599vd)−(−vd+450 mV)1.4 kΩ+(2.969−6.599vd)−A×vd1 Ω=0−6.599vd1.4 kΩ+vd1.4 kΩ−6.599vd1 Ω−A×vd1 Ω+2.9691.4×103 Ω−450×10−3 V1.4×103 Ω+2.9691 Ω=0vd(−6.5991.4 kΩ+11.4 kΩ−6.5991 Ω−A1 Ω)+2.121×10−3−3.214×10−4+2.969=0

vd(−4.713×10−3+7.14×10−4−6.599−A)+2.970=0 (7)

Substitute 2×105 V/V for A in the equation (7).

vd(−4.713×10−3+7.14×10−4−6.599−2×105)+2.970=0vd(−200006.603)+0.0349=0vd=2.970200006.603 Vvd=1.485×10−5 V

Substitute 1.485×10−5 V for vd in equation (3).

1.485×10−5 V=(id×1 TΩ)1.485×10−5 V=(id×1×1012 Ω) {∵1 TΩ=1×1012 Ω}

Rearrange for id.

id=1.485×10−17 A

Conclusion:

Thus, the differential voltage vd of LF411 is 1.485×10−5 V and input bias current id is 1.485×10−17 A.

(c)

Expert Solution

To determine

Find the differential input voltage and input bias current of AD549K op amp.

Answer to Problem 50E

The differential voltage vd of AD549K is 1.99×10−7 V and input bias current id is 1.996×10−20 A.

Explanation of Solution

Given Data:

Differential op amp is AD549K.

Calculation:

Refer to the TABLE 6.3 in the textbook.

Substitute 1.4 kΩ for Rf, 250 Ω for R1 and 10 TΩ for Ri in the equation (4).

−vd(10 TΩ×1.4 kΩ+250 Ω×10 TΩ+250 Ω×1.4 kΩ10 TΩ×1.4 kΩ×250 Ω)=−450 mV250 Ω+vout1.4 kΩ−450 mV1.4 kΩ−vd(14000 TΩ+2500 TΩ+0.00035 TΩ3500000 TΩ)=−450×10−3V250 Ω+vout1.4×103 Ω−3.214×10−4−vd(4.714×10−3)=−1.8×10−3+7.143×10−4vout−3.214×10−4

Rearrange for vout.

−vd(4.714×10−3)=7.143×10−4vout−2.121×10−37.143×10−4vout=2.121×10−3−vd(4.714×10−3)vout=2.121×10−3−vd(4.714×10−3)7.143×10−4

vout=2.969−6.599vd

Substitute 2.969−6.599vd for vout in equation (2).

(2.969−6.599vd)−(−vd+450 mV)1.4 kΩ+(2.969−6.599vd V)−A×vd15 Ω=0−6.599vd1.4 kΩ+vd1.4 kΩ−6.599vd15 Ω−A×vd15 Ω+2.9691.4×103 Ω−450×10−3 V1.4×103 Ω+2.96915 Ω=0vd(−6.5991.4 kΩ+11.4 kΩ−6.59915 Ω−A15 Ω)+2.121×10−3−3.214×10−4+0.198=0

vd(−4.713×10−3+7.14×10−4−0.439−A)+0.1997=0 (8)

Substitute 106 V/V for A in equation (8).

vd(−4.713×10−3+7.14×10−4−0.439−106)+0.1997=0vd(−1000000.443)+0.1997=0vd=0.19971000000.443 Vvd=1.99×10−7 V

Substitute 1.99×10−7 V for vd in equation (3).

1.99×10−7 V=(id×10 TΩ)1.99×10−7 V=(id×10×1012 Ω) {∵1 TΩ=1×1012 Ω}

Rearrange for id,

id=1.996×10−20 A

Conclusion:

Thus, the differential voltage vd of AD549K is 1.99×10−7 V and input bias current id is 1.996×10−20 A.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

Find the differential input voltage and input bias current of op amp.

Answer to Problem 50E

The differential voltage vd of μA741 is 1.548×10−5 V and input bias current id is 7.743×10−12 A.

Explanation of Solution

Given Data:

Differential op amp is μA741.

Calculation:

The redrawn circuit of op amp is shown in Figure 1 as follows:

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 6, Problem 50E

Refer to the redrawn Figure 1:

The expression for the nodal analysis at the nodal voltage vd is:

−vd+450 mVR1+−vd+450 mV−voutRf+−vdRi=0 (1)

Here,

vd is the differential voltage at node terminal of input across op amp,

vin is the applied input across op amp,

R1 is the input resistance of op amp,

Rf is the feedback resistance of op amp and

vout is the output voltage of μA741 op amp.

The expression for nodal analysis at nodal voltage vout is:

vout−(−vd+450 mV)Rf+vout−AvdR0=0 (2)

Here,

A is the open loop gain of op amp and

R0 is the output resistance of op amp.

The expression for the bias input current of differential op amp is:

vd=(id×Ri) (3)

Here,

id is the bias input current of practical op amp.

Simplify the equation (1) for vd.

−vd+450 mVR1+−vd+450 mV−voutRf+−vdRi=0−vdR1−vdRf−vdRi−voutRf+450 mVRf=−450 mVR1−vd(1R1+1Rf+1Ri)−voutRf+450 mVRf=−450 mVR1

−vd(RiRf+R1Ri+R1RfR1RfRi)=−450 mVR1+voutRf−450 mVRf (4)

Refer to the TABLE 6.3 in the textbook.

Substitute 1.4 kΩ for Rf, 250 Ω for R1 and 2 MΩ for Ri in the equation (4).

−vd(2 MΩ×1.4 kΩ+250 Ω×2 MΩ+250 Ω×1.4 kΩ2 MΩ×1.4 kΩ×250 Ω)=−450 mV250 Ω+vout1.4 kΩ−450 mV1.4 kΩ−vd(2800 MΩ+500 MΩ+0.35 MΩ700000 MΩ)=−450×10−3V250 Ω+vout1.4×103 Ω−3.214×10−4−vd(4.714×10−3)=−1.8×10−3+7.143×10−4vout−3.214×10−4

Rearrange for vout.

−vd(4.714×10−3)=−1.8×10−3+7.143×10−4vout−3.214×10−4−vd(4.714×10−3)=7.143×10−4vout−2.121×10−37.143×10−4vout=2.121×10−3−vd(4.714×10−3)vout=2.121×10−3−vd(4.714×10−3)7.143×10−4

vout=2.969−6.599vd (5)

Substitute 2.969−6.599vd for vout from equation (5) in equation (2).

(2.969−6.599vd)−(−vd+450 mV)1.4 kΩ+(2.969−6.599vd)−A×vd75 Ω=0−6.599vd1.4 kΩ+vd1.4 kΩ−6.599vd75 Ω−A×vd75 Ω+2.9691.4×103 Ω−450×10−3 V1.4×103 Ω+2.96975 Ω=0vd(−6.5991.4 kΩ+11.4 kΩ−6.59975 Ω−A75 Ω)+2.121×10−3−3.214×10−4+0.0395=0

vd(−4.713×10−3+7.14×10−4−0.0879−A75 Ω)+0.0413=0 (6)

Substitute 2×105 V/V for A in the equation (6).

vd(−4.713×10−3+7.14×10−4−0.0879−2×105 V/V75 Ω)+0.0413=0vd(−2666.758)+0.0413=0vd=0.04132666.758 Vvd=1.548×10−5 V

Substitute 1.548×10−5 V for vd in equation (3).

1.548×10−5 V=(id×2 MΩ)1.548×10−5 V=(id×2×106Ω) {∵1 MΩ=1×106Ω}

Rearrange for id.

id=7.743×10−12 A

Conclusion:

Thus, the differential voltage vd of μA741 is 1.548×10−5 V and input bias current id is 7.743×10−12 A.

(b)

Expert Solution

To determine

Find the differential input voltage and input bias current of op amp.

Answer to Problem 50E

The differential voltage vd of LF411 is 1.485×10−5 V and input bias current id is 1.485×10−17 A.

Explanation of Solution

Given Data:

Differential op amp is LF411.

Calculation:

Refer to the TABLE 6.3 in the textbook.

Substitute 1.4 kΩ for Rf, 250 Ω for R1 and 1 TΩ for Ri in the equation (4).

−vd(1 TΩ×1.4 kΩ+250 Ω×1 TΩ+250 Ω×1.4 kΩ1 TΩ×1.4 kΩ×250 Ω)=−450 mV250 Ω+vout1.4 kΩ−450 mV1.4 kΩ−vd(1400 TΩ+250 TΩ+0.00035 TΩ350000 TΩ)=−450×10−3V250 Ω+vout1.4×103 Ω−3.214×10−4−vd(4.714×10−3)=−1.8×10−3+7.143×10−4vout−3.214×10−4

Rearrange for vout.

−vd(4.714×10−3)=7.143×10−4vout−2.121×10−37.143×10−4vout=2.121×10−3−vd(4.714×10−3)vout=2.121×10−3−vd(4.714×10−3)7.143×10−4

vout=2.969−6.599vd

Substitute 2.969−6.599vd for vout in equation (2).

(2.969−6.599vd)−(−vd+450 mV)1.4 kΩ+(2.969−6.599vd)−A×vd1 Ω=0−6.599vd1.4 kΩ+vd1.4 kΩ−6.599vd1 Ω−A×vd1 Ω+2.9691.4×103 Ω−450×10−3 V1.4×103 Ω+2.9691 Ω=0vd(−6.5991.4 kΩ+11.4 kΩ−6.5991 Ω−A1 Ω)+2.121×10−3−3.214×10−4+2.969=0

vd(−4.713×10−3+7.14×10−4−6.599−A)+2.970=0 (7)

Substitute 2×105 V/V for A in the equation (7).

vd(−4.713×10−3+7.14×10−4−6.599−2×105)+2.970=0vd(−200006.603)+0.0349=0vd=2.970200006.603 Vvd=1.485×10−5 V

Substitute 1.485×10−5 V for vd in equation (3).

1.485×10−5 V=(id×1 TΩ)1.485×10−5 V=(id×1×1012 Ω) {∵1 TΩ=1×1012 Ω}

Rearrange for id.

id=1.485×10−17 A

Conclusion:

Thus, the differential voltage vd of LF411 is 1.485×10−5 V and input bias current id is 1.485×10−17 A.

(c)

Expert Solution

To determine

Find the differential input voltage and input bias current of AD549K op amp.

Answer to Problem 50E

The differential voltage vd of AD549K is 1.99×10−7 V and input bias current id is 1.996×10−20 A.

Explanation of Solution

Given Data:

Differential op amp is AD549K.

Calculation:

Refer to the TABLE 6.3 in the textbook.

Substitute 1.4 kΩ for Rf, 250 Ω for R1 and 10 TΩ for Ri in the equation (4).

−vd(10 TΩ×1.4 kΩ+250 Ω×10 TΩ+250 Ω×1.4 kΩ10 TΩ×1.4 kΩ×250 Ω)=−450 mV250 Ω+vout1.4 kΩ−450 mV1.4 kΩ−vd(14000 TΩ+2500 TΩ+0.00035 TΩ3500000 TΩ)=−450×10−3V250 Ω+vout1.4×103 Ω−3.214×10−4−vd(4.714×10−3)=−1.8×10−3+7.143×10−4vout−3.214×10−4

Rearrange for vout.

−vd(4.714×10−3)=7.143×10−4vout−2.121×10−37.143×10−4vout=2.121×10−3−vd(4.714×10−3)vout=2.121×10−3−vd(4.714×10−3)7.143×10−4

vout=2.969−6.599vd

Substitute 2.969−6.599vd for vout in equation (2).

(2.969−6.599vd)−(−vd+450 mV)1.4 kΩ+(2.969−6.599vd V)−A×vd15 Ω=0−6.599vd1.4 kΩ+vd1.4 kΩ−6.599vd15 Ω−A×vd15 Ω+2.9691.4×103 Ω−450×10−3 V1.4×103 Ω+2.96915 Ω=0vd(−6.5991.4 kΩ+11.4 kΩ−6.59915 Ω−A15 Ω)+2.121×10−3−3.214×10−4+0.198=0

vd(−4.713×10−3+7.14×10−4−0.439−A)+0.1997=0 (8)

Substitute 106 V/V for A in equation (8).

vd(−4.713×10−3+7.14×10−4−0.439−106)+0.1997=0vd(−1000000.443)+0.1997=0vd=0.19971000000.443 Vvd=1.99×10−7 V

Substitute 1.99×10−7 V for vd in equation (3).

1.99×10−7 V=(id×10 TΩ)1.99×10−7 V=(id×10×1012 Ω) {∵1 TΩ=1×1012 Ω}

Rearrange for id,

id=1.996×10−20 A

Conclusion:

Thus, the differential voltage vd of AD549K is 1.99×10−7 V and input bias current id is 1.996×10−20 A.

Answer 8

(a)

Expert Solution

To determine

Find the differential input voltage and input bias current of op amp.

Answer to Problem 50E

The differential voltage vd of μA741 is 1.548×10−5 V and input bias current id is 7.743×10−12 A.

Explanation of Solution

Given Data:

Differential op amp is μA741.

Calculation:

The redrawn circuit of op amp is shown in Figure 1 as follows:

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 6, Problem 50E

Refer to the redrawn Figure 1:

The expression for the nodal analysis at the nodal voltage vd is:

−vd+450 mVR1+−vd+450 mV−voutRf+−vdRi=0 (1)

Here,

vd is the differential voltage at node terminal of input across op amp,

vin is the applied input across op amp,

R1 is the input resistance of op amp,

Rf is the feedback resistance of op amp and

vout is the output voltage of μA741 op amp.

The expression for nodal analysis at nodal voltage vout is:

vout−(−vd+450 mV)Rf+vout−AvdR0=0 (2)

Here,

A is the open loop gain of op amp and

R0 is the output resistance of op amp.

The expression for the bias input current of differential op amp is:

vd=(id×Ri) (3)

Here,

id is the bias input current of practical op amp.

Simplify the equation (1) for vd.

−vd+450 mVR1+−vd+450 mV−voutRf+−vdRi=0−vdR1−vdRf−vdRi−voutRf+450 mVRf=−450 mVR1−vd(1R1+1Rf+1Ri)−voutRf+450 mVRf=−450 mVR1

−vd(RiRf+R1Ri+R1RfR1RfRi)=−450 mVR1+voutRf−450 mVRf (4)

Refer to the TABLE 6.3 in the textbook.

Substitute 1.4 kΩ for Rf, 250 Ω for R1 and 2 MΩ for Ri in the equation (4).

−vd(2 MΩ×1.4 kΩ+250 Ω×2 MΩ+250 Ω×1.4 kΩ2 MΩ×1.4 kΩ×250 Ω)=−450 mV250 Ω+vout1.4 kΩ−450 mV1.4 kΩ−vd(2800 MΩ+500 MΩ+0.35 MΩ700000 MΩ)=−450×10−3V250 Ω+vout1.4×103 Ω−3.214×10−4−vd(4.714×10−3)=−1.8×10−3+7.143×10−4vout−3.214×10−4

Rearrange for vout.

−vd(4.714×10−3)=−1.8×10−3+7.143×10−4vout−3.214×10−4−vd(4.714×10−3)=7.143×10−4vout−2.121×10−37.143×10−4vout=2.121×10−3−vd(4.714×10−3)vout=2.121×10−3−vd(4.714×10−3)7.143×10−4

vout=2.969−6.599vd (5)

Substitute 2.969−6.599vd for vout from equation (5) in equation (2).

(2.969−6.599vd)−(−vd+450 mV)1.4 kΩ+(2.969−6.599vd)−A×vd75 Ω=0−6.599vd1.4 kΩ+vd1.4 kΩ−6.599vd75 Ω−A×vd75 Ω+2.9691.4×103 Ω−450×10−3 V1.4×103 Ω+2.96975 Ω=0vd(−6.5991.4 kΩ+11.4 kΩ−6.59975 Ω−A75 Ω)+2.121×10−3−3.214×10−4+0.0395=0

vd(−4.713×10−3+7.14×10−4−0.0879−A75 Ω)+0.0413=0 (6)

Substitute 2×105 V/V for A in the equation (6).

vd(−4.713×10−3+7.14×10−4−0.0879−2×105 V/V75 Ω)+0.0413=0vd(−2666.758)+0.0413=0vd=0.04132666.758 Vvd=1.548×10−5 V

Substitute 1.548×10−5 V for vd in equation (3).

1.548×10−5 V=(id×2 MΩ)1.548×10−5 V=(id×2×106Ω) {∵1 MΩ=1×106Ω}

Rearrange for id.

id=7.743×10−12 A

Conclusion:

Thus, the differential voltage vd of μA741 is 1.548×10−5 V and input bias current id is 7.743×10−12 A.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

Find the differential input voltage and input bias current of op amp.

Answer 13

Answer to Problem 50E

The differential voltage vd of μA741 is 1.548×10−5 V and input bias current id is 7.743×10−12 A.

Answer 14

Explanation of Solution

Given Data:

Differential op amp is μA741.

Calculation:

The redrawn circuit of op amp is shown in Figure 1 as follows:

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 6, Problem 50E

Refer to the redrawn Figure 1:

The expression for the nodal analysis at the nodal voltage vd is:

−vd+450 mVR1+−vd+450 mV−voutRf+−vdRi=0 (1)

Here,

vd is the differential voltage at node terminal of input across op amp,

vin is the applied input across op amp,

R1 is the input resistance of op amp,

Rf is the feedback resistance of op amp and

vout is the output voltage of μA741 op amp.

The expression for nodal analysis at nodal voltage vout is:

vout−(−vd+450 mV)Rf+vout−AvdR0=0 (2)

Here,

A is the open loop gain of op amp and

R0 is the output resistance of op amp.

The expression for the bias input current of differential op amp is:

vd=(id×Ri) (3)

Here,

id is the bias input current of practical op amp.

Simplify the equation (1) for vd.

−vd+450 mVR1+−vd+450 mV−voutRf+−vdRi=0−vdR1−vdRf−vdRi−voutRf+450 mVRf=−450 mVR1−vd(1R1+1Rf+1Ri)−voutRf+450 mVRf=−450 mVR1

−vd(RiRf+R1Ri+R1RfR1RfRi)=−450 mVR1+voutRf−450 mVRf (4)

Refer to the TABLE 6.3 in the textbook.

Substitute 1.4 kΩ for Rf, 250 Ω for R1 and 2 MΩ for Ri in the equation (4).

−vd(2 MΩ×1.4 kΩ+250 Ω×2 MΩ+250 Ω×1.4 kΩ2 MΩ×1.4 kΩ×250 Ω)=−450 mV250 Ω+vout1.4 kΩ−450 mV1.4 kΩ−vd(2800 MΩ+500 MΩ+0.35 MΩ700000 MΩ)=−450×10−3V250 Ω+vout1.4×103 Ω−3.214×10−4−vd(4.714×10−3)=−1.8×10−3+7.143×10−4vout−3.214×10−4

Rearrange for vout.

−vd(4.714×10−3)=−1.8×10−3+7.143×10−4vout−3.214×10−4−vd(4.714×10−3)=7.143×10−4vout−2.121×10−37.143×10−4vout=2.121×10−3−vd(4.714×10−3)vout=2.121×10−3−vd(4.714×10−3)7.143×10−4

vout=2.969−6.599vd (5)

Substitute 2.969−6.599vd for vout from equation (5) in equation (2).

(2.969−6.599vd)−(−vd+450 mV)1.4 kΩ+(2.969−6.599vd)−A×vd75 Ω=0−6.599vd1.4 kΩ+vd1.4 kΩ−6.599vd75 Ω−A×vd75 Ω+2.9691.4×103 Ω−450×10−3 V1.4×103 Ω+2.96975 Ω=0vd(−6.5991.4 kΩ+11.4 kΩ−6.59975 Ω−A75 Ω)+2.121×10−3−3.214×10−4+0.0395=0

vd(−4.713×10−3+7.14×10−4−0.0879−A75 Ω)+0.0413=0 (6)

Substitute 2×105 V/V for A in the equation (6).

vd(−4.713×10−3+7.14×10−4−0.0879−2×105 V/V75 Ω)+0.0413=0vd(−2666.758)+0.0413=0vd=0.04132666.758 Vvd=1.548×10−5 V

Substitute 1.548×10−5 V for vd in equation (3).

1.548×10−5 V=(id×2 MΩ)1.548×10−5 V=(id×2×106Ω) {∵1 MΩ=1×106Ω}

Rearrange for id.

id=7.743×10−12 A

Conclusion:

Thus, the differential voltage vd of μA741 is 1.548×10−5 V and input bias current id is 7.743×10−12 A.

Answer 15

(b)

Expert Solution

To determine

Find the differential input voltage and input bias current of op amp.

Answer to Problem 50E

The differential voltage vd of LF411 is 1.485×10−5 V and input bias current id is 1.485×10−17 A.

Explanation of Solution

Given Data:

Differential op amp is LF411.

Calculation:

Refer to the TABLE 6.3 in the textbook.

Substitute 1.4 kΩ for Rf, 250 Ω for R1 and 1 TΩ for Ri in the equation (4).

−vd(1 TΩ×1.4 kΩ+250 Ω×1 TΩ+250 Ω×1.4 kΩ1 TΩ×1.4 kΩ×250 Ω)=−450 mV250 Ω+vout1.4 kΩ−450 mV1.4 kΩ−vd(1400 TΩ+250 TΩ+0.00035 TΩ350000 TΩ)=−450×10−3V250 Ω+vout1.4×103 Ω−3.214×10−4−vd(4.714×10−3)=−1.8×10−3+7.143×10−4vout−3.214×10−4

Rearrange for vout.

−vd(4.714×10−3)=7.143×10−4vout−2.121×10−37.143×10−4vout=2.121×10−3−vd(4.714×10−3)vout=2.121×10−3−vd(4.714×10−3)7.143×10−4

vout=2.969−6.599vd

Substitute 2.969−6.599vd for vout in equation (2).

(2.969−6.599vd)−(−vd+450 mV)1.4 kΩ+(2.969−6.599vd)−A×vd1 Ω=0−6.599vd1.4 kΩ+vd1.4 kΩ−6.599vd1 Ω−A×vd1 Ω+2.9691.4×103 Ω−450×10−3 V1.4×103 Ω+2.9691 Ω=0vd(−6.5991.4 kΩ+11.4 kΩ−6.5991 Ω−A1 Ω)+2.121×10−3−3.214×10−4+2.969=0

vd(−4.713×10−3+7.14×10−4−6.599−A)+2.970=0 (7)

Substitute 2×105 V/V for A in the equation (7).

vd(−4.713×10−3+7.14×10−4−6.599−2×105)+2.970=0vd(−200006.603)+0.0349=0vd=2.970200006.603 Vvd=1.485×10−5 V

Substitute 1.485×10−5 V for vd in equation (3).

1.485×10−5 V=(id×1 TΩ)1.485×10−5 V=(id×1×1012 Ω) {∵1 TΩ=1×1012 Ω}

Rearrange for id.

id=1.485×10−17 A

Conclusion:

Thus, the differential voltage vd of LF411 is 1.485×10−5 V and input bias current id is 1.485×10−17 A.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

Find the differential input voltage and input bias current of op amp.

Answer 20

Answer to Problem 50E

The differential voltage vd of LF411 is 1.485×10−5 V and input bias current id is 1.485×10−17 A.

Answer 21

Explanation of Solution

Given Data:

Differential op amp is LF411.

Calculation:

Refer to the TABLE 6.3 in the textbook.

Substitute 1.4 kΩ for Rf, 250 Ω for R1 and 1 TΩ for Ri in the equation (4).

−vd(1 TΩ×1.4 kΩ+250 Ω×1 TΩ+250 Ω×1.4 kΩ1 TΩ×1.4 kΩ×250 Ω)=−450 mV250 Ω+vout1.4 kΩ−450 mV1.4 kΩ−vd(1400 TΩ+250 TΩ+0.00035 TΩ350000 TΩ)=−450×10−3V250 Ω+vout1.4×103 Ω−3.214×10−4−vd(4.714×10−3)=−1.8×10−3+7.143×10−4vout−3.214×10−4

Rearrange for vout.

−vd(4.714×10−3)=7.143×10−4vout−2.121×10−37.143×10−4vout=2.121×10−3−vd(4.714×10−3)vout=2.121×10−3−vd(4.714×10−3)7.143×10−4

vout=2.969−6.599vd

Substitute 2.969−6.599vd for vout in equation (2).

(2.969−6.599vd)−(−vd+450 mV)1.4 kΩ+(2.969−6.599vd)−A×vd1 Ω=0−6.599vd1.4 kΩ+vd1.4 kΩ−6.599vd1 Ω−A×vd1 Ω+2.9691.4×103 Ω−450×10−3 V1.4×103 Ω+2.9691 Ω=0vd(−6.5991.4 kΩ+11.4 kΩ−6.5991 Ω−A1 Ω)+2.121×10−3−3.214×10−4+2.969=0

vd(−4.713×10−3+7.14×10−4−6.599−A)+2.970=0 (7)

Substitute 2×105 V/V for A in the equation (7).

vd(−4.713×10−3+7.14×10−4−6.599−2×105)+2.970=0vd(−200006.603)+0.0349=0vd=2.970200006.603 Vvd=1.485×10−5 V

Substitute 1.485×10−5 V for vd in equation (3).

1.485×10−5 V=(id×1 TΩ)1.485×10−5 V=(id×1×1012 Ω) {∵1 TΩ=1×1012 Ω}

Rearrange for id.

id=1.485×10−17 A

Conclusion:

Thus, the differential voltage vd of LF411 is 1.485×10−5 V and input bias current id is 1.485×10−17 A.

Answer 22

(c)

Expert Solution

To determine

Find the differential input voltage and input bias current of AD549K op amp.

Answer to Problem 50E

The differential voltage vd of AD549K is 1.99×10−7 V and input bias current id is 1.996×10−20 A.

Explanation of Solution

Given Data:

Differential op amp is AD549K.

Calculation:

Refer to the TABLE 6.3 in the textbook.

Substitute 1.4 kΩ for Rf, 250 Ω for R1 and 10 TΩ for Ri in the equation (4).

−vd(10 TΩ×1.4 kΩ+250 Ω×10 TΩ+250 Ω×1.4 kΩ10 TΩ×1.4 kΩ×250 Ω)=−450 mV250 Ω+vout1.4 kΩ−450 mV1.4 kΩ−vd(14000 TΩ+2500 TΩ+0.00035 TΩ3500000 TΩ)=−450×10−3V250 Ω+vout1.4×103 Ω−3.214×10−4−vd(4.714×10−3)=−1.8×10−3+7.143×10−4vout−3.214×10−4

Rearrange for vout.

−vd(4.714×10−3)=7.143×10−4vout−2.121×10−37.143×10−4vout=2.121×10−3−vd(4.714×10−3)vout=2.121×10−3−vd(4.714×10−3)7.143×10−4

vout=2.969−6.599vd

Substitute 2.969−6.599vd for vout in equation (2).

(2.969−6.599vd)−(−vd+450 mV)1.4 kΩ+(2.969−6.599vd V)−A×vd15 Ω=0−6.599vd1.4 kΩ+vd1.4 kΩ−6.599vd15 Ω−A×vd15 Ω+2.9691.4×103 Ω−450×10−3 V1.4×103 Ω+2.96915 Ω=0vd(−6.5991.4 kΩ+11.4 kΩ−6.59915 Ω−A15 Ω)+2.121×10−3−3.214×10−4+0.198=0

vd(−4.713×10−3+7.14×10−4−0.439−A)+0.1997=0 (8)

Substitute 106 V/V for A in equation (8).

vd(−4.713×10−3+7.14×10−4−0.439−106)+0.1997=0vd(−1000000.443)+0.1997=0vd=0.19971000000.443 Vvd=1.99×10−7 V

Substitute 1.99×10−7 V for vd in equation (3).

1.99×10−7 V=(id×10 TΩ)1.99×10−7 V=(id×10×1012 Ω) {∵1 TΩ=1×1012 Ω}

Rearrange for id,

id=1.996×10−20 A

Conclusion:

Thus, the differential voltage vd of AD549K is 1.99×10−7 V and input bias current id is 1.996×10−20 A.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

Find the differential input voltage and input bias current of AD549K op amp.

Answer 27

Answer to Problem 50E

The differential voltage vd of AD549K is 1.99×10−7 V and input bias current id is 1.996×10−20 A.

Answer 28

Explanation of Solution

Given Data:

Differential op amp is AD549K.

Calculation:

Refer to the TABLE 6.3 in the textbook.

Substitute 1.4 kΩ for Rf, 250 Ω for R1 and 10 TΩ for Ri in the equation (4).

−vd(10 TΩ×1.4 kΩ+250 Ω×10 TΩ+250 Ω×1.4 kΩ10 TΩ×1.4 kΩ×250 Ω)=−450 mV250 Ω+vout1.4 kΩ−450 mV1.4 kΩ−vd(14000 TΩ+2500 TΩ+0.00035 TΩ3500000 TΩ)=−450×10−3V250 Ω+vout1.4×103 Ω−3.214×10−4−vd(4.714×10−3)=−1.8×10−3+7.143×10−4vout−3.214×10−4

Rearrange for vout.

−vd(4.714×10−3)=7.143×10−4vout−2.121×10−37.143×10−4vout=2.121×10−3−vd(4.714×10−3)vout=2.121×10−3−vd(4.714×10−3)7.143×10−4

vout=2.969−6.599vd

Substitute 2.969−6.599vd for vout in equation (2).

(2.969−6.599vd)−(−vd+450 mV)1.4 kΩ+(2.969−6.599vd V)−A×vd15 Ω=0−6.599vd1.4 kΩ+vd1.4 kΩ−6.599vd15 Ω−A×vd15 Ω+2.9691.4×103 Ω−450×10−3 V1.4×103 Ω+2.96915 Ω=0vd(−6.5991.4 kΩ+11.4 kΩ−6.59915 Ω−A15 Ω)+2.121×10−3−3.214×10−4+0.198=0