EXEMRIE (1) Determine the output of the DAC in Figure (a) if the waveforms representing a sequence of 4-bit numbers in Figure (b) are applied to the inputs. Input Do is the least significant bit (LSB). 200 kn Da o R 0123 456 7 89 1011 1213 14 15 +5 V Do +5V 100 kn D, o 10 kn 50 k D, O- +5 V 25 k +5 V Dy D, O (a) (b) Re solving this question and show me step step pls Solution First, determine the current for each of the weighted inputs. Since the inverting |(-) input of the op-amp is at 0 V (virtual ground) and a binary 1 corresponds to +5 V, the current through any of the input resistors is 5 V divided by the resistance value. 200 kn 5 V -0.025 mA 200 k 100 kn D, o WM 10 kt - 0.05 mA 100 k 50 kn SV -0.1 mA 1 - 50 KN 25 k SV -0,2 mA 25 k D, o (a) Solution Almost no current goes into the inverting op-amp input because of its extremely high impedance. Therefore, assume that all of the current goes through the feedback resistor Ry. Since one end of Ry is at 0 V (virtual ground), the drop across R, equals the output voltage, which is negative with respect to virtual ground. 200 k SV -0.025 mA 200 k 100 kn SV Voun - (10 kf-0.025 mA)- -0.25 V Vun (10 kt0x-0.05 mA)- -0.5 V Veun (10 kf-0.1 mA)= -IV VounD - (10 kx-0.2 mA)--2 V = 0.05 mA 100 K 5V = (0.1 mA 25 ka 50 K SV -0.2 mA 25 k (a) Solution
EXEMRIE (1) Determine the output of the DAC in Figure (a) if the waveforms representing a sequence of 4-bit numbers in Figure (b) are applied to the inputs. Input Do is the least significant bit (LSB). 200 kn Da o R 0123 456 7 89 1011 1213 14 15 +5 V Do +5V 100 kn D, o 10 kn 50 k D, O- +5 V 25 k +5 V Dy D, O (a) (b) Re solving this question and show me step step pls Solution First, determine the current for each of the weighted inputs. Since the inverting |(-) input of the op-amp is at 0 V (virtual ground) and a binary 1 corresponds to +5 V, the current through any of the input resistors is 5 V divided by the resistance value. 200 kn 5 V -0.025 mA 200 k 100 kn D, o WM 10 kt - 0.05 mA 100 k 50 kn SV -0.1 mA 1 - 50 KN 25 k SV -0,2 mA 25 k D, o (a) Solution Almost no current goes into the inverting op-amp input because of its extremely high impedance. Therefore, assume that all of the current goes through the feedback resistor Ry. Since one end of Ry is at 0 V (virtual ground), the drop across R, equals the output voltage, which is negative with respect to virtual ground. 200 k SV -0.025 mA 200 k 100 kn SV Voun - (10 kf-0.025 mA)- -0.25 V Vun (10 kt0x-0.05 mA)- -0.5 V Veun (10 kf-0.1 mA)= -IV VounD - (10 kx-0.2 mA)--2 V = 0.05 mA 100 K 5V = (0.1 mA 25 ka 50 K SV -0.2 mA 25 k (a) Solution
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
Related questions
Question
![E混mpie (1)
Determine the output of the DAC in Figure (a) if the waveforms representing a
sequence of 4-bit numbers in Figure (b) are applied to the inputs. Input Do is
the least significant bit (LSB).
200 kn
Doo M
R
100 kn
D o M-
0123 4567 89 1011 1213 14 15
+5V U
Do
+5 V
10 kn
50 kn
D, O
+5 V
25 kf?
+5 V
D, o W-
D,
(а)
(b)
Re solving this question and
show me step step pls
Solution
First, determine the current for each of the weighted inputs. Since the inverting
(-) input of the op-amp is at 0 V (virtual ground) and a binary 1 corresponds to
+5 V, the current through any of the input resistors is 5 v divided by the
resistance value.
200 k
D, o M
5V
-0.025 mA
200 k
5V
4-
100 kn
D, o W-
10 ka
0.05 mA
100 k
50 kn
Dy o W-
25 k
D, o W
5V
= 0.1 mA
50 k
SV
-0,2 mA
25 k
(a)
Solution
Almost no current goes into the inverting op-amp input because of its
extremely high impedance. Therefore, assume that all of the current goes
through the feedback resistor Rr. Since one end of R, is at 0 V (virtual ground),
the drop across R, equals the output voltage, which is negative with respect to
virtual ground.
5V
200 k
D W-
-0.025 mA
200 k
100 kn
D, o W-
S0 k
SV
Voun - (10 kf-0.025 mA) = -0.25 V
VnD (10 ktnN-0.05 mA)- -0.5 V
Ve (10 k-0.1 mA)=-IV
VounD (10 kfX-0.2 mA)- -2 V
= 0,05 mA
100 kn
25 k
=0.1 mA
50 kN
SV
-0.2 mA
(a)
25 k?
Solution
From Figure (b), the first binary input code is 0000, which produces an output voltage of 0 V.
The next input code is 0001, which produces an output voltage of -0.25 V. The next code is
0010, which produces an output voltage of -0.5 v. The next code is 0011, which produces an
output voltage of -0.25 V + -0.5 V = -0.75 V. Each successive binary code increases the output
voltage by -0.25 V, so for this particular straight binary sequence on the inputs, the output is
a stairstep waveform going from 0 V to -3.75 V in -0.25 V steps. This is shown in Figure c.
0123456789 1011 121314 15
+sV
D. 0-
29
200
D. C
2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F772c4772-f1b0-4a5a-9ec9-1fc873927841%2F3c729f13-d172-4e40-abfd-511eddea34d8%2Fsw1e7vc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:E混mpie (1)
Determine the output of the DAC in Figure (a) if the waveforms representing a
sequence of 4-bit numbers in Figure (b) are applied to the inputs. Input Do is
the least significant bit (LSB).
200 kn
Doo M
R
100 kn
D o M-
0123 4567 89 1011 1213 14 15
+5V U
Do
+5 V
10 kn
50 kn
D, O
+5 V
25 kf?
+5 V
D, o W-
D,
(а)
(b)
Re solving this question and
show me step step pls
Solution
First, determine the current for each of the weighted inputs. Since the inverting
(-) input of the op-amp is at 0 V (virtual ground) and a binary 1 corresponds to
+5 V, the current through any of the input resistors is 5 v divided by the
resistance value.
200 k
D, o M
5V
-0.025 mA
200 k
5V
4-
100 kn
D, o W-
10 ka
0.05 mA
100 k
50 kn
Dy o W-
25 k
D, o W
5V
= 0.1 mA
50 k
SV
-0,2 mA
25 k
(a)
Solution
Almost no current goes into the inverting op-amp input because of its
extremely high impedance. Therefore, assume that all of the current goes
through the feedback resistor Rr. Since one end of R, is at 0 V (virtual ground),
the drop across R, equals the output voltage, which is negative with respect to
virtual ground.
5V
200 k
D W-
-0.025 mA
200 k
100 kn
D, o W-
S0 k
SV
Voun - (10 kf-0.025 mA) = -0.25 V
VnD (10 ktnN-0.05 mA)- -0.5 V
Ve (10 k-0.1 mA)=-IV
VounD (10 kfX-0.2 mA)- -2 V
= 0,05 mA
100 kn
25 k
=0.1 mA
50 kN
SV
-0.2 mA
(a)
25 k?
Solution
From Figure (b), the first binary input code is 0000, which produces an output voltage of 0 V.
The next input code is 0001, which produces an output voltage of -0.25 V. The next code is
0010, which produces an output voltage of -0.5 v. The next code is 0011, which produces an
output voltage of -0.25 V + -0.5 V = -0.75 V. Each successive binary code increases the output
voltage by -0.25 V, so for this particular straight binary sequence on the inputs, the output is
a stairstep waveform going from 0 V to -3.75 V in -0.25 V steps. This is shown in Figure c.
0123456789 1011 121314 15
+sV
D. 0-
29
200
D. C
2
Expert Solution
![](/static/compass_v2/shared-icons/check-mark.png)
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps with 3 images
![Blurred answer](/static/compass_v2/solution-images/blurred-answer.jpg)
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.Recommended textbooks for you
![Introductory Circuit Analysis (13th Edition)](https://www.bartleby.com/isbn_cover_images/9780133923605/9780133923605_smallCoverImage.gif)
Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:
9780133923605
Author:
Robert L. Boylestad
Publisher:
PEARSON
![Delmar's Standard Textbook Of Electricity](https://www.bartleby.com/isbn_cover_images/9781337900348/9781337900348_smallCoverImage.jpg)
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:
9781337900348
Author:
Stephen L. Herman
Publisher:
Cengage Learning
![Programmable Logic Controllers](https://www.bartleby.com/isbn_cover_images/9780073373843/9780073373843_smallCoverImage.gif)
Programmable Logic Controllers
Electrical Engineering
ISBN:
9780073373843
Author:
Frank D. Petruzella
Publisher:
McGraw-Hill Education
![Introductory Circuit Analysis (13th Edition)](https://www.bartleby.com/isbn_cover_images/9780133923605/9780133923605_smallCoverImage.gif)
Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:
9780133923605
Author:
Robert L. Boylestad
Publisher:
PEARSON
![Delmar's Standard Textbook Of Electricity](https://www.bartleby.com/isbn_cover_images/9781337900348/9781337900348_smallCoverImage.jpg)
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:
9781337900348
Author:
Stephen L. Herman
Publisher:
Cengage Learning
![Programmable Logic Controllers](https://www.bartleby.com/isbn_cover_images/9780073373843/9780073373843_smallCoverImage.gif)
Programmable Logic Controllers
Electrical Engineering
ISBN:
9780073373843
Author:
Frank D. Petruzella
Publisher:
McGraw-Hill Education
![Fundamentals of Electric Circuits](https://www.bartleby.com/isbn_cover_images/9780078028229/9780078028229_smallCoverImage.gif)
Fundamentals of Electric Circuits
Electrical Engineering
ISBN:
9780078028229
Author:
Charles K Alexander, Matthew Sadiku
Publisher:
McGraw-Hill Education
![Electric Circuits. (11th Edition)](https://www.bartleby.com/isbn_cover_images/9780134746968/9780134746968_smallCoverImage.gif)
Electric Circuits. (11th Edition)
Electrical Engineering
ISBN:
9780134746968
Author:
James W. Nilsson, Susan Riedel
Publisher:
PEARSON
![Engineering Electromagnetics](https://www.bartleby.com/isbn_cover_images/9780078028151/9780078028151_smallCoverImage.gif)
Engineering Electromagnetics
Electrical Engineering
ISBN:
9780078028151
Author:
Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher:
Mcgraw-hill Education,