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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)
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...
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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
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
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