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AND/OR Gates Determine Vo for the given applied input voltages. Si E1 E2 Vo E1 1 OV O V Si E2 o Vo O V 10 V D2 10 V OV R 1 ΚΩ 10 V 10 V OR gate - The output will be 1 if either or both inputs are 1 Positive logic OR gate

AND/OR Gates Determine Vo for the given applied input voltages. Si E1 E2 Vo E1 1 OV O V Si E2 o Vo O V 10 V D2 10 V OV R 1 ΚΩ 10 V 10 V OR gate - The output will be 1 if either or both inputs are 1 Positive logic OR gate

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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I've already asked this question, and the answer is wrong. Use the formula given to solve for the output voltage. Subtract the voltage drop due to the diode.

AND/OR Gates Determine Vo for the given applied input voltages. Si E1 E2 Vo E1 DI O V O V Si E2 O V 10 V D2 10 V O V R ►1 kQ 10 V 10 V OR gate - The output will be 1 if either or both inputs are 1 Positive logic OR gate AND/OR Gates Determine Vo for the given applied input voltages. Si E1 E1 E2 Vo 1 D1 OV O V Si D2 10 V R21 kQ 10 V OV E 10 V 10 V 10 V For an AND gate, The output will be O if either one of the inputs is 0 Positive logic AND gate
Transcribed Image Text:AND/OR Gates Determine Vo for the given applied input voltages. Si E1 E2 Vo E1 DI O V O V Si E2 O V 10 V D2 10 V O V R ►1 kQ 10 V 10 V OR gate - The output will be 1 if either or both inputs are 1 Positive logic OR gate AND/OR Gates Determine Vo for the given applied input voltages. Si E1 E1 E2 Vo 1 D1 OV O V Si D2 10 V R21 kQ 10 V OV E 10 V 10 V 10 V For an AND gate, The output will be O if either one of the inputs is 0 Positive logic AND gate
Re-drawn network: AND/OR Gates 0.7V D1 Si + (1) E= 10 Vo 1 E 10 V E – 10 V D2 Si R '1 kQ 1 kQ (0) 0 Vo D2 O V R '1 kQ V, = E - VK = 9.3 V (a 1 level) OR gate - The output will be 1 if either or both Е - Vк 10V — 0.7 V = R 1 kΩ inputs are 1 Positive logic OR gate = 9.3 mA
Transcribed Image Text:Re-drawn network: AND/OR Gates 0.7V D1 Si + (1) E= 10 Vo 1 E 10 V E – 10 V D2 Si R '1 kQ 1 kQ (0) 0 Vo D2 O V R '1 kQ V, = E - VK = 9.3 V (a 1 level) OR gate - The output will be 1 if either or both Е - Vк 10V — 0.7 V = R 1 kΩ inputs are 1 Positive logic OR gate = 9.3 mA
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