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1₁-1A-1₂2=0 (3.64) However, an ideal op amp draws negligible current (A=0) and, therefore, I₁=I₂ and the currents through the two resistors are equal. We can write expressions for I and I₂ by using Ohm's law and dividing the respective voltage drop by the resistance ein (1) - eA R₁ e₁ - eo R₂ The left-side term in Eq. (3.65) is current I, and the right-side term is current I₂. We can rewrite Eq. (3.65) as R₂(ein (1) - eA) = R₁ (e₁-eo) Rearranging Eq. (3.66) with the op-amp input voltage e, on the right-hand side, we obtain R₂ein (1) + R₁eo= (R₁ + R₂)еA or, solving for e R₂ R₁ + R₂ Next, substitute Eq. (3.68) into the amplifier gain equation (3.63) KR₂ R₁ + R₂ ein (1) R₁ www eo = K(eBeA) = + R₂ www 12 K ein (1) + eo R₁ R₁ + R₂ KR₁ R₁ + R₂, eo KR₁ R₁ + R₂ Note that eg = 0 as the positive op-amp input terminal in Fig. 3.13 is directly connected to the ground. Equation (3.69) is rearranged with all output voltage terms on the left-hand side to yield ∙eo KR₂ R₁ + R₂ in (1) (3.65) Figure 3.13 Op-amp circuit for Example 3.6. (3.66) (3.67) (3.68) (3.69) (3.70) The output voltage eo of the op amp shown in Fig. 3.12 is eo = K(eB - eA) (3.63)

1₁-1A-1₂2=0 (3.64) However, an ideal op amp draws negligible current (A=0) and, therefore, I₁=I₂ and the currents through the two resistors are equal. We can write expressions for I and I₂ by using Ohm's law and dividing the respective voltage drop by the resistance ein (1) - eA R₁ e₁ - eo R₂ The left-side term in Eq. (3.65) is current I, and the right-side term is current I₂. We can rewrite Eq. (3.65) as R₂(ein (1) - eA) = R₁ (e₁-eo) Rearranging Eq. (3.66) with the op-amp input voltage e, on the right-hand side, we obtain R₂ein (1) + R₁eo= (R₁ + R₂)еA or, solving for e R₂ R₁ + R₂ Next, substitute Eq. (3.68) into the amplifier gain equation (3.63) KR₂ R₁ + R₂ ein (1) R₁ www eo = K(eBeA) = + R₂ www 12 K ein (1) + eo R₁ R₁ + R₂ KR₁ R₁ + R₂, eo KR₁ R₁ + R₂ Note that eg = 0 as the positive op-amp input terminal in Fig. 3.13 is directly connected to the ground. Equation (3.69) is rearranged with all output voltage terms on the left-hand side to yield ∙eo KR₂ R₁ + R₂ in (1) (3.65) Figure 3.13 Op-amp circuit for Example 3.6. (3.66) (3.67) (3.68) (3.69) (3.70) The output voltage eo of the op amp shown in Fig. 3.12 is eo = K(eB - eA) (3.63)

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Need assistance with solving equation (3.69) to (3.70).  I'm not understanding why output voltage is outside of parenthesis in equation 3.70 and the # 1  is on inside of parenthesis? 

This is not part of a graded assignment, I'm reading through the chapters and solving examples. 

1₁-1A-1₂2=0 (3.64) However, an ideal op amp draws negligible current (A=0) and, therefore, I₁=I₂ and the currents through the two resistors are equal. We can write expressions for I and I₂ by using Ohm's law and dividing the respective voltage drop by the resistance ein (1) - eA R₁ e₁ - eo R₂ The left-side term in Eq. (3.65) is current I, and the right-side term is current I₂. We can rewrite Eq. (3.65) as R₂(ein (1) - eA) = R₁ (e₁-eo) Rearranging Eq. (3.66) with the op-amp input voltage e, on the right-hand side, we obtain R₂ein (1) + R₁eo= (R₁ + R₂)еA or, solving for e R₂ R₁ + R₂ Next, substitute Eq. (3.68) into the amplifier gain equation (3.63) KR₂ R₁ + R₂ ein (1) R₁ www eo = K(eBeA) = + R₂ www 12 K ein (1) + eo R₁ R₁ + R₂ KR₁ R₁ + R₂, eo KR₁ R₁ + R₂ Note that eg = 0 as the positive op-amp input terminal in Fig. 3.13 is directly connected to the ground. Equation (3.69) is rearranged with all output voltage terms on the left-hand side to yield ∙eo KR₂ R₁ + R₂ in (1) (3.65) Figure 3.13 Op-amp circuit for Example 3.6. (3.66) (3.67) (3.68) (3.69) (3.70)
Transcribed Image Text:1₁-1A-1₂2=0 (3.64) However, an ideal op amp draws negligible current (A=0) and, therefore, I₁=I₂ and the currents through the two resistors are equal. We can write expressions for I and I₂ by using Ohm's law and dividing the respective voltage drop by the resistance ein (1) - eA R₁ e₁ - eo R₂ The left-side term in Eq. (3.65) is current I, and the right-side term is current I₂. We can rewrite Eq. (3.65) as R₂(ein (1) - eA) = R₁ (e₁-eo) Rearranging Eq. (3.66) with the op-amp input voltage e, on the right-hand side, we obtain R₂ein (1) + R₁eo= (R₁ + R₂)еA or, solving for e R₂ R₁ + R₂ Next, substitute Eq. (3.68) into the amplifier gain equation (3.63) KR₂ R₁ + R₂ ein (1) R₁ www eo = K(eBeA) = + R₂ www 12 K ein (1) + eo R₁ R₁ + R₂ KR₁ R₁ + R₂, eo KR₁ R₁ + R₂ Note that eg = 0 as the positive op-amp input terminal in Fig. 3.13 is directly connected to the ground. Equation (3.69) is rearranged with all output voltage terms on the left-hand side to yield ∙eo KR₂ R₁ + R₂ in (1) (3.65) Figure 3.13 Op-amp circuit for Example 3.6. (3.66) (3.67) (3.68) (3.69) (3.70)
The output voltage eo of the op amp shown in Fig. 3.12 is eo = K(eB - eA) (3.63)
Transcribed Image Text:The output voltage eo of the op amp shown in Fig. 3.12 is eo = K(eB - eA) (3.63)
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