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Consider the non-inverting amplifier of Figure 2. e(t) Oy(t) O r(t) R, y(t), R1 Figure 2: Non-Inverting Amplifier The resistors values are: R, =1 [KQ]; R, = 4 [KQ]. As seen in Electronics I, the gain of this amplifier is y(t) R, + R, = 5 r(t) R, The equations that govern the behavior of this circuit are: y(1) = Ae(t) e(t) = r(t)– y,(t) R, (Voltage Divider Rule) R, + R, y,(1) = By(t) where ß=- Notice that the Voltage Divider Rule is “perfect", as there in no bleeding current at the junction R,,R, , since the Op Amp does not take any current at the -input (and neither at the + input). a) Express the circuit governing equations in block diagram form. This will make apparent that the circuit has feedback b) Using A = 100, and the CL gain equation y(t) find the exact r(t) value of the output voltage when the input voltage is r(t) =1 [V] c) Obtain the gain that relates r(t) to e(t), that is, express e(t) =(GAIN) r(t). Y(1) -, together with the CL gain equation y(t) ). r(t) (Hint: Use the “trick" : e(t)= A Find the error when the input voltage is r(t) =1[V] d) What happens to the gains y(t) e(t) and when A→0? Remember that we r(t) implement A by using non-linear semiconductors. Therefore, the resulting A will not be a perfect constant. What is the idea behind making (even the smallest value of) A as large r(t) as possible?

Consider the non-inverting amplifier of Figure 2. e(t) Oy(t) O r(t) R, y(t), R1 Figure 2: Non-Inverting Amplifier The resistors values are: R, =1 [KQ]; R, = 4 [KQ]. As seen in Electronics I, the gain of this amplifier is y(t) R, + R, = 5 r(t) R, The equations that govern the behavior of this circuit are: y(1) = Ae(t) e(t) = r(t)– y,(t) R, (Voltage Divider Rule) R, + R, y,(1) = By(t) where ß=- Notice that the Voltage Divider Rule is “perfect", as there in no bleeding current at the junction R,,R, , since the Op Amp does not take any current at the -input (and neither at the + input). a) Express the circuit governing equations in block diagram form. This will make apparent that the circuit has feedback b) Using A = 100, and the CL gain equation y(t) find the exact r(t) value of the output voltage when the input voltage is r(t) =1 [V] c) Obtain the gain that relates r(t) to e(t), that is, express e(t) =(GAIN) r(t). Y(1) -, together with the CL gain equation y(t) ). r(t) (Hint: Use the “trick" : e(t)= A Find the error when the input voltage is r(t) =1[V] d) What happens to the gains y(t) e(t) and when A→0? Remember that we r(t) implement A by using non-linear semiconductors. Therefore, the resulting A will not be a perfect constant. What is the idea behind making (even the smallest value of) A as large r(t) as possible?

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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Consider the non-inverting amplifier of Figure 2. e(t) Oy(t) r(t) R2 R, Figure 2: Non-Inverting Amplifier The resistors values are: R, =1 [KQ]; R, = 4 [KQ]. As seen in Electronics I, the gain of y(t) R, + R, this amplifier is -= 5 r(t) R, The equations that govern the behavior of this circuit are: y(1) = Ae(t) e(t) = r(t)– y,(t) y,(t) = By(t) where ß= R, (Voltage Divider Rule) R, + R, Notice that the Voltage Divider Rule is “perfect", as there in no bleeding current at the junction R, R, , since the Op Amp does not take any current at the -input (and neither at the + input). a) Express the circuit governing equations in block diagram form. This will make apparent that the circuit has feedback b) y(t) Using A = 100, and the CL gain equation find the exact r(t) value of the output voltage when the input voltage is r(t) =1 [V] c) Obtain the gain that relates r(t) to e(t), that is, express e(t) =(GAIN) r(t). y(t) ', together with the CL gain equation A (Hint: Use the “trick" : e(t) = r(t Find the error when the input voltage is r(t) =1 [V] d) What happens to the gains y(t) and r(t) e(t) when A→o ? Remember that we r(t) implement A by using non-linear semiconductors. Therefore, the resulting A will not be a perfect constant. What is the idea behind making (even the smallest value of) A as large as possible?
Transcribed Image Text:Consider the non-inverting amplifier of Figure 2. e(t) Oy(t) r(t) R2 R, Figure 2: Non-Inverting Amplifier The resistors values are: R, =1 [KQ]; R, = 4 [KQ]. As seen in Electronics I, the gain of y(t) R, + R, this amplifier is -= 5 r(t) R, The equations that govern the behavior of this circuit are: y(1) = Ae(t) e(t) = r(t)– y,(t) y,(t) = By(t) where ß= R, (Voltage Divider Rule) R, + R, Notice that the Voltage Divider Rule is “perfect", as there in no bleeding current at the junction R, R, , since the Op Amp does not take any current at the -input (and neither at the + input). a) Express the circuit governing equations in block diagram form. This will make apparent that the circuit has feedback b) y(t) Using A = 100, and the CL gain equation find the exact r(t) value of the output voltage when the input voltage is r(t) =1 [V] c) Obtain the gain that relates r(t) to e(t), that is, express e(t) =(GAIN) r(t). y(t) ', together with the CL gain equation A (Hint: Use the “trick" : e(t) = r(t Find the error when the input voltage is r(t) =1 [V] d) What happens to the gains y(t) and r(t) e(t) when A→o ? Remember that we r(t) implement A by using non-linear semiconductors. Therefore, the resulting A will not be a perfect constant. What is the idea behind making (even the smallest value of) A as large as possible?
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