Outline: |
Dependent Sources |
Op Amps |
Behavior of Op Amps with Negative Feedback |
Basic Op Amp Circuits |
More analysis of Op Amp and Dependent Source Circuits |
This is a VCVS:This is a dependent voltage source. The voltage across the source depends on Vx. For example, if Vx=5V and u=3 then the voltage across the source is uVx or 15V. The value of u is unitless and is determined by how the dependent source is designed. |
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This is a CCVS:This too is a dependent voltage source. The voltage across the source depends on the current Ix. For example, if Ix=2A and u=10 then the voltage across the source is uIx or 20V. The value of u is determined by how the dependent source is designed. Since the dependent source puts out voltage, its value is always |
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This is a VCCS:This is a dependent current source. The current through the source depends on the voltage Vx. For example, if Vx=10V and g=0.5 then the current through the source is gIx or 5A. The value of g is determined by how the dependent source is designed. Since the dependent source puts out current, its value is always |
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This is a CCCS:This is a dependent current source. The current through the source depends on Ix. For example, if Ix=3A and B=2 then the current through the source is BIx or 6A. The value of B is unitless and is determined by how the dependent source is designed. |
This a simple Op Amp circuit: Using the Ideal op amp model, find Vout. The first step is to replace the Op Amp with the model above. |
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This is the same circuit with the Op Amp replaced with the Ideal Model: We now will do analysis using KVL/KCL and Ohm's Laws to find Vout. By inspection we can see that Vout = 10000(Vin), therefore all we need to find is Vin. Also by inspection we see that V+ = 1V because V+ is connected to the source. The 100 and 50 Ohm resistors are in series because they carry the same current (note that no current goes into the op amp at V+ or V-). Applying Voltage Division V50 = 1V[50/(50+100)] V50 = 1V(1/3) = 0.333V V- = V50 = 0.333V Vin = (V+ - V-) = 1V - 0.333V = 0.67V Vout = 10000Vin = 10000*(0.67V) = 6700V |
Here is an op amp circuit with negative feedback: We will first analyze this circuit using the Ideal Voltage Op Amp Model shown in the last section. |
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Here is the same circuit with the Op Amp replaced with the Ideal Model: Now we will perform analysis - KVL on the outside: (Note that the current in the two resistors is the same) -V1 + iR1 + iR2 + Vout = 0 Vout = V1 - i(R1 + R2) Now we will do KVL starting at Vin going clockwise: +Vin + iR2 + AVin = 0 Solving for i: i = [-Vin(1 + A)]/R2 Since A = 10000 or more, then A+1=A Therefore i = [-Vin*A]/R2 Now we plug i into the first equation: Vout = V1 - (-Vin*A/R2)(R1 + R2) Vout = V1 + (Vin*A)(R1/R2 + 1) By Inspection we know AVin = Vout Vout = V1 + (Vout)(R1/R2 + 1) Vout = V1 + Vout*R1/R2 + Vout Vout - Vout - Vout*R1/R2 = V1 (The first two terms cancel out) Vout = -(R2/R1)*V1 |
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Note that Vout does not depend on A.
Below we show this new op amp model!
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This circuit was analyzed in the last section using the Ideal Op Amp Model. Now we will analyze it with the other op amp model: First we assume that V+ = V- (i.e. Vin = 0) and that I1 = 0 and I1 = 0 Since I1 = 0 then IR1 = IR2 By Inspection V- is connected to the ground node, therefore V- = 0 Since V+ = V- then V+ = 0 KVL loop equation 1: -V1 + IR1R1 + Vin = 0 V1 = IR1R1 IR1 = V1/R1 KVL loop equation 2: - Vin + IR2R2 + Vout = 0 But recall that IR1 = IR2, therefore IR2 = V1/R1 Plugging IR2 into the equation gives: - Vin + (V1/R1)R2 + Vout = 0 Recall that Vin = 0, therefore the equation reduces to: (V1/R1)R2 + Vout = 0 Vout = - V1(R2/R1)
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Here we will analyze the circuit to find Vout/V1: Since I2 = 0 then i through R1 equals i through R2 KVL loop equation 1: -V1 + Vin + iR1 = 0 Recall that Vin = 0 Therefore V1 = iR1 i = V1/R1 KVL loop equation 2: -iR1 - iR2 + Vout = 0 -i(R1 + R2) + Vout = 0 Vout = i(R1 + R2) Now we can plug in i = V1/R1 to get: Vout = V1/R1(R1 + R2) Vout = V1(1 + R2/R1)
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Here we will analyze the circuit to find Vout/V1: Here i = 0 because I2 = 0 V+ = V1 because they are the same node Since V+ = V-, the V- = V1 But Vout = V- because they are the same node Therefore Vout = V- = V1 Vout = V1 |