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. 

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 

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 

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. 

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 V_{out}. By inspection we can see that V_{out} = 10000(V_{in}), therefore all we need to find is V_{in}. 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 V_{50} = 1V[50/(50+100)] V_{50} = 1V(1/3) = 0.333V V = V_{50} = 0.333V V_{in} = (V+  V) = 1V  0.333V = 0.67V V_{out} = 10000V_{in} = 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. 

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) V_{1} + iR_{1} + iR_{2} + V_{out} = 0 V_{out} = V_{1}  i(R_{1} + R_{2}) Now we will do KVL starting at V_{in} going clockwise: +V_{in} + iR_{2} + AV_{in} = 0 Solving for i: i = [V_{in}(1 + A)]/R_{2} Since A = 10000 or more, then A+1=A Therefore i = [V_{in}*A]/R_{2} Now we plug i into the first equation: V_{out} = V_{1}  (V_{in}*A/R_{2})(R_{1} + R_{2}) V_{out} = V_{1} + (V_{in}*A)(R_{1}/R_{2} + 1) By Inspection we know AV_{in} = V_{out} V_{out} = V_{1} + (V_{out})(R_{1}/R_{2} + 1) V_{out} = V_{1} + V_{out}*R_{1}/R_{2} + V_{out} V_{out}  V_{out}  V_{out}*R_{1}/R_{2} = V_{1} (The first two terms cancel out) V_{out} = (R_{2}/R_{1})*V_{1} 

Note that V_{out} does not depend on A.
Below we show this new op amp model!

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. V_{in} = 0) and that I_{1} = 0 and I_{1} = 0 Since I_{1} = 0 then I_{R1} = I_{R2} By Inspection V is connected to the ground node, therefore V = 0 Since V+ = V then V+ = 0 KVL loop equation 1: V_{1} + I_{R1}R_{1} + V_{in} = 0 V_{1} = I_{R1}R_{1} I_{R1} = V_{1}/R_{1} KVL loop equation 2:  V_{in} + I_{R2}R_{2} + V_{out} = 0 But recall that I_{R1} = I_{R2}, therefore I_{R2} = V_{1}/R_{1} Plugging I_{R2} into the equation gives:  V_{in} + (V_{1}/R_{1})R_{2} + V_{out} = 0 Recall that V_{in} = 0, therefore the equation reduces to: (V_{1}/R_{1})R_{2} + V_{out} = 0 V_{out} =  V_{1}(R_{2}/R_{1})


Here we will analyze the circuit to find V_{out}/V_{1}: Since I_{2} = 0 then i through R_{1} equals i through R_{2} KVL loop equation 1: V_{1} + V_{in} + iR_{1} = 0 Recall that V_{in} = 0 Therefore V_{1} = iR_{1} i = V_{1}/R_{1} KVL loop equation 2: iR_{1}  iR_{2} + V_{out} = 0 i(R_{1} + R_{2}) + V_{out} = 0 V_{out} = i(R_{1} + R_{2}) Now we can plug in i = V_{1}/R_{1} to get: V_{out} = V_{1}/R_{1}(R_{1} + R_{2}) V_{out} = V_{1}(1 + R_{2}/R_{1})


Here we will analyze the circuit to find V_{out}/V_{1}: Here i = 0 because I_{2} = 0 V+ = V_{1} because they are the same node Since V+ = V, the V = V_{1} But V_{out} = V because they are the same node Therefore V_{out} = V = V_{1} V_{out} = V_{1} 