20.309:Recitation 092107: Difference between revisions
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#Apply KCL at the <math>V_-</math> node | #Apply KCL at the <math>V_-</math> node | ||
#Solve one equation for <math>V_x</math> and substitute into the other | #Solve one equation for <math>V_x</math> and substitute into the other | ||
#Rewrite in the form of a transfer function <math>V_o / V_i</math> | #Rewrite the result in the form of a transfer function <math>V_o / V_i</math> | ||
===The gruesome details=== | ===The gruesome details=== | ||
==Apply the Golden Rules== | ====Apply the Golden Rules=== | ||
In the Sallen Key circuit, a wire connects <math>V_-</math> to <math>V_+</math>. Therefore, <math>V_- = V_+ = V_o</math> | In the Sallen Key circuit, a wire connects <math>V_-</math> to <math>V_+</math>. Therefore, <math>V_- = V_+ = V_o</math>. This will be a useful substitution when applying KCL. | ||
====Apply KCL at the <math>V_x</math> node | ====Apply KCL at the <math>V_x</math> node>==== | ||
=====KCL===== | |||
:<math> | :<math> | ||
(V_i-V_x)/{R_1} + (V_o-V_x)/R_2 + C_1 s (V_o - V_x) = 0 | (V_i-V_x)/{R_1} + (V_o-V_x)/R_2 + C_1 s (V_o - V_x) = 0 | ||
| Line 43: | Line 45: | ||
<br/> | <br/> | ||
=====Multiply by <math>R_1 R_2</math>===== | |||
:<math> | :<math> | ||
R_2(V_i-V_x) + R_1(V_o-V_x) + R_1 R_2C_1 s (V_o - V_x) = 0 | R_2(V_i-V_x) + R_1(V_o-V_x) + R_1 R_2C_1 s (V_o - V_x) = 0 | ||
| Line 48: | Line 51: | ||
<br/> | <br/> | ||
====Apply KCL at the | =====Gather terms===== | ||
:<math> | |||
R_2(V_i-V_x) + R_1(V_o-V_x) + R_1 R_2C_1 s (V_o - V_x) = 0 | |||
</math> (3) | |||
<br/> | |||
====Apply KCL at the <math>V_-</math> node==== | |||
=====KCL===== | |||
:<math> | :<math> | ||
(V_x-V_o)/R_2 - V_o C_2 s = 0 | (V_x-V_o)/R_2 - V_o C_2 s = 0 | ||
</math> | </math> (3) | ||
<br/> | <br/> | ||
=====Multiply by <math>R_2</math>===== | |||
:<math> | :<math> | ||
V_x=V_o (1 + R_2 C_2 s) | V_x=V_o (1 + R_2 C_2 s) | ||
</math> | </math> (4) | ||
<br/> | |||
====Substitute | ====Substitute for <math>V_x</math> (equation 4 into equation 2)==== | ||
</div> | </div> | ||
Revision as of 07:44, 22 September 2007
Review of ideal circuit elements
Linear passive
Independent sources
Dependent sources
Nonlinear
Modeling real components with ideal elements
Modeling a battery
Modeling an op amp
Sallen Key circuit
Approach to solving the Sallen Key circuit
- Apply the Golden Rules
- Apply KCL at the [math]\displaystyle{ V_x }[/math] node
- Apply KCL at the [math]\displaystyle{ V_- }[/math] node
- Solve one equation for [math]\displaystyle{ V_x }[/math] and substitute into the other
- Rewrite the result in the form of a transfer function [math]\displaystyle{ V_o / V_i }[/math]
The gruesome details
=Apply the Golden Rules
In the Sallen Key circuit, a wire connects [math]\displaystyle{ V_- }[/math] to [math]\displaystyle{ V_+ }[/math]. Therefore, [math]\displaystyle{ V_- = V_+ = V_o }[/math]. This will be a useful substitution when applying KCL.
Apply KCL at the [math]\displaystyle{ V_x }[/math] node>
KCL
- [math]\displaystyle{ (V_i-V_x)/{R_1} + (V_o-V_x)/R_2 + C_1 s (V_o - V_x) = 0 }[/math] (1)
Multiply by [math]\displaystyle{ R_1 R_2 }[/math]
- [math]\displaystyle{ R_2(V_i-V_x) + R_1(V_o-V_x) + R_1 R_2C_1 s (V_o - V_x) = 0 }[/math] (2)
Gather terms
- [math]\displaystyle{ R_2(V_i-V_x) + R_1(V_o-V_x) + R_1 R_2C_1 s (V_o - V_x) = 0 }[/math] (3)
Apply KCL at the [math]\displaystyle{ V_- }[/math] node
KCL
- [math]\displaystyle{ (V_x-V_o)/R_2 - V_o C_2 s = 0 }[/math] (3)
Multiply by [math]\displaystyle{ R_2 }[/math]
- [math]\displaystyle{ V_x=V_o (1 + R_2 C_2 s) }[/math] (4)
