20.309:Recitation 092107: Difference between revisions
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=====KCL===== | =====KCL===== | ||
:<math> | :<math> | ||
\frac{V_i-V_x}{R_1} + \frac{V_o-V_x}{R_2} + C_1 s (V_o - V_x) = 0 | |||
</math> (1) | </math> (1) | ||
<br/> | <br/> | ||
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=====KCL===== | =====KCL===== | ||
:<math> | :<math> | ||
\frac{V_x-V_o}{R_2} - V_o C_2 s = 0 | |||
</math> (4) | </math> (4) | ||
<br/> | <br/> |
Revision as of 07:48, 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{ \frac{V_i-V_x}{R_1} + \frac{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_2 C_1 s (V_o - V_x) = 0 }[/math] (2)
Gather terms
- [math]\displaystyle{ R_2 V_i + (-R_2-R_1-R_1 R_2 C_1 s) V_x + (R_1 + R_1 R_2 C_1 s)V_o = 0 }[/math] (3)
Apply KCL at the [math]\displaystyle{ V_- }[/math] node
KCL
- [math]\displaystyle{ \frac{V_x-V_o}{R_2} - V_o C_2 s = 0 }[/math] (4)
Multiply by [math]\displaystyle{ R_2 }[/math]
- [math]\displaystyle{ V_x=V_o (1 + R_2 C_2 s) }[/math] (5)