20.309:Recitation 092107: Difference between revisions
From OpenWetWare
Jump to navigationJump to search
Line 28: | Line 28: | ||
*Because of the buffer amplifer at its output, the Sallen Key circuit has low output impedence. | *Because of the buffer amplifer at its output, the Sallen Key circuit has low output impedence. | ||
*The circuit implements a ''second order'' low pass filter. (There are also band and high pass versions.) | *The circuit implements a ''second order'' low pass filter. (There are also band and high pass versions.) | ||
*The cutoff frequency, <math>F_c</math>, is defined to be where the magnitude of the transfer funcion falls to <math>1/\sqrt{2}</math>. This occurs where the real and imaginary parts of the denominaor are equal. | *The cutoff frequency, <math>F_c</math>, is defined to be where the magnitude of the transfer funcion falls to <math>1/\sqrt{2}</math>. This occurs at the frequency where the real and imaginary parts of the denominaor are equal. | ||
*Above the cutoff frequency, the second order transfer function falls off twice as fast as the first order — 40 dB per decade versus 20. | *Above the cutoff frequency, the second order transfer function falls off twice as fast as the first order — 40 dB per decade versus 20. | ||
Revision as of 16:25, 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
Key concepts
- Even though the circuit is complicated, solving it requires only two techniques: the golden rules and Kirchoff' Current Law.
- Because of the buffer amplifer at its output, the Sallen Key circuit has low output impedence.
- The circuit implements a second order low pass filter. (There are also band and high pass versions.)
- The cutoff frequency, [math]\displaystyle{ F_c }[/math], is defined to be where the magnitude of the transfer funcion falls to [math]\displaystyle{ 1/\sqrt{2} }[/math]. This occurs at the frequency where the real and imaginary parts of the denominaor are equal.
- Above the cutoff frequency, the second order transfer function falls off twice as fast as the first order — 40 dB per decade versus 20.
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_o }[/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)
Substitute for [math]\displaystyle{ V_x }[/math] (equation 5 into equation 3)
Sustitute
- [math]\displaystyle{ R_2 V_i - (R_2 + R_1 + R_1 R_2 C_1 s) (1 + R_2 C_2 s) V_o + (R_1 + R_1 R_2 C_1 s)V_o = 0 }[/math] (6)
Multiply and collect terms
- [math]\displaystyle{ R_2 V_i + (-R_2 -R_1 - R_1 R_2 C_1 s - R_2^2 C_2 s - R_1 R_2 C_2 s - R_1 R_2^2 C_1 C_2 s^2 + R_1 + R_1 R_2 C_1 s)V_o = 0 }[/math] (7)
- [math]\displaystyle{ V_i = \{ R_1 R_2 C_1 C_2 s^2 + C_2(R_1 + R_2) s + 1\}V_o }[/math] (8)
Rewrite as a transfer function
- [math]\displaystyle{ \frac{V_o}{V_i} = \frac{1}{ R_1 R_2 C_1 C_2 s^2 + C_2(R_1 + R_2) s + 1} }[/math] (9)