20.309:Recitation 092107: Difference between revisions
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The magnitude of both functions is plotted below. Although they have similar cutoff frequencies, the Sallen Key function falls off twice as fast as the passive RC filter (40 dB/decade versus 20 dB/decade). | Because its transfer function includes an <math>s^2</math> term, the Sallen Key filter is called a second order filter. The magnitude of both functions is plotted below. Although they have similar cutoff frequencies, the Sallen Key function falls off twice as fast as the passive RC filter (40 dB/decade versus 20 dB/decade). | ||
[[Image:First Order vs Second Order Low Pass.jpg|700 px|center]] | [[Image:First Order vs Second Order Low Pass.jpg|700 px|center]] |
Revision as of 19:35, 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]
Use this same approach for the op amp question on Homework #1
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)
Sallen Key transfer function
Compare the Sallen Key transfer function with the first order low pass filter we considered in class:
- [math]\displaystyle{ \frac{V_o}{V_i} = \frac{1}{ R C s + 1} }[/math]
Because its transfer function includes an [math]\displaystyle{ s^2 }[/math] term, the Sallen Key filter is called a second order filter. The magnitude of both functions is plotted below. Although they have similar cutoff frequencies, the Sallen Key function falls off twice as fast as the passive RC filter (40 dB/decade versus 20 dB/decade).
Here is the Matlab code that generated the plot:
%Clear the plot axes clf; %Component values for Sallen Key circuit (gives Fc = 1) R1=1; R2=1; C1=1; C2=1; %Generate a vector for the frequency axis with exponential spacing w = -2:.01:2; w = 10 .^ w; s = Fc * w * j; % Sallen Key transfer function Vsk = 1 ./ (R1 * R2 * C1 * C2 * s .^2 + (R1 + R2) * C2 * s + 1); %Plot the magnitude of the transfer function on a log-log scale vs. %frequency loglog( abs(s), abs(Vsk), 'b', 'LineWidth', 3); %Maintain the current plot hold on %Adjust axes for a little headroom around the plot axis([ min(abs(s)), max(abs(s)), 1.2 * min(abs(Vsk)), 1.2 * max(abs(Vsk))]) %Component values for 1st order circuit R = 1; C = 1; % Single order low pass transfer function Vlp = 1 ./(R * C * s + 1); %Plot the magnitude of the transfer function on athe same axes loglog(abs(s), abs(Vlp), 'r', 'LineWidth', 3); %Add a chart title, legend, and axis labels title('First and Second Order Filter Responses', 'FontSize', 18) legend('Second Order', 'First Order', 'Location', 'SouthWest'); ylabel('|H(s)|', 'FontSize', 14); xlabel('Frequency', 'FontSize', 14);