20.309:Recitation 092107

20.309 Fall Semester 2007 Recitation Notes for 9/21/2007'''

Modeling real components

 * The ideal elements comprise a vocabulary of mathematical relationships that are useful for modeling real systems.
 * In many situations, a single ideal element models the behavior of a real component well.
 * At other times, such as high frequency, high load, or high gain, it may be necessary to create a more sophisticated model

Example: modeling a battery
Consider models for common AAA and D cell batteries.



AAA and D cell

Simple model: ideal voltage source

 * The voltage output of both the AAA and the D cell is 1.6 V
 * A simple model for both batteries is a voltage source with a value of 1.6V
 * This model works well with small loads
 * This model does not capture any difference between the two sizes



Simple Battery Model

Measuring output resistance

 * Output resistance can be estimated from open circuit voltage and short circuit current.

Better model: output resistance

 * Output resistance is simply add in series with the voltage source.



Battery Model including Output Resistance

Comparing the two models under heavy loading (0.2 &Omega;)

 * The improved model predicts the output voltage and current flow under a load of 0.2 Ohms
 * This model is much more accurate under heavy load conditions.

Op amp model

 * Just as with the battery, it is possible to make a more accurate op amp model by adding elements
 * Offset voltage, input bias current, input offset current can be modeled by adding simple souces to the ideal op amp
 * Unfortunately, these parameters are temperature dependent
 * Non-infinite gain and frequency dependence can be modeled by cascading a transfer function at the output of the op amp
 * Regardless of the connection, the product of gain and bandwidth is constant. This parameter is usually specified in the datasheet.



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, &omega;c, occurs where the magnitude of the transfer funcion falls to $$1/\sqrt{2}$$.
 * Above the cutoff frequency, the second order transfer function falls off twice as fast as the first order &mdash; 40 dB per decade versus 20.

Approach to solving the Sallen Key circuit

 * 1) Apply the Golden Rules
 * 2) Apply KCL at the $$V_x$$ node
 * 3) Apply KCL at the $$V_-$$ node
 * 4) Solve one equation for $$V_x$$ and substitute into the other
 * 5) Rewrite the result in the form of a transfer function $$V_o / V_i$$

ALthough the details are different, the same approach will work to solve the op amp question on Homework #1

Apply the Golden Rules
In the Sallen Key circuit, a wire connects $$V_-$$ to $$V_o$$. Therefore, $$V_- = V_+ = V_o$$. This will be a useful substitution when applying KCL.

KCL


\frac{V_i-V_x}{R_1} + \frac{V_o-V_x}{R_2} + C_1 s (V_o - V_x) = 0 $$     (1)

Multiply by $$R_1 R_2$$


R_2(V_i-V_x) + R_1(V_o-V_x) + R_1 R_2 C_1 s (V_o - V_x) = 0 $$     (2)

Gather terms


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 $$     (3)

KCL


\frac{V_x-V_o}{R_2} - V_o C_2 s = 0 $$     (4)

Multiply by $$R_2$$


V_x=V_o (1 + R_2 C_2 s) $$     (5)

Sustitute


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 $$     (6)

Multiply and collect terms


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 $$     (7)



V_i = \{ R_1 R_2 C_1 C_2 s^2 + C_2(R_1 + R_2) s + 1\}V_o $$     (8)

Rewrite as a transfer function


\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} $$     (9)

Comparison of first and second order transfer functions
Compare the Sallen Key transfer function with the first order low pass filter we considered in class:



\frac{V_o}{V_i} = \frac{1}{ R C s + 1} $$

Because its transfer function includes an $$s^2$$ 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);