Part 13 - Op-Amps and Filters

In this part we’ll cover circuit response, filters and amplifiers.

Transfer Function

Recall (Transfer function)

Recall from the last part that, the transfer function is defined as,

H(f) = \dfrac{V_{out}}{V_{in}}.

Let’s find the transfer function of this circuit:

Let’s find $V_{out}$ ,

\begin{align*} V_{out} & = I \cdot Z_C \newline & = \left(\dfrac{V_{in}}{Z_R + Z_C}\right) Z_C \newline & = \left(\dfrac{V_{in}}{R + \dfrac{1}{j2\pi fC}}\right) \left(\dfrac{1}{j2\pi fC}\right). \end{align*}

Which means that $H(f)$ ,

\begin{align*} H(f) & = \dfrac{V_{out}}{V_{in}} \newline & = \dfrac{\left(\dfrac{V_{in}}{R + \dfrac{1}{j2\pi fC}}\right) \left(\dfrac{1}{j2\pi fC}\right)}{V_{in}} \newline & = \cdots \newline & = \dfrac{1}{1 + j2\pi fRC}. \end{align*}

We can define $f_B$ as,

f_B = \dfrac{1}{2\pi RC}.

Which means,

H(f) = \dfrac{1}{1 + j \left( \dfrac{f}{f_B} \right)}.

Therefore, the magnitude of $H(f)$ is,

|H(f)| = \dfrac{1}{\sqrt{1 + \left( \dfrac{f}{f_B} \right)^2}}.

The angle,

\angle{H(f)} = -\arctan \left(\dfrac{f}{f_B} \right).

Decibels

We all have heard (yes, pun intended) of decibels. Let’s properly define what decibels are:

Definition 1 (Decibel)

|H(f)|_{\mathrm{dB}} = 20 \log_{10}(|H(f)|)

Amplifier

Is just as the name suggests,

V_{0}(t) = A_{v}V_{i}(t).

Note that, $A_{v}$ , can be of any sign, which means we also have inverting amplifiers!

A typical model would look something like:

Operational Amplifier (Op-Amp)

An Op-Amp looks like:

We use the Op-Amp like:

Definition 2 (Ideal Op-Amp)

An Ideal Op-Amp has:

Infinite input impedance
Infinite gain for differential signal
Zero gain for the common-mode signal
Zero output impedance
Infinite bandwidth

Differential signal,

V_d = V_1 - V_2.

Common-mode signal,

V_{cm} = \dfrac{1}{2} \left(V_1 + V_2 \right).

Let’s take a look at this Op-Amp circuit,

I_1 = I_2 = \dfrac{V_{in}}{R_1}.

KVL,

\begin{align*} 0 + I_2 R_2 + V_{o} & = 0, \newline I_2 = -\dfrac{V_{o}}{R_2} & = \dfrac{V_{in}}{R_1}. \end{align*}

Which means,

A_{v} = \dfrac{V_{o}}{V_{in}} = \boxed{-\dfrac{R_2}{R_1}}.

Let’s take a look at this Op-Amp:

Figure 6: Op-Amp with positive feedback.

Here we’ll quickly discover that we have positive feedback. Positive feedback saturates the output - which means we can not use the derived formulas from above!

Always check if there is positive or negative feedback.

For non-inverting amplifiers, we get,

A_{v} = \dfrac{V_{o}}{V_{in}} = \boxed{1 + \dfrac{R_2}{R_1}}.

So, steps to analyze an ideal Op-Amp circuit:

Algorithm 1: Analyzing an ideal Op-Amp circuit

Verify that negative feedback is present.
Assume that the voltage between the terminals and input current are forced to 0.
Apply standard circuit analysis principles (KCL, KVL, and Ohm’s Law).