Operational Amplifiers

While real operational amplifiers often require a lot of complicated components to work, we can use a dependent-source model to approximately model the linear behavior of an op-amp:

Here, \(V_{\text{in}} = v_p - v_n\), and \(A\) is our open-loop gain.

To determine input impedance, we can:

1. Leave the output open.
2. Apply a test voltage, \(v_x\), to the input.
3. Measure the current, \(i_x\), going out of \(v_x\).
4. Compute \(R_i = \frac{v_x}{i_x}\).

Or, we can use the voltage divider:

\begin{align} v_p &= v_s\frac{R_i}{R_s+R_i} \notag \\ \frac{v_p}{v_s} &= \frac{R_i}{R_s+R_i} \notag \\ v_pR_s + v_pR_i &= v_sR_i \notag \\ (v_s-v_p)R_i &= v_pR_s \notag \end{align}

Then, the input impedance is:

\begin{align} \boxed{R_i = \frac{v_p}{v_s-v_p}R_s} \end{align}

To determine output impedance, we can:

1. Replace the input with a short to ground.
2. Attach a test voltage \(v_x\) to the output.
3. Measure the resulting current \(i_x\).
4. Compute \(R_o = \frac{v_x}{i_x}\).

Practically, in a real op-amp, we can make \(A\) very large, but not very precise. Our goal is to design an OPA circuit that is “immune” to the precise value of \(A\). Consider the following circuit that incorporates a negative feedback loop:

Here, \(x_o = Ax_i\), \(x_i = x_s-x_f\), and \(x_f = \beta x_0\). Then, we want to find the gain \(G = \frac{x_0}{x_s}\):

\begin{align} x_0 &= A(x_s-\beta x_o) \notag \\ \frac{x_0}{A} &= x_s-\beta x_0 \notag \\ \frac{x_0}{A}+\beta x_0 &= x_s \notag \\ \left(\frac{A\beta + 1}{A}\right)x_0 &= x_s \notag \end{align}

Then, the gain can be found by:

\begin{align} G = \frac{x_0}{x_s} = \frac{A}{1+A\beta} = \frac{1}{\frac{1}{A} + \beta} \approx \frac{1}{1 + \beta} \notag \end{align}

This approximation works for sufficiently large \(A\).

The following circuit is an inverting amplifier:

KCL tells us that \(i_1 + i_2 - i_n = 0\). Recall that for an ideal OPA, \(i_n=i_p=0\), which means that \(i_1+i_2=0\). Then, expanding \(i_1\) and \(i_2\) using voltages, we get:

\begin{align} \frac{v_i-v_n}{R_i} + \frac{v_o-v_n}{R_f} = 0 \notag \end{align}

Since \(v_p = 0\), we know that by virtual ground \(v_n=0\). Then,

\begin{align} \frac{v_i}{R_i} + \frac{v_o}{R_f} &= 0 \notag \\ \frac{v_o}{R_f} &= -\frac{v_i}{R_i} \notag \end{align}

Then, our gain is:

\begin{align} G = -\frac{R_f}{R_i} \notag \end{align}

Notice that since our gain is negative, the circuit flips the polarity of the input signal, which is why this circuit is called an inverting amplifier.

The following circuit is a non-inverting amplifier:

Recall that for an ideal op-amp, \(i_p=0\), which means that \(v_p=v_s\). Additionally, \(v_p=v_n\), which means that \(v_n=v_s\). By the voltage divider, we know that:

\begin{align} v_n = v_o\frac{R_2}{R_1+R_2} = v_s \notag \end{align}

Since gain \(G\) is \(\frac{v_o}{v_s}\), we have:

\begin{align} G = \frac{R_1+R_2}{R_2} = 1 + \frac{R_1}{R_2} \notag \end{align}

Notice that for this amplifier, this gain must be greater than one.

The following circuit is a unity-gain amplifier, which is also known as a buffer or follower:

Analyzing with an ideal op-amp, we realize that this ampilfier satisfies:

\begin{align} v_o=v_n=v_p=v_i \notag \end{align}

This circuit is often used as a buffer with cascaded filters. Consider the following circuit with a low-pass filter cascaded with a high-pass filter:

We want \(v_{\text{LPF}} = \frac{Z_{C1}}{R_1+Z_{C1}}v_i\). But, the later high-pass filter means that the math changes for every component that we add thereafter. We can solve this by adding a buffer between the two filters.

We can add two voltages \(v_1\) and \(v_2\) with a summing amplifier circuit:

Using superposition and considering each voltage source separately, we get the following two equations:

\begin{align} v_o &= -\frac{R_f}{R_1}v_1 \notag \\ v_o &= -\frac{R_f}{R_2}v_2 \notag \end{align}

Combining, we have:

\begin{align} v_o = -\frac{R_f}{R_1}v_1 - \frac{R_f}{R_2}v_2 \notag \end{align}

Realize that this builds off of the inverting amplifier. Specifically, if \(R_f = R_1 = R_2\), we have:

\begin{align} v_o = -(v_1 + v_2) \notag \end{align}

which is the inverted sum of the two voltages.

We can also build a differentiator circuit, which takes the derivative of a signal:

We know that since \(v_n=v_p=0\), we have:

\begin{align} \frac{v_n-v_o}{R_f} &= i_2 \notag \\ v_o &= -i_2R_f \notag \end{align}

Additionally, \(i_1=i_2 = C\frac{\text{d}v_i}{\text{d}t}\). Since \(v_o = -i_2R_f\), we have:

\begin{align} v_o = -R_fC\frac{\text{d}v_i}{\text{d}t} \notag \end{align}

We can also take the integral of a signal with the following circuit:

We know that \(v_n=v_p=0\). Then,

\begin{align} i_2 &= C\frac{\text{d}}{\text{d}t}(v_n-v_o) \notag \\ &= -C\frac{\text{d}v_o}{\text{d}t} \notag \end{align}

We also know that:

\begin{align} i_1 = \frac{v_i-v_n}{R} = \frac{v_i}{R} \notag \end{align}

Since \(i_1=i_2\), we have:

\begin{align} \frac{v_i}{R} &= -C\frac{\text{d}v_o}{\text{d}t} \notag \\ \frac{\text{d}v_o}{\text{d}t} &= -\frac{v_i}{RC} \notag \end{align}

Thus, we find that the output voltage is:

\begin{align} v_o = -\frac{1}{RC} \int v_i \text{ d}t \notag \end{align}

The instrumentation amplifier (INA) circuit is used to amplify the difference between two signals:

Using superposition, we get the following two equations:

\begin{align} v_o &= -\frac{R_4}{R_3}v_2 \notag \\ v_n &= v_p = \frac{R_2}{R_1+R_2}v_1 \notag \end{align}

We know that \(i_3=i_4\) and \(i_n=i_p\), so we have:

\begin{align} \frac{v_2-v_n}{R_3} = \frac{v_n - v_o}{R_4} \Rightarrow v_2 = \frac{R_3+R_4}{R_4}v_n - \frac{R_3}{R_4}v_o \notag \end{align}

But \(v_2 = 0\), so we have:

\begin{align} v_o &= \frac{R_3+R_4}{R_3} v_n \notag \\ &= \frac{R_3+R_4}{R_3} \cdot \frac{R_2}{R_1+R_2} v_1 \notag \end{align}

Combining superposition, we get:

\begin{align} v_o = \frac{R_3+R_4}{R_3} \cdot \frac{R_2}{R_1+R_2} v_1 - \frac{R_4}{R_3}v_2 \notag \end{align}

If we equate the coefficients, we find that if \(R_2R_3 = R_1R_4\), we get:

\begin{align} v_o = \frac{R_4}{R_3}(v_1-v_2) \notag \end{align}