Op amp circuit analysis - System Analysis and Control

In the previous section, we learned how to analyze simple op amp circuits using realistic or idealized op amp models. In this section, we will build on that knowledge to analyze more complicated op amp circuits.

Ideal op amp circuit analysis¶

All the circuits in this section use negative feedback, so we assume the ideal op amp assumptions hold. As a reminder, the op amp has three terminal:

Figure 1:Ideal op amp with input and output terminals labeled.

The ideal op amp assumptions are:

\boxed{\begin{aligned} I_\textsf{in} &= 0 &&\textsf{(no input current)} \\ V_+ &= V_- &&\textsf{(equal input voltages)} \end{aligned}}

(1)

We wrote the currents and voltages with upper-case letters because we will be using the impedance method to analyze circuits from now on.

Let’s take a look at some common op amp circuits, starting with amplifiers.

Example: non-inverting amplifier¶

A non-inverting amplifier circuit is shown below.

To analyze the circuit, we first label the nodes. Let the voltage at the inverting input terminal be $V_A$ , and let the current through resistors $R_1$ and $R_2$ be $I$ (the same current flows through both resistors since no current flows into the op amp input terminal). We now write the constitutive equations for the circuit elements and for the op amp:

\begin{aligned} V_A &= R_1 I &&\textsf{(Resistor $R_1$)} \\ V_\textsf{out}-V_A &= R_2 I &&\textsf{(Resistor $R_2$)} \\ V_A &= V_\textsf{in} &&\textsf{(ideal op amp assumption)} \end{aligned}

(2)

We have three equations and three unknowns: $V_A$ , $I$ , and $V_\textsf{out}$ . We can first eliminate $V_A$ by substituting $V_\textsf{in}$ for $V_A$ in the first two equations:

\begin{aligned} V_\textsf{in} &= R_1 I\\ V_\textsf{out}-V_\textsf{in} &= R_2 I \end{aligned}

(3)

Next, we can solve the first equation for $I$ and substitute it into the second equation:

\begin{aligned} V_\textsf{out}-V_\textsf{in} &= \frac{R_2}{R_1} V_\textsf{in} \end{aligned}

(4)

Finally, we can solve for $V_\textsf{out}$ :

\boxed{V_\textsf{out} = \left(1 + \frac{R_2}{R_1}\right) V_\textsf{in}}

(5)

This circuit amplifies the input voltage by a factor of $1 + \frac{R_2}{R_1}$ . Of course, we remain limited by the op amp’s supply voltages, so the output voltage cannot exceed those limits.

Picking $R_1$ and $R_2$ appropriately allows us to set the gain of the amplifier to any desired value greater than or equal to 1. If we want a gain between 0 and 1, we can use a buffered voltage divider circuit such as the one shown in the previous section in Figure 8 (right) but with two different resistor values.

Example: inverting amplifier¶

If we want to amplify and invert a signal, we can use an inverting amplifier circuit such as the circuit shown below.

We can analyze this circuit using the same approach as before. Label the voltages and currents as before and write the constitutive equations:

\begin{aligned} V_\textsf{in} - V_A &= R_1 I &&\textsf{(Resistor $R_1$)} \\ V_A - V_\textsf{out} &= R_2 I &&\textsf{(Resistor $R_2$)} \\ V_A &= V_+ &&\textsf{(ideal op amp assumption)} \\ V_+ &= 0 &&\textsf{(non-inverting terminal grounded)} \end{aligned}

(6)

The last two equations force $V_A=0$ . In such situations, we call node $A$ a virtual ground. This is different from simply grounding node $A$ , since here we can do it without drawing any current from ground. Eliminating $I$ as before and solving for $V_\textsf{out}$ gives:

\boxed{V_\textsf{out} = -\frac{R_2}{R_1} V_\textsf{in}}

(7)

This time, picking $R_1$ and $R_2$ appropriately allows us to set the gain of the amplifier to any negative value, again subject to the limits imposed by the op amp’s supply voltages.

Amplifiers with impedances

Although the circuits of Figure 2 and Figure 3 use resistors, we can replace the resistors with arbitrary impedances (e.g., capacitors, inductors, or combinations of these elements) and analyze the circuits in the same way. In other words, the formulas for the output voltage still hold if we replace $R_1$ and $R_2$ with impedances $Z_1$ and $Z_2$ :

\begin{aligned} V_\textsf{out} &= \left(1 + \frac{Z_2}{Z_1}\right) V_\textsf{in} &&\textsf{(non-inverting amplifier)} \\ V_\textsf{out} &= -\frac{Z_2}{Z_1} V_\textsf{in} &&\textsf{(inverting amplifier)} \end{aligned}

(8)

Example: more complex amplifier¶

We can also analyze more complex op amp circuits once we recognize the basic building blocks. Consider the following circuit.

Figure 4:More complex amplifier circuit.

Our goal is twofold:

Find an expression relating the output voltage $V_\textsf{out}$ to the input voltage $V_\textsf{in}$
Interpret the behavior of the circuit when $R_2$ is disconnected.

Solution: To analyze this circuit, we could apply our usual approach: label all the node voltages and branch currents, write the constitutive equations for each element, write KCL at all junctions, and solve the resulting system of equations.

However, we can simplify our work by making some observations.

The resistor $R_g$ can be ignored because no current flows into the non-inverting input terminal, so no current flows through $R_g$ . Thus, there is no voltage drop across $R_g$ , and it’s as if the non-inverting input terminal is directly connected to ground.
We can now see that the rest of the circuit is just an inverting amplifier, where the feedback impedance is $(C \parallel R_2)$ and the input impedance is $R_1$ .

Combining these observations, we have the following equivalent circuit.

More complex amplifier of with aggregated impedances. — Figure 5:More complex amplifier of Figure 4 with aggregated impedances.

And now we can directly apply the inverting amplifier formula from (8):

\boxed{V_\textsf{out} = -\frac{Z_2}{Z_1} = -\left(\frac{\frac{1}{R_1}}{Cs + \frac{1}{R_2}}\right) V_\textsf{in}}

(9)

We can also write the formula above as a differential equation by multiplying it out and recognizing that multiplication by $s$ corresponds to differentiation in the time domain:

C\, \dot v_\textsf{out} + \frac{1}{R_2}v_\textsf{out} = -\frac{1}{R_1} v_\textsf{in}

(10)

Now to answer the second part of the question, we consider what happens when we disconnect $R_2$ . This is the same as taking $R_2 \to \infty$ , so the ODE simplifies to:

C\, \dot v_\textsf{out} = -\frac{1}{R_1} v_\textsf{in}

(11)

Or, written another way:

v_\textsf{out}(t) = - \frac{1}{R_1 C} \int_0^t v_\textsf{in}(\tau) \, \dd\tau

(12)

In other words, this circuit is an inverting integrator!

Cascaded op amp circuits¶

We can also analyze circuits with multiple op amps connected in series (cascaded). Consider the following circuit.

The first step is recognizing that each op amp stage is just an inverting amplifier with impedances. Thus, we can analyze each stage separately using the formula from (8). Let’s start by aggregating the impedances in each stage. We will also label the intermediate voltage between the two stages as $V_A$ .

Cascaded op amp circuit of with aggregated impedances. — Figure 7:Cascaded op amp circuit of Figure 6 with aggregated impedances.

Since we have two inverting amplifier stages, we can write:

\begin{aligned} V_A &= -\frac{Z_2}{Z_1} V_\textsf{in} &&\textsf{(first stage)} \\ V_\textsf{out} &= -\frac{Z_4}{Z_3} V_A &&\textsf{(second stage)} \end{aligned}

(13)

Where the impedances are:

\begin{aligned} Z_1 &= \frac{1}{C_1 s + \frac{1}{R_1}}, & Z_2 &= \frac{1}{C_2 s} + R_2, & Z_3 &= R_3, & Z_4 &= R_4 \end{aligned}

(14)

Substituting the first stage equation into the second stage equation gives:

V_\textsf{out} = \left(-\frac{Z_4}{Z_3}\right) \left(-\frac{Z_2}{Z_1} \right)V_\textsf{in} \\ = \frac{\left(\frac{1}{C_2 s} + R_2\right) R_4}{\left(\frac{1}{C_1 s + \frac{1}{R_1}}\right) R_3} V_\textsf{in}

(15)

which simplifies to:

\boxed{V_\textsf{out} = \frac{R_4 (R_1C_1 s+1) (R_2C_2 s+1)}{ R_1 R_3 C_2s} V_\textsf{in}}

(16)

Another way to write this is in terms of the transfer function from input to output:

\frac{V_\textsf{out}}{V_\textsf{in}} = \left(\frac{V_A}{V_\textsf{in}}\right) \left(\frac{V_\textsf{out}}{V_A}\right)

(17)

When we know the transfer function for each stage of the cascade, we can multiply them together to get the overall transfer function.

Cascaded circuits and loading effects

When analyzing cascaded circuits, we can multiply transfer functions from each stage to get the overall transfer function if the act of connecting the stages does not change either stage’s transfer function. In other words, the stages do not load previous stages.

For example, consider a simple voltage divider. We can view this as a system with input voltage $V_\textsf{in}$ and output voltage $V_\textsf{out}$ , with the following circuit diagram:

Figure 8:Voltage divider circuit.

If we just look at voltages (assume nothing is connected to the output), we have:

\left.\begin{aligned} V_\textsf{in} - V_\textsf{out} &= R_1 I \\ V_\textsf{out} &= R_2 I \end{aligned}\;\;\right\} \implies \frac{V_\textsf{out}}{V_\textsf{in}} = \frac{R_2}{R_1 + R_2}

(18)

What if we connect two such stages together in a cascade?

Figure 9:Cascaded voltage divider circuit.

Again assuming nothing is connected to the output, we have:

\left.\begin{aligned} V_\textsf{in} - V_A &= R_1 I \\ V_A &= R_2 I_2 \\ I_A + I_2 &= I \\ V_A - V_\textsf{out} &= R_1 I_A \\ V_\textsf{out} &= R_2 I_A \end{aligned}\;\;\right\} \implies \frac{V_\textsf{out}}{V_\textsf{in}} = \frac{R_2^2}{R_1^2 +3R_1R_2 +R_2^2}

(19)

This is different from simply multiplying the individual transfer functions together:

\left(\frac{R_2}{R_1 + R_2}\right) \left(\frac{R_2}{R_1 + R_2}\right) = \frac{R_2^2}{R_1^2 + 2R_1R_2 + R_2^2}

(20)

What’s going on here? When we connect the two voltage dividers together, there is current flowing between the stages, which alters the voltage at node $A$ .

This phenomenon is called a loading effect. It doesn’t happen with op amp circuits because drawing additional current from the output of an op amp simply draws it from the internal op amp voltage source; it does so without changing any voltages.

For example, if we put a voltage follower (buffer) stage between the two voltage dividers, the op amp would prevent any current from flowing out of the first stage into the second stage, eliminating the loading effect and allowing us to multiply the transfer functions together as in Eq. (20).

Test your knowledge¶

Solution to Exercise 1 #

We begin by labeling any relevant node voltages and branch currents. Since no current can flow in or out of the op amp input terminals, the capacitor and $R_1$ share the same current, as do the two resistors $R_2$ and $R_3$ . Here is our labeled circuit:

Op amp circuit from with annotated currents . — Figure 11:Op amp circuit from Figure 10 with annotated currents .

Next, we write the constitutive equations for each element and for the op amp:

\begin{aligned} V_\textsf{in} - V_+ &= \frac{1}{Cs} I_1 &&\textsf{(capacitor $C$)} \\ V_+ &= R_1 I_1 &&\textsf{(resistor $R_1$)} \\ V_\textsf{out} - V_- &= R_2 I_2 &&\textsf{(resistor $R_2$)} \\ V_- &= R_3 I_2 &&\textsf{(resistor $R_3$)} \\ V_+ &= V_- &&\textsf{(ideal op amp)} \end{aligned}

(21)

We have five equations and five unknowns: $V_+$ , $V_-$ , $I_1$ , $I_2$ , and $V_\textsf{out}$ . We can first eliminate $V_+$ and $V_-$ by substituting the second and fourth into the remaining equations:

\begin{aligned} V_\textsf{in} &= \left( \frac{1}{Cs} + R_1 \right) I_1 \\ V_\textsf{out} &= \left( R_2 + R_3 \right) I_2 \\ R_1 I_1 &= R_3 I_2 \end{aligned}

(22)

Dividing the second equation by the first and substituting the third equation gives:

\frac{V_\textsf{out}}{V_\textsf{in}} = \frac{(R_2 + R_3) I_2}{\left( \frac{1}{Cs} + R_1 \right) I_1} = \frac{(R_2 + R_3)}{\left( \frac{1}{Cs} + R_1 \right)} \cdot \frac{R_1}{R_3}

(23)

Simplifying, we obtain our final answer:

\boxed{\frac{V_\textsf{out}}{V_\textsf{in}} = \frac{R_1 (R_2 + R_3) Cs}{R_3 (R_1 Cs + 1)}}

(24)

Although this circuit isn’t a standard amplifier configuration, its output $V_\textsf{out}$ is connected to the output of the op amp, so if we wanted to drive a load, we could do so without affecting the transfer function. In other words, if we cascaded several such stages, the overall transfer function would simply be the product of the individual transfer functions.