Inverting and Non-Inverting Op Amps
Introduction: This article focuses on the concept of op amps. This article also covers op amps in inverting and non-inverting modes.
The Essentials
The operational amplifier is a circuit element that amplifies a given differential voltage input to a single voltage output. Figure 1. contains a schematic of the op amp element.
Figure 1: Op Amp Schematic
There are multiple nodes to consider here. \( V_A \) and \( V_B \) are the input nodes for the op amp. These are also referenced as the differential voltage nodes. The \( V_+ \) ans \( V_- \) nodes are the power supply for the op amp. The op amp, like any other electronic device, needs power to operate. These can also be considered as the range for which the op amp operates "linearly". \( V_O \) is the output voltage of the op amp based on the differential input.
Op Amps involve multiple circuit elements such as diodes and transistors, but that is really getting down to the nitty gritty and for the purpose of this class we do not need to get into that. We are mainly focused on three main features of the op amp; 1) the operation of the op amp, 2) the current through the input terminals, and 3) the voltage across the input terminals.
1.) Let's consider the operation of the op amp. With the op amp, we apply voltages \( V_+ \) and \( V_- \) to power the op amp. This can also be considered the rails. The following graph in Figure 2. describes the relationship of the output voltage to the input differential voltage.
Figure 2: Input Differential vs. Output
There are three portions to notice here. First is the negative saturation section. This shows that from a given input voltage from negative infinity to \( - \frac{V_{CC}}{A} \), the output voltage will be \( -V_{CC} \) or \( V_- \). A similar situation happens when the input voltage is \( \frac{V_{CC}}{A} \). For all values of input from \( \frac{V_{CC}}{A} \) to infinity, the output voltage is \( V_{CC} \) or \( V_+ \). These two scenarios are described as the op amp operating in "saturation mode". The last region is one that we are most interested in, which is when the op amp is operating linearly. This linear relationship is shown in the middle of the graph, going from input voltage being \( - \frac{V_{CC}}{A} \) to \( \frac{V_{CC}}{A} \). In this middle region the output voltage increases linearly depending on its input differential voltage. Essentially, in quick summary, depending on the input differential voltage and the voltage railing values, the op amp may operate linearly or in saturation.
2.) The current through the op amp. Mentioned in the textbook, the ideal resistance, seen from the input terminals, in the op amp is infinite. Since that resistance is infinite, we are safe to assume that the current through the input terminals is 0. So,
3.) When the op amp is operating linearly, both of the input terminals are the same voltage. So,
These three are assumptions we make when it comes to analyzing a circuit with an op amp. While they are not the most accurate, the error is so small that it essentially becomes negligible.
An application of op amps in real life circuit is inverting and non-inverting amplifiers. Figure 3. shows an example of both inverting and non-inverting amplifiers.
Figure 3: Inverting (top) and Non-inverting (bottom) amplifier circuits
A general equation to describe the output voltage in an inverting amplifier circuit is the following;
In this equation, the fraction \( \frac{R_f}{R_s} \) is the gain of this inverting amplifier circuit.
And for a non-inverting amplifier circuit.
In this equation, the fraction \( \frac{R_s + R_f}{R_s} \) is the gain of its amplifier circuit.
Example
To help us understand the amplifiers, we'll do an example of one of each kind. Firstly, the inverting amplifier circuit. Figure 4. has an example of one such amplifier.
Figure 4: Inverting Amplifier Example
For the purpose of this circuit, let's say that the op amp is operating under \( V_+ = 100 \, \text{V} \) and \( V_- = -100 \, \text{V} \).Now using equation 3, we can solve for the gain and \( V_O \) of this circuit.
\[ V_O = -\frac{R_f}{R_s}V_s = -\frac{10000}{2000} \cdot 15 = -75 \, \text{V}\]
The gain of this circuit is shown as the fraction, \( \frac{10000}{2000} \), so the gain of the circuit is 5 V/V. So as the op amp is operating in linear mode, every volt presented into the op amp, 5 Volts will come out.
Let's do another example with non-inverting amplifiers. Figure 5. has an example of such an amplifier.
Figure 5: Non-Inverting Amplifier Example
Let's say for this circuit, the op amp is powered by \( V_{+/-} = \pm 50 \, \text{V} \). Note how the orientation of the op amp is reversed from before, where the voltage power source is attached to the positive terminal of the op amp. To find the gain and \( V_O \) in the circuit, let's use equation 4.
\[ V_O = \frac{R_f + R_s}{R_s}V_s = \frac{3000+9000}{3000}\cdot12 = 48 \, \text{V}\]
The gain, in this scenario, is 4 V/V. That's also given from the fraction \( \frac{3000+9000}{3000} \).
Practice
- Given that the op amp operates at \( V_{+/-} = \pm 50\, \text{V} \), find \( V_O \).
Figure 6: Circuit for practice problem #1.
- Given that the op amp operates at \( V_{+/-} = \pm 20 \, \text{V} \), find \( V_O \).
Figure 7: Circuit for practice problem #2.
- Given that the op amp operates at \( V_{+/-} = \pm 75 \, \text{V} \), find \( V_O \).
Figure 8: Circuit for practice problem #3.
- Given that the op amp operates at \( V_{+/-} = \pm 25 \, \text{V} \), find \( V_O \).
Figure 9: Circuit for practice problem #4.
Solutions:
Problem 1 Solution.) \( V_O = -41.2 \, \text{V} \). Using equation 3, we get the following:
\[ V_O = -\frac{7000}{1700}\cdot10 = -41.2 \, \text{V}\]
Problem 2 Solution.) \( V_O = -20 \, \text{V} \). With two resistors being used in parallel for our effective \( R_f \), so find the resistance equivalent to the two resistors in parallel. Then using equation 3, we get the following:
\[ V_O = -\frac{3878.98}{1700}\cdot10 = -22.82 \, \text{V}\]
Since the value of \( V_O \) exceeds the railing specified in the beginning of the problem, this op amp is operating in saturation mode, so the final \( V_O = -20 \, \text{V} \).
Problem 3 Solution.)\( V_O = 60 \, \text{V} \). Using equation 4,
\[ V_O = \frac{10000+5000}{5000}\cdot20 = 60 \, \text{V}\]
Problem 4 Solution.)\( V_O = 25 \, \text{V} \). With resistors in series and parallel, we can use simplify the resistors into single \( R_s \) and \( R_f \) values. Then using equation 4
\[ V_O = \frac{10600+3333.33}{10600}\cdot20 = 26.29 \, \text{V}\]
Since this op amp is operating at \( \pm 25 \), this op amp is saturated at 25 V, so that is our output voltage.