Next: Analysis of Op-Amp Circuits Up: Chapter 5: Operational Amplifiers Previous: Chapter 5: Operational Amplifiers

## Operational Amplifier

The circuit schematic of the typical 741 op-amp is shown below:

A component-level diagram of the common 741 op-amp. Dotted lines outline:

Like all op-amps, the circuit basically consists of three stages:

• Differential amplifier with high input impedance that generates a voltage signal, the amplified voltage difference .
• Voltage amplifier (class A amplification) with a high voltage gain to further amplify the voltage.
• Output amplifier (class AB push-pull emitter follower) with low output impedance and high current driving capability.
The op-amp requires two voltage supplies of both polarities (typically V).

Although the op-amp circuit may look complicated, the analysis of its operation and behaviors can be simplfied based on the following assumptions:

• The huge input resistance can be treated as infinity .
• The input current drawn by an op-amp is small ( ), and could be approximated to be zero .
• The small output impedance can be treated as zero , i.e., the output is not affected by the load (so long as it is much greater than ).
• The bandwidth is large (), i.e., the property of the op-amp remain unchanged for all frequencies of interest.

Based on these approximations, an op-amp can be modeled in terms of the following three parameters:

• Input impedance : very large, typically a few mega-Ohms or higher ( , e.g., 741 ), depending on the frequency and specific components used (e.g., BJT or FET).

• Output resistance :, very small, typically a few tens of ohms, e.g., 75 .

• Open-circuit gain :, based on both the inverting input and the non-inverting input :

where is the differential-mode gain and is the common-mode gain. It is desired that and , i.e., the output is only proportional to the difference between the two inputs. The common-mode rejection ratio (CMRR) is defined as the ratio between differential-mode gain and common-mode gain:

Also, as the output is in the range between and and is large, is small (in the micro-volt range), i.e., . If, as in some op-amp circuits, is grounded, then is very close to zero, i.e., it is almost the same as ground, or virtual ground. The analysis of various op-amp circuits can be much simplified by this virtual ground assumption.

As is large, is usually saturated, equal to either or (called the rails''), depending on whether or not is greater than . For to be meaningful, some kind of negative feedback is needed. In the following, we consider some typical op-amp circuits to show how to analyze an Op-amp circuit to find its input resistance , output resistance , and open-circuit voltage gain .

• Voltage follower: The input is connected to the positive input while the output is directly connected to the negative input (100% negative feedback), as shown in (A) in the figure below. The op-amp can be modeled by its input impedance , output impedance and voltage gain , as shown in (B). Then the voltage follower can be modeled by its input impedance , output impedance , and voltage gain , as shown in (C).

Specifically, , and can be found below. Here the voltage source in the op-amp is .

• Input impedance : Applying KVL to the loop we get

Dividing both sides by we get the input impedance:

• Open-circuit voltage gain : The open-circuit output voltage is

The open-circuit gain is

The approximation is due to the fact that and . We therefore have

• Output impedance : With a short-circuit load, we have , and the short-circuit current can be found by superposition:

As we also know , we get the output impedance (Thevenin's model):

The approximation is due to the fact that

In summary, we see that the voltage follower has a unit voltage gain, but much increased input resistance (e.g., ) and much reduced output resistance (e.g., ). In practice we could simply assume and .

Example:

The figure on the left shows a circuit represented by an ideal voltage source in series with an internal resistance (Thevenin's theorem), with a load . The voltage delivered to the load by this non-ideal source is

We see that the output voltage across the load is only a fraction of the voltage due to the voltage drop across the internal resistance . If it is desired for the output voltage to be as close to the source as possible, the internal resistance needs to be small compared to the load resistance .

Next consider inserting a voltage follower (buffer), in between the source and the load, as shown in the middle figure. The follower is modeled by its input and output resistances and , as well as its voltage gain , as shown in the right figure. The output voltage can be obtained after two levels of voltage dividers:

We see that the output voltage across can be very close to source voltage, i.e., , due to the nature of the op-amp:

• Inverting Amplifier

As the analysis of the circuit using full model of the op-amp is very involved, certain approximation is made to simplify the analysis.

• Open-circuit voltage gain :

As and , we approximate and . Also, we have , and . Now we have

Solving for we get

The output voltage is:

Now we have the open-circuit voltage gain:

The approximation is due to the fact that .

• Input resistance:

We assume , and find the input resistance as the ratio of and the input current . By KCL applied to the node of :

Solving for :

The input current is

Dividing by we get

Note that this input resistance is significantly smaller than that of the voltage follower with !

• Output resistance: Here we assume to simplify the analysis.
• Find short-circuit output current: Applying KCL to the output node we get

but as , the above becomes

• Find open-circuit output voltage: Applying KCL to the node of we get

Solving for we get

The open-circuit output voltage can be found to be (voltage divider)

• Find output resistance:

The approximation is based on and .

In summary,

• open-circuit voltage gain:

• input resistance:

• output resistance:

• Non-Inverting Amplifier (Homework)

The three parameters of this non-inverting amplifier can be found to be (see here):

• open-circuit voltage gain:

• input resistance:

• output resistance:

Comparing these results with those of the voltage follower, we see that is a little better, but both and are a little worse. In particular if , this non-inverting amplifier becomes a voltage follower with , , and .

Next: Analysis of Op-Amp Circuits Up: Chapter 5: Operational Amplifiers Previous: Chapter 5: Operational Amplifiers
Ruye Wang 2019-07-07