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:

- Current mirrors/current source (red)
- Differential amplifier (blue)
- Class A gain stage (magenta)
- Voltage level shifter (green);
- output stage (cyan)

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.

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 .

- Find short-circuit output current: Applying KCL to the
output node we get

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:

- open-circuit voltage gain: