To simplify the analysis of the op-amp circuits, we further make 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 ).
- Based on the assumptions that
is in the range between
the positive and negative voltage supplies (e.g., , the
*rails*) and , we can assume , i.e., . We can further assume if one of the two inputs is grounded, the other one is*virtually grounded*. Based on this virtual ground assumption, the analysis of all op-amp circuits is significantly simplified. - The bandwidth is large (), i.e., the property of the op-amp remain unchanged for all frequencies of interest.

**Voltage follower (buffer)**

As the output is the same as the input, why can't we replace this op-amp circuit by a piece of wire?**Inverter**Current into the op-amp is negligible, and . Applying KCL to the node of , we have

In general, and of the inverter can be replaced by two networks (with impedances and respectively) containing resistors and capacitors and the analysis of the circuit can be carried out easily in frequency domain:

This is a convenient way to design filters of various frequency characteristics.**Non-Inverting Amplifier**

**Summer-inverter**Apply KCL to :

**Differential amplifier**We first define , and then apply KCL to both and to get:

and

Substituting the second to the first we get:

If one of the two inputs, e.g., is connected to a constant voltage treated as a reference voltage, then the differential amplifier can also be used as a level shifter. The output is times the input , shifted by a constant value

Alternatively, if , the output is some scaled version of shifted by

Consider some special cases:

- If and , then we get

- If (open circuit, and can be any value), then ,
and the circuit is a combination of inverter and a non-inverter amplifiers:

- If
, then , the circuit becomes the
follower:

- If and , then the circuit becomes the inverter:

- If , , and , then the circuit becomes the
non-inverter:

It is likely that both inputs are subjected to some common noise (e.g., the interference of 60Hz power supply):

In this case the output is

not affected by the common noise at all, i.e., the differential amplifier can suppress*common-mode signal*(e.g., the noise signal ) while amplify the*differential-mode signal*(e.g., and ).- If and , then we get
**Instrumentation Amplifier**The main drawback of the differential amplifier is that its input impedance () may not be high enough if the output impedance of the source is high. To overcome this problem, two non-inverting amplifiers with high input resistance are used each for one of the two inputs to the differential amplifier. The resulting circuit is called the instrumentation amplifier.

The analysis of this circuit is very simple. As the output impedance resistance of the non-inverting amplifiers is very low, the output voltage of the source circuit is not affected by the load circuit, and the input voltage received by the load circuit is the same as the open-circuit output voltage produced by the source circuit. Therefore the outputs of the two non-inverters are:

The output voltage of the differential amplifier is:

Of course the two resistors can be combined to become , i.e., , then the output can be written as:

Alternatively, we consider the current going from to through , , and :

From the equation of the first two terms we get:

From the equation of the second two terms we get:

Using the equation of the differential amplifier above, we get the same result as above:

In the instrumentation circuit AD623, , (open-circuit), i.e., the circuit has a unit voltage gain. However, if an external resistor () is connected to the circuit, the gain can be greater up to 1000.**Algebraic summer (inputs of different signs)**It can be shown that (see here) the output is some algebraic sum of the inputs with both positive and negative coefficients:

**Square Wave converter**Without feedback, the output of an op-amp is . As is large, is saturated, equal to either the positive or the negative voltage supply, depending on whether or not is greater than . When an input of any waveform is compared with a reference voltage, the output is a square wave:

**A/D converter**These two possible outputs, positive and negative, can be treated as ``1'' and ``0'' of the binary system. The figure shows an A/D converter built by three op-amps to measure voltage from 0 to 3 volts with resolution 1 V.

Due to the voltage divider, the input voltages to the three op-amps are, respectively, 2.5V, 1.5V and 0.5V. The output of these op-amps are listed below for each of the input voltage levels. A digital logic circuit (a decoder) can convert the 3-bit output of the op-amps to the 2-bit binary representation.

**Integrator and differentiator****Integrator**In time domain, as and , we have (KCL)

where . In frequency domain, we have:

**Differentiator**If we swap the resistor and the capacitor, we get in time domain:

In frequency domain, we have:

**PID controller**A proportional-integral-derivative (PID) controller can be implemented as shown. The output of the circuit is a linear combination of the signal together with its integral and derivative:

**Howland Current Source**Assuming , we can show that the output current through the load is a constant determined by the input voltages and , as well as the circuit parameters (see here):

**Logarithmic and Exponential Amplifiers**Based on the relationship between the current through and voltage across a diode and the virtual ground assumption, we can show that the output voltage of the exponential amplifier (left) is approximately an exponential function of the input voltage, and the output voltage of the logarithmic amplifier (right) is approximately a logarithmic function of the input voltage:

(Homework: Derive this relationship and determine the coefficients and .)