Electronics Reference
Operational Amplifiers (Special-Purpose Circuits)
| Overall Closed-Loop Gain |  where R1 = R2 = R |
| External Gain-Setting Resistor |  |
| Output Voltage |  |
An instrumentation amplifier is basically composed of three internal op-amps and few internal resistors. It is a differential voltage-gain amplifier which has a primary function of amplifying small signals riding on a large common-mode voltages. An instrumentation amplifier amplifies the difference between the voltages present at its two inputs. To calculate the external gain-setting resistor, the overall closed-loop gain, and the output voltage of an instrumentation amplifier, you can use the equations above.
| Input Stage Gain |  |
| Output Stage Gain |  |
| Total Amplifier Gain |  |
An isolation amplifier is a device that can amplify an input signal but at the same time electrically isolates the input from the output. It uses an isolation barrier to put an isolation between the input and output stages. An isolation amplifier could use a capacitive, optical, or a transformer coupling for isolation. It is usually used in medical equipment and applications such as chemical processing, nuclear processing, and metal processing. The equations above can be used to determine the total gain of an isolation amplifier that uses transformer coupling as its isolation barrier.
Note: IR is a constant for a given diode
A log amplifier is an op-amp configuration in which its output voltage is proportional to the logarithm of the input voltage. It's usually used for signal compression applications. What produces the logarithmic response is the feedback element of the op-amp that demonstrates logarithmic characteristic. In this equation, the feedback element used is a diode and the output voltage can be calculated using the equation above.
Note: IEBO is the emitter-to-base leakage current
This circuit also functions as a log amplifier just like the log amplifier with a diode feedback element. However, the feedback element used here is a bipolar junction transistor (BJT). The base-emitter junction of a BJT responds similarly to a diode, so the analysis for this circuit is similar to the diode log amplifier. The equation above can be used to determine the output voltage.
While a log amplifier outputs a voltage that is proportional to the logarithm of the input voltage, the antilog amplifier acts like the inverse of log amplifier. In mathematics, you’ll get the antilogarithm of a number by raising the base of the logarithm to a power equal to the logarithm of that number. The antilogarithm amplifier operates that way. As you can see in the diagram, instead of putting the transistor in the feedback element like in the log amplifier, the transistor is placed at the input of the antilog amplifier. To calculate the output voltage, you can use the equation above.
An op-amp can be configured to be a constant-current source. As you can see in the circuit diagram, the op-amp will provide a constant current to the load even if the resistance of the load will vary. All of the input current is considered to flow to the load since the input impedance of the op-amp is considered to be infinite. To calculate the current that flows through the load, you can use the equation above.
Another application of an op-amp is converting current to voltage. An example of that is the circuit that you see above. It has a light sensing device that produces current which flows through the feedback resistor. The output voltage depends on the amount of current that flows from the input to the feedback resistor and the resistance of the feedback resistor. To calculate the output voltage you can simply use the equation above.
