Operational Amplifier - Operation

Operation

The amplifier's differential inputs consist of a V+ input and a V input, and ideally the op-amp amplifies only the difference in voltage between the two, which is called the differential input voltage. The output voltage of the op-amp is given by the equation:

where V+ is the voltage at the non-inverting terminal, V is the voltage at the inverting terminal and AOL is the open-loop gain of the amplifier (the term "open-loop" refers to the absence of a feedback loop from the output to the input).

The magnitude of AOL is typically very large—100,000 or more for integrated circuit op-amps—and therefore even a quite small difference between V+ and V drives the amplifier output nearly to the supply voltage. Situations in which the output voltage is equal to or greater than the supply voltage are referred to as saturation of the amplifier. The magnitude of AOL is not well controlled by the manufacturing process, and so it is impractical to use an operational amplifier as a stand-alone differential amplifier. Without negative feedback, and perhaps with positive feedback for regeneration, an op-amp acts as a comparator. If the inverting input is held at ground (0 V) directly or by a resistor, and the input voltage Vin applied to the non-inverting input is positive, the output will be maximum positive; if Vin is negative, the output will be maximum negative. Since there is no feedback from the output to either input, this is an open loop circuit acting as a comparator. The circuit's gain is just the AOL of the op-amp.

If predictable operation is desired, negative feedback is used, by applying a portion of the output voltage to the inverting input. The closed loop feedback greatly reduces the gain of the amplifier. When negative feedback is used, the circuit's overall gain and response becomes determined mostly by the feedback network rather than by the op-amp itself. If the feedback network is made of components with relatively constant, stable values, the variability of the op-amp's open loop response does not seriously affect the circuit's performance. The response of the op-amp circuit with its input, output and feedback circuits to an input is characterized mathematically by a transfer function. Designing an op-amp circuit to have a desired transfer function is in the realm of electrical engineering. The transfer functions are important in most applications of op-amps, such as in analog computers. High input impedance at the input terminals and low output impedance at the output terminal(s) are particularly useful features of an op-amp.

For example, in a non-inverting amplifier (see the figure on the right) adding a negative feedback via the voltage divider Rf, Rg reduces the gain. Equilibrium will be established when Vout is just sufficient to reach around and "pull" the inverting input to the same voltage as Vin. The voltage gain of the entire circuit is determined by 1 + Rf/Rg. As a simple example, if Vin = 1 V and Rf = Rg, Vout will be 2 V, the amount required to keep V at 1 V. Because of the feedback provided by Rf, Rg this is a closed loop circuit. Its overall gain Vout / Vin is called the closed-loop gain ACL. Because the feedback is negative, in this case ACL is less than the AOL of the op-amp.

Another way of looking at it is to make two relatively valid assumptions.
One, that when an op-amp is being operated in linear (not saturated) mode, the difference in voltage between the non-inverting (+) pin and the inverting (−) pin is so small as to be considered negligible.
The second assumption is that the input impedance at both (+) and (−) pins is extremely high (at least several megohms with modern op-amps).
Thus, when the circuit to the right is operated as a non-inverting linear amplifier, Vin will appear at the (+) and (−) pins and create a current i through Rg equal to Vin/Rg. Since Kirchoff's current law states that the same current must leave a node as enter it, and since the impedance into the (−) pin is near infinity, we can assume the overwhelming majority of the same current i travels through Rf, creating an output voltage equal to Vin + i × Rf. By combining terms, we can easily determine the gain of this particular type of circuit.

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