An Op Amp Gain Bandwidth Product

I can see some chat on internet about the operational amplifier gain bandwidth product. People are interested in having a better understanding of this parameter, as it appears in any op amp datasheet and it is used in many articles and books. In this article I will describe this parameter and show you an example with Analog Devices’ ADA4004, which is a precision amplifier.

The Gain Bandwidth Product describes the op amp gain behavior with frequency. Manufacturers insert a dominant pole in the op amp frequency response, so that the output voltage versus frequency is predictable. Why do they do that? Because the operational amplifier, which is grown on a silicon die, has many active components, each one with its own cutoff frequency and frequency response. Because of that, the operational amplifier frequency response would be random, with poles and zeros which would differ from op amp to op amp even in the same family. As a consequence, manufacturers thought of introducing a dominant pole in the schematic, so that the op amp response becomes more predictable. It is a way of “standardizing” the op amp frequency response. At the same time, it makes the op amp more user friendly, because its stability in a schematic becomes more predictable.

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A Summing and Differential Amplifier with One Op Amp

In a comment, one of my readers asked me what is the transfer function of the non-inverting summing amplifier in Figure 1, when R3 is connected to a reference voltage instead of ground.  Well, this is a summing amplifier with a differential configuration.

Figure 1

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The Virtual Ground

In my articles I talked about the op amp virtual ground and sometimes I wrote a brief explanation of this concept. In this article I will show you why an op amp input can be considered at a zero potential, without being galvanically connected to ground. Let’s take a simple circuit, the inverting amplifier.

inverting_amplifier_1Figure 1

In MasteringElectronicsDesign.com : How to Derive the Inverting Amplifier Transfer Function I showed the proof of its formula by using the virtual ground. The inverting input is at a zero potential, therefore virtual ground, which is a direct consequence of the feedback provided by R2 and the op amp high gain. Let’s see why.

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