A Pragmatic Introduction to the Art of Electrical by Paul H. Dietz

By Paul H. Dietz

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Here, G = 1. This is often referred to as a unity gain buffer. You may have thought that there wasn’t much point to an amplifier with a gain of one, but if you examine the figure closely, you’ll notice that it has a dramatic effect. If the load resistor, Rl, had been placed across V1, a voltage divider would have A Pragmatic Introduction to the Art of Electrical Engineering 39 Guess the Number Rs Vs V1 GV1 Rl been formed, and the load would have seen only a fraction of the voltage Vs. With the buffer, the load is directly driven by the dependent source, bypassing the source impedance, Rs.

V out = A ( V + – V - ) = A ( V in – V out ) Collecting the Vout terms on the left yields: V out ( 1 + A ) = AV in This reduces to: AV in V out = ------------------ ≅ V in (1 + A) for sufficiently large gain, A. Thus, the circuit is a unity gain, noninverting, buffer that is independent of A, so long as A is large. Let’s consider another op amp circuit. Shown below is a more general noninverting amplifier. In this case, instead of feeding back the output to the negative input, we Vout = A (V+ - V-) V+ A V- Rf Vin Rs first reduce the output with a voltage divider, and then feed back the reduced voltage.

Before you can wire this up in the lab, you need to know a little bit more about the components. Typically, LEDs have two leads, one longer than the other, and one side of the case is flattened slightly. These are used to indicate which are the positive and negative (or, as they are called in the case of diodes, the anode and the cathode, respectively). Unfortunately, not all manufacturers have uniformly agreed upon which is which. Typically, the lead near the flattened side is the negative one.

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