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A short description of the workings of an op-amp

Op-amp

The op-amp has four terminals to which connections can be made. Inputs attach to nodes a and b , and the output is node c . As the circuit model on the right shows, the op-amp serves as an amplifier for the difference ofthe input node voltages.

Op-amps not only have the circuit model shown in [link] , but their element values are very special.

  • The input resistance , R in , is typically large , on the order of 1 MΩ.
  • The output resistance , R out , is small , usually less than 100 Ω.
  • The voltage gain , G , is large , exceeding 10 5 .
The large gain catches the eye; it suggests that an op-amp could turn a 1 mV input signal into a 100 V one. If you were to buildsuch a circuit--attaching a voltage source to node a , attaching node b to the reference, and looking at the output--you would be disappointed. In dealing with electronic components, you cannotforget the unrepresented but needed power supply.
It is impossible for electronic components to yield voltagesthat exceed those provided by the power supply or for them to yield currents that exceed the power supply's rating.
Typical power supply voltages required for op-amp circuits are ± 15 V . Attaching the 1 mv signal not only would fail to produce a 100 V signal, the resulting waveform would beseverely distorted. While a desirable outcome if you are a rock&roll aficionado, high-quality stereos should not distort signals. Another consideration in designing circuits withop-amps is that these element values are typical: Careful control of the gain can only be obtained by choosing a circuitso that its element values dictate the resulting gain, which must be smaller than that provided by the op-amp.

Op-amp

The top circuit depicts an op-amp in a feedback amplifier configuration. On the bottom is the equivalent circuit, andintegrates the op-amp circuit model into the circuit.

Inverting amplifier

The feedback configuration shown in [link] is the most common op-amp circuit for obtaining what is knownas an inverting amplifier .

R F R out R out G R F 1 R out 1 R in 1 R L 1 R 1 R in 1 R F 1 R F v out 1 R v in
provides the exact input-output relationship. In choosing element values with respect to op-amp characteristics, we can simplify theexpression dramatically.
  • Make the load resistance, R L , much larger than R out . This situation drops the term 1 R L from the second factor of [link] .
  • Make the resistor, R , smaller than R in , which means that the 1 R in term in the third factor is negligible.
With these two design criteria, the expression( [link] ) becomes
R F R out G R F 1 R 1 R F 1 R F v out 1 R v out
Because the gain is large and the resistance R out is small, the first term becomes 1 G , leaving us with
1 G 1 R 1 R F 1 R F v out 1 R v in
  • If we select the values of R F and R so that G R R F , this factor will no longer depend on the op-amp's inherentgain, and it will equal 1 R F .
Under these conditions, we obtain the classic input-outputrelationship for the op-amp-based inverting amplifier.
v out R F R v in
Consequently, the gain provided by our circuit is entirely determined by our choice of the feedback resistor R F and the input resistor R . It is always negative, and can be less than one or greaterthan one in magnitude. It cannot exceed the op-amp's inherent gain and should not produce such large outputs that distortionresults (remember the power supply!). Interestingly, note that this relationship does not depend on the load resistance. Thiseffect occurs because we use load resistances large compared to the op-amp's output resistance. Thus observation meansthat, if careful, we can place op-amp circuits in cascade, without incurring the effect of succeeding circuits changing the behavior (transfer function)of previous ones; see this problem .

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Source:  OpenStax, Fundamentals of electrical engineering i. OpenStax CNX. Aug 06, 2008 Download for free at http://legacy.cnx.org/content/col10040/1.9
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