Summing Amplifier

3 Key excerpts on "Summing Amplifier"

• 

  eBook - ePub

  Electronics

  • David Crecraft(Author)
  • 2018(Publication Date)
  • CRC Press

    (Publisher)

  These ideas are put into practice in the following sections which describe some of the applications of feedback circuits. In each case the simplest possible approach is used to arrive at an understanding of the circuit operation. However, since the loop gain of any feedback circuit falls off at high frequencies, there will always be a frequency beyond which simplifying assumptions no longer apply. It is in this regime that CAD packages prove to be invaluable. With computer assistance you can quickly set up circuit models and evaluate their performance over a wide frequency range without resorting to tedious calculations. However, do remember that the use of the CAD software only makes sense when it is guided by a basic understanding of the underlying principles of the circuit under investigation.

  4.5.1 Summing Amplifiers

  Figures 4.19 and 4.20 show two amplifier circuits, each with two inputs. These circuits can be modified to have any number of inputs simply by adding more resistors, one for each input. When the resistors are chosen correctly, the first circuit has an output voltage which is the sum of its input voltages. Alternatively, the output can be the sum multiplied by a gain factor, or the output can be the weighted sum of its inputs (some inputs are amplified more than others).

  Fig 4.19 A non-inverting Summing Amplifier.

  Fig 4.20 An inverting Summing Amplifier.

  The same remarks apply to the second circuit, but its output voltage is inverted too.

  The resistive summing network

  To understand how these circuits work, first consider the resistive summing network of Fig. 4.18 . We need to know its Thévenin and Norton equivalent circuits:

  Fig 4.18 A resistive summing network.

  (i) Short-circuit current. If the output terminals are shorted, the current through the short-circuit is simply

  I sc

  =

  V 1

  /

  R 1

  +

  V 2

  /

  R 2

  =

  V 1

  /

  G 1

  +

  V 2

  /

  G 2

  This represents a weighted sum of the input voltages.

  (ii) Output resistance (and conductance). This is found by looking into the network’s output, with the voltage sources replaced by short-circuits. Thus

  R o

  =

  R 1

  //

  R 2

  or

  G o

  =

  G 1

  +

  G 2

  (iii) Open-circuit output voltage. This is simply

  V oc

  =

  I sc

  R o

  =

  I sc

  /

  G o

              =

  (

  V 1

  G 1

  +

  V 2

  G 2

  )

  /

  G o

  This network is used in both the Summing Amplifiers. Let us look at the first one. The non-inverting Summing Amplifier

  In Fig. 4.19

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  A Practical Introduction to Electrical Circuits

  • John E. Ayers(Author)
  • 2024(Publication Date)
  • CRC Press

    (Publisher)

  Figure 3.8 , allows us to reproduce a voltage from a source node to a target node without drawing current from the source node. For this reason, it is called a “buffer amplifier” or a “unity-gain buffer.”

  Long Description for Figure 3.8

  A buffer amplifier circuit configured using a single op amp, with a single input voltage. The power supply connections for the op amp are not shown explicitly. A voltage source with value VIN has the negative terminal connected to ground and the positive terminal connected to the noninverting input terminal. A wire is connected between the inverting input terminal and the output. The voltage at the output is labeled VOUT.

  FIGURE 3.8 Buffer amplifier.

  3.7 Summing Amplifier

  A Summing Amplifier allows us to scale and add two or more voltage signals. A two-input Summing Amplifier is illustrated in Figure 3.9 . Each input voltage is applied to the inverting input terminal through a resistor. The non-inverting input is grounded and there is a feedback resistor

  R F

  .

  Long Description for Figure 3.9

  A Summing Amplifier circuit constructed using an op amp and three resistors, with two input voltages. The power supply connections for the op amp are not shown explicitly. The noninverting input terminal is grounded. A voltage source with value V1 has the negative terminal connected to ground and the positive terminal connected to a resistor R1, which is then connected to the inverting input terminal. The current leaving the inverting input in R1 is labeled I1. A second voltage source with value V2 has the negative terminal connected to ground and the positive terminal connected to a resistor R2, which is then connected to the inverting input terminal. The current leaving the inverting input in R2 is labeled I2. The voltage at the inverting terminal is labeled vn. The current flowing from the inverting input in the resistor R1 is labeled I1. The current entering the inverting input is labeled in. There is a resistor RF connected between the inverting input terminal and the output. The current leaving the inverting input through this resistor is labeled IF. The voltage at the output is labeled VOUT.

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  Introduction to Energy, Renewable Energy and Electrical Engineering

  Essentials for Engineering Science (STEM) Professionals and Students

  • Ewald F. Fuchs, Heidi A. Fuchs(Authors)
  • 2020(Publication Date)
  • Wiley

    (Publisher)

  (8.21) one gets 8.7 Summing Networks The summing circuit of Figure 8.8 is similar to the inverting amplifier shown in Figure 8.6. The only difference is that Figure 8.8 has two input voltages v inA and v inB and two input resistors R 1A and R 1B. The output voltage v out is the weighted sum of the values of the two input voltages. Assuming ideal OP amplifier with v + = v − = 0 and i + = i − = 0, one can write for the two input currents (8.23a) (8.23b) and for the feedback (F) current Figure 8.8 Summing network with two inputs. (8.24) With i A + i B = − i F one can write or (8.25) This method of summing the weighted input voltages can be applied to three, four, and a higher number of input voltages. Application Example 8.5 Output voltage for summing network Determine the output voltage v out (t) for the summing network of Figure 8.8 provided the input voltages are v inA (t) = 10 V sin ωt, v inB = 5 V cos ωt, the resistors [ 3, 4 ] are R 1A =. 1 kΩ, R 1B = 422 Ω, R 2 = 562 Ω, and the power supply voltages are V DD = 10 V and V SS = −10 V. Solution According to Eq. (8.25) one gets 8.8 Integrating and Differentiating Networks Integrating and differentiating networks are frequently applied in control [ 1 ] networks. Integrating networks eliminate/reduce electric noise by smoothing the signal, while differentiating networks are applied to anticipate the change (increase or decrease) of the control variable, making the control faster as compared with the integrating network. However, the differentiating approach increases existing electric noise in the signal, while the integration reduces signal noise

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