Skip to content Skip to navigation

Connexions

You are here: Home » Content » Inverting Amplifiers with Impedances

Navigation

Content Actions
  • Download module PDF
  • Add to ...
    Add the module to:
    • My Favorites
    • A lens
    • An external social bookmarking service
    • My Favorites (What is 'My Favorites'?)
      'My Favorites' is a special kind of lens which you can use to bookmark modules and collections directly in Connexions. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need a Connexions account to use 'My Favorites'.
    • A lens (What is a lens?)

      Definition of a lens

      Lenses

      A lens is a custom view of Connexions content. You can think of it as a fancy kind of list that will let you see Connexions through the eyes of organizations and people you trust.

      What is in a lens?

      Lens makers point to Connexions materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

      Who can create a lens?

      Any individual Connexions member, a community, or a respected organization.

      What are tags? tag icon

      Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

    • External bookmarks
  • E-mail the author
  • Rate this module (How does the rating system work?)

    Rating system

    Ratings

    Ratings allow you to judge the quality of modules. If other users have ranked the module then its average rating is displayed below. Ratings are calculated on a scale from one star (Poor) to five stars (Excellent).

    How to rate a module

    Hover over the star that corresponds to the rating you wish to assign. Click on the star to add your rating. Your rating should be based on the quality of the content. You must have an account and be logged in to rate content.

    		(0 ratings)

Recently Viewed

This feature requires Javascript to be enabled.

Inverting Amplifiers with Impedances

Module by: Don Johnson

Summary: Inverting amplifiers with impedances.

Note: Your browser may not currently support MathML. See our browser support page for additional details. You can always view the correct math in the PDF version.

As long as design requirements are met, the input-output relation for the inverting amplifier also applies when the feedback and input circuit elements are impedances (resistors, capacitors, and inductors).

Figure 1: V out V in =- Z F Z V out V in Z F Z
opamp
opamp (opamp11.png)

Example 1

Let's design an op-amp circuit that functions as a lowpass filter. We want the transfer function between the output and input voltage to be

Hf=K1+ⅈ2πf f c H f K 1 2 f f c (1)
where K K equals the passband gain and f c f c is the cutoff frequency. Let's assume that the inversion (negative gain) does not matter. With the transfer function of the above op-amp circuit in mind, let's consider some choices.
  • Z F =K Z F K , Z=1+ⅈ2πf f c Z 1 2 f f c . This choice means the feedback impedance is a resistor and that the input impedance is a series combination of an inductor and a resistor. In circuit design, we try to avoid inductors because they are physically bulkier than capacitors.
  • Z F =11+ⅈ2πf f c Z F 1 1 2 f f c , Z=1K Z 1 K . Consider the reciprocal of the feedback impedance (its admittance): Z F -1=1+ⅈ2πf f c Z F 1 2 f f c . Thus, the admittance is a sum of admittances, suggesting the parallel combination of a resistor (value = 1 Ω 1Ω ) and a capacitor (value = 1 f c F 1 f c F ). We have the right idea, but the values (like 1 Ω) are not right. Consider the general RC R C  parallel combination; its admittance is 1 R F +ⅈ2πfC 1 R F 2 f C . Letting the input resistance equal R R, the transfer function of the op-amp inverting amplifier now is Hf=- R F R1+ⅈ2πf R F C H f R F R 1 2 f R F C
Thus, we have the gain equal to R F R R F R and the cutoff frequency 1 R F C 1 R F C .

Thus, creating a specific transfer function with op-amps does not have a unique answer. As opposed to design with passive circuits, electronics is more flexible (a cascade of circuits can be built so that each has little effect on the others) and gain (increase in power and amplitude) can result. To complete our example, let's assume we want a lowpass filter that emulates what the telephone companies do. Signals transmitted over the telephone have an upper frequency limit of about 3 kHz. For the second design choice, we require R F C=3.3×10-4 R F C 3.3 10 4 . Thus, many choices for resistance and capacitance values are possible. A 1 μF capacitor and a 330 Ω resistor; 10 nF and 33 kΩ, and 10 pF and 33 MΩ would all theoretically work. Let's also desire a voltage gain of ten: R F R=10 R F R 10 , which means R= R F 10 R R F 10 . Recall that we must have R≪ R in ≪ R R in . As the op-amp's input impedance is about 1 MΩ we don't want R R too large, and this requirement means that the last choice for resistor/capacitor values won't work. We also need to ask for less gain than the op-amp can provide itself. Because the feedback "element" is an impedance (a parallel resistor capacitor combination), we need to examine the gain requirement more carefully. We must have | Z F |R≪105 ≪ Z F R 10 5 for all frequencies of interest. Thus, R F |1+ⅈ2πf R F C|R<105 R F 1 2 f R F C R 10 5 . As this impedance decreases with frequency, the design specification of R F R=10 R F R 10 means that this criterion is easily met. Thus, the first two choices for the resistor and capacitor values (as well as many others in this range) will work well. Additional considerations like parts cost might enter into the picture. Unless you have a high-power application (this isn't one) or ask for high-precision components, costs don't depend heavily on component values as long as you stay close to standard values. For resistors, having values r10d r 10 d , easily obtained values of r r are 1, 1.4, 3.3, 4.7, and 6.8, and the decades span 0-8.

Exercise 1

What is special about the resistor values; why these rather odd-appearing values for r r?

Solution

The ratio between adjacent values is about 2 2 .

Comments, questions, feedback, criticisms?

Send feedback