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Inverting vs. Noninverting

-December 14, 2013

Picking up from my introductory blog on negative feedback [1], I now wish to discuss its two most common op amp implementations, the inverting and noninverting configurations.  These popular circuits exhibit similarities/differences, some tricky, that I will try to clarify. 

 


Figure 1 - (a) Basic negative-feedback block diagram, and (b) the noninverting amplifier as its most direct implementation. 


Let’s first reexamine the block diagram of Fig. 1a in more detail.  Writing

 


 

expanding, collecting, and solving for the ratio vO/vI, which we shall call the closed-loop gain A, gives

 


 

where

 


 

We observe that as the error signal vE propagates around the loop, it is first amplified by aε, then is attenuated to 1/Aideal, and finally is inverted at the summer, for an overall gain of –aε/Aideal.  Its negative is called – if improperly – the loop gain T.  As we move along, we’ll see that T provides all relevant information about a negative-feedback circuit, so it pays to think of T as the circuit’s DNA. 

For large T (that is, for aε >> Aideal) we can approximate Eq. (2) as

 



so here is the first piece of DNA information: T tells you how close your actual gain A is to Aideal.  You want a gain error not exceeding 0.1%?  Then you need T >103, which you achieve by using an error amplifier with aε >103Aideal.  Equivalently, T tells you how much open-loop gain aε you need to “throw away” in order to approach Aideal within a given percentage error.  (All this is familiar stuff, but I wanted to put it down explicitly to prepare the framework for the rest of this blog and the blogs to follow.) 

   

The Noninverting Configuration

 

The noninverting amplifier of Figure 1b fits the block diagram of Figure 1a to a T (pun intended).  The idea here is that if you want to set it up for a given Aideal, you use a voltage divider such that

 


 

The combined role of input summer and error amplifier is played by the op amp, whose voltage gain we denote as av, and error signal as vD.  Recycling Eq. (2), we thus write

 


 

indicating a loop gain of T = av/(1 + R2/R1).

   

The Inverting Configuration

 

Let us now turn to the noninverting amplifier’s closest sibling, the inverting amplifier of Figure 2a.  This configuration does not conform to the block diagram of Fig. 1a as obviously as the noninverting configuration, but we can still analyze it by applying the superposition principle,

 


 

Again expanding, collecting, and solving for the closed-loop gain, we get



 

Figure 2 - (a) The inverting amplifier, and (b) source transformation to show it in its most natural form. 

 

 

The loop gain is still T = av/(1 + R2/R1), and this makes sense because as vD propagates around the loop, it still undergoes magnification by av and then attenuation to 1/(1 + R2/R1).  However, we now have Aideal = –R2/R1, so we can no longer claim that T = av/Aideal.  What gives?  This is the first potential source of confusion between the inverting and noninverting configurations, which students have rightly pointed out in my classes countless times. 

            The confusion stems from the fact that even though we are driving the circuit with a voltage, its natural input is really a current [2], as evidenced by the source transformation of Figure 2b.  Nobody in his or her right mind would drive an op amp’s inverting input pin with a voltage source, as this would prevent the op amp from exercising its negative-feedback action.  So, we decouple vD from vI by means of R1, and in so doing we convert vI to iI (= vI/R1), making the op amp act as a current-to-voltage (I-V) converter.  This, even though the op amp is inherently a voltage-to-voltage (V-V) converter.  Replacing vI with R1iI and manipulating, we put Equation (7) in the form of Equation (1)

 


 

Comparison with Equation (1) indicates that the circuit realizes a negative-feedback system with

 

 


 

where both the error gain aε (≠av!) and the I-V converter gain Aiv are in V/A.  The ideal voltage gain from vI and vO is then Aideal = vO/vI = (vO/iI)×(iI/vI) = Aiv(ideal)/(R1) = –R2/R1, thus confirming Equation (8).  Summarizing, while sharing the same expression for T, the inverting and noninverting amplifiers differ both in the magnitudes and polarities of their voltage gains. 

The quantity 1 + R2/R1, common to both amplifiers, is called the noise gain because this is the gain with which either circuit will amplify any noise present between the op amp’s input pins, such as the input offset voltage VOS.  Noise gain and signal gain coincide in the noninverting configuration, but differ in the inverting configuration, as the latter’s signal gain is –R2/R1.

   

Comparing the Frequency Responses

 

Most op amps are designed for a frequency response av(jf) of the type of Fig. 3.  Gain starts out high and then rolls off with frequency until it drops to unity, or 0-dB, at a frequency aptly called the transition frequency ft.  We wonder how this affects the closed-loop gain A(jf).  For both amplifiers this gain takes on the common form

 


 

where

 


 




Figure 3 - Graphical method for the visualization of the discrepancy function |D(jf)|.


 

is the discrepancy function, so called because it tells you by how much your actual gain departs from the ideal.  Since T(jf) = av(jf)/(1 + R2/R1), you can visualize the decibel plot of |T| as the difference between the plot of |av| and the noise-gain curve 1 + R2/R1 (this, because the log of a ratio equals the difference of the logs).  As depicted in Figure 3, |T| is large at low frequencies, making |D| approach 0 dB, or unity there.  However, |T| itself rolls off with frequency until it drops to 0 dB, or unity, at the crossover frequency

 



 
The phase angle of av at fx is close to –90°, so we can write T(jfx) ≈ 1∠ -90o = -j.  Then, D(jfx) = 1/[1 + 1/(–j)] = 1/(1 + j), so  |D(jfx)| = 1/√2 = -3 dB.  Clearly, the crossover frequency fx represents the closed-loop bandwidth (this is the second important piece of DNA information).  Above fx, |D| itself rolls off with frequency, so D(jf) is just the ordinary low-pass function with fx as the –3-dB frequency,

 


 

With identical R2/R1 ratios, the inverting and noninverting amplifiers exhibit the same noise gain and closed-loop bandwidth, but dc signal gains of –R2/R1 and 1 + R2/R1, respectively, so 


The difference in their magnitudes is minor for high R2/R1 ratios, but becomes more pronounced at low gains.  The greatest difference occurs when both amplifiers are configured for unity gain, as depicted in Figure 4.  In this case we have 1 + R2/R1 = 1 + 0/∞ = 1 for the follower, so fx = ft, but 1 + R2/R1 = 1 + 1 = 2 (or 6 dB) for the inverter, so fx = ft/2.  Clearly, the –3-dB frequency of the inverter is half as large as that of the follower. 

The two circuits exhibit also a marked difference in the input resistance Ri.  So long as vD is vanishingly small, the current drawn from the input source in Fig. 1b will also be vanishingly small (this current is vD/rd, rd being the op amp’s internal input resistance, usually large).  Consequently, the noninverting amplifier presents a very high input resistance Ri (ideally Ri→∞), thus eliminating loading at the input (in fact, this is the very reason for using the follower as a buffer).  In Fig. 2a, however, a vanishingly small vD will make the inverting-input node approach a virtual ground, so this circuit presents a non-infinite input resistance of Ri = R1, which is likely to cause input loading. 

   
 

Figure 4 - Comparing unity-gain (a) the voltage follower and (b) inverting amplifier.

Compared with the voltage follower, the unity-gain inverter has (a) lower input resistance, (b) half as much bandwidth, (c) twice as much noise gain, and (d) it even requires two resistors (compared to the follower’s plain feedback wire) to achieve less.  So, why use it at all?  Your answer here:__________.

 

 

References

 

[1]http://www.edn.com/electronics-blogs/analog-bytes/4424393/The-magic-of-negative-feedback

 

[2] http://online.sfsu.edu/sfranco/BookAnalog/AnalogJacket.pdf

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