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Chapter 3.5.6 of Analysis and Design of Analog Integrated Circuits by Gray, Hurst, Meyer, and Lewis (GHLM) discusses the effect of device mismatches on the behavior of emitter- or source-coupled differential pairs. The discussion is here:

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I am confused about two things in particular:

(1) What is the argument for why Figs 3.62a and 3.62b are equivalent. In particular, the claim seems to be made that they are equivalent, but no proof is given. In particular, I cannot follow or understand the discussion beginning at "For example..." and perhaps that is the crux of my issues.

(2) Question (1) begets a second question, which is what the nature of the differential amp without mismatches should be. Does one use average values of the corresponding components of that in the mismatched amplifier?

Edit The references 11 and 12 seem to be rigorous analyses of the differential amplifier from the late 60s. I've only started to look at them and haven't really understood that which I've read, but I suspect the gist of what GHLM are using from these references is that as a (network?) theorem, for every real differential differential amplifier there exists a unique (balanced amplifier, \$V_{os}\$, \$I_{os}\$) triple which is such that it reproduces the real (unbalanced) amplifier in terms of its terminal voltage characteristics. As the references seem rather involved, I would accept as an answer simply confirmation that I have understood the claim of this "theorem" correctly, as well as any comment as to how one obtains said triple.

The references 11 and 12 are both texts called Differential Amplifiers, by Middlebrook and then Giacoletto, respectively.

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    \$\begingroup\$ Current and voltage sources are added to include any type of input drive condition. The voltage offset is useless if you drive the amplifier with a current source. And the current offset is useless if driven by a voltage source. \$\endgroup\$
    – sarthak
    Commented Feb 2 at 8:50

6 Answers 6

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...In particular, the claim seems to be made that they are equivalent, but no proof is given. ...

It is more a definition than a proof. Ideally (treating the amplifier as a black box) if \$V_{in}=0\$ then \$V_{out}=0\$ can be expected. If \$V_{out}\$ is different from zero, then this quantity is referred to the input by dividing by the gain of the amplifier to get \$V_{os}\$. This so-called "input referred offset" can now be included as if a real source in series with the input voltage for analysis and design.

Another way to understand it is \$V_{os}\$ is the input voltage required to make the output voltage zero.

The offset current \$I_{os}/2\$ can be treated in much the same way except the current requires a trans-resistance to make an output voltage. The input resistance of the amplifier in parallel with the source resistance will provide a resistance to make the conversion to a voltage.

If the input circuit is open then the output voltage should be zero. If it is not then \$I_{os}/2\$ is the current required to be applied to bring the output voltage to zero.

I cannot follow or understand the discussion beginning at "For example..."

The example demonstrates that both \$I_{os}/2\$ and \$V_{os}\$ are required to fully specify the voltage offset at the output when the amplifier gain is included.

The authors demonstrates that \$V_{os}\$ is the source for offset error while \$I_{os}\$ has no influence for a source voltage with zero resistance.

Then they demonstrate that for a current source with infinite resistance, \$V_{os}\$ has no influence, but \$I_{os}\$ does.

This implies that both offset sources must be considered when the source resistance is significant.

...what the nature of the differential amp without mismatches should be.

Essentially it is being treated as a black box with gain and possibly input resistance, though it is not specified. If the input voltage is zero and if the input current is zero, then the output voltage is zero.Nothing more.

Does one use average values of the corresponding components of that in the mismatched amplifier?

No. This is strictly a model to explain a physical obseration. That is why it is included in a specification sheet for op-amps for example. It varies from device to device during the manufacturing process, so trying to calculate it would be fruitless.

Any design must allow for offset variation or provide a means to "zero-out" for each system.

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  • \$\begingroup\$ Thanks so much for this answer. I think it starts getting me towards understanding, but I want to check a few things that I didn't follow. (1) Zooming all the way out, it seems like you agree with the assessment that the figure is saying that a real (unbalanced) differential amplifier (Fig 3.62a) can be modelled as a balanced differential amplifier with sources as shown attached. What I don't follow is how (even conceptually) we think about the referral. You say "this quantity is referred to the input by dividing by the gain of the amplifier to get 𝑉𝑜𝑠" which intuitively makes sense, but... \$\endgroup\$
    – EE18
    Commented Feb 1 at 1:43
  • \$\begingroup\$ ...what is "the amplifier" here? Presumably the balanced differential amplifier of Fig 3.62b since that will be amplifying whatever \$V_{os}\$ we use at the input, but this seems to me like an incompletely specified situation since there are infinitely many \$V_{os}\$ and balanced amplifier pairs which will yield the correct zero-input output voltage! (2) With all of this said in (1), it seems like \$I_{os}\$ has not been spoken of. Perhaps there is a unique (balanced amplifier, \$V_{os}\$, \$I_{os}\$ ) triple which is such that it reproduces the real (unbalanced) amplifier? \$\endgroup\$
    – EE18
    Commented Feb 1 at 1:46
  • \$\begingroup\$ I've placed a summary of a "lesser" question in the OP, perhaps that will be helpful in understanding what I am not understanding :) \$\endgroup\$
    – EE18
    Commented Feb 1 at 1:51
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(1) What is the argument for why Figs 3.62a and 3.62b are equivalent. In particular, the claim seems to be made that they are equivalent, but no proof is given. In particular, I cannot follow or understand the discussion beginning at "For example..." and perhaps that is the crux of my issues.

This is a theoretical solution to a practical problem. Input offset voltage is intrinsic to real amplifiers. So is the input bias currents. In the first example, a "real" amplifier is shown. It has input offset voltage and input bias currents that make it non-ideal (there are other properties that make an amplifier non-ideal, but the scope of your question is limited to Voffset and IinBias). So for a given device, operating at a known temperature, working at a fixed common-mode operating point, the actual input offset voltage and currents are known.

Figure 2 shows that if you put a theoretical voltage source that exactly cancels out the offset voltage in the practical example, and you do the same with the input current in that same vein, you end up with an amplifier that has no mismatches (from the standpoint of the input signals).

No proof is given cause the result should be self-evident - adding the external offsets nulls the intrinsic offsets of the amplifier precisely to make it ideal.

The "for example..." language that follows is basically describing how an ideal differential amplifier should behave with regard to DC performance - voltages that are equal in magnitude get canceled precisely; there is no offset voltage and therefore you have no static DC bias voltage on the input. The same could be said for the input current.

(2) Question (1) begets a second question, which is what the nature of the differential amp without mismatches should be. Does one use average values of the corresponding components of that in the mismatched amplifier?

Without mismatches, the idea is you treat the amplifier as if it's ideal. In fact, you can usually do this in practice with real amplifiers assuming the offsets are negligible with respect to the signals' magnitudes and the accuracy of the DC performance required.

You don't use average values in the mismatched amplifier - you use the actual offsets. Ios/2 isn't implying an average. What that means is that half the offset current is split between both inputs: half the current in one input and half the current out of the other input amounts to the total Input offset current. If it was drawn as just Ios and say that's 1mA, well than that's 1mA into one input and 1mA out of the other input which would be 2mA, which is double the input offset.

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Circuit analysis of device mismatches with equivalent sources has strong parallels with the analysis of noise in circuits--- an often-referenced comparison is that offsets can be treated as "very low-frequency noise" in circuit analysis. Both circuit noise and mismatch can be thought of as the result of internal stochastic processes in the circuit--- "noise" usually refers to stochastic processes generated by the underlying device physics whilst "mismatch" often refers to imbalances caused by manufacturing that can be modeled stochastically. Since both are usually modeled as a stochastic quantity, they can be treated equivalently in circuit analysis.

A linear, noisy, N-port network will have N open-circuit noise voltages, \$V_n\$ , present at its N ports. These noise voltages are the result of the internal noise generators in the circuit. The entire network can be completely described by it's \$Z\$ matrix and the complex \$V_n\$'s which have an associated correlation matrix that defines the correlation between the \$V_n\$'s. Since the circuit is linear, we can use the \$Z\$ matrix to transform \$V_n\$ into an equivalent \$V_n'\$ which will result in the same N open-circuit noise voltages present at the ports, but may be represented by a different set of noise sources. The total noise sources will remain as N due to the linearity of the transformation.

For the two-port case, as shown in the figures you reference, we could represent the circuit with the impedance matrix \$Z_2\$, as well as the noise sources \$V_{n2}\$, where there is a noise source at each port. A common linear transformation is to reflect the output noise source to the input, resulting in a new noise source vector, \$V_{n2}'\$, which consists of two noise sources at the input. Note that in this formalism the noise source vector contains the necessary information for computing the correlations between the noise sources. Also, we used the impedance formalism with \$Z\$ and \$V_n\$ but could have carried out this analysis in other network formalisms, i.e. admittance, hybrid parameters etc.

This formalism can naturally be extended to mismatch by considering the open-circuit noise voltages as arising from mismatches in the circuit. In this case, we look at the \$V_n\$, find the \$Z\$ matrix describing the circuit, and then massage the noise sources (now mismatch sources) into the form we want. With two-port networks we usually just skip right ahead to computing the sources when they are both referred to the input, i.e. an input-referred mismatch voltage and current. Gray and Meyer present the intuition for why we need two input-referred sources by presenting the limiting cases of a zero-impedance and infinite-impedance source impedance, and how the current source and voltage source would have no effect on the output respectively, but we can also view it in the general case as the result of a series of linear transformations from the original situation we described above.

The \$Z\$ matrix is computed in the presence of no sources, which in the mismatch case corresponds to no mismatch sources--- i.e. a perfectly balanced amplifier. So when computing the amplifier to be used for the input-referred sources, we assume no mismatch.

In summary, the "triple" which you mention is in general the N-port \$Z\$ matrix, and the N open-circuit noise sources. It reduces to a "triple" for the two-port case, i.e. an amplifier two-port description (\$Z\$) and two noise sources.

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  • \$\begingroup\$ Thank you for this answer! I'm afraid there's still a few things I don't follow but I think this line of connecting it to noise is very helpful for me. I'll go ahead and list my follow-ups if it's OK: 1) how does this referring actually happen? Would it be possible to show an example with a two port? 2) Supposing that I understand the referring, I'm afraid I still don't follow how the current source emerges from this argument. 3) In the case of noise, I can imagine how one would compute the \$Z\$ matrix and I guess one can do the same for these mismatched amps. But is there a way to look... \$\endgroup\$
    – EE18
    Commented Feb 4 at 22:04
  • \$\begingroup\$ ...internally to figure out the values of components/circuit elements for a corresponding balanced amplifier? That seems to be what Gray and Meyer allude to but I can't figure out the argument. \$\endgroup\$
    – EE18
    Commented Feb 4 at 22:04
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If you accept that an amplifier will have offset (the DC bias at the input will have a difference WRT the output), then you know that this is only visible at the output. However, different amplifiers have different DC gains. Therefore, how can we quantify the offset of many several amplifiers? how could we say one has less offset than the other?

The answer lies in abstracting away this offset and refer it to the input. If the input stage of an amplifier were a diff. pair made of NMOSes, then, you should know that these NMOS need a input bias current (however small they are) to work, and will have deviations in their threshold voltages \$V_{th}\$, they cannot be perfectly equal due to systematic and random mismatches, these mismatches causes their gate voltages to be different in order to have the same current running through both branches of the diff. pair. These 2 issues, among others, can be modeled as an input current source and a voltage source in series with the input, respectively.

We could theoretically model every source of mismatch in all the stages, though the 1st stage will most likely dominate. These mismatches get transferred to the output via some amplification or attenuation factor across the subsequent stages of the amplifier. However, we are only able to measure this offset at the output. We only see the net effect, and we're not able to discern what contributed where (unless you have the amplifier built out of its transistor models, then you could get that information). Therefore, we'd like to abstract this net effect into a lumped voltage and/or current source.

Why are these current and voltage sources enough to model such effects? If we assume that every amplifier has actually 4 types of transfer (V->V, V->I, I->V and I->I), then there's only 2 types of input signals we can use, V (voltage) or I (current). Obviously, a single feedback network determines only one transfer, while, ideally, making the rest of them very small or close to 0.

Needless to say, these current and voltage offset sources, as notated in op-amp datasheets, are only valid under set of conditions. They are a function of the amplifier's environment, such as supply voltage, biasing voltage, temperature, etc.

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There's no proof. There's no need for a proof. It's a model that does a good job of representing nonidealities for the purpose of design (and op amp selection).

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It is a bit hard to understand - neither did I in the first glimpse

In figure (a) the arrow has a text of "Differential amplifier with mismatches", so the "mismatches" are "hidden" inside the symbol of a differential amplifier.

If the two mismatch sources are moved out of the symbol (unhidden) like in figure (b), the symbol of a differential amplifier now really represents a "Differential amplifier without (these 2) mismatches".

(these 2) <-- I added after some comments. Because the 2 mentionned mismatches are the two most disturbing in opamps in most application cases. Of course there are a lot other mismatches like - already someone mentionned probably:

  • common mode input current
  • noise
  • drift by age, temperature of all/some parameters of the IC
  • influences by supply voltage, ... to all/some/none of these mismatch values
  • ...
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  • \$\begingroup\$ Sure, to some extent. But I think the point being made in the transformation is that this ideal voltage and current source as offsets are sufficient to model all of the mismatches in the diff amp in a). It's not clear to me a priori why this should be? \$\endgroup\$
    – EE18
    Commented Dec 4, 2023 at 2:13
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    \$\begingroup\$ @EE18 Practically there are most probably more effects too, but these are most probably far less influencing the result. These effects are not drawn: just guessing: common mode input current, parasite capacitance, input resistors, noise sources, gain change, ... and all perhaps influenced by temperature, age, supply voltage, ... \$\endgroup\$ Commented Dec 7, 2023 at 22:01
  • \$\begingroup\$ @BitLauncher: To continue the explanation in your answer, the "differential amplifier without mismatches" is only without mismatches, it still can and does have all the other non-ideal behavior including the ones you mentioned. \$\endgroup\$
    – Ben Voigt
    Commented Feb 5 at 22:31

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