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In any normal op amp, if it was ideal it would have a gain such that the output voltage would be

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However in reality this is not the case as there is a common mode voltage gain and hence in reality the gain is different and the output voltage would be:

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If we look at a simple long tail differential amplifier which uses BJTs how could this be explained? On a transistor level how can you show where this comes from?

What actually causes this non ideal behavior of common mode voltage gain ?

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fred
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    There are multiple reasons for non-ideality resulting in common-mode gain (more usually referred to as its reciprocal - Common mode rejection ratio - CMRR). For example if the current source in the emitter was not perfect and the current changed with the voltage across it the output would depend upon the level of the input voltage. Early effect or thermal drift also can affect the output causing a reduction in CMRR. – Kevin White Mar 19 '19 at 01:14
  • Suppose the current source in the tail doesn't have infinite output impedance. – The Photon Mar 19 '19 at 01:14
  • @ThePhoton

    So lets say we have a current mirror as a source and a Wilson current mirror then this error would be reduced with a Wilson current mirror as the output impedance would be higher. But why exactly would this lead to the common mode voltage gain? how does it increase Vout if output impedance is not infinite?

     – fred Mar 19 '19 at 01:17
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    @fred There are lots of sources for common mode gain: Early Effect, Late Effect, as well as non-ideal behavior when operating in region I or region III (for a few notes on the regions and a chart to illustrate, see: BJT operating regions.) Also, resistor value variation, I suppose. And there's more too, as BJTs are complex devices. (Note that rather than using resistors as collector loads, results may be often improved by using a current mirror for both loads, instead.) – jonk Mar 19 '19 at 01:26
  • @jonk is the Late Effect solely a play on words, or is it an actual effect that received a tongue-in-cheek name? – nanofarad Mar 19 '19 at 01:41
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    @fred I forgot to mention variation between BJT devices, too. Even on an IC, they aren't the same as each other. They might even not be at the same temperature, depending on how the IC is laid out. All of the various effects sum together. – jonk Mar 19 '19 at 01:41
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    @AndreyAkhmetov No, it's separate. There is basewidth modulation based upon $V_\text{BC}$ and separate basewidth modulation based upon $V_\text{BE}$. Of course, it was facetiously named. No argument there. But the effect was uncovered later and documented by Gummel and Poon in their work modifying the Ebers-Moll model. – jonk Mar 19 '19 at 01:49
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    @AndreyAkhmetov I'm asking Ian Getreu where the actual name comes from. (It may not have been named, as such, by Gummel/Poon.) I just wrote him today and I'll let you know what he says about it. – jonk Mar 19 '19 at 01:55
  • I think Late leakage is the inverse to Dr Early's effect and it applies tongue in cheek to the reversed CE operation which does not apply here – Tony Stewart EE75 Mar 19 '19 at 01:57
  • if everything is balanced, then common-mode inputs may not produce differential-outputs. – analogsystemsrf Mar 19 '19 at 02:36
  • R Ratio tolerance errors can often contribute the most to CMRR error – Tony Stewart EE75 Mar 19 '19 at 04:38
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    @AndreyAkhmetov Oh, that's funny. Ian Getreu wrote back, this evening: "Guilty as charged. Play of words on the Early effect. Jim Early was not amused :-)" So I guess Ian is why I'm so aware of it as a separate name. Ian created the term (would have been circa 1976, or so, when he published it.) – jonk Mar 19 '19 at 05:01

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Let`s calculate the common mode gain Acm (real case with a finite dynamic resistance re in the common emitter leg):

(1) Forgetting - for the first moment - the second transistor Q2, the first transistor Q1 is operated in common emitter configuration (with current-controlled voltage feedback provided by a finite dynamic resistance re in the emitter leg): Hence, the gain is

A=-gm * Rc/(1+gm * re).

(2) The same gain applies to Q2 (without Q1). Now - when both transistors are active and excited with the same input voltage (common mode), the current through re will be doubled (compared with case 1) causing a doubling of the feedback voltage. This can be incorporated into the given gain formula by replacing re with 2re. Hence, the common mode gain expression is:

Acm=A=-gm * Rc/(1+gm * 2re).

This expression shows that the common mode gain will be zero for an ideal current source (re approachung infinite) only.

Note: The above (rough) calculation is accurate enough to demonstrate the systematic common mode effect caused by the a finite re. Other influences (asymmetry effects, Early-effect,...) are neglected.

LvW
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