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Assuming the following:

  1. The voltage across the base-emitter junction of a BJT is constant in forward-active and saturation mode
  2. The BJTs in a long-tailed pair are normally forward-active or saturated

(If any of these assumptions are incorrect, read not further and correct me)

then how is the difference between Vin+ and Vin- not fixed to the difference between the base-emitter voltage drops of the two transistors? (Zero, if the transistors are identical)

long tailed pair

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4 Answers 4

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This is a classic misconception: "VBE is constant, BJT transistors are current-controlled devices." Nope, wrong.

So, your #1 above is wrong. Long-tailed pairs are based on changes to VBE. The VBE becomes the transistor's voltage-input.

In BJT transistors, the collector current is determined by the potential-barrier of the EB junction. (This is the same way that diodes work, with an exponential V-I curve where the VF voltage-drop determines diode current.) The base current of a BJT actually has no direct effect on the collector current. Transistors are voltage-input devices, so the IB and the beta are mostly irrelevant to explaining their internal workings.

However, much technician-level training material (especially in the military) teaches that BJTs are current-input devices. They assume that VBE is fixed at 0.7V. This oversimplification is fine at the black-box level, and works for the tech-students who never will become engineers or scientists. (It's a "lie to children," like teaching kids that electrons orbit inside atoms just like little planets.) The full engineering version of BJT explanation is too complicated. It involves the built-in voltage-fields of the PN junction, depletion zones and the diode exponential V-I curve, and the Ebers-Moll equation IC = IS*(eVBE/VT -1).

In many circuits, hfe is irrelevant, and Ib is treated as an insignificant leakage. The classic BJT transistor circuits which rely on direct VBE voltage-input are: DC amplifiers (op amps, long-tailed pair,) current-mirrors, cascode circuits.

Suppose a student had been taught that "hfe is primary, VBE is irrelevant," and has never even encountered the idea that Ic is actually controlled by VBE ...then that student will have no hope of understanding the inside of an IC op-amp. Those ICs are composed of voltage-input transistor circuits: current-mirrors, long tailed pairs, cascode stages!

In other words, if the internal schematic of an LM741 amplifier looks utterly incomprehensible, as if it was designed by alien minds ...it just means that you still believe that BJT transistors are current-input devices. Technicians were taught this mistake, and the vast majority still believe it. Even some engineers have this "newbie misconception." They missed the part of their coursework where it was all debunked, when hfe was replaced by Ebers-Moll equation.

For a good textbook which directly attacks this "hfe mistake," see Art of Electronics by Horowitz and Hill. Their lab-manual especially delves into all the problems caused by hfe-based thinking in bipolar transistor design. Win Hill on the C4 forum even discusses how he first encountered this issue.


[Note that hfe and base current can be a helpful concept, a simple rule-of-thumb. After all, base current can be used to determine VBE, and then the EB-junction's potential-barrier will control the collector current ...so if we oversimplify and remove the middle part, we can pretend that base current can have an effect upon collector current. Because both currents are linked to VBE, the two currents are roughly proportional, and this is a useful concept. But it can become a mental stumbling-block for students encountering BJT circuits which harness VBE directly, where the base current plays no role except as unwanted leakage. ]


enter image description here Figure 1 internal schematic of 741 op amp [ Texas Instruments ] Wikimedia Commons

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    \$\begingroup\$ Ah, now I see that it was the PN juction that I misunderstood, when I saw it as the mathematical ideal of perfectly conductive after a certain voltage and as an open circuit below that voltage. In this sense, it could be said that the op-amp takes advantage of a PN junction's imperfection. \$\endgroup\$
    – nc404
    Commented May 3, 2018 at 1:49
  • \$\begingroup\$ Should the gm be a part of (this already excellent) answer? \$\endgroup\$ Commented May 3, 2018 at 5:16
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    \$\begingroup\$ nc404, to further explain the background of your false assumption keep in mind the following: During calculation of the DC bias point of a BJT amplifier stage it is common practice to ASSUME a fixed value for Vbe (0.6 or 0.7 volts). However, this happens only because we do not know the exact value (somewhere between 0.6 and 0.7 V). The resulting error due to this rough assumption is acceptable in most cases, because we apply DC feedback which makes the DC bias point rather insensitive to such uncertainties... \$\endgroup\$
    – LvW
    Commented May 3, 2018 at 9:52
  • \$\begingroup\$ Designs that are based only on the assumptions about β/hFE are not great anyway, since in practice it can vary so much. Good designs will perform reasonably over quite a wide range of β. E.g. a 741 would work probably within spec for current gains of all transistors between 50 and infinity. Most bipolar op-amps would, in fact. They may work a bit less admirably at lower β, but work they will. Classroom "designs" that use a fixed β or are tweaked for the β of transistor on hand work with selected parts and nothing else... β is usually a limiting factor, not a constant, in a design. \$\endgroup\$ Commented Jan 10, 2023 at 19:57
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The long-tailed pair is interesting enough that I should probably post a mathematical solution, using the original Ebers-Moll equation for the active region of operation for the two BJTs. No Early Effect (basewidth modultion) included.

Assumed: Identical (matching) BJTs whose saturation currents are the same and \$\beta\approx \infty\$ is assumed.

Knowing:

$$\begin{align*} I_{\text{C}_1}&=I_\text{SAT}\:\left(e^\frac{V_{\text{B}_1} -V_\text{E}}{V_T}-1\right)\\\\ I_{\text{C}_2}&=I_\text{SAT}\:\left(e^\frac{V_{\text{B}_2} -V_\text{E}}{V_T}-1\right) \end{align*}$$

Nodal analysis provides:

$$\begin{align*} V_\text{E}&=V_\text{EE} + V_T\:\operatorname{LambertW}\left[\frac{I_\text{SAT}\:R_\text{E}}{V_T}\left(e^\frac{V_{\text{B}_1}}{V_T} + e^\frac{V_{\text{B}_2}}{V_T}\right)\:e^{\frac{2\:I_\text{SAT}\:R_\text{E} - V_\text{EE}}{V_T}}\right]-2\:I_\text{SAT}\:R_\text{E} \end{align*}$$

It's probably better to focus on \$\Delta V_\text{B}=\frac{V_{\text{B}_1}-V_{\text{B}_2}}{2}\$ and \$V_{\text{B}}=\frac{V_{\text{B}_1}+V_{\text{B}_2}}{2}\$. Then set:

$$\begin{align*} V_{\text{B}\eta} &=V_\text{B} + 2\: I_\text{SAT}\: R_\text{E} - V_\text{EE}\\\\ \eta&=\operatorname{LambertW}{\left [\frac{I_\text{SAT}\: R_\text{E}}{V_T} \left(e^{\frac{2\:\Delta V_\text{B}}{V_T}} + 1\right) e^{\frac{V_{\text{B}\eta}- \Delta V_\text{B}}{V_T}} \right ]}\end{align*}$$

(Clearly, \$V_{\text{B}\eta}\$ is a voltage. In this case, one referred to the most negative rail but that also takes into account a small voltage drop due to the very tiny saturation currents of the two BJTs through \$R_\text{E}\$. Also, \$\eta\$ is a unitless ratio of the voltage drop caused by the sum of both emitter currents flowing through \$R_\text{E}\$, with respect to the thermal voltage used to compute those currents.)

Which makes it a little simpler to write out:

$$\begin{align*} V_\text{E}&=V_\text{EE} + \eta\:V_T-2\:I_\text{SAT}\:R_\text{E} \end{align*}$$

And of course:

$$\begin{align*} V_{\text{C}_1}&=V_\text{CC} - R_{\text{C}_1}\:I_\text{SAT}\:\left(e^\frac{V_{\text{B}_1} -V_\text{E}}{V_T}-1\right)=V_\text{CC} - R_{\text{C}_1}\:I_{\text{C}_1}\\\\ V_{\text{C}_2}&=V_\text{CC} - R_{\text{C}_2}\:I_\text{SAT}\:\left(e^\frac{V_{\text{B}_2} -V_\text{E}}{V_T}-1\right)=V_\text{CC} - R_{\text{C}_2}\:I_{\text{C}_2} \end{align*}$$

Assuming saturation is avoided, that's all there is to it. As you can see, it begins with figuring out the value of \$V_\text{E}\$. The rest is just \$V_\text{CC}\$ less the voltage drop across the appropriate resistor times the appropriate collector current.


To test the values, assume \$I_\text{SAT}=10\:\text{fA}\$, \$V_\text{CC}=+10\:\text{V}\$, \$V_\text{EE}=-10\:\text{V}\$, \$V_T=26\:\text{mV}\$, and all three resistors are \$1\:\text{k}\Omega\$. If \$V_{\text{B}_1}=+10\:\text{mV}\$ and \$V_{\text{B}_1}=+20\:\text{mV}\$, then ignoring \$\beta\$ one would get: \$V_\text{E}=-683\:\text{mV}\$, \$V_{\text{C}_1}=6.227\:\text{V}\$, and \$V_{\text{C}_2}=4.457\:\text{V}\$. By comparison, with \$\beta=150\$ included these would be: \$V_\text{E}=-683\:\text{mV}\$, \$V_{\text{C}_1}=6.251\:\text{V}\$, and \$V_{\text{C}_2}=4.493\:\text{V}\$.

The small signal voltage gain would be about \$\mid\: A_v\mid\:\approx 365\$.


The current ratios now look like:

$$\begin{align*} \frac{I_{\text{C}_2}}{I_{\text{C}_1}}&=\frac{1-e^{\left(\frac{V_{\text{B}\eta}-\Delta V_\text{B}}{V_T} -\eta\right)}}{1-e^{\left(\frac{V_{\text{B}\eta}+\Delta V_\text{B}}{V_T} -\eta\right)}}\end{align*}$$

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  • \$\begingroup\$ I am learning long-tailed pair too and I find it so complicated to understand how a long-tailed pair works. In your question, what is LambertW? \$\endgroup\$
    – Dat
    Commented Nov 2, 2021 at 10:13
  • \$\begingroup\$ @Dat It's also called a product-log function. Google it. Or look it up on Wolfram's site. \$\endgroup\$
    – jonk
    Commented Nov 2, 2021 at 10:27
  • \$\begingroup\$ @Dat If you have \$u e^u=z\$ where \$z\$ is known then \$u=\operatorname{LambertW}(z)\$. \$\endgroup\$
    – jonk
    Commented Nov 2, 2021 at 10:38
  • \$\begingroup\$ ok, I understand all the math now. In the "Practical electronics for inventors" book, the author writes: \$ V_{out} = \frac{R_c}{r_{tr}}(V1-V2) \$. Could you show me the proof of this simpler equation? \$\endgroup\$
    – Dat
    Commented Nov 4, 2021 at 10:06
  • \$\begingroup\$ In other texts, they usually said the current through Re is constant. Why is that? \$\endgroup\$
    – Dat
    Commented Nov 4, 2021 at 10:08
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Base-Emitter junction is a PN junction, and works like a diode would, absent current gain. So, for an initial analysis, replace the transistors with diodes from base to emitter, and collectors open, and see what happens. When the B-E diode is reverse-biased, the transistor it's part of becomes an open circuit - until the B-C junction starts conducting, that is.

schematic

simulate this circuit – Schematic created using CircuitLab

So, in a long-tailed pair, if Vin+>Vin- (by more than a few hundred mV), then Q1 conducts, but Q2's B-E diode is reverse biased and that transistor "vanishes" from the circuit - the circuit works as-if the transistor wasn't there, just its parasitic capacitances and leakage currents.

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The voltage across the base-emitter junction of a BJT is constant in forward-active and saturation mode.

The trick is that the base-emitter voltage is not absolutely constant but varies within small limits (a few hundred mV). This means that the differential voltage is also within such limits, i.e. it is very small compared to the common-mode voltage.

The BJTs in a long-tailed pair are normally forward-active or saturated.

If the differential voltage changes outside these limits, one of the transistors continues to be active but the other goes "off" and releases the emitter of the first transistor. Because of the negative feedback ("emitter degeneration"), the active transistor cannot be saturated. This situation can be observed in ECL logic gates.

So the transistors of the differential pair can be only in one of two possible states - active and off; they cannot be saturated.

Figuratively speaking, in normal operation, the differential pair consists of two interacting emitter followers. When the input differential voltage becomes unacceptably high, one of the emitter followers ceases to function and only the other follower remains.

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