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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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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).

The classic circuits which rely on direct VBE voltage-input are: DC amplifiers (op amps, long-tailed pair,) current-mirrors, cascode circuits.

In other words, 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 a modern op-amp. Those ICs are composed of voltage-input transistor circuits: current-mirrors, long tailed pairs, cascode stages!

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 beta-based thinking in transistor design. Win Hill on C4 even discusses how he first encountered this issue.


[Note that hfe and base current can be a helpful concept. After all, base current can be used to determine VBE, and then the EB junction potential-barrier will control the collector current ...so if we oversimplify and remove the middle part, we can pretend that base current can affect 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. ]

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  • 1
    \$\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 May 3 '18 at 1:49
  • \$\begingroup\$ Should the gm be a part of (this already excellent) answer? \$\endgroup\$ – analogsystemsrf May 3 '18 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 May 3 '18 at 9:52
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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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