# How does the resistor at the tail act as a current source in a differential pair?

Here is a version of differential pair:

I can understand the working principle of differential pair above. When Ic1 increases Ic2 decreases that's the thing. And for that to happen the tail current must be constant. And the above differential pair example is using current source at the tail which makes this idea work. But have a look at below:

In the above version there's no current source but just a resistor at the tail. How is the tail current kept constant by that resistor Re? Or am I misunderstanding something here?

Edit:

Finally after reading τεκ's answer I started to understand. Below is an example how we can replace a "voltage source in series with a resistor" with a "current source in parallel with the same resistor". I added disturbance to see how much the current is regulated. I and II shows that source transformation which was the keyword to make things sense. Black plot of the current for I and II is identical which proves this. In III I've increased the source impedance of that equivalent current source and that shows us that moving to an ideal current source makes the current variation much smaller as shown in the red plot.

An ideal current source has infinite output impedance (i.e. $dV/dI = \infty$). A real current source is equivalent to an ideal current source in parallel with some finite output impedance Ro.

A current source in parallel with a resistor is equivalent to a voltage source in series with a resistor (source transformation).

simulate this circuit – Schematic created using CircuitLab

So the long-tail resistor in your circuit is just a non-ideal current source.

simulate this circuit

In the traditional vacuum tube version, V- would be a very large voltage and Re would be a very large resistance, so that the resulting current was correct but it was closer to the ideal of infinite resistance.

• Thanks a lot finally things made sense to me. Please also see my edit. – floppy380 Apr 5 '18 at 18:04

First of all, the 'tail' resistor is a current sink, not a source. No need to get into semantics over this.

In many low cost designs resistors are used with the understanding that they limit dynamic range by not regulating the current, which also limits the working voltage range. The constant current sink you saw in the first version is typical of op-amps and other IC's where it is easy to implement current sources and sinks all over the place. Being on the same die helps a lot to avoid temperature drift and external noise. A medium value resistor in the 10K to 50K is treated as if it were a current sink, because so little current flows to begin with. It cannot be more than the sum of Q1 current and Q2 current.

However, for boards made with discrete parts it becomes a tyranny of numbers, because each source or sink must be made with temperature stable parts. Special 5 pin dual transistors can be used as differential inputs or current mirrors, but it starts to consume space and drives up the cost.

Yes there are amplifiers made by Crown and Cerwin Vega that are built like that, but they are thousands of dollars each, just to get a 70V/uS slew rate and 130dB dynamic range, for 500 to 5,000 watts.

It comes down to cost vs. required or mandated performance. In IC's it is easy to just lithograph all the extra goodies that make an excellent op-amp or other analog IC.

Note that many RF IC's in the GHZ range avoid current sinks or sources as they have more capacitive loading then a resistor, but they usually consume more current than audio IC's or slow micro-power IC's.

• But I'm not asking why resistor is used instead of current source. I'm asking how does that resistor keep the current roughly/almost constant? How does it act as if it is a current source? – floppy380 Mar 23 '18 at 2:28
• Oh so its purpose is not to keep the current constant at all? I really thought it was. But what is its real purpose? I thought the idea of diff pair is to decrease Ic1 when Ic2 increases or the other way around. But that can only be done by a current source. Could you explain how Re makes things work here? Thanks – floppy380 Mar 23 '18 at 2:34
• No forget about the current source version; Im wondering how does the pair with resistor work. – floppy380 Mar 23 '18 at 2:35
• Hmm I still dont understand how this works. I thought the whole idea is to make Ic1+Ic2 = constant – floppy380 Mar 23 '18 at 2:37
• @pipe What Im trying to understand is not that imperfection. Im trying to understand how does it even mimic a "constant current sink" even though it is not ideal. How does a resistor act like a "sort of non ideal constant current sink" in that tail. Even though it doesn't regulate the current perfect, something make it enough to be used as a current regulator. I dont know what. We dont put an inductor there but a resistor. Why a resistor? I needed an analytic step by step explanation. – floppy380 Mar 23 '18 at 16:27

The resistor keeps the tail current fairly constant, constant enough.

If the magnitude of V- is much more than the of Vin range, then the current is fairly constant indeed.

There is a common mode gain, RC/2RE, but it's much smaller than the differential gain, RC/2(Qx intrinsic Re), and for many purposes, this degree of CMRR is sufficient.

You'll notice there's no resistor used between Q1/2 emitters to reduce the differential gain, so this gain stays very high. Where this degeneration is used, the CMRR is much less, and a current source may be used in the tail to increase it again.

The emitter resistor behaves approximately like a current source provided the base emitter resistance is much smaller and that you only consider the small signal model. Forget about the differential pair for now and consider the common emitter amplifier. The emitter current is $$I_\mathrm{e} = \frac{V_\mathrm{in}}{r_\mathrm{e} + R_\mathrm{E}} \approx \frac{V_\mathrm{in}}{R_\mathrm{E}},$$ using the fact that the base emitter resistance is small. Under the small signal assumption the voltage across the emitter resistor will remain relatively constant. Therefore the emitter resistor behaves approximately like a constant current source.

"How is the tail current kept constant by that resistor Re?"

No - it cannot be kept constant. However, the total current change can be kept very low. Just imagine that the long-tailed pair is excited with a small signal voltage Vin(+) only (Unsymmetrical operation with Vin-=0). In this case, the diff. amplifier can be seen as a common collector-common base series connection.

Now, let`s write down the collector current changes in both transistors, caused by Vin(+). Of course, Ic1=Ic2 with gm1=gm2=gm.

(1) d(Ic1)=gm[Vin(+)-d(Ve)] with d(Ve): Emitter voltage variation.

(2) d(Ic2)=gm[0-d(Ve)].

The small voltage variation d(Ve) is found to be (emitter follower loaded by 1/gm of the second stage): d(Ve)=Vin(+)[gm/(gm + 1/re)] .

Here the diff. resistance re is the effective dynamic resistance at the emitter node (with the input resistance 1/gm of T2): re=(1/gm)||RE.

Therefore,

(1) d(Ic1)=gm[Vin(+)][1-gm/(gm + 1/re)]

(2) d(Ic2)=-gm[Vin(+)][gm/(gm + 1/re)].

Now, for the condition d(Ic1)+d(Ic2)=0 we have

[1-gm/(gm + 1/re)]=[gm/(gm + 1/re)].

Solving this, we see that this equality is possible for re=1/gm only.

This means: Both variations of the collector currents compensate each other only (constant current in the common emitter path) if the differential resistance of the common emitter path re=(1/gm)||RE reduces to re=1/gm which is equivalent to RE>>infinity (Currrent source).

Comment: Therefore, of course both current variations will NOT cancel each other for a finite value of RE. This effect is expressed using the term "Common mode gain" (which is not zero for a finite RE).

A few days ago I decided to see what happens in the forum... and I came across this question. It seemed simple... but actually, the whole philosophy of the differential pair was hidden behind its answer.

Obviously, the HelpMee's “naive” question (persistently repeated several times) is about the nature of the “tail” - he/she wants to know WHAT, HOW and WHY is done there. I perfectly know the answer… but how do I explain it in the possibly simplest way?

OK, let’s start with a short answer: The differential pair consists of two emitter followers connected in parallel to a common load Re so that each of them tries to make the common emitter voltage equal to its input voltage.

At differential mode, they oppose each other by changing their currents in different directions so both voltage across and current through Re stay constant. In this case, the humble resistor does a good job. It should be resistor (not an inductor or something else) to set the common emitter current according to Ohm's law. This current stays constant; only the partial collector currents change (crossfade, redistribute or are steered). We take one of them, convert to voltage by the corresponding collector resistor and use as a single-ended output.

At common mode, the two emitter followers contribute each other by changing their currents in the same direction. As a result, the common voltage across Re and current through Re, partial collector current and output voltage change as well… which is undesirable. To suppress these current variations, we can make Re dynamic. Then when the emitter voltage increases, the Re resistance increases with the same extent and v.v. By this trick we artificially keep constant the ratio Ve/Re in Ohm's law, which is the common emitter current Ie.

See also this RG discussion - Can we reveal the basic idea behind the long-tailed pair and explain its operation in an intuitive way?

(Another clever trick would be to leave the constant ohmic resistor and make Vee variable. When the emitter voltage increases, the Vee magnitude decreases with the same extent and v.v. Thus we artificially keep constant both the voltage across Re - the numerator in Ohm's law, and common current Ie. See the RG discussion Are there situations in which the common emitter current source of the long-tailed pair can be replaced with a humble resistor?)

Now I will answer the question in detail as HelpMee's wants (and I like) - using “an analytic step by step explanation”. I will do it as a fictional story about the invention of the “long-tailed pair”. I will refrain from using special terms that discourage understanding.

1. Single common-emitter amplifier. In this configuration, we apply the input voltage to the transistor base and take the collector current (converted to voltage by the collector resistor) as an output signal. We can change the gain by inserting various “things” with different (differential) resistance dRe between the emitter and ground. Thus the stage acts as an emitter follower loaded with some “load”... but surprisingly, we do not use the voltage drop across this load as an output… instead we use the collector current that creates this voltage drop. Here are three typical cases:

dRe = 0. If we insert a voltage source, voltage stabilizing element (e.g. Zener diode)... or simply ground the emitter, the emitter voltage will be fixed. Figuratively speaking, the emitter is “stiff”, “immovable”. The whole input voltage is applied to the base-emitter junction and the gain is maximum.

dRe = Re. If we insert an ordinary ohmic resistor with static resistance Re, the emitter voltage will begin “moving” in the same direction as the input voltage. Now the emitter is “soft”, “movable”. A part of the input voltage is compensated and the gain is decreased (as they say, there is a negative feedback or emitter degeneration).

dRe -> infinity. If we insert a current source, or more likely, current stabilizing element (usually, a collector-emitter junction of a transistor), the emitter will become extremely “soft” and will exactly follow the input voltage at the base. As a result, the input voltage is fully compensated and the gain is zero.

The conclusion is that we can increase the gain by “hardening” the common emitter voltage and decrease it by “loosening” this voltage. This trick will lead us to the differential pair...

2. Paired common-emitter amplifier. To make a differential amplifier, it is not enough just to take two single common-emitter amplifiers for at least two reasons. First, we want to have a single-ended output but here we have a differential one. Second, they will amplify both input signals - differential and common-mode. Somehow we have to make them amplify as much as possible the differential signal but do not amplify (and even attenuate) the common-mode signal. According to the trick above, this means to “harden” the emitter voltages at differential mode and “soften” them at common mode.

We can do this magic by connecting their emitters to a common current stabilizing resistor (“current source” or a resistor in the simplest case). Now, at differential mode, each of them will fix the other emitter voltage thus providing maximum gain (the other transistor will have the “feeling” that a voltage source is connected in its emitter). At common-mode, both will work on a common current-stabilizing load that provides a minimum gain (both transistors will have the “feeling” that a current source is connected in their emitters).

See also my answer to the question Why the common-mode gain of the differential pair is almost zero?