BJT Differential Amplifier problem

Question

I have tried to reproduce and understand the BJT differential amplifier, and unfourtunetely I have encountered some issues. The circuit is from Art of Electronics, 3rd edition.

First, my question is on what basis does the circuit function? I would guess that the 0 degrees phase shift sinusoidal wave of Q1 acts on Q2 to amplify the voltage difference Vq1-Vq2 by a factor G. but how does this work exactly? Q1 and Q2 aren't in direct contact with one another's terminals. They only share the same voltage rail supplies, but other than this I don't see how does the amplifier amplify the difference between the two waves. The common mode gain I would guess that is near 0 because as Vq1 increases or decreses so does q2 which wouldn't give a voltage difference too big to amplify as Vout equals nearely (vq1-Vq2)G. Plus if the signals are equal in amplitude, the amplification would be 0. I only want an explanation regarding the amplifier process of this circuit and how does it work.

My other problem is recreating this circuit in circuitlab. I tried to build a similar one and understand better this way how the circuit works. The only problem is that...... the circuit behaves differently behind my knowledge of what goes wrong.

^{simulate this circuit – Schematic created using CircuitLab}

The differential amplifier transforms the sinusoidal inputs into some sort of cracky square wave. My first thought was that the amplitude of the voltage AC sources was too high and that the RMS voltage saturated the transistors for too long of a time. Nothing happened as expected once I made the modification. Then I reduced frequency and increased rail negative and positive voltage swings. I have modified the paramaters for resistance and the same result appears. I begin to believe that it is circuitlab's fault, but who knows? I sure don't. Oh and as a note, I have removed Re from both emitters as of lack of space on circuitlab. But this I don't think that should modify too much the linearity as circuitlab doesn't count for temperature change. What is the problem? Plese anyone explain. I understand the gain functions, I just do not understand how this circuit works and what is the problem with my reproduction of the circuit. Thanks!

Answer 1

The motive

Transistor differential amplifier (aka "transistor differential pair", "emitter-coupled pair", "long-tailed pair", etc.) is an ingenius circuit solution that has puzzled human minds for more than a century (me from the early 90s); the OP's question is the most recent example of this. I was moved by the effort, desire and enthusiasm with which OP seeks the answer to basic questions... and decided to write this incredible story about the famous circuit. It has nothing to do with the boring, banal and trite textbook "explanations" that actually do not explain but only say. This is a fictional story, a kind of "circuit fairy tale"... and only a possible scenario for the "invention" of the transistor pair.

Basic idea

I would formulate it like this: Gain control of a common-emitter stage by inserting various sources into the emitter.

Reinventing the circuit

Let's now build the circuit step by step following the basic idea above. For the purposes of this intuitive explanation (and to the dismay of conventionally thinking minds) I will frequently use figurative names in quotes.

"Stiff" emitter

Common-emitter stage with grounded emitter and current output: A BJ transistor has a voltage input (base-emitter junction) and a current output (collector-emitter junction). The simplest way to drive it is to ground the emitter (Ve = 0 V) and apply the input voltage Vin to the base. Figuratively speaking, the emitter is "immovable" and we "move" the base.

^{simulate this circuit – Schematic created using CircuitLab}

In the CircuitLab simulation, open Vin's parameters window and, looking at the ammeter Iout, carefully set the input voltage (658 mV in the schematic) so that a (quiescent) current of 1 mA flows through the collector. This initial input voltage is known as a "bias voltage". You can then wiggle Vin around that value and watch Iout wiggle too around 1 mA.

Common-emitter stage with grounded emitter and voltage output: But our devices "prefer" to work with voltage. So let's insert a resistor Rc in the collector circuit to convert the collector current Iout into an output voltage VRc = Iout.Rc. For convenience, let's use the complementary (to Vcc) voltage Vout which is relative to ground.

^{simulate this circuit}

Now adjust the resistance Rc so that the initial (quiescent) output voltage is equal to half of the supply voltage (5 kΩ in the schematic). Wiggle Vin and watch Vout; the circuit has maximum gain.

Common-emitter stage with "lifted" emitter: The circuit will behave exactly the same (have maximum gain) if we "shift" the emitter and base voltages by, say, 1 V.

^{simulate this circuit}

Conclusion: A transistor stage with a voltage source in the emitter (fixed emitter voltage) has maximum gain because the entire input voltage is applied to the base-emitter junction.

"Soft" emitter

Common-emitter stage with a resistor in the emitter: If when, figuratively speaking, we "move" the base, the emitter also "moves" (in the same direction but with a smaller amplitude), the input voltage applied to the base-emitter junction and, accordingly, the gain decreases... as they say, because of the negative feedback aka "emitter degeneration". We can make the transistor do this by itself if we insert a resistor Re into the emitter (in place of the voltage source).

^{simulate this circuit}

"Very soft" emitter

Common-emitter stage with a current source in the emitter: We can greatly enhance this effect almost zeroing the gain if instead of a resistor we include a current source Ibias or, more precisely, a current-stabilizing non-linear (dynamic) resistor. As a result, the emitter voltage will follow the input voltage exactly and the difference applied to the base-emitter junction will be zero.

^{simulate this circuit}

"Hardened soft" emitter

So these are the rules:

• If we want the transistor stage to amplify as much as possible, we must include a voltage source in the emitter.
• If we want it not to amplify, we must include a current source.

Voltage source in parallel to the current source: We can do this switching easily by connecting the voltage source in parallel without disconnecting the current source; the voltage source overrides the current source. As you can see, this schematic is equivalent to the third schematic above.

^{simulate this circuit}

Variable "softness"

And how do we do the opposite - disconnect the voltage source? Well, here is the biggest trick... The transistor is actually a voltage source (follower) and we have included a second voltage source in parallel which above was causing conflict and high gain. But what if we keep this voltage equal to the emitter (input) voltage? No current will flow between them as if there is no second source. The name of this technique is "bootstrapping".

To implement this idea, we can connect a second emitter follower Q2 in parallel to Q1 and drive its input by an equal input voltage Vin2 = Vin1. As they say, this is the so-called "common mode" input signal.

^{simulate this circuit}

As a result, in the common mode, both base and emitter voltages will change equally and there will be no gain. If we stop the voltage from changing or start changing it opposite to the first voltage (differential mode), the gain becomes maximum.

Now it remains only to draw the schematic more beautifully to get the classic "long-tailed pair".

^{simulate this circuit}

Interacting voltage sources

But now we notice something even more interesting - the Q2's emitter is not only an output (of an emitter follower) but also an input (of a common-base stage). This is because the base-emitter voltage can be varied by both Vb and Ve (the so-called common-base stage). The same goes for Q1. So this is a symmetrical configuration where the two transistors interact (the OP's "interference").

Helping voltage sources (differential mode): When both input voltages change in the same direction, the common emitter voltage follows them, and the two interacting voltage sources "help" each other. Both base-emitter voltages and collector currents do not change; the output voltage also does not change, and the gain is zero.

Opposing voltage sources (common mode): When both input voltages change in the opposite direction, the emitter voltage does not change ("virtual ground"). The base-emitter voltages change with the maximum input voltage amplitude, collector currents change and the gain is maximum. Figuratively speaking, a kind of "tug of war" is observed where one transistor "pulls up" the emitters and the other "relaxes the pull" (at the same time the current source "pulls down" the emitters). Two "pull-up" elements (transistors) and one "pull-down" element (current source) are connected to the common point of the emitters.

Movable ground

I will take advantage of the OP's mistake of connecting the emitters to ground (the second schematic in the body of the question) to show the genius schematic idea here (by the way, when I was doing my thesis in 1980, I also wondered why the emitters were not connected through separate resistors to ground but through a common one).

Long-tailed pair. The clever trick here is that the emitters are connected to a movable but not static ground. Thus, when common-mode (same) input voltages are applied, the ground "moves" with them, and they are not amplified. When differential (opposite) input voltages are applied, the ground does not "move" with them, and their difference is amplified.

OP's pair. In contrast, the OP's transistor emitters are permanently connected to ground. As a result, they amplify equally both common and differential input voltages... but the main requirement of a differential amplifier is to amplify only differential signals and suppress common-mode signals.

Why emitter current source?

This is the second main point in this circuit that remains to be fully clarified. The problem is that the circuit has a differential input but we want its output to be "single-ended" (with respect to ground). A simple ohmic resistor with as much resistance as possible connected between the emitter and ground (or negative power supply) has somewhat solved the problem in the past (and in the H&H bestseller). But when the common input voltages and, accordingly, the emitter voltage Ve vary, the emitter current Ie varies accordingly to Ohm's law - Ie = Ve/Re. So the collector currents and the output voltage vary as well. The problem is because the emitter resistor is constant. If it was a "dynamic resistor" whose resistance increases when the voltage inreases and v.v., their ratio Ve/Re (the current) would stay constant. Accordingly, Ic and Vout would not change.

So, the common device between the emitters and ground (the circle with an arrow inside) is not a true current source (producing a current) but a constant-current dynamic resistor (setting the current). It is usually implemented by another transistor with a constant input voltage.

Limitations

Maximum common-mode voltage: Due to the significant gain, the input differential voltage is insignificant against the background of the common-mode voltage, which varies widely; let's see in what limits.

When both input voltages simultaneously increase (common mode), the emitter current "source" increases its internal resistance with the same rate. In the same time, the collector-emitter jubctions decrease their "resistance" with the same rate. As a result, the total resistance between the supply rails, quescient collector current and output voltage do not change. Only there is a "small"problem.

The voltage drops across the collector-emitter junctions continuosly decrease and when approach zero, transistors exhaust their supply of "resistance" and as they say, they saturate. So, the common-mode voltage is limited above by the quiescent collector voltage.

Answer 2

There's a couple of things. First of all, the circuit is not the same as that from Fig. 2.63. Both transistor's emitters are connected to ground, and the negative power supply does nothing besides dissipating power in R1.

Let's draw the actual circuit from Fig. 2.63:

^{simulate this circuit – Schematic created using CircuitLab}

The circuit must be symmetric to work as a normal differential amplifier.

Let's see how the outputs look for VIN = 2Vp-p, 0.2Vp-p, and 0.02Vp-p respectively:

Clearly, at 2Vp-p input the circuit has way too much gain for the output signal to fit between about -0.5V and +5V - the output range of the differential stage here.

At 0.2Vp-p we can see that the output waveforms clearly resemble the input waveforms, but the input voltage is still too large.

At 0.02Vp-p, we get the desired clean action of a differential amplifier.

I have tested this on the bench, and the performance is very close to the simulation results. The transistors in my case were matched to have the same offset voltage to within 0.1mV and were super-glued together to maintain thermal bond. Both outputs have to be loaded the same - the circuit misbehaves if there's a scope probe connected only to one output, as that loads the collector of one transistor differently from the other. In the simulation everything in each half of the circuit is completely identical, so the bench model had to be as close as possible to that.

In a practical, easy-to-put-together breadboard variant you could use transistors without matching the V(be) offsets. Then, one of the AC sources would need to have its DC offset adjusted +/-20mV from 0V to bring the DC component of the output signal within the output range of the amplifier.

If the input was truly floating - no galvanic connection to the supply rails or ground, say a transformer winding - then the inputs could be self-biased, and the output offset could be nulled with a potentiometer in the emitter leg:

^{simulate this circuit}

The outputs for Rbal at either extreme and the center position are:

Self biasing would be used for free-potential AC inputs - say if this was a differential amplifier front-end for a balanced XLR line audio input.

A simple floating AC source is a speaker, used in this case as a microphone. Works great as shown. Rbal may need to subsume more of the Re, so for simplicity you could have RBal=1kΩ, and Re1=Re2=499Ω.

Otherwise, the input potential common mode would be expected to be within the operating range of the input stage. For a stage with resistive collector and emitter loads, the gain varies with common mode, so that usable common mode range is narrow. For a stage with current source emitter load and a current mirror collector load, the gain is much higher and has less dependence on the common mode voltage, but much higher dependence on temperature. Thus, such high-gain input stages only make sense when negative feedback is available to control the gain - as is the case in an op-amp, for example.

Answer 3

I got the circuit right and now I have tested my hypothesis with the interference between the two voltages at Q2's emitter.

Again this is the modified amplifier with .03V amplitude at the input base of Q1 and Q2, and with the ground detached from the emitters of the transistors (it mostly disturbed the linearity, opsie doosy). And now I have plotted the voltage at the bases versus the voltage at Q2's emitter.

where Vp and Vn are the voltages at the bases and Ve is the voltage at the emitter of Q2. It seems like the voltages of the transistors do interfere at the emitter, and that it get's amplified as a result of the voltage difference. Now let's plot Ve versus Vout (at Q2's collector) and see what happens

the voltage entering q2 get's amplified by the respective voltage gain, as described by a differential amplifier. the difference of voltages at Q2's emitter between Vq1 and Vq2 get's amplified by G and the resulting voltage is observed at Q2's collector. If we would merge the two collectors together to see their respective voltages we would get a higher amplification resulting from the difference of the two Vcollector voltages. From all the gathered infromation I will state this as the final answer. Any other additional comments are welcomed.Thanks!