What's inside an op-amp?
The internals of operational amplifiers are typically not explained well. Real talk about the differential pair.
In an earlier article on signal amplification, I discussed a couple of basic transistor amplifier circuits and then introduced op-amps as a better-behaved version of the same. This left one question unanswered: what’s inside an op-amp, anyway?
It’s easy to give a superficial answer. A typical operational amplifier has three main parts: a differential input stage, a voltage amplifier, and a push-pull output circuit. We covered the latter two in the previous article, and it’s tempting to wave away the first bit too.
Yet, the differential stage is a pretty interesting animal. At its heart lies an arrangement of transistors known as the differential pair (or “long-tailed pair”), and an accurate explanation of its mechanics is hard to come by. I’ll try to address this below — but let’s start with a brief detour.
The I-V curve of a field effect transistor
Let’s consider a typical n-channel MOSFET. We know that it’s a voltage-controlled device: that is, a voltage signal applied to the gate terminal (Vgs) modulates the flow of current between the two other terminals — drain and the source (Id).
From this description, one could assume that FET is just a variable resistor: a device where the gate voltage dials in a specific drain-source resistance, and where that resistance stays roughly the same no matter what voltage (Vds) appears across the other two terminals. If that model is true, the drain-source I-V plot (Id-Vds) for a given gate voltage should look about the same as it does for a resistor:
The diagram, captured for a 10 kΩ resistor, makes perfect sense: I = V/R, so if R is constant, the current through a resistive load should be directly proportional the applied voltage.
But that’s not at all what happens with a FET! Don’t take my word for it. I recorded the following curve for a common BS170 transistor at Vgs = 2.25 V. I selected this voltage empirically just to maintain the same scale for both plots:
As it turns out, the resistance of a FET is strongly dependent not only on the Vgs input signal, but also on the potential across its drain and source terminals. In particular, for any given Vds past a modest threshold, the transistor appears to be a (nearly) constant-current device. In the plot, the current stays at 800 µA +/- 10% for Vds all the way from 0.5 to 9 V.
The long-tailed pair
Keeping this detail in mind, let’s get back to op-amps. As mentioned earlier, the essential building block of most differential amplifiers is the long-tailed pair; its simplest variant, drawn with n-channel MOSFETs for consistency with the earlier article, looks the following way:
The circuit consists of a pair of voltage followers sharing a common resistor on their low side. Each transistor will conduct only if the voltage on its gate terminal is sufficiently higher than the voltage on the source terminal (Vgs > Vth). Because the flow of current through the transistor causes a voltage to build up across R2, the transistor will admit only as much current as needed for the resistor voltage to reach Vin - Vth.
Let’s make a couple of assumptions about component values and voltages and look at what happens if A > B:
In this scenario, the equilibrium state is a 200 µA current is flowing entirely through the left branch of the circuit; the transistor on the right is not conducting because its source terminal is sitting at 2 V and the gate is only 1 V higher than that (i.e., Vgs < Vth). And because this transistor is off, the voltage at the output node is equal to Vdd.
In the inverse situation (A < B), the branch on the right ends up doing all the work:
Because there is 200 µA flowing through the upper resistor, there is a V = IR = 2 V drop at the output pin; in other words, Vout = 15 V - 2 V = 13 V.
Finally, if A and B are both at 4 V, both branches are handling 50% of the current, so the voltage drop across the upper resistor is halved, putting Vout at 14 V.
At first glance, this might sound like the behavior of an op-amp: we have three distinct output voltages for A < B, A = B, and A > B. Indeed, many online sources try to pass it off as a differential amplifier. That said, the circuit has a fatal flaw: its output voltage doesn’t depend only on the difference between A and B. To illustrate, let’s consider a scenario where A = B, but their common voltage is a bit higher — say, 6 V:
In this scenario, a drop of 4 V needs to develop across the bottom resistor before the transistors reach their cut-off points. This results in a combined current of 400 µA divided equally across both branches. In the right branch, the 50% current (200 µA) causes a 2 V drop across the output resistor, putting Vout at 13 V — exactly the same as in the A < B scenario before!
In other words, the differential amplifier is a dud. It is sensitive to common-mode inputs, making it impossible to reliably distinguish between differential voltages.
A workaround is hinted at by the name of the circuit: it’s called a “long tail” because in order for the amplifier to behave correctly, you need to maintain ample distance between the permissible input voltages and the bottom supply rail. For example, in a 15 V circuit, we could constrain the inputs to a range between 14 and 15 V. This way, a significant voltage drop would always be present across the “tail” resistor, and the overall current would vary only a tiny bit — making it easy to distinguish between 0%, 50%, and 100%.
Toward a shorter-tailed pair
Of course, this input constraint is rather severe, especially in the age of single-supply electronics that commonly run off as little as 3.3 volts.
To improve this circuit without sacrificing its input range, it would be necessary to replace R2 with a device that can recreate a wide range of voltage drops while requiring only fairly modest current swings. This would make the current flowing R1 fairly insensitive to any common-mode artifacts.
Well — that’s our cue to wheel out the previously-discussed FET I-V curve:
As seen in the graph, a spectrum of voltage drops ranging from 0.5 to 9 V (x scale) can be created by varying current by as little as +/- 10% (y scale).
A single FET with a carefully trimmed gate voltage could be used as a drop-in replacement for R2. That said, its characteristics will drift in response to factors such as ambient temperature, so it’s best to add some sort of a feedback loop. One such circuit — known as the current mirror — is shown below:
For a moment, let’s focus only on the part on the left. At first, there’s no current flowing through the transistor, so the drain terminal quickly shoots toward Vdd; so does the gate terminal, prompting the transistor to start conducting. This, in turn, creates a voltage drop across R, eventually lowering the gate voltage toward the cut-off region. This is a negative feedback loop; an equilibrium is reached when the transistor is admitting current equal to I = (Vdd - Vth) / R.
On its own, this is worthless, but there is a carefully matched transistor on the right. It has no feedback mechanism, but is supplied with the same gate voltage — and therefore, should admit the same current through the connected load, with relatively little dependence on what the load is.
This particular current mirror is controlled by a fixed reference — R — so it becomes what’s known as a “current source”. This term is somewhat sketchy because the device is not a source of any electromotive force in the sense that a battery would be; rather, it’s just a current-limiting contraption. But I digress.
Current sources have their own (uncommon) circuit symbol, so a revised version of the not-so-long-tailed pair might look this way:
And that, in a nutshell, is a very basic but reasonably well-behaved differential stage.
Because the Vgs < Vth dead zone near 0 V is typically undesirable, the circuit is often constructed as an upside-down version with p-channel MOSFETs. It’s also possible to employ JFET transistors to get rid of the threshold altogether. In all cases, the operating principle is the same.
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Minor nitpick: The arrow in the current source symbol represents the direction of current flow. The second long-tailed pair diagram has it the wrong way around.