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enter image description here

What is the purpose of Q1? Is it just to amplify the current? What region will Q1 be operating in and how do I determine this?

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    \$\begingroup\$ please cite the source of the diagram, unless it's your own work. \$\endgroup\$
    – Neil_UK
    Commented Jul 2 at 13:07
  • \$\begingroup\$ @Neil_UK This circuit was provided by university with no source unfortunately. \$\endgroup\$
    – Harrison
    Commented Jul 2 at 13:16
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    \$\begingroup\$ As this is a homework question, what work have you done on this problem so far? What step are you stuck at? What background have you learned that might be applied to solve it? \$\endgroup\$ Commented Jul 2 at 15:04
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    \$\begingroup\$ If there's no power supplies connected to the op-amp it probably won't do much of value. \$\endgroup\$ Commented Jul 2 at 15:08
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    \$\begingroup\$ Note that this power supply lacks a "current" protection ... which can be very useful. \$\endgroup\$
    – Antonio51
    Commented Jul 2 at 16:07

5 Answers 5

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Hmm. Okay. Let's redraw the schematic. Perhaps that will help you think about it:

schematic

simulate this circuit – Schematic created using CircuitLab

That should get you started.

Just think about each of the separate sections show above.

The left-most accepts an unregulated voltage and, by using a Zener's breakdown voltage, creates a simple reference voltage that doesn't vary much regardless of the unregulated voltage source or temperature variations. The opamp doesn't load it down, so the reference voltage should be fairly predictable.

The next part is a simple resistor voltage divider that divides down the output voltage by some specified fraction determined by the values chosen for \$R_2\$ and \$R_3\$. The resistor values should be low enough that the input bias and offset variations don't significantly impact the resulting feedback voltage developed by the divider. But not so low that it is wasteful of power.

These two inputs are then supplied to an opamp, which compares the two values and attempts to drive its output (to the BJT base) such that the divider's fraction of the output voltage matches the Zener's reference voltage. The opamp has the ability to supply whatever recombination current the BJT requires, so if the load current needed, the load attached to the output voltage node, varies in its needs the opamp can adjust itself in just the right way that the BJT (\$Q_1\$) still performs in just the right way to keep the output voltage relatively stable.

The collector of the BJT is directly connected to the unregulated supply. This just allows it access to the most-positive source so that its emitter can be held stable, in response to the opamp's attempt to keep the divided output voltage (due to the resistor divider already mentioned) equal to the Zener voltagе.

You should be able to see that there are three sections to the circuit and that two of them are dead-easy to understand. The only "difficult" part is understanding how the opamp observes differences in its inputs and uses negative feedback methods to keep the output voltage stable. So focus on that last part to understand this in detail.

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    \$\begingroup\$ Flow from above is always good. It gives much better perspective. \$\endgroup\$
    – Electronx
    Commented Jul 4 at 6:29
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    \$\begingroup\$ This does not answer the question, what is the purpose of the BJT? As far as I can tell, just connecting the opamp's output to the voltage divider would achieve the exact same result. \$\endgroup\$
    – Aetol
    Commented Jul 4 at 8:32
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    \$\begingroup\$ @Circuitfantasist Oh, I didn't mean to close their question but instead to select an answer, closing it that way. Regardless, that's up to Harrison to work out, I guess. I'd expand on the details of the BJT if Harrison expands on what's remaining to be said. I'm sure you would do the same! ;) And perhaps better than I would! In any case, Harrison needs to say something or do something. Or do nothing, I suppose. :) Otherwise, none of us have better clues than we already have. It's nice, I think, that you write with many different ways of seeing things! \$\endgroup\$ Commented Jul 5 at 10:32
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    \$\begingroup\$ periblepsis, ah, no... you're a specialist in transistors, I'm in resistors :-) I'm stopping here because we're being asked to chat ... \$\endgroup\$ Commented Jul 5 at 10:52
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    \$\begingroup\$ One thing to consider is that many op amps will have a circuit either functionally identical (i.e. a voltage buffer with low output impedance), or actually identical (i.e. NPN emitter follower) on chip. It is just obscured by the op amp symbol—we would see it if the transistorized amplifier was drawn. I would consider first explaining the function of the circuit with an internal BJT, i.e., just the op amp. Then we see the real purpose of moving the BJT off chip which is for heat dissipation and power handling. \$\endgroup\$
    – DavidG25
    Commented Jul 5 at 16:17
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Suggestion

Since this is homework, you will have to make an effort to understand this famous circuit solution. Okay, let's do it this way:

I have broken down the sterile university schematic into individual steps, and you will try to understand and explain what is behind them. Are you okay with doing it this way?

Implementation

OK, let's do it anyway...

15 V -> 5 V?

There are strange things in analog circuits. For example, to make an amplification in amplifiers, we introduce attenuation. Here we do something similar - to get a constant output voltage VRL (5 V) across the load, we reduce the input "unregulated" voltage Vin (15 V). Simply put, we decrease the voltage so we can increase it. The problem is how to do it.

schematic

simulate this circuit – Schematic created using CircuitLab

Note that the load is represented by an imperfect voltmeter with 1 kΩ internal resistance (you can change it in the parameters window).

Resistor in series

The simplest possible way (known since the 19th century) is to connect a 2 kΩ resistor R in series. A 10 V voltage drop occurs across it and the load voltage drops to 5 V; hence the name "ballast resistor".

schematic

simulate this circuit

But if we change the load resistance (from 1 kΩ to 10 kΩ), the output voltage significantly varies from 5 V to about 13 V. We can do this experiment by the CircuitLab DC Sweep Simulation.

STEP 2

Static voltage divider

To reduce the influence of the load on the voltage, we can shunt it with a resistor R2. Thus we get a voltage divider.

schematic

simulate this circuit

As a result, the output voltage changes "only" from 5 V to about 7 V but that is still a pretty significant variation.

STEP 3

Dynamic voltage divider

To reduce it, we can apply a clever trick - when the voltage tries to increase, the voltage divider decreases its "gain" with the same rate and vice versa. To make such a "dynamic divider", we can replace the bottom resistor R2 with a "dynamic resistor", e.g. a Zener diode. For convenience, as such I have used a 5 V "ideal" diode D from the CircuitLab library (you can change its forward voltage in the parameters window).

schematic

simulate this circuit

As we can see in the graph below, Vout is exactly 5 V.

STEP 4

Transistor buffer added

But if the load current becomes too high, the Zener diode will cease to function as a voltage stabilizer (aka "regulator"). To solve the problem, we can add a buffer (emitter follower Q) to supply the significant load current directly from the input source.

According to the negative feedback principle, the transistor compares (by its base-emitter junction) the output voltage with the reference Zener voltage, and changes its collector current to make them almost equal.

schematic

simulate this circuit

We see that there is a "small" problem - about 0.7 V are lost across the Q's base-emitter junction, and Vout is lower.

STEP 5

Error amplifier added

To solve the problem, we can connect an op-amp that amplifies the Vout - Vref difference. Note that the negative feedback is closed not to the op-amp output but to the circuit output. Thus the op-amp compenates the undesired Vbe error (disturbance) by increasing its output voltage with Vbe. The op-amp compares (by its differential input) the output voltage with the reference Zener voltage, and changes its output voltage to make them almost equal.

schematic

simulate this circuit

Now the circuit output voltage is exactly equal to the Zener voltage.

STEP 6

Amplification added

To be able to adjust the output voltage to the desired value (different from the Zener voltage), we can use the same trick as in negative feedback amplifiers take a portion of the output voltage instead the full voltage. For this purpose, we can use a voltage divider R3-R4 and compare its output voltage with Vref...

Transistor implementation: ... by a Q2 transistor amplifier.

schematic

simulate this circuit

Thus the Zener voltage will be multiplied (R3 + R4)/R4 times.

STEP 7

Op-amp implementation: Finally, we can use an op-amp amplifier to make a perfect voltage regulator.

schematic

simulate this circuit

As we can see from the graph below, the circuit is perfect.

STEP 8

The phenomenon: In both transistor and op-amp circuit, the voltage divider acts as a disturbance. The amplifier responds to this by increasing its output voltage/current proportionally and thus overcomes the disturbance. This is a fundamental property of negative feedback circuits, which is why they are so widely used.

How to understand negative feedback circuits

To visualize the operation of the negative feedback circuits above, it is necessary to think of the amplifier element (transistor or op-amp) as some kind of slow-thinking "creature" that compares, analyzes and changes the output value. If we think of it as an instantaneously acting amplifier element, we fall into a vicious circle and come to the paradoxical conclusion that the circuit cannot work.

And yet, what is Q1's role?

This question has been asked several times already, so it makes sense to answer it in more depth.

The idea

Indeed, at first glance, there is no need for a transistor because the op-amp performs its function (there is a transistor inside it, and not just one). But the op-amp is "weak", and cannot deliver the necessary current that a low resistance load would require. That is why we do as in life - we support the weak op-amp with a powerful transistor "in parallel". Its big collector current is added to the weak base current provided by the op-amp and it does the hard work. Let's take a look at how the two devices interact.

Implementation

Now we will do a few sophisticated experiments with which we will not only illustrate this question live (as opposed to dead academic drawings) but also show a more interesting and intuitive way to use the simulator:

  • For our experiment to be impressive, we need a "weak" op-amp (to need help and justify the use of the transistor). Fortunately for us, CircuitLab allows us to "weaken" a normal op-amp by going into parameters and setting a high output resistance, e.g. 1 kΩ.

  • In addition, we will increase the load (reduce its resistance to 100 Ω).

  • Then, to show the role of the transistor, we will do two experiments - with the collector on and with the collector off.

  • Finally, we will replace the collector-emitter part of the transistor with an ordinary resistor with equivalent resistance to show how simple the idea is.

The operation

Q's collector OFF: At power-up, the op-amp tries to power the load itself through the base-emitter junction, but fails; the load voltage "moves" slightly with about 1 V.

schematic

simulate this circuit

STEP 9.1

Q's collector ON: The transistor "senses" this and starts passing its powerful collector current (in parallel to the base current) through the load, thus forcing the voltage across it to rise. The operational amplifier monitors the circuit output voltage through the feedback loop and when it becomes equal to the setpoint it stops increasing its output voltage.

schematic

simulate this circuit

STEP 9.2

Conceptual circuit: Although the transistor is presented as something mystical, for the purpose of intuitive understanding, we can think of it as an electrically controlled resistor. Its resistance is non-linear but for our purposes we can model it as linear with the caveat that this will only apply to the current operating point. Its value can be either calculated from the above schematic or adjusted in the simulator; it turns out to be 48 Ω.

schematic

simulate this circuit

The graph is not true, but still interesting to see...

STEP 9.3

Even more conceptual circuit: After all, two voltage sources (weak and strong) are connected in parallel through resistors (high and low resistance) to the load, so that their voltages add up with (very) different weighting factors.

schematic

simulate this circuit

Not true but interesting to see...

STEP 9.4

Concepts seen

In the specific circuit implementations above, we can see important concepts and conceptual devices:

  • "Ballast" resistor
  • Static voltage divider
  • Dynamic voltage divider
  • Negative feedback
  • Emitter follower
  • Error amplifier
  • Voltage multiplier

They can help us to explain the operation of other circuits as well.

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The external transistor boosts the output current capability of the regulator. Without it, the current is limited by what the op-amp can deliver.

Put simply, the Q1 NPN is a high-current voltage follower, or a current buffer. Meanwhile, the op-amp is connected as a non-inverting amplifier with a gain of Vref * (1 + R2/R3), with Vref being generated by the Zener diode. Because R2 and R3 provide feedback, the circuit has the same gain regardless of whether there is a Q1 or not: the R2/R3 feedback cancels out any difference.

And what about Q1? It operates in its linear region until the regulator reaches its minimum possible Vin-Vout (the condition known as the dropout voltage), at which point Q1 saturates and the op-amp begins to source a lot of current into the base. (more about that below.)

Here's a sim to play with (simulate it here):

enter image description here

You'll see that most of the Vout current flows from the NPN collector to emitter, with the op-amp contributing the base current only. The collector and base (from op-amp) currents add together to drive Iout.

That is,

  • Ie = Ic + Ib
  • Ic = \$\beta\$ * Ib

Where \$\beta\$ (beta) is the transistor current gain, Ic/Ib.

Substituting for Ic, we have:

  • Ie = (Ib*\$\beta\$) + Ib = (\$\beta\$ + 1)*Ib

In this example I've set the regulator to 6.1V and inserted a 6.1 ohm load to get close to 1A Vout emitter current. With a transistor \$\beta\$ of 100, the 1A emitter current (Ie) would be the sum of 990mA (Ic) from the collector and 9.9mA (Ib) from the base.

Thus, for the simulation we get:

  • 1A = (100 + 1)*9.9mA

What else is going on? Because Q1 is an NPN follower, its emitter output Vout will be offset -0.7V (one Vbe drop) below its base, which in turn is driven by the op-amp output. The op-amp compensates for this offset and any other nonlinearity so Vout tracks the reference vs. the feedback.

(Exercise for the student: try the Vin slider. Bring Vin closer to Vout to simulate the dropout case. What happens, and why? How to fix it? Hint: see the comments.)

While this works, it has some serious drawbacks:

  • High drop-out (op-amp max swing and Vbe)
  • Gain limited by single NPN beta

The first drawback means that the dropout voltage, that is, lowest possible Vin-Vout, is at best case 0.7V if the op-amp can swing all the way to Vin; it gets even worse if it can't. For example, if the op-amp can only swing to Vin-2V, the total dropout voltage will be 2V+0.7V = 2.7V.

The dropout issue can be overcome if the op-amp bias runs off a separate, higher voltage. With the increased op-amp swing we can drag the NPN base up higher than Vin. Then as Vin approaches Vout, dropout becomes only the collector-emitter saturation drop (Vce(sat)) of about 0.45V.

Then there's \$\beta\$. I used a beta of 100, but good design implies that you assume much lower betas than typical, especially for big power transistors. If the worst-case design beta were only 20, we'd need be sourcing 50mA from the op-amp to get 1A at the emitter. That's kind of a lot.

(Side note: what if you used an NMOS pass? You have the same issue: needing to overcome the Vgs threshold with a higher-than Vin gate drive. There are regulators that do this, including TI's 'Cap Free' family that offer good stability and fast transient response. They use an internal charge pump to make the gate drive bias. And because it's MOS, the op-amp only needs to provide voltage, not current.)

The dropout and beta issues can be reduced by using an enhancement: PNP bypass instead of NPN, with an NPN buffer (simulate it here):

enter image description here

With this arrangement the op-amp only needs to swing positive enough to turn on the NPN. The NPN collector then sinks current from the PNP base. The dropout is reduced to just the PNP's Vce(sat), again about 0.45V.

We also get much greater gain due to the combined betas of the PNP and NPN, which form a Sziklai pair. We get higher currents with less op-amp drive.

(Again, try the Vin slider and see what happens as the circuit approaches dropout. How would you fix it?)

A final note. I've used a 5.6V Zener reference which limits the minimum Vout to 5.6V. Most linear (and switching) regulators will use a bandgap reference of about 1.25V. This is why many regulators have that as a minimum voltage. It's possible to go lower by either adding gain to the feedback path or by scaling down the reference.

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  • \$\begingroup\$ Your answer is very well and concisely written and provides useful information. Just to point out in good faith two things I noticed: 1) In your last simulation there should be resistors in series to the bases of transistors; 2) In the web page about the Sziklai pair, in the first simulation the emitter and collector of the PNP transistor are swapped (this is of course not your problem, but it still bothers you a bit because you provide a link). \$\endgroup\$ Commented Jul 5 at 9:53
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    \$\begingroup\$ The resistors are unnecessary so long as the op-amp is in loop closure. It does become an issue as Vin approaches Vout and the transistor goes into saturation, then some current limiting is needed. That said, OP’s diagram doesn’t have base resistors either, so I left them out in my sims to avoid confusion. \$\endgroup\$ Commented Jul 5 at 16:46
  • \$\begingroup\$ Interesting... I hadn't thought about that. I wonder if it's a general principle for negative feedback circuits. I can't think of another configuration like this right now. But can you always rely on a properly functioning feedback loop? For example, if the output is short-circuited, the op-amp will raise its output voltage to its maximum. Also, the op-amp output voltage will stay very close (about 0.7V) to ground; so the amplifier must be a rail-to-rail op-amp. In the OP's circuit, the transistor is connected in an emitter follower that does not require a base resistor... \$\endgroup\$ Commented Jul 5 at 16:51
  • \$\begingroup\$ Now I realize that in a simple emitter follower without a base resistor, the emitter voltage automatically limits the current, and the problem of shorting the output is still there. So a small base resistor would limit the current in this case. But still, these are details that are not so important at this stage. \$\endgroup\$ Commented Jul 5 at 17:02
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    \$\begingroup\$ What ends up happening with an NPN emitter follower as shown is that as Vc approaches Ve, the op-amp starts injecting a lot of current into the base to make up the difference - that is, the op-amp is back to sourcing big current into the load. With the PNP the op-amp does the same thing, and the NPN current sink also maxes out. Neither situation is good, in a real design it would let the magic smoke out. In fact, the sim also gives up as the currents turn stupid. So, yes, base current limiting is needed to allow the regulator to behave itself as it enters into dropout. \$\endgroup\$ Commented Jul 5 at 18:37
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Diode Z1 is producing a fixed potential, which for argument's sake, let's call \$V_{REF}=+3V\$. This will be used as a reference to control the output potential \$V_{OUT}\$.

You might use only the op-amp (and resistors R2 and R3) to produce any \$V_{OUT}\$ you desire, just by changing R2 and R3. You're already familiar with the non-inverting amplifier configuration:

schematic

simulate this circuit – Schematic created using CircuitLab

We have gain \$1+\frac{R_2}{R_1}\$, and as you can see on the left, without any load connected to the op-amp's output we obtain the expected output:

$$ \begin{aligned} V_{OUT} &= V_{REF}\left( 1+\frac{R_2}{R_1} \right) \\ \\ &= 3V \left( 1+\frac{10k\Omega}{5k\Omega} \right) \\ \\ &= 3V \times 3 \\ \\ &= 9V \end{aligned} $$

On the right above, though, I have installed a modest load that would normally draw \$I=\frac{V}{R}=\frac{9V}{90\Omega}=100mA\$, but here draws only 15mA. The output impedance of the op-amp is so high that it is unable to provide 100mA, and maintain the required 9V, and output voltage falls dramatically. Really we shouldn't ask for more than 10mA or so from a typical op-amp.

The purpose of the BJT is to improve the ability to source current. The transistor is able to pass much more than 15mA. For instance using a transistor with current gain \$\beta=100\$, we could theoretically get \$10mA \times \beta=1A\$ or so.

The transistor is employed as an emitter-follower (common collector), for which the emitter follows the base, but 0.7V lower (approximately; that value varies with base current):

schematic

simulate this circuit

In this configuration, the transistor is able to source 100mA (and probably much more) from the collector, while base current remains a factor of \$\beta\$ smaller. The only problem is that the base needs to be held 0.8V higher in potential than the emitter, for the output to be 9V exactly. Ignoring that inconvenience for the moment, note that an op-amp would be easily able to source the 1mA of base current required here.

We can combine the non-inverting amplifier arrangment with this emitter-follower, as follows:

schematic

simulate this circuit

If you look closely, that's exactly the same schematic as yours, only drawn slightly differently to help understand its function.

In this arrangement, negative feedback is not taken from the op-amp's own output at B, but rather from the overall output OUT. This ensures that the op-amp adjusts its own output B to whatever voltage is required to satisfy the condition of its own two inputs having equal potential, which happens when \$V_{OUT}=+9V\$. In other words, the base-emitter junction of transistor Q1 has been inserted into the feedback path, so overall gain remains exactly \$1+\frac{R_2}{R_1}\$, in spite of the 0.8V "drop".

You can see that the op-amp has found the exact base voltage \$V_B=+9.785V\$ required for the emitter to be at \$V_{OUT}=+9.0V\$ as before, but this time the op-amp is not overloaded, and the transistor is easily able to supply the 100mA demanded by the 90Ω load.


Emitter followers operate exclusively in the active region, never saturated. Technically, with no load, they could be in cut-off, but that's not the case here, especially considering that the transistor is also providing the ever-present feedback current via R1 and R2.

Since emitter-followers have near-unity gain, the only thing being amplified by that transistor is current, and you could say that it is a current amplifier only.

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  • \$\begingroup\$ As always, I enjoy reading your answer because you very well explain key points of circuit solutions and illustrate them with "live" CircuitLab schematics. Regardless of the fact that we do not keep in touch, I am happy to have at least one more like-minded person here. \$\endgroup\$ Commented Jul 5 at 10:08
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    \$\begingroup\$ @Circuitfantasist Thanks! You've surprised me with a few insights, always worth a read \$\endgroup\$ Commented Jul 5 at 11:01
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There are many good answers already, but I think one key point is missing: an external BJT is used because it has better heat dissipation and power handling.

The purpose of the Zener voltage reference has already been explained well, so I will leave it out—the output of the voltage reference will be our input signal. Now, consider the circuit without the external BJT: we are looking at a non-inverting amplifier.

schematic

simulate this circuit – Schematic created using CircuitLab

Now consider that you have implemented this circuit with a 741 op amp. Re-drawing the circuit using the actual op amp schematic rather than the op amp symbol, we see that we already have a BJT at the output, connected as an emitter follower, just like the circuit in the question. You can ignore the resistor and BJT right next to it—that is an over-current protection circuit, and it is effectively not there during normal operation.

enter image description here

So we already have an emitter-follower in this circuit, whether we put one there ourselves externally or not. The purpose of the emitter follower is explained by several of the answers in terms of current gain of a BJT, a low impedance output stage, etc. But if we already have an emitter follower in the op amp, why do we add another one outside of the amplifier?

The answer is because the external BJT can dissipate much more heat and therefore handle more power. In the op amp, the physical dimensions of all devices are minimized, because silicon area is directly proportional to cost: the smaller a chip, the less it costs to make. The tradeoff is that small physical dimensions result in poor power dissipation: a small amount of current will lead to a large temperature rise, and the op amp will break once a certain temperature limit is reached. In contrast, the external BJT has large metal leads which act like a heat sink. Also, part of the marketing of the external BJT is how much power it can handle, so the devices are made larger even though it increases the silicon area. High power BJTs have tabs for connecting to an even larger, separate heat sink.

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  • \$\begingroup\$ Nicely written answer, David! This explanation was needed. Your answer made me want to think more about this configuration. Looking at the output stage of the famous op-amp with the transistor outlined by you, an idea occurred to me - does it not form a Dallington pair together with the external transistor? Just another perspective on this configuration... \$\endgroup\$ Commented Jul 5 at 17:20
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    \$\begingroup\$ It does form a Darlington pair. This just means we don't have a lot of current in the op amp output stage. You can take this concept to the extreme by replacing the external BJT with a MOSFET which has infinite current gain at DC. Then we can talk about AC current gain of MOS vs BJT, but I think it is beyond the scope of the question. \$\endgroup\$
    – DavidG25
    Commented Jul 5 at 17:25
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    \$\begingroup\$ You can choose the input to the output device be current or voltage. Independent of the underlying physics, we can easily switch between the two by multiplying or dividing by the input impedance of the device. The key is that MOSFETs have a near infinite input impedance at DC (the gate oxide is an insulator). The input impedance at higher frequencies may be low, especially for big power FETs, because the gate capacitance is large. \$\endgroup\$
    – DavidG25
    Commented Jul 5 at 17:45
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    \$\begingroup\$ So the FET is Iout = Gm*Vin. Vin = Iin*Zin. So then Iout = gm*Iin*Zin. Zin=1/(s*Cin), and yes, the DC value is limited by leakage paths. \$\endgroup\$
    – DavidG25
    Commented Jul 5 at 18:00
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    \$\begingroup\$ Well, usually there is another shunt element to define the impedance at the gate. In the regulator example, the impedance at the gate would be the output impedance of the emitter follower. In other words, only the bias current flows in the op amp output stage. \$\endgroup\$
    – DavidG25
    Commented Jul 5 at 18:09

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