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I am trying to understand the functioning of this circuit since I have read that it is appropriate for power regulation:

enter image description here

Here is the explanation that I have found:

The zener diode provides the reference voltage to the positive input of the opamp. Note that any other kind of voltage reference, such as a bandgap voltage reference would also work (it is actually better). A voltage divider composed by R1 and R2 provides a measure of the output voltage suitable for the opamp, i.e., if we want the output voltage to be k times higher than the voltage reference, than the voltage divider should divide the output voltage by k so that both inputs of the opamp can be equal for the desired output voltage. Then, the op amp supplies a voltage to the gate of the MOSFET transistor such that its negative input (the output voltage) follows its positive input (the reference), due to the beautiful effect of negative feedback. Any perturbation of the input current or the load current (or even others) will affect the output voltage, but the feedback loop will quickly adjust the driving of the BJT base to stabilize the output voltage at the desired reference.

But I pretty much am lost after they talk about the output voltage being divided down and then compared to a reference voltage. My question is, how does the comparing of \$V_\text{n} = V_{\text{out}}\times\frac{R_2}{R_1+R_2}\$ to \$V_{\text{ref}}\$ drive this circuit? I just find it so complex even though I know how to apply the equations to see which components I need, and I want to understand it conceptually.

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  • \$\begingroup\$ Do you know and understand how a noninverting amplifier works? \$\endgroup\$
    – G36
    Commented Feb 15, 2021 at 18:18
  • \$\begingroup\$ @G36 I believe I do? I should know that it takes in the input voltage at the positive terminal and amplifies the signal. I also know the inout terminals have ideally, infinite impedance, and that the voltage of both positive and negative terminals is equal. I am more unsure about how it interacts with the transistor. \$\endgroup\$
    – Bee
    Commented Feb 15, 2021 at 18:20
  • \$\begingroup\$ Notice that we have a noninverting amplifier here as well, Vin is a Vref and noninverting gain is set by the voltage divider ratio Vout = Vref * (1 + R1/R2) \$\endgroup\$
    – G36
    Commented Feb 15, 2021 at 18:20
  • \$\begingroup\$ And the BJT is here to provide a "current gain" only. \$\endgroup\$
    – G36
    Commented Feb 15, 2021 at 18:22
  • \$\begingroup\$ I believe this is a MOSFET right? Since it's voltage-driven \$\endgroup\$
    – Bee
    Commented Feb 15, 2021 at 18:28

3 Answers 3

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enter image description here

First note that the feedback from the output is connected to the inverting input of the op-amp. That means were dealing with negative feedback and the op-amp output will stabilise when the non-inverting input (1) and inverting input (2) are at the same voltage.

  • At power-up Vout is zero as Vin begins to rise.
  • The voltage at (1) will rise to Vref.
  • Since the feedback signal (2) is close to zero and the reference voltage (1) > 0 the op-amp output (3) will start to swing towards the positive supply.
  • The transistor is wired as an emitter follower so the voltage at (4) will be V4 = V3 - 0.7 V due to the base-emitter voltage drop.
  • The voltage at (5) will rise and this feeds back to (2).
  • As V2 gets very close to V1 the output of the op-amp will start to fall until V2 = V1. At this point V5 = Vref.

So if V5 = Vref then \$ V_{\text{out}} = \frac {R_1 + R_2}{R_2}V_{\text{ref}} \$.

You've changed your schematic to a MOSFET since I started my answer but the principle is the same.

From the comments:

How does the Zener diode fit into this?

Note that the Zener diode is pointing towards the positive supply so that it is reverse biased. A regular diode will not pass any current in reverse mode1 unless you exceed the reverse breakdown voltage at which point you have probably destroyed it.

Zener diodes are designed to breakdown in a controlled manner at a specific voltage. All that is required is to feed it a voltage higher than it's breakdown voltage with some form of current limiting, R3 in this case. The voltage at (1) will be reasonably steady for a wide range of voltages Vin > Vref. In practice you'll need a little "headroom" of a couple of volts at least.

1 Not quite true. There is a little leakage current and you can get its value from the datasheet.

So ... what heavy lifting does MOSFET do?

Op-amps typically have an output capability of a few tens of mA max. The addition of a BJT or MOSFET increases the output current capability. It's like power steering; the op-amp is the little old lady at the wheel and the transistor is the hydraulic amplifier moving the steering rack.

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  • \$\begingroup\$ Thank you, I appreciate the walk-through of the circuit. How does the Zener diode fit into this? \$\endgroup\$
    – Bee
    Commented Feb 15, 2021 at 18:59
  • \$\begingroup\$ How does the Zener diode fit into this? I suggest you look up what a zener diode does and how it can be used to make a constant (reference) voltage. Because that is what it is doing here, R3 and the Zener diode generate \$V_{ref}\$ \$\endgroup\$
    – Bimpelrekkie
    Commented Feb 15, 2021 at 19:57
  • \$\begingroup\$ @Bee, see the update. \$\endgroup\$
    – Transistor
    Commented Feb 15, 2021 at 20:05
  • \$\begingroup\$ Thanks, your explanation helps a lot and watching a few videos about Zener diodes also helped. So then, in the context of putting everything together (and I'm aware that you answered my first question that had the BJT and not the MOSFET, which I'm interested in using): - Vin is the battery power, might drain over time - R3 + Z diode maintain Vref stable as long as Vin > Vref - Opamp takes the difference between positive and negative terminals, and outputs the voltage required to bring V(1) and V(2) together, all regulated through negative feedback. So... what heavy lifting does MOSFET do? \$\endgroup\$
    – Bee
    Commented Feb 15, 2021 at 20:14
  • \$\begingroup\$ I'd expect this to be less stable with a bipolar transistor than with a Mosfet. I will try have to check the transfer function of the linearization. \$\endgroup\$
    – Ben
    Commented Feb 15, 2021 at 20:26
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If you remove the transistor and add power rails to the op-amp and let the op-amp drive the load resistor and potential divider, do you see how it works: -

enter image description here

If the answer is "yes, I see how that works" then, adding a MOSFET (or BJT) as a source (or emitter) follower is only a means of letting the transistor do the heavy lifting. Do you see that now?

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  • \$\begingroup\$ Thank you. I follow this until the point of knowing what to do with the Zener diode (not super familiar with them). My understanding is that it guarantees that Vp = Vref at all times, otherwise it drains until this happens. But I don't fully understand this thought either. \$\endgroup\$
    – Bee
    Commented Feb 15, 2021 at 18:57
  • \$\begingroup\$ Your explanation is very clear, however, I thought that we wanted Vout to be a A*Vz, where A is the gain of the op-amp, 1 + R1/R2, then the voltage divider would take Vn = (R2/(R1+R2))*Vref*(1 + R1/R2) = Vref. Then, is there is a mismatch, then Z drains current until Vn = Vref again? But what happens if the voltage across it is smaller than Vz, not greater? \$\endgroup\$
    – Bee
    Commented Feb 15, 2021 at 19:11
  • \$\begingroup\$ Sorry, I made a mistake that I was in the process of correcting in the answer @Bee i.e. hmm do I correct the comment or add more details to the answer (dilemma). \$\endgroup\$
    – Andy aka
    Commented Feb 15, 2021 at 19:12
  • \$\begingroup\$ Thanks! So, just to sum up from your schematic: the op-amp compares Vn to Vz continuously. In the event that they differ (e.g. if RL pulls more current), then Zener diode will readjust (by shorting out or becoming an open circuit (??)) until Vp = Vref again and then it will output an amplified response to Vout? \$\endgroup\$
    – Bee
    Commented Feb 15, 2021 at 19:57
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    \$\begingroup\$ The voltage on the zener diode can be regarded as being constant for all time. If the feedback voltage from the resistor divider gets smaller, the opamp readjusts its output to maintain equilibrium. \$\endgroup\$
    – Andy aka
    Commented Feb 15, 2021 at 22:19
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To understand the whole circuit, you have to break it down into its subcircuit building blocks and understand the operation of each of them (this approach applies to any unknown circuit; master it and apply it in the future). Here you can recognise three such sub-circuits: passive voltage regulator (RD and Z), op-amp non-inverting amplifier (op-amp, voltage divider R1 and R2) and powerful source follower (MOSFET).

As you can see, the source follower is put into the feedback loop of the op-amp follower. With this clever trick the Vgs voltage of the transistor is compensated by the op-amp that "lifts" its output voltage with this value. As a result, the output voltage is the same as in the case of the ordinary op-amp noninverting amplifier (as though there is no transistor inserted). This is a fundamental property of negative feedback circuits that you will see in many other circuit solutions... and can apply in your projects.

As a conclusion, you can think of the whole circuit as a power amplifier that amplifies a constant voltage.

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  • \$\begingroup\$ Thanks for the breakdown. I am a bit lost when you say: "With this clever trick the Vgs voltage of the transistor is compensated by the op-amp that "lifts" its output voltage with this value. As a result, the output voltage is the same as in the case of the ordinary op-amp noninverting amplifier (as though there is no transistor inserted)". Is this what @Transistor was saying in their answer -- that the output voltage through VGS is the same as the Opamp output voltage, but the current flowing through is increased? \$\endgroup\$
    – Bee
    Commented Feb 16, 2021 at 14:51
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    \$\begingroup\$ @Bee, OK, I will explain it here and then in my answer... The phenomenon of negative feedback is extremely important because not only most electronic circuits but the whole living world is based on it. This is due to the unique property of negative feedback systems to compensate for side effects (interference, disturbance). To see the effect of the negative feedback, let's consider two ways of applying the negative feedback - 'before' and 'after' the disturbance. To apply the negative feedback before the disturbing Vgs, disconnect R1 from the source and connect it to the gate... \$\endgroup\$
    – Circuit fantasist
    Commented Feb 16, 2021 at 19:01
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    \$\begingroup\$ In this case, two circuits with negative feedback are cascaded - an 'op-amp non-inverting amplifier' and a 'source follower'. The first is perfect but the second is imperfect because of the voltage drop Vgs between the gate and source... and the output voltage across the load will be Vout - Vgs instead Vout. To eliminate this voltage lost, we have to close the negative feedback after it. So, disconnect R1 from the op-amp output and connect it again to the circuit output. Now the op-amp "observes" the voltage across the load… "sees" that it is less than the input voltage with Vgs… \$\endgroup\$
    – Circuit fantasist
    Commented Feb 16, 2021 at 19:20
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    \$\begingroup\$ … and begins compensating for it (as we all do in life when something tries to stop us from reaching our goal). For this purpose, the op-amp begins increasing its output voltage until it becomes Vout + Vgs; as a result, the source (circuit output) voltage will be Vout + Vgs - Vgs = Vout. This was an example of how an undesired disturbance is neutralized. But in some cases, the disturbance is useful... and we intentionally introduce it. E.g., in the non-inverting amplifier shown in your picture, the voltage divider R1-R2 is a useful disturbance since it makes the op-amp behave as an amplifier. \$\endgroup\$
    – Circuit fantasist
    Commented Feb 16, 2021 at 19:32
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    \$\begingroup\$ Understood, thank you!! \$\endgroup\$
    – Bee
    Commented Feb 16, 2021 at 20:19

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