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I'm looking for some help with understanding (in fact, correcting or confirming my understanding) of how the following circuit works:

enter image description here

U1 here is TL431 voltage regulator, for example from Texas Instruments. This circuit is intended to provide a regulated voltage over Vout pins, and I'm looking for proper explanation how that is achieved. Here is what I managed to figure out:

At some moment, U1 is closed and current is flowing from Vin through R1 resistor directly to the Q1 gate, which opens the transistor at some voltage. In that way, we achieve some voltage at Vout, but we want some regulated, i.e. constant voltage, more or less independent of a load resistance being connected to the output.

When the volate at the source pin of Q1 reaches some point, determined by resistor divider on R2, R3 and R4, U1 opens and current starts to flow through R1, then U1, reducing the voltage applied to Q1 and thus partially closing it, which reduce the voltage at Q1 source pin.

If we remove C2 and C3 capacitors from the circuit, we should see some the pulsations at Vout pin, caused by this transistor opening/closing.

Please let me know if my understanding is correct and if not, how this circuit achieves regulation.

Also if this question is not quite suitable for this site, I do apologise and will remove it.

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    \$\begingroup\$ Question is entirely valid for this site :-). The following is NOT criticism - just advice. | Note that your open/close terminology is opposite to normal for electrical purposes. That is not wrong as such but will tend to be misunderstood by people. ie You say "open" as in "open a tap" and in close a valve so there is no flow, BUT in electrical discussions people usually say close as in close a switch = flow happens and open = open a switch so flow stops. \$\endgroup\$ – Russell McMahon Jun 4 '15 at 14:20
  • \$\begingroup\$ @RussellMcMahon Actually I kind-of translated terms from my native language, russian. We usually say like "a transistor is open" when current flows through it. But thanks for the advise, I'll take that into account in future. \$\endgroup\$ – Alexey Malev Jun 4 '15 at 19:29
  • \$\begingroup\$ Please notice the Q1 is a depletion-type MOSFET. These are normally-"on" and need to be turned off with negative Vgs. As such, it can be low-drop regulator. \$\endgroup\$ – Martin Feb 19 '18 at 12:48
  • \$\begingroup\$ I don't mean to be critical, but shouldn't Q1 be an enhancement mode mosfet instead of a depletion mode device. The schematic symbol is a depletion mode device. That might be confusing for some. \$\endgroup\$ – Techman_222 Dec 28 '18 at 5:16
  • \$\begingroup\$ If you use a depletion mode MOSFET, the circuit works as in the case an enhancement mode MOSFET (or a BJT) is used, but you have the plus of a lower voltage dropout, since you do not have to worry to make \$V_\mathrm{S}<V_\mathrm{G}\$. \$\endgroup\$ – Daniele Tampieri Dec 28 '18 at 6:53
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This is a linear regulator, not a switching regulator, so you should not see pulsations.

Other than that, your understanding is fairly accurate. The TL431 will draw more and more current through the cathode as the voltage on the sense terminal exceeds its internal reference voltage of about 2.5V. The MOSFET is wired as a source follower, so it has a voltage gain of about 1. C2 is there to make sure the feedback to U1 is not unduly delayed, which could cause oscillation.

So U1 will maintain the voltage at the gate of Q1 in order to have the divided output voltage (R3/R4/C2- node) equal roughly 2.5V.

Because Q1 is used as a source follower, this regulator will not be low drop-out, however it's much easier to assure stability with various load capacitances because Q1 is not adding voltage gain. There is also a lower limit on output voltage of ~2.5V- achieved when R2+R3 = 0 ohms.

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  • \$\begingroup\$ Well, even linear regulators will oscillate if their output filtering is inappropriate. \$\endgroup\$ – Ignacio Vazquez-Abrams Jun 4 '15 at 13:55
  • \$\begingroup\$ @IgnacioVazquez-Abrams Hence "should not" rather than "will not", but this configuration (source follower) is relatively stable against output loading. Like an LM7805 emitter follower, and unlike most LDOs which have the output connected to the collector or drain of the pass transistor. \$\endgroup\$ – Spehro Pefhany Jun 4 '15 at 14:01
  • \$\begingroup\$ @SpehroPefhany Can you please explain what LDO is? I'm currently investigating the options to build a power supply unit and this might be helpful too. \$\endgroup\$ – Alexey Malev Jun 4 '15 at 19:32
  • \$\begingroup\$ Low Drop Out (regulator). It means that the regulator will work with very low Vin - Vout differential, usually well under 1 volt. There are advantages and disadvantages to such regulators. \$\endgroup\$ – Spehro Pefhany Jun 4 '15 at 19:44
  • \$\begingroup\$ @SpehroPefhany Ah, thanks. Didn't guess myself) \$\endgroup\$ – Alexey Malev Jun 4 '15 at 21:03
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The actions of U1 the TL431 was not well explained here. This is an adjustable Voltage Zener diode, the Voltage of it is based on the Voltage fed into it's "Ref" terminal. As the output Voltage of the regulator drops (like when the load current has increased) the K to A (top/bottom) terminal Voltage of U1 rises causing the gate and source Voltage to rise by turning the MOSFET harder on, returning the output Voltage to the original value. Changing the Voltage ratio fed back to U1's Ref terminal allows the output Voltage of this regulator to be set where it is desired, here by changing R3. TL431s can be used as a comparator but it is not doing that here.

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