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I'm not sure how to title this question, any suggestions are welcome.

I want to analyze this circut, it's a zener regulator and I am following this video that does the analysis

enter image description here

The proposed method is to sum \$V_{zener}+V_{be}\$ and then use the voltage divider (I will make it equal for easy calculations) to calculate the output voltage.

enter image description here

I feel this is very arbitrary, since it does not seem to take the other transistor into consideration, if it where to be replaced with a resistor or a wire, this analysis will not work. It seems there are some assuptions that are not clear, but I'm not sure what they are.

here for example I have substituted the transistor that was not used in the analysis with a resistor, changing the behaviour of the voltage divider completely. while the new behaviour is not unexpected when analyzed separately it was not obvious at all that this could happen on the first analysis, my guess is that something was assumed but I cant quite understand what it is.

enter image description here

could somebody clarify it? what is the assumption for the first analysis to work? why is there no mention of the other transistor if it's so necessary?.

EDIT: I have had some awesome answers and thanks to them I have gained new insights to how this circuit works, however my question is related to this partcular analysis approach and the assumptions that have to be made to make this approach viable, because I want to perfect my analysis technique. However the answers here so far are awesome and I REALLY appreciate them.

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  • \$\begingroup\$ Joaquin, Congratulations for the attractive voltage visualization in the second picture! I will be happy if others like you on the web use this visualization technique! Now all we have to do is convince the creators of Falstad to build it into their simulator. I have only one remark to the voltage bar representing VBE - it would be good if it was also vertical... \$\endgroup\$ – Circuit fantasist May 8 at 17:33
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Here's the first schematic, redrawn:

schematic

simulate this circuit – Schematic created using CircuitLab

\$Q_1\$ is a pass transistor and \$R_1\$ provides the base current needed to keep it active. If you got rid of the POT, \$Z\$, and \$Q_2\$, then the load would see \$V_\text{CC}\$, less a little voltage. But there wouldn't be any way to set the load voltage to some particular, different value.

\$Q_2\$ is added as a comparator. It compares its base voltage with its emitter voltage and adjusts its collector current so that they are different by about one diode drop. Whatever it takes. Since \$Z\$'s voltage is supposed to be fairly constant, the base voltage is what's under observation by \$Q_2\$. If the base voltage increases, then the \$V_\text{BE}\$ difference also increases and this causes \$Q_2\$ to pull much harder downward on its collector (sinking more current into \$Z\$.) But that causes the voltage drop across \$R_1\$ to increase and that causes both the base and the emitter voltages of \$Q_1\$ to decline in response. So, if \$V_\text{REG}\$ momentarily increases, then \$Q_1\$ moves back downward and therefore decreases \$V_\text{REG}\$. That's negative feedback in action.

Ideally, you'd ignore \$I_{B_1}\$ and \$I_{B_2}\$ and treat them as nil. But a good design really can't ignore them and must work out a way to make their effects small enough that they can be mostly ignored.

The worst case error will be when the POT is in its middle, halfway position as that will maximize the voltage drop due to \$I_{B_2}\$. But that may not be all that important, because you can always just turn the pot a little more in one or another direction to compensate. So that's not critical. But it's worth keeping in the back of the mind.

For a back of the envelope design and assuming that power dissipation issues can be largely set aside (it's a relatively low power circuit), you would start with the worst case load current that may be needed: \$I_{\text{LOAD}_\text{MAX}}\$. You'd assume a BJT for \$Q_1\$ that can handle that current (and then some) but defer locating it for a later time and instead assume that it can probably support \$\beta\approx 100\$ in active mode. (Again, if this is a power device that won't be true. But I'm stating at the outset this is not a power device because then you'd need a bigger envelope.)

Now, this means \$I_{B_1}\approx \frac{I_{\text{LOAD}_\text{MAX}}}{\beta\approx 100}\$. To be safe, you'll want \$I_\text{BIAS}\$ to be at least \$10\times\$ this much, though that's not cast in stone.

Now this is the point where you want to compare \$I_\text{BIAS}\approx 10\cdot\frac{I_{\text{LOAD}_\text{MAX}}}{\beta\approx 100}\$ with appropriate zener currents. If they are in a close match with each other, great. If not, and the zener current is a lot higher, then you may need to let the required zener current replace your estimate for \$I_\text{BIAS}\$ and take the hit on \$I_{B_2}\$ and the implications for \$I_\text{DIV}\$. If not, and the zener current is a lot lower, then you may want to find a different zener family because very low zener currents imply very high zener impedances and that's going to mess things up for the regulator.

A final redrawing may be helpful:

schematic

simulate this circuit

Here, I've divided this up into two sections. An upper section which moves up and down, relative to ground, and a lower section which stays solid, relative to ground. Kind of like a giraffe neck going higher or lower to reach an apple in a tree (that no giraffe is really lucky enough to eat), the upper section moves up and down based upon the voltage drop across \$R_1\$. That voltage drop is controlled by the collector current of \$Q_2\$. But that collector current is controlled by the base-emitter voltage difference of \$Q_2\$, which is measuring the output and comparing a fraction of it with the zener voltage.

And at last, we come to \$I_\text{DIV}\$. (Note that I've added \$R_2\$??) This current should also be about \$10\times\$ that needed by \$I_{B_2}\$ (worst case, again.) I've added \$R_2\$ since there's no point in losing a lot of range for the POT just because of that zener voltage. So you will want to drop about one diode drop more than the zener voltage across \$R_2\$.


There's a simple modification you can add if you want to provide a current limit capability, too.

schematic

simulate this circuit

Now you can set a current limit. Here \$Q_3\$ measures the voltage drop across IPOT. If it exceeds a certain value, \$Q_3\$ starts sinking more current via \$R_1\$ and around and behind \$Q_2\$ and into the load path. This increases the drop across \$R_1\$ and lowers the output voltage in response.

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  • \$\begingroup\$ Thank you, seeing Q2 as a comparator was enlightening!!. However I'm still a bit stuck with my original question, I added an edit to clarify what my intent was when asking. I'm not looking for new or different ways of seeing or understanding this circuit, I want to understand why this particular analysis approach (that is kind of systematic instead of intuitive) seemingly fails when chainging the circuit. I suppose that the method is correct and I'm skipping some unsaid assumption. \$\endgroup\$ – Joaquin Brandan May 8 at 1:44
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    \$\begingroup\$ @JoaquinBrandan Well, it's a little bit different than that. It's not so much that the base current of \$Q_1\$ is limited by \$Q_2\$ action. In fact, \$Q_1\$'s base current might well increase. The main point is that \$Q_2\$'s collector current increases the voltage drop across \$R_1\$ and pulls down the voltage at the base of \$Q_2\$. Focusing on the base current of \$Q_1\$ is a distraction of sorts. It's real, but it's not the main thrust. The voltage regulation occurs because the voltage at the base of \$Q_1\$ is regulated. Not because its base current is regulated. \$\endgroup\$ – jonk May 8 at 2:36
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    \$\begingroup\$ I see, thank you!. Your answer was very helpful in clearing up my misunderstanding. Sorry I could not be more clear in my question, I found it very hard to articulate where exactly my confusion was. \$\endgroup\$ – Joaquin Brandan May 8 at 2:44
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    \$\begingroup\$ @JoaquinBrandan I think we all get focused on one particular thing, at times, and it gets in our way of seeing other things also going on. We learn from how others think. That's why this is such a great place. But also keep in mind it is NOT so much the conclusions we have (me or others) and push on you, but instead much much more about how we think about something. We all get things wrong. And most of our conclusions are probably wrong in a number of ways, even from the best of us here or elsewhere. What you want to get isn't the conclusion, but the methods we apply in getting there. \$\endgroup\$ – jonk May 8 at 2:58
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    \$\begingroup\$ @JoaquinBrandan Listen for the approaches, what key ideas are brought into a problem and why. But do NOT get caught up in memorizing the results of that process. The best of us will have found a few extremely general and extremely powerful mental tools in thinking about the world around us. That doesn't mean we apply them correctly. But it does mean that we've uncovered some really good ways to think about the world. Those who, instead, only have lots of scattered and mostly unconnected "rules" should not be listened to, so much. Just FYI. \$\endgroup\$ – jonk May 8 at 3:02
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This part does two things: first, when no current is pulled from the base of Q1, R1 supplies current to turn Q1 on, roughly to the point where the emitter of Q1 is one volt below the collector; second, it presents a load to whatever is pulling current from Q1's base. I can't say what that load is because it depends on the selection of Q1 and the nature of the stuff hanging off of its emitter, but it's less than \$1000\Omega\$.

schematic

simulate this circuit – Schematic created using CircuitLab

This part controls Q1. It does so because when the voltage on the base of Q2 exceeds the zener voltage plus the base voltage, Q2 pulls current away from the base of Q1. This tends to turn Q1 off (remember that by itself, Q2 plus R1 just passed the input voltage to the output).

If the output rises, more current is stolen from Q1, which causes the output to fall. If the output falls, less current is stolen from Q1, which causes the output to rise. If you're lucky (or a good circuit designer), then the output will go where it belongs and stay there without oscillating.

Note that -- because there's a finite load presented to the collector of Q2 -- simple regulator circuits like this do a sorta-kinda-good job of regulating their output voltage, but they're not perfect. As you pull more and more current from the output, it will drop, and as you remove load from the output it will rise. That, plus the amount of components used and the power demands of this circuit, is why people buy integrated circuits that do the same thing, only better.

schematic

simulate this circuit

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The zener —for reasonable currents, with reasonable linearity— forces the voltage at the lower transistor's emitter to be its zener voltage. And thus it forces the base voltage to be 0.7V higher. And the potentiometer scales that up.

That correct "upscaling" depends on the output voltage to be close-loop controlled of course. A resistor divider cannot upscale a voltage for real. It can only downscale a voltage. And that's where the series transistor comes into play. If the output voltage is higher than calculated, a higher current will flow trough the lower transistor, taking base current from the series transistor so the output voltage will be lowered. And vice versa.

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