I'm studying the following series voltage regulator, and I'd like to know how to "prove" the last two sentences by using equations.

For example, in the case of the 2nd sentence, the first part says that if \$ V_o \$ increases, the \$V_{BE}\$ decreases; this is quite easy to see, indeed:

$$ V_{BE} = V_Z - V_o $$

Which equations should I consider to see the rest of the sentence (Q1 conducts less, thereby \$V_o\$ decreases and returns to the prevoius value)?

enter image description here

EDIT: thanks to @jonk comment I understood another piece; by using the Ebers-Moll equations in active mode, it's visible that when \$V_{BE}\$ decreases, the \$I_B\$, \$I_C\$, \$I_E\$ decrease:

enter image description here

Now, to see that \$V_o\$ decreases, it's sufficient to consider the following equation:

$$ V_o = R_L \, I_E $$

  • 2
    \$\begingroup\$ The book likely has explained how transistors work, and particularly how they work in an emitter follower circuit, before moving on to more advanced concepts. The equations should be around that area. \$\endgroup\$
    – Justme
    Jun 27, 2022 at 21:48
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    \$\begingroup\$ If you claim that the book has no equations, how do you explain that the example right below figure 15.13 solves an identical circuit shown in figure 15.14 with equations? I see a lot of other equations in the book too. \$\endgroup\$
    – Justme
    Jun 27, 2022 at 21:54
  • \$\begingroup\$ @Justme the example you mentioned doesn't contain the equations that explain what I'd like to know, but only the first part that I reported. \$\endgroup\$ Jun 27, 2022 at 21:56
  • \$\begingroup\$ @Justme If you found the right equations, can you post them please? \$\endgroup\$ Jun 27, 2022 at 22:10
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    \$\begingroup\$ @Gennaro You do follow that an active-mode BJT's collector current is directly related to its \$V_{_\text{BE}}\$ through an exponential function; which means that a tiny increasing change of 60 mV or so will yield a huge 10X collector current change? \$\endgroup\$
    – jonk
    Jun 28, 2022 at 0:55

1 Answer 1


Opening Summary

The circuit is the following (you really should learn to redraw schematics for better reading):


simulate this circuit – Schematic created using CircuitLab

If you are willing to assume that the series current through \$R\$ and the zener diode is very large with respect to the base current required by \$Q_1\$, then it follows that the voltage at the base of \$Q_1\$ is fixed and determined by the zener diode. So we can ignore the details of the behavior of the left-hand side and just claim that \$Q_1\$'s base voltage is fixed for the purposes of analysis.

That leaves the right-hand side to worry over.

Solving for \$V_{_\text{OUT}}\$

For simplification purposes, we can assume that the emitter and collector currents are equal. (They aren't. But for reasonable values of \$\beta\$ they are close enough that we can consider them equal.) And we can ignore the -1 term in the Shockley diode model to get:

$$V_{_\text{OUT}}\approx R_{_\text{LOAD}}\,I_{_\text{C}}=R_{_\text{LOAD}}\,I_{_\text{SAT}}\,e^{^{\left[\frac{V_{_\text{Z}}-V_{_\text{OUT}}}{V_T}\right]}}$$

This solves out as:

$$V_{_\text{OUT}}\approx V_T\cdot \operatorname{LambertW}\left(\frac{I_{_\text{SAT}}\,R_{_\text{LOAD}}}{V_T}\cdot e^{^{\left[\frac{V_{_\text{Z}}}{V_T}\right]}}\right)$$

We know that \$V_T\approx \frac1{40}\$ so it follows that \$\operatorname{LambertW}\left(\frac{I_{_\text{SAT}}\,R_{_\text{LOAD}}}{V_T}\cdot e^{^{\left[\frac{V_{_\text{Z}}}{V_T}\right]}}\right)\approx 40\,V_{_\text{OUT}}\$.


The concept of a sensitivity equation starts with recognizing that a %-change in something (using finite values) would be \$\frac{\Delta x}{x}\$. But using infinitesimal values this is still more precisely stated as \$\frac{\text{d}\, x}{x}\$. A sensitivity equation relates the %-change of one thing to the %-change of another thing, so the expression might look like \$\frac{\frac{\text{d}\, y}{y}}{\frac{\text{d}\, x}{x}}=\frac{\text{d}\, y}{\text{d}\, x}\cdot\frac{x}{y}\$, which gives you the %-change of y as compared to the %-change of x.

In this case, we may want to know what %-change in \$V_{_\text{OUT}}\$ results from some given %-change in \$R_{_\text{LOAD}}\$. Performing the required calculus, this sensitivity equation is then:

$$\frac{\% V_{_\text{OUT}}}{\% R_{_\text{LOAD}}}=\frac{V_T}{V_{_\text{OUT}}\left(1+\frac1{\operatorname{LambertW}\left(\frac{I_{_\text{SAT}}\,R_{_\text{LOAD}}}{V_T}\,\cdot\, e^{^{\left[\frac{V_{_\text{Z}}}{V_T}\right]}}\right)}\right)}$$

But we already determined that the LambertW function is going to be rather large (about \$40\cdot V_{_\text{OUT}}\$.) So the last term in the denominator therefore fades to zero.

The simplified result is then:

$$\frac{\% V_{_\text{OUT}}}{\% R_{_\text{LOAD}}}=\frac{V_T}{V_{_\text{OUT}}}$$

Or, in more useful terms:

$$\% V_{_\text{OUT}}=\left[\frac{V_T}{V_{_\text{OUT}}}\right]\,\% R_{_\text{LOAD}}$$

Final Summary

Roughly speaking this means, taking everything into account, we should find that the %-change in \$V_{_\text{OUT}}\$ will be about \$\frac1{40\,\cdot\, V_{_\text{OUT}}}\$ of the %-change in \$R_{_\text{LOAD}}\$.

This pretty much tells you that the %-change in \$V_{_\text{OUT}}\$ is very, very, very much less than the %-change in \$R_{_\text{LOAD}}\$ for all practical circumstances. (Only a few parts per thousand.)

Or, put again in other more qualitative terms, the output voltage won't budge much with respect to load changes.

And yes, stuff like this can be quantified (as shown above) and not just hand-waved-at. You want numbers? You got numbers. But when you get done with all that quantifying work, you'll once again find (in this case) that qualitative hand-waving is good enough for all intents and purposes.

Final Note: All of the above work assumes that \$V_{_\text{Z}}\$ is rock solid and doesn't itself move in response to load changes. The fact is, it does change because the base current is a function of the load current and a zener diode is specified for some operating current that is often quite limited. The above is good enough where the zener current is, as earlier stated, very much greater than the worst case expected base current. But it fails when that is no longer true and then a deeper analysis is required. Just keep it in mind.

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    \$\begingroup\$ I appreciate this final formula. It is so concise! +1 I'd like to comment on the qualitative summary assessment, that the output voltage doesn't change "much" depending on load. This kind of regulator is hardly a good solution for cases, where one actually needs "low output impedance regulation", especially at >10 kHz where analog circuitry PSRR fades away. It's good enough not to violate input voltage ranges of ICs but feedback-based regulation is needed for cases where voltage stability is critical. \$\endgroup\$
    – tobalt
    Jun 29, 2022 at 6:08
  • \$\begingroup\$ @tobalt Technically, I call what happens in this case "local NFB." It happens as a result of the topology and device behavior. This isn't unlike the "local NFB" one also finds in a degenerated BJT CE amplifier stage used to teach BJT circuits at school. There is NFB. But it's not global, but instead local to the subcircuit topology. As far as the final formula, I just enjoy showing that theory and math do apply if you care to. Things can be quantified. It bothers me, seeing 'hand-waving'. I do understand that sometimes that is all someone can gather. But math shines and it's nice to see. \$\endgroup\$
    – jonk
    Jun 29, 2022 at 6:18

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