This diagram is from https://www.homemade-circuits.com/simple-voltage-regulator-circuits-using-transistor-and-zener-diode/

The article says that when the input voltage exceeds the zener diode's rating, current will pass through the zener diode and that this will somehow cause the input voltage to drop and therefore cause the output voltage to do the same.

Why does the voltage drop? There is a current limiting resistor stopping the input source from being overburdened, and if there wasn't, that would be a short circuit.

Also, how does this circuit relate to a chopper circuit?

  • 1
    \$\begingroup\$ Something is wrong or I couldn't understand as my English is terrible. I think that the article somehow assumes that there's no limiting resistor. If the load resistor is zero Ohms and the input voltage source has a non-zero output impedance then yes, when the input voltage reaches the Zener voltage then the input voltage will drop. \$\endgroup\$ Sep 8, 2020 at 5:20
  • 1
    \$\begingroup\$ The input voltage never drops any extra input voltage is found across across the resistor. This is a linear regulator, so it does not directly relate to a chopper, which is a switching regulator. \$\endgroup\$
    – User
    Sep 8, 2020 at 5:20
  • 8
    \$\begingroup\$ The article is not very good. The schematics are fine, but the verbal descriptions are quite poor. \$\endgroup\$ Sep 8, 2020 at 5:40
  • 5
    \$\begingroup\$ @SpehroPefhany Seems Sam Gibson recognized where the schematics are stolen from. Shame that the website authors didn't also steal the text, it would have less terrible, wild guess. \$\endgroup\$ Sep 8, 2020 at 14:54
  • 3
    \$\begingroup\$ @MarcusMüller Unfortunately there is a thriving business in "creating content" by re-writing existing articles for very small sums of money (in order to technically avoid copyright infringement). The English in the writing has tell-tale characteristics of English as spoken in certain countries where that makes economic sense. SEO methods can be used to promote the crummy "content". The hapless learner has difficulty telling the difference, particularly if their technical knowledge is limited and if their first language is not English. \$\endgroup\$ Sep 8, 2020 at 15:36

3 Answers 3


Easy one first:

One more question: How does this circuit relate to a chopper circuit?

Unrelated. The chopper chops, i.e., there's some kind of control that turns a switch (e.g. a transistor) on and off. There's nothing like that in here, it's not a switch-mode power supply.

I don't understand why the voltage drops.

The current through a diode grows very quickly if you increase the voltage after you've crossed the breakdown voltage. A little more voltage, way, way, way more current.

Way, way, way more current has to come from somewhere: it must flow through your orange resistor. That means way, way, way more voltage drop across that resistor, so that the voltage across the zener doesn't increase very much. That's the "regulated voltage".

(It's not a great regulator. If I see a design where anyone uses this, unless it's a very special case, it's probably a very bad design. Don't do this in the wild, there's always a better way. All the circuits on the page you've linked to have been obsolete since ca the early 1970's. Also, they are colorized scanned photocopies of photocopies of copies from copies ... maybe find a better source for schematics, honestly.)

There is a current limiting resistor stopping the input source from being overburdened and if there wasn't that would be a short circuit.

Well, that resistor is half of the regulating mechanic, the other half being the Zener diode.

  • 1
    \$\begingroup\$ A linear regulator can be all right if you know the input voltage is very close to the output voltage. For one obvious example, if you have a circuit that's sensitive to noise on its power supply, a switching supply followed by a linear regulator can work quite nicely (and while I might buy that being a special case, it doesn't strike me as rare enough to qualify as "very special"). \$\endgroup\$ Sep 8, 2020 at 5:46
  • 3
    \$\begingroup\$ @JerryCoffin: A linear regulator following a switching regulator to reduce noise is only "special," not "very special. A Zener diode regulator is more wasteful than a regular linear regulator and has really crummy regulation. Any circumstance that makes a Zener regulator preferable to a regular linear regulator qualifies as "very special." The most common reason for using a Zener regulator these days is someone working from a textbook from 1975, or following a schematic from the same time frame. \$\endgroup\$
    – JRE
    Sep 8, 2020 at 5:55
  • 2
    \$\begingroup\$ Correction: "way more current drop across that resistor" should be "way more voltage drop across that resistor", right? \$\endgroup\$
    – RJR
    Sep 8, 2020 at 10:24
  • 6
    \$\begingroup\$ @MarcusMüller - Hi, Regarding: "All the circuits on the page you've linked to [...] are colorized scanned photocopies of photocopies [...]". Indeed. Just FYI, I recognise that almost all the schematics on the webpage linked by the OP (before they were colorized and had a website URL put on them) have been copied from the book: "Power Supply Projects" by R. A. Penfold, first published in paperback by Bernard Babani in 1980 (ISBN 0 900162 96 1). \$\endgroup\$
    – SamGibson
    Sep 8, 2020 at 14:48
  • 4
    \$\begingroup\$ @SamGibson and the "copycat" / intellectual property problem here aside, it would be better if the authors of the websites actually understood that 1980 publication, presumably. \$\endgroup\$ Sep 8, 2020 at 14:49

The article gives an incorrect description of how a zener operates. The circuit shown is a simple potential divider, the bottom arm of which is the zener diode.

Below the zener voltage, the zener dynamic impedance is high and the output voltage very nearly equals the input voltage. Above the zener voltage, its dynamic resistance drops such that the voltage across the zener is very nearly constant. The current through the zener therefore varies, such that the output of the potential divider equals the zener voltage. There is no rapid up and down movement of the voltage as the article implies. Hence it has no relation to a chopper circuit.

  • 2
    \$\begingroup\$ Given the question is asked by a beginner, it might be clearer to define your terms, what the difference is between V/I and dynamic resistance and why the article gets that wrong. \$\endgroup\$ Sep 8, 2020 at 15:27
  • \$\begingroup\$ I was not sure the OP is a beginner. Even so I thought it best to use the correct term in order to encourage the OP to research it himself, I just wanted to make it clear that the explanation given was incorrect. \$\endgroup\$
    – Ian Bell
    Sep 9, 2020 at 11:33

How does this circuit compare to a voltage divider? Are there any conceptual parallels/similarities?

Right observation... Yes, there are conceptual similarities. More precisely speaking, both circuits are specific implementations of a more general arrangement consisting of two elements in series (I don't exactly know but maybe, 90% of electronic circuits are based on this connection).

While the ordinary voltage divider is linear this is a "dynamic voltage divider" implemented by a "dynamic resistor" (Zener diode) connected to ground.

The ordinary voltage divider has a constant transfer ratio of R1/(R1 + R2). The ratio of this "dynamic voltage divider" K = Rdyn/(Rdyn + R2) is dynamic - it changes in an opposite direction regarding the input voltage variations in such a way that its output voltage stays constant.

As an example, if Vin increases, Rdyn decreases -> K decreases -> Vout stays unchanged... and v v., , if Vin decreases, Rdyn increases -> K increases -> Vout stays unchanged again.

You can demonstrate the operation of this electronic circuit by replacing the Zener diode by a variable resistor (rheostat). If you change its resistance, when the input voltage varies, so that to keep a constant output voltage, the variable resistor (and you:) will act as a "Zener diode".


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.