I’m a newbie to the topic of circuits so sorry if this seems a stupid question. enter image description here

This is the picture given in my textbook of a Zener circuit.

First of all, isn’t Vz = the fluctuating DC input voltage? As far as I understood, potential difference across parallel circuits is the same so it should be same across both the diode and the load right?

Secondly, how does the diode work? Once it’s crossed its ‘breakdown’ stage, what exactly happens when the voltage fluctuates? How does the diode regulate it? My guess is that it ensures the same current flows through the load no matter how much current in the circuit changes (due to change in Voltage) and since V=IR (to reason mathematically) and current and resistance are same, V is also same. Am I right?

  • $\begingroup$ That is a really bad way to try and make a ‘constant DC’ output by the way. $\endgroup$
    – Jon Custer
    Nov 26, 2019 at 13:35
  • $\begingroup$ @JonCuster, that way specifically, yes. But there are shunt regulators and references that behave like zeners (just with better accuracy) that are entirely acceptable in many situations. $\endgroup$
    – The Photon
    Nov 26, 2019 at 17:22
  • $\begingroup$ Kathikeya, The "Fluctuating DC input voltage" is not the same as Vz in your drawing. They're separated by the resistor labelled "R". $\endgroup$
    – The Photon
    Nov 26, 2019 at 17:22

1 Answer 1


Think about how the zener diode works. It is conducting for voltages above 0, and for voltages below -$V_z$. When the voltage is above $0$ the diode will conduct a current and the voltage over $R_L$ will be $0$. The load will however experience a voltage when the value is $V_z < V < 0$ because then the diode is not conducting and we have a voltage over it, the current goes through $R_L$. $V_z$is the max voltage $R_L$ will experience. If this is known the current going through the closed loop can be calculated and thus you can calculate other parameters as well. If you plot the voltage over $R_L$ you should get a signal where we only have the lower part of a sinusoid that is clipped at $V_Z$ (given that the input voltage is a sinusoid).

A Zener diode is a semiconductor that permits zener breakdown at a specific voltage, and this means that a diode will conduct a current even in the backwards direction. In practice this current is dependent on the voltage applied, but given the simplicity of the circuit above, it should be enough to use the ideal diode functionality where the current is the same no matter the voltage applied.

When the Zener-voltage is passed, the voltage over the circuit will be $V_z$, thus the current/voltage on $R$ will be constant as long as the diode is conducting.

Something that'd help you could be to redraw the circuit for the scenarios when the diode is and is not conducting, replacing the diode with an open circuit and a short circuit. And keep in mind that current seeks the path of lowest resistance.

  • $\begingroup$ But since the two are connected in parallel, shouldn't the load have the same V as the fluctuating source and therefore fluctuate? Or does that rule not apply here? $\endgroup$ Nov 26, 2019 at 12:45
  • $\begingroup$ @KarthikeyaKaza When the voltage is above 0 the diode is open, and thus the current will prefer to travel the way of least resistance, and thus the diode can be replaced with a short circuit. In the case of negative voltage, the voltage over the load is the same is the negative voltage up until the zener voltage, at which point this is the maximum voltage it will experience. $\endgroup$
    – DakkVader
    Nov 26, 2019 at 12:49

Your Answer

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

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