The following diagram is a trace of a Zener diode in avalanche. I can understand the explanations behind shapes A - D, but what causes shape E?


A is the accumulating charge within the junction's capacitance, and thus follows the common exponential form. But no charge seems to accumulate immediately following breakdown (shape B).

And from my own experiment, this time with a 24V diode at 30V:-


My theory is the electrons are cascading so fast that they're exceeding the current flow through the 100k resistor, spurred on by impact ionisation. So no charge can build. The avalanche gradually stops between ~26.4V and ~28V, and only then does charge rebuild > 28V.

Is this correct?


1 Answer 1


The breakdown event involves a very fast pulse of current. Any circuit of nonzero size also has nonzero inductance. That voltage "spike" you're pointing to is the result of that current pulse interacting with that inductance.

$$V = L\frac{d i(t)}{d t}$$

  • \$\begingroup\$ Ah! You don't really see inductance mentioned in the diode data sheets. Is there anyway to estimate it? Can \$dt\$ be measured on a faster scope? \$\endgroup\$
    – Paul Uszak
    Commented Nov 9, 2019 at 16:15
  • \$\begingroup\$ Yes, but any measurement you make will be "filtered" through the parasitic inductances, capacitances and resistances of the test setup and the instrumentation. \$\endgroup\$
    – Dave Tweed
    Commented Nov 9, 2019 at 16:18
  • \$\begingroup\$ @DaveTweed it's not carrier recombination? \$\endgroup\$
    – TimWescott
    Commented Nov 9, 2019 at 17:40
  • \$\begingroup\$ The dominant factor is the capacitance of the reverse biased junction (not any inductance factor). Once ionization occurs avalanche will continue to create hole pairs until the capacitance is discharged to a point where insufficient field exists to continue the process. The capacitance of the junction then starts to charge again. \$\endgroup\$ Commented Nov 9, 2019 at 18:08
  • \$\begingroup\$ @JackCreasey: That's the explanation for the "B)" part of the curve. The OP is asking specifically about the "E)" part that he marked. How do you explain the rapid rise following the most negative point on the waveform? \$\endgroup\$
    – Dave Tweed
    Commented Nov 9, 2019 at 21:03

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.