# Why is Zener avalanche noise saw tooth shaped?

I have the following schematic of a Zener based noise source:- simulate this circuit – Schematic created using CircuitLab

When built, an oscilloscope reveals a saw tooth noise signal at the "Noise" node, like: The time base is 1us/div. Can anyone explain why the signal is saw tooth shaped? Initially I expected a triangular, or even sine shaped wave form. I think that it's something to do with the impedance of the Zener in conjunction with the much higher 100 kOhm resistor. The electrons cascade freely across the junction, but the resistor restricts current flow when the avalanche stops. We're talking 60uA. The result being slower charge build up than when current flows during avalanche.

This wave form isn't particular to my set up. There are other examples elsewhere on the Interweb when people have really zoomed in on the signal, one being https://youtu.be/CAas_kbTW3Q?t=714. Also there's a good chart here showing the rising edge to be slightly curved. It's probably unfamiliar as it's usually shown with a much slower time base. Am I right about the resistance/impedance explanation?

• Is there a capacitor you aren't telling us about? Or, what is the junction capacitance of the diode? – Brian Drummond Mar 2 '18 at 13:54
• @BrianDrummond Nope, just what's shown and a 50 Ohm lead direct to the scope. It was soldered bug style (clearly excluding the 30V psu). – Paul Uszak Mar 3 '18 at 23:50

Consider that you effectively have this: simulate this circuit – Schematic created using CircuitLab

where C is the junction capacitance, plus any external capacitance (leads, breadboard, etc). Some of the current from R1 leaks through D1, but the rest charges C. Once the voltage reaches a certain level, avalanche breakdown occurs and current flows from C until the avalanche stops. Then the current begins charging C again.

To calculate C you first need to know the leakage. Decrease V1 until the noise disappears. Then measure the current. Then increase V1 back to 30V. Measure the rising slope of the noise dV/dt. Measure the average value of V. The current through R1 is approximately constant at (30V - V)/100kohm. Subtract the leakage current from this, then use I=C dV/dt to calculate the capacitance.

• How did you arrive at the 10pF value please? And do you see breakdown as effectively being a short then across the Zener? – Paul Uszak Mar 2 '18 at 12:35
• @PaulUszak I've added a bit on how to calculate the capacitance (which also includes the breadboard capacitance, if you're building this on a breadboard). Typical values for the junction capacitance of zener diodes seem to be in the range 10-100pF. Look at the graph on page 6 of this datasheet. They also do make low-capacitance zener diodes. – τεκ Mar 2 '18 at 15:55
• @PaulUszak When breakdown occurs it's the zener impedance (70 ohms-ish). If you zoomed in enough you could probably estimate that from the slope too, especially if you added more capacitance to slow it down. – τεκ Mar 2 '18 at 15:59
• Could this setup be used in a comparator for a PWM controller designed for a DC-DC converter, just as a hobby project? – Daniel Tork Mar 2 '18 at 19:35
• @DanielTork to get a random pulse time? – τεκ Mar 2 '18 at 20:00

The random discharges near breakdown are from random crystal dielectric charges breaking down under a high E field producing a pulse current that drops the voltage with an RC fall time. If you could measure how small the fall time was , you could estimate the size of the C in that charged particle.

If I guess each particle sees at least 50kV/mm or 50V/um or 50mV/nm so the charge size may be about 10 to 20 nm to get 500 to 1000 mV. This can be scaled according to the epixtaxial particle sizes in the Si crystal lattice.

Like a Unijunction oscillator except with random thresholds in a limited range, the C charges up and the Zener voltage collapses rapidly 1~5% just below the breakdown threshold at very low currents.

From looking at the waveform I expect the rise/fall time ratio to be ~ 100 or less in this sawtooth.