4
\$\begingroup\$

What exactly are the Zeners doing in this circuit?

I've been trying to improve my understanding of BJT's, by trying to 'improve' the classic astable multivibrator. In an effort to 'sharpen' the transistions, I included a Zener diode between the capacitor and the base of the transistor. My thinking was to 'protect' the base from the capacitor charge curve until it was well past the ~0.7V threshold. What ended up happening instead was an increase in oscillation frequency, and an increase in oscillation stability with very small capacitance values.

With the 'classic' multivibrator circuit, I can only get to about ~100kHz. With the Zeners (as well as R7,R8 and R9), I see stable oscillations well into the low MHz range.

Can someone help me understand how to analyze how the Zeners are working?

[The breadboard circuit performs very similarly to the simulation in LTSPice, in this case, about 3MHz with rise/fall times <10ns]

super multivibrator Zener current trace

\$\endgroup\$
0

1 Answer 1

2
\$\begingroup\$

I believe it's providing a path for the stored base charge to discharge. Basically turning the capacitors into "speed up capacitors" in addition to timing caps. It could be good to plot the current through the zeners to confirm.

  1. a cap starts off discharged and starts charging up towards VCC when its transitor is an open switch

  2. the base of its transistor eventually rises to turn on (thanks to r9), and closes its switch.

  3. the diode side of the cap suddenly gets pulled low, making the other side a negative voltage.

  4. the zener is now in reverse breakdown and pulls the attached base negative - sucking out all the base charge for a rapid turn off.

  5. Repeat

The next question could be, "does it have to be a zener?" I keep oscillating back and forth as to whether a regular diode would do the same thing. Since you don't use the reverse breakdown part of the zener diode, maybe it doesn't need to be a zener diode. Plotting current through it will help illustrate that.

EDIT ha! I originally wrote that it had to be a zener, and as I was editing my answer to say it doesn't have to be, you went and showed that it does get faster with lower voltage zeners. Ok, so there's still something to think about!

\$\endgroup\$
5
  • \$\begingroup\$ I think you're on to something.... though it DOES have to be a zener. Conducting in both directions seems to be important (at least, for high frequencies. I didn't try it with RC values for <1MHz)..... I added a screenshot of the sim w/I(D2), and it does do things in each direction at different points in the cycle. The traces for both zeners are identical in all but phase \$\endgroup\$
    – Brad Roy
    Mar 11, 2021 at 22:29
  • \$\begingroup\$ You're right - lower values seems to be better, to an extent. 3v3 and 3v6 work better than 5v1 on the breadboard. (the only values I have handy) Higher values increase the oscillation frequency, but also make it more likely for the circuit to fail to start up. \$\endgroup\$
    – Brad Roy
    Mar 11, 2021 at 22:35
  • \$\begingroup\$ Thanks for plotting current - it's possible that the excess current from the capacitor discharge is going through the other transistor's zener as well. That speeds up the capacitor discharge rate, increasing frequency too. The lower the zener voltage, the more total charge will flow through it - so lower voltage would be faster. \$\endgroup\$
    – KD9PDP
    Mar 11, 2021 at 22:44
  • \$\begingroup\$ also re: your point 2: r9 is optional - the circuit works without it. I discovered the effect when I noticed that the frequency increased when my finger touched the anodes of the zeners \$\endgroup\$
    – Brad Roy
    Mar 11, 2021 at 22:46
  • 1
    \$\begingroup\$ Awesome, that "proves" that the zeners are controlling timing too. When a transistor is "off", the base current is zero, base voltage is low (or negative, based on how the cap just "sucked out" all the base charge). VCC then charges up the reverse biased zener (like it's a tiny capacitor), and the capacitor charges up again (from when it was negative). Eventually the combined voltage turns on the transistor again, causing the other capacitor to turn off the other transistor. If R9 wasn't doing anything, then the zeners + caps controlled timing. \$\endgroup\$
    – KD9PDP
    Mar 11, 2021 at 23:29

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.