I have the following circuit and I have to find the duty cycle of the output signal and its frequency:

relaxation oscillator

I figured out that if \$v_{cond} > 0\$, then D4 will be forward biased and D3 reverse biased, so it's basically a relaxation oscillator with the time constant \$T_{b} = R_{1b}*C_{1}\$, and if \$v_{cond} < 0\$, then D3 will be forward biased and D4 will be reverse biased and the time constant will be \$T_{a} = R_{1a}*C_{1}\$.

The threshold value is 4V, so if the capacitor is fully charged at 4V and it begins to discharge, then it does so with the time constant \$T_{b}\$ until the voltage reaches 0 and then it continues to discharge with the time constant \$T_{a}\$ until it reaches -4V, then it starts charging again and so on.

If my way of thinking until now is not flawed, then on half of the period of the signal I have a capacitor charging/discharging at different speeds and so I'm stuck at finding the duty cycle. It would be quite easy if I could find out how much time is elapsed from the moment the capacitor is fully charged until it reaches zero, but I cannot use the usual relationship \$ V_{cond} = (V_{c_{stop}} - V_{c_{start}}) * (1 - e^{-\frac{t}{R C}})\$, so I'm obviously missing something. After I clear this thing out I think finding out the frequency of the output signal will be quite simple.

Any help is greatly appreciated.

  • 1
    \$\begingroup\$ Homework with some effort. Extremely rare. +1. \$\endgroup\$
    – dim
    Jun 10, 2016 at 19:17
  • \$\begingroup\$ yes, once you determine the charging and discharging times, getting the duty cycle and freqiency should be easy. \$\endgroup\$ Jun 11, 2016 at 4:11
  • \$\begingroup\$ Thanks, but it's not for a homework. I'm preparing for an exam and trying to understand some basic principles. \$\endgroup\$
    – Laura_F
    Jun 11, 2016 at 9:21

1 Answer 1


Any help is greatly appreciated.

It charges between -4V and +4V with one time constant and discharges back to -4V with the other time constant. 0V (midrail is of no consequence).

  • \$\begingroup\$ Doesn't the time constant change when the voltage crosses 0V (either when the capacitor charges or discharges)? I assumed that at this threshold one diode is forward biased, while the other is reverse biased, so the resistor through which the capacitor is charging/discharging is changing. Is my assumption flawed? \$\endgroup\$
    – Laura_F
    Jun 11, 2016 at 9:24
  • \$\begingroup\$ The schmitt trigger threshold points are at -4V and +4V and nothing happens of much consequence in between except that the capacitor is ramping one way or the other. That ramp direction is determined ONLY by the output polarity which is (due to the zener diode clamps) +6V or -6V. So if the output is +6V then the cap charges up via D3 until it reaches the +4V threshold then the output switches to -6V and discharges via D4. Nothing to do with 0V or midrail. \$\endgroup\$
    – Andy aka
    Jun 11, 2016 at 9:43

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