I’ve a few problems and questions to the principle of monostable multivibrator’s work. Basic schematic is as below: monostable multivibrator basic schematic

  1. What is the reason for D1, C1, R3 elements at input? Why I can’t just tie a transistor base to the trigger input, without these elements?
  2. I've read that Vcc voltage can’t be as high as you would and it has some max value, but I can’t determine what it is and where it comes from.
  • \$\begingroup\$ I'd say the input circuit is to condition an AC waveform without reverse-biasing the B-E junction, but I can't be sure without context. The main limitation on Vcc is going to be the Vce spec on the transistor (maximum collector-emitter voltage). \$\endgroup\$ Dec 5, 2018 at 14:59
  • \$\begingroup\$ I think that schematic has a mistake (no feedback making it more of a pulse shaper than a one-shot), but in general the C/R/D on the input is a differentiator that forms a well defined pulse for the one-shot to work properly. If you DC couple the input then it will become dependent on when the input "releases" the input stage. Whether you want this or not depends on your intended use for the one-shot. In most applications it's used as a pulse-stretcher, but there are other uses. \$\endgroup\$
    – user201365
    Dec 5, 2018 at 15:05
  • \$\begingroup\$ @isdi you got a point. I've fixed it \$\endgroup\$ Dec 5, 2018 at 17:18
  • \$\begingroup\$ The circuit is designed to respond to the rising edge and not the falling edge. R3 is needed to provide a falling edge path for the charge on C1 (D1 and T2's base-emitter and Rs provide paths for the rising edge.) As for the power supply limit? The BJT's have a limit with respect to their Vce, probably? Oh... T1 should have its base protected with a diode to ground, though that will affect the timing. \$\endgroup\$
    – jonk
    Dec 5, 2018 at 17:41

1 Answer 1


The steady-state of this circuit, removing C2 to find that steady-state, is with T1 (the right bipolar) on, and that collector voltage low, thus taking away any bias current into T2.

Steady-state is with rightmost transistor on/saturated and left most transistor off(cutoff).

Now bring in a rising edge to C1. (negative edges are blocked by the diode). If your input amplitude is at least 2 diode drops, and the pulse source resistance << Rs (the feedback resistor), then left transistor turns on and drives its collector near to ground.

You are injecting a trigger pulse, thru the high-pass-filter of C1 and R3. There is a propagation delay around the feedback loop; the rightmost transistor needs time to flush out charges from the collector junction, so that transistor can exit the saturation-mode and that collector can rise enough to provide current through Rs to base of left transistor, and replace the pulse-current flowing through C1/D1.

Your input pulse width must exceed T_on of the left transistor, and T_delay+T_off of the right transistor; plan on at least 50nS to 200nS.

Notice the high quiescent (steady-state) voltage across C2: Vdd - 0.6 volts, or about 4.4 volts.

If the VDD > emitter-base breakdown voltage, the negative swing on base of right transistor is clamped at the Zener (VbeReverse) voltage, and your computed time-constant is no longer valid.


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