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Q&A Change of pins in monostable multivibrator

The circuit you show doesn't make any sense: Start by examining the steady state condition. In steady state, C1 is effectively an open, so you ignore it. R4 keeps Q1 on. That means the collec...

posted 2y ago by Olin Lathrop‭

Answer
#1: Initial revision by user avatar Olin Lathrop‭ · 2021-09-18T16:36:20Z (over 2 years ago)
The circuit you show doesn't make any sense:

<img src="https://electrical.codidact.com/uploads/rMiB5SbBoBT67CrRnNC6Cmz2">

Start by examining the steady state condition.  In steady state, C1 is effectively an open, so you ignore it.  R4 keeps Q1 on.  That means the collector of Q1 will be low, which keeps the base of Q2 low, which means Q2 is off.

The diode doesn't do anything useful, and its cathode is certainly not a "trigger".  Driving <tt>trigger</tt> high does nothing, since the diode is reverse biased.  Driving <tt>trigger</tt> low only lowers the base of Q2 more.  That still keeps Q2 off.  Driving even lower will only exceed the reverse voltage capability of Q2's B-E junction, causing damage.

One way to trigger this one-shot would be to momentarily hold the base of Q1 low.  That allows the collector of Q1 to go high, which turns on Q2, which causes the collector of Q2 to go low.  C1 then provides a positive feedback path to continue keeping Q1 off.  Eventually C1 charges up to the point where Q1 is no longer held off.  Now there is positive feedback to flip Q1 on and Q2 off, ending the pulse.  Since the circuit is stable in the Q1-on Q2-off state, it will remain that way until externally perturbed again.