# collector coupled monostable multivibrator circuit operation

Discussion in 'LEDs and Optoelectronics' started by amateur_101, Dec 1, 2012.

1. ### amateur_101

1
0
Dec 1, 2012
hi. i find it hard to understand how the Transistor 1 gets a -0.7 volts when the trigger is applied. and so is the Ct and then discharges the -0.7V to transistor 2, making it in its cut-off state and produces an Vcc output. can someone explain this to me? thanks

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2. ### Harald KappModeratorModerator

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Nov 17, 2011
Me too - because it doesn't.
C1 + R4 make a differentiator. Only the edge of the trigger signal generates a voltage across R4. Of that voltage only the positive pulses get throughthe diode to Tr1.

In the absence of a trigger, TR2 is open, biased by Rt, therefore the voltae at the collector of TR2 is ~0V. The collector voltage of TR2 is base voltage for TR1 and therefore TR1 is off. Ct is fully charged ("+" side = Vcc, "-" side = Vbe(Tr2)). This is the stable state of the circuit.

On a positive pulse at the base of TR1, Tr1 opens. The collector voltage of TR1 becomes becomes ~0V. Since Ct cannot be discharged instantaneously, it keeps its charge and therefore the voltage across it. The "+" side of Ct is connected to TR1's collector and therefore is ~0V. Since the voltage across Ct doesn't change immediately, the "-" side becomes ~-Vcc. The "-" side is connected to the base of TR2 which is therefore negatively biased and turned off. This is the astable state. Rt will discharge Ct to 0V and then start charging it with reverse polarity (remember: the "+" side is still held to ~0V by TR1) until the "-" side reaches ~0.7V. At that moment TR2 becomes conductive and via R3 draws TR1's base to 0V. TR1 is thus turned off.

Therefore the time constant RT*Ct determines the astable on-time of the monoflop.

Having written all that sermon, I Googled "monostable transistor multivibrator" and guess what I found? This. You could have saved me all that work by giving your source. The circuit is fully explained there!

The only thing is that on that page the explanation
is not correct. As stated above this should be -(Vcc-0.7V). We could have discusses that issue without the ballast of the full circuit's operation.

Next time, please give us all information you have. It helps.

Harald