Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Monostable Multivibrator Circuit - NPN Transistors

Status
Not open for further replies.

Zener_Diode

New Member
Hey,

Monostable Multivibrators (One-Shot)

1.
look at the "Monostable Multivibrator Circuit" at the link above. As I understand there is a mistake in the explanation "The Collector of TR1 which was previously at Vcc drops quickly to below zero volts effectively giving capacitor C2 a reverse charge of -0.6v across its plates. This results in transistor TR2 now having a Base voltage of -0.6v holding the transistor fully "OFF" at point VbTR2".
I clame that voltage on C2 is -(Vcc-0.6) and not -0.6 as was written, am I right?


2. I can not understand if there is a current into the base of TR2 while voltage on C2 is less than 0.6V. if the answer is YES explain why? if the answer is NO explain why too.


Thanks ahead
 

dougy83

Well-Known Member
look at the "Monostable Multivibrator Circuit" at the link above. As I understand there is a mistake in the explanation "The Collector of TR1 which was previously at Vcc drops quickly to below zero volts effectively giving capacitor C2 a reverse charge of -0.6v across its plates. This results in transistor TR2 now having a Base voltage of -0.6v holding the transistor fully "OFF" at point VbTR2".
I clame that voltage on C2 is -(Vcc-0.6) and not -0.6 as was written, am I right?
The explanation is bs. The voltage at the collector of TR1 will go from vcc to ~0V. The voltage of TR2 base will go from +0.6 to ~-(Vcc-0.6).
 

MrAl

Well-Known Member
Most Helpful Member
Hi there,


Im not sure if this helps or not but...

That circuit works mostly on the basis of which cap is charged and which cap
is discharged for a given half cycle of the output. When the input pulse goes
positive one cap charges up to +Vpulse, then when the input pulse goes back
to ground the cap that charged up during +Vpulse now puts a negative voltage
at the base of that transistor effectively turning it off and that turns the other
transistor on. This means the base of each transistor goes from about 0.7v
down to -Vpulse for a short time period. The collector voltage however goes from
about 0.1v up to about +Vcc yes, and never goes negative.

The result on a pulsing drive signal of frequency F is of course an output
frequency of one half that or F/2.

The current into the base of either transistor depends on the other transistors
collector resistor and the base series resistor and the base ground resistor and
whether or not it is going through a switching cycle and whether or not it was
on or off previously. If the transistor was on and it is being turned off, then
the base current could possibly be negative for a short time duration while the
base charge is being swept out.
 
Last edited:
Status
Not open for further replies.

EE World Online Articles

Loading
Top