help needed understanding monostable circuit

Thread Starter

daz E

Joined Nov 7, 2017
47
http://www.electronics-tutorials.ws/waveforms/monostable.html

The text underneath "Monostable Multivibrator Circuit" states, in para 2,

"If a negative trigger pulse is now applied at the input, the fast decaying edge of the pulse will pass straight through capacitor, C1 to the base of transistor, TR1 via the blocking diode turning it “ON”. The collector of TR1 which was previously at Vcc drops quickly to below zero volts effectively giving capacitor CT a reverse charge of -0.7v across its plates"

I'm having real difficulty understanding this. Please help and explain in layman terms ....

1.If a -ve voltage is applied to the input pin, then how does a -ve current pass through what would be a reverse biased diode given the anode is now -ve? The diaode is forward biased when +ve from R4/C1 through to TR1b. So how does a -ve current flow in that direction to effect a reverse charge at Ct

2. C1 is 100nf so what effect does this have, other than to smooth out the pulse ?

3. How does Ct get a reverse charge of -0.7v because R1 and R5 are connected to TRb and not TR1b where the -ve charge is.

Ive set this up in falstad circuit applet and applied -5v to the input and it does seem to work as described, but I just cant get my head around how??

Thank you in advance
(we all learned somewhere/somehow !)
 
Last edited:

AlbertHall

Joined Jun 4, 2014
9,869
It is not well explained.
Look at the waveform given at the anode of the diode. It is a series of positive and negative spikes. It is the positive spikes which pass through the diode and turn on TR1. If it was a positive going trigger pulse then the decaying edge of that pulse would be its negative going edge - the end of the pulse. So for a negative pulse its decaying edge is positive going.

The voltage across a capacitor cannot change quickly so when a fast transition is applied to one plate the voltage on the other plate will change by the same amount in the same direction.
 

Thread Starter

daz E

Joined Nov 7, 2017
47
It is not well explained.
Look at the waveform given at the anode of the diode. It is a series of positive and negative spikes. It is the positive spikes which pass through the diode and turn on TR1. If it was a positive going trigger pulse then the decaying edge of that pulse would be its negative going edge - the end of the pulse. So for a negative pulse its decaying edge is positive going.

The voltage across a capacitor cannot change quickly so when a fast transition is applied to one plate the voltage on the other plate will change by the same amount in the same direction.
Thank you Sir for your reply

Although, My head is spinning

Yes, I get the diode issue now, I thought it must be a +ve influx to TR1b that switched it on.

But can you just explain a bit more simply, :) , the signal bit (going +ve, going -ve, edge) - im in a spin here having read your words over and over.

and

I thought I understood capacitors, but I dont, so what is the significance of your words around plates of the capacitor changing in the same direction by the same amount? As I understand, a plate will accumulate charge of electrons and at full capacity will allow current to flow. When the charge is removed the capacitor discharges with the same polarity, so a -ve charge plate releasing its -ve charge. This much I understand; what I do not understand is what is happening through the capacitor and wires each side when the pulse is +ve and when the pulse is -ve.

??? this is fundemental stuff I guess so thank oyu for your patience with me.,

i would add that I studied Computer Science at Uni but dealt with MCUs , ALUs , CPUs and so on... 0 and 1s .... leading edge, trailing edge, pulse width, mark, train, period, etc. , this anlaogue stuff is another world !!
 
Last edited:

Thread Starter

daz E

Joined Nov 7, 2017
47
Have a read here and see if it helps.
Yes, i have read three times trying to get my head round it all to understand the physics and importantly the basics. THANK YOU for the link. I'll need to read and digest some more for it to be etched to memory.

So the capacitor yields a transient response whereupon the leading edge positive gets through the diode?well, it can only be a +ve current through the forward biased diode to TR 1 can't it. So.... What i don't get now...

Is, once the (surely +ve) pulse gets through the diode,...

1. how can TR1b that was at vcc now at below zero volts (according to the article) if the current through diode is +ve.
2. Even if TR1b is at v<0 then that would shut off TR1?
3.H ow does Ct then get reversed charged from the pulse, its not even connected to TR1b and the supposed -ve volts now at TR1b

Oohhhhh my head
 

AlbertHall

Joined Jun 4, 2014
9,869
TR1 collector was at VCC. After the trigger it will be close to 0V.
Ct was charged to near VCC. When the collector of TR1 falls by VCC volts, both sides of Ct will fall by VCC volts. So the base of TR2 will now be negative by approximately VCC volts. Hence TR2 will be off and its collector voltage will rise and supply current to TR1 base keeping it turned on.
Until Ct discharges (the voltage across it falls) and TR2 can turn on again...
 

Thread Starter

daz E

Joined Nov 7, 2017
47
TR1 collector was at VCC. After the trigger it will be close to 0V.
Ct was charged to near VCC. When the collector of TR1 falls by VCC volts, both sides of Ct will fall by VCC volts. So the base of TR2 will now be negative by approximately VCC volts. Hence TR2 will be off and its collector voltage will rise and supply current to TR1 base keeping it turned on.
Until Ct discharges (the voltage across it falls) and TR2 can turn on again...
I do thank you for taking the time to reply, however, one can only perceive in your response, a recital of the orginal article. There is nothing in your reply to explain the WHY?

Again, let one if I may, put the questions once more for clarity , this time with a few more words appended:

1. how can TR1b that was at vcc now at below zero volts (according to the article) at trigger pulse, if the current through diode is +ve.
Surely TR1b has a +ve base current because only +ve current can move through the forward biased diode?
If not WHY?


2. Even if TR1b is / was at v<0 at trigger, then that would shut off TR1 no?
If not then WHY?

3.How does Ct then get reversed charged from the pulse, Ct is not even connected to TR1b and thus the supposed -ve volts now at TR1b, Ct is always charged via R1/Rt and always at Vcc from the Vcc rail?
If not WHY?
 

AlbertHall

Joined Jun 4, 2014
9,869
1. TR1 base (I'm assuming that is what you mean by TR1b) is never at VCC. It is either 0V or 0.7V.
2. Before the trigger TR1 base is at 0V. The trigger takes it positive - up to 0.7V - and turns it on.
3. In the stable state, with TR1 off and Tr1 on, TR1 collector is at VCC and TR2 base is at 0.7V so Ct is charged with Vcc-0.7V across it.
When triggered, TR1 collector will be at 0V and as the voltage across Ct has not yet had a chance to change then TR2 base will be at -(VCC-0.7V).
Now Ct begins to discharge and the base voltage of TR2 slowly rises. When that voltage has risen to +0.7V TR2 will turn on, which turns off TR1 again.
 

Thread Starter

daz E

Joined Nov 7, 2017
47
1. TR1 base (I'm assuming that is what you mean by TR1b) is never at VCC. It is either 0V or 0.7V.
2. Before the trigger TR1 base is at 0V. The trigger takes it positive - up to 0.7V - and turns it on.
3. In the stable state, with TR1 off and Tr1 on, TR1 collector is at VCC and TR2 base is at 0.7V so Ct is charged with Vcc-0.7V across it.
When triggered, TR1 collector will be at 0V and as the voltage across Ct has not yet had a chance to change then TR2 base will be at -(VCC-0.7V).
Now Ct begins to discharge and the base voltage of TR2 slowly rises. When that voltage has risen to +0.7V TR2 will turn on, which turns off TR1 again.
Yes, sorry, I have overlooked the 10k dropping all but the charge to TR1b (yes , base of TR1, I cant get subscript on the 'b' until now!)

So TR1b at 0 at stable state until +ve pulse at the base turning it on and allowing TR1ce , which in turn allows Ct discharge .

But where is the change on C
t ? Is this through a +ve charge now through Rt ? which makes TR1b go -700mv ?
 

AlbertHall

Joined Jun 4, 2014
9,869
But where is the change on Ct ? Is this through a +ve charge now through Rt ? which makes TR1b go -700mv ?
After TR1 switches on, current flows through Rt and the voltage on TR2 base begins to rise from -(VCC - 0.7) up to +0.7V.
The only time TR1 base will be negative is during the negative edge of the trigger pulse.
 
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