negative feedback loop at -180 and A*B less than 1 stability?

[TicTek]

Hello everyone,
SO I am reading this article on negative feedback and the criteria to be stable.
I have not fully understood why if the loop gain at -180degree if the loop gain is less than 1 can be stable. why? because for me when the feedback has a shift phase of -180 and even the loop gain is less than 1,lets suppose 0.5 it will still add to the V+ since Vin=V(+)-V(-) and V(-)=A*B*Vo, A*B=-0.5(A=-1,B=0.5), so if I can run a simulation now as following
1. V(+)=1, V(-)=0, Vin=1, Vo=A*Vin=-1
2.V(+)=1,V(-)=(0.5)*(-1)=-0.5, Vin=1.5, Vo=-1.5
3.V(+)=1,V(-)=(0.5)*(-1.5)=-0.75, Vin=1.75, Vo=-1.75

so as you can see even it is increasing not so much but still it is increasing and not stable.
I know I have a lack in my logic here, so where I did not get it right?
Introduction to stability

 

[scopeprobe]

The way i understand it, if the gain is less than 1 then any ring or oscillation has not enough gain so therefore unsustainable so will eventually die away and settle at a steady state, too high a gain and you will hit the rails of the amplifier or the system could go bang, a gain of one will mean the oscillation is sustained.

 

[danadakk]

If you satisfied the angle criteria, 0, or N x 360 degrees (N is integer), but had a fdbk factor of < 1,
each time signal traverses the loop it is multiplied by < 1 so it gets smaller, until finally its 0,
eg. no signal left in the loop.


Regards, Dana.

 

[TicTek]

The way i understand it, if the gain is less than 1 then any ring or oscillation has not enough gain so therefore unsustainable so will eventually die away and settle at a steady state, too high a gain and you will hit the rails of the amplifier or the system could go bang, a gain of one will mean the oscillation is sustained.

Yes, this is what they say but why I cannot prove it by

 

[TicTek]

View attachment 176475

If you satisfied the angle criteria, 0, or N x 360 degrees (N is integer), but had a fdbk factor of < 1,
each time signal traverses the loop it is multiplied by < 1 so it gets smaller, until finally its 0,
eg. no signal left in the loop.


Regards, Dana.

Thanks for your answer, Dana.
I am still confused about it.
First, if Vs=0 then Vo =Vs*A/(1-AB) which means A/(1-AB) must be different from 0,which means that AB #1!! so where I am getting wrong?
second, quote "
If you satisfied the angle criteria, 0, or N x 360 degrees (N is integer), but had a fdbk factor of < 1,
each time signal traverses the loop it is multiplied by < 1 so it gets smaller, until finally its 0,
eg. no signal left in the loop." lets suppose as you wrote angle is 0 and AB < 1, so Verr=Vs+Vf
if Vs=1 at T=0,Vf=A*B*Verr=AB/(1-AB)*Vs, if AB=0.25, we end up with Verr=1/3Vs, so Vf is alwas has the sign of the Vs ,which makes Verr increasing

 

[danadakk]

There are two cases, one where Vs is 0 and the other non zero.

G = AB

Case one Vs = 0. So lets start by using Vs to inject a signal into the loop, then
turn it off. So signal, if loop G < 1, slowly keeps adding to Vs = 0, so its irrelevant.
and so it makes it first pass around the loop itself as it transverses the loop.
Now assume G is .9, so it passes thru the G elements (AB) but now is smaller, and that
process iteratively each pass keeps reducing signal until its 0. Corollary if G = 1 the loop
is sustained with the signal when Vs was reduced to 0. If G > 1 then it keeps growing
until physical circuit limits saturate.

Case two Vs is non zero. So Vs flows into summer and is is added to the fdbk
portion. If G >=1 then each transversal of loop the output of the summer grows
until physical limitsof the circuits saturate, typical the supply V on the circuits as
an approximation. Corollary, G < 1 then each time thru loop less and less is added
to Vs coming out of the summer, and ultimately the loop simple settles to Vs flowing
around loop, eg. stable.

Of course the above is not discussing the phase impact, but you can walk the loop and
see the impact of how phase affects loop. Keep in mind phase affects at the summer,
where you can experience less or more added to loop depending on phase, can even
experience sign reversal where summer actually becomes subtractor.

Regards, Dana.

 

[TicTek]

There are two cases, one where Vs is 0 and the other non zero.

G = AB

Case one Vs = 0. So lets start by using Vs to inject a signal into the loop, then
turn it off. So signal, if loop G < 1, slowly keeps adding to Vs = 0, so its irrelevant.
and so it makes it first pass around the loop itself as it transverses the loop.
Now assume G is .9, so it passes thru the G elements (AB) but now is smaller, and that
process iteratively each pass keeps reducing signal until its 0. Corollary if G = 1 the loop
is sustained with the signal when Vs was reduced to 0. If G > 1 then it keeps growing
until physical circuit limits saturate.

Case two Vs is non zero. So Vs flows into summer and is is added to the fdbk
portion. If G >=1 then each transversal of loop the output of the summer grows
until physical limitsof the circuits saturate, typical the supply V on the circuits as
an approximation. Corollary, G < 1 then each time thru loop less and less is added
to Vs coming out of the summer, and ultimately the loop simple settles to Vs flowing
around loop, eg. stable.

Of course the above is not discussing the phase impact, but you can walk the loop and
see the impact of how phase affects loop. Keep in mind phase affects at the summer,
where you can experience less or more added to loop depending on phase, can even
experience sign reversal where summer actually becomes subtractor.

Regards, Dana.

sorry for my late answer. Was very busy.
in the second case, you mentioned that :
Corollary, G < 1 then each time thru loop less and less is added
to Vs coming out of the summer, and ultimately the loop simple settles to Vs flowing
around loop, eg. stable.
I do not understand this correctly since it is still adding and the voltage increases with tiny increases.

 

[danadakk]

Start the loop with Vs = 0. sig in loop = 0, so output of loop summer = 0, and so
0 is added back to 0, and loops, stays off.

Start loop with Vs = 1, G = 0, sig in loop = 0, loop stays off.

Start loop with Vs = 1, G = 1, then take away Vs, so loop oscillates and stays at
1 signal in the loop.

Start loop with Vs = 1, G > 1, then take away Vs, so loop oscillates and signal
in loop grows without bounds.

Start loop with Vs = 1, G = .9, then take away Vs, so first pass thru loop signal = .9,
then .81, then .72 and falls to 0.

Alternate view, loop but no Vs ever. Noise in loop, G < 1, loop never grows.
Now make G > 1 so each time the sig transverses G > 1 it keeps growing.
Now make G = 1, so noise into gain block = out of gain block, so only
energy in loop is the noise and stays that way, neither growing or shrinking.

Phase of course contributes to stability, you can work the numbers for a
signal at each instant of time to see its affects.

Use a couple of OpAmps and set this up on bench, check it out....just altering G, then
do same for phase.....or in Spice....or Falstad

Regards, Dana.