Ian Thompson-Bell wrote:
>
> Patrick Turner wrote:
> >
> > Ian Thompson-Bell wrote:
> >> Patrick Turner wrote:
> >>> Alex wrote:
> >>>> "Ian Thompson-Bell" <ruffrecords@[EMAIL PROTECTED]
> wrote in message
> >>>> news:fukkiv$qch$1@[EMAIL PROTECTED]
> >>>>> I have just been working through the math for shunt derived shunt
> >>>>> applied NFB around an amp and Ican't get the expected result.
> >>>>>
> >>>>> Imagine an inverting amp with an open loop gain of -Ao with a
feedback
> >>>>> resistor from output to input of Rf and an input resistor from
signal
> >>>>> source to the input of Ri. I get a closed loop gain of:
> >>>>>
> >>>>> An = Ao/(1+(Ri/Rf)+(Ri.Ao/Rf)) when I expected some thing of the
form:
> >>>>>
> >>>>> An = Ao/(1+ß.Ao) where ß = Ri/Rf
> >>>>>
> >>>>> but instead I get Ao/(1+ß+ß.Ao)
> >>>>>
> >>>>> I have checked the math several times and cannot see where I have
gone
> >>>>> wrong. Anyone throw any light on this?
> >>>>>
> >>>>> Cheers
> >>>>>
> >>>>> Ian
> >>>> Hello, Ian.
> >>>> Your equations are perfectly correct.
> >>>>
> >>>> To test them use an (imaginery) ridiculous amplifier with very low
gain, say
> >>>> Ao=0.01. Also let us assume beta=1, i.e. Ri = Rf. Well, because of
> >>>> ridiculously low gain the feedback is virtually inoperative. What
will the
> >>>> gain An be?
> >> Patrick replied.
> >>
> >>> But ß cannot be 1.0 where R1 = R2 ( Ri = Rf ).
> >>>
> >>> ß is the fraction of the output voltage fed back to the input.
> >>>
> >>> Consider the input voltage terminal to be grounded.
> >>>
> >>> A noise signal within the amp appears at the output, Vn.
> >>> The fed back Vn at the grid, or inverting input = Vn x R1 / ( R1 +
R2 ).
> >>>
> >>> The Vn at the grid is thus 0.5Vn, and is amplified by the OLG to
oppose
> >>> its own production.
> >>>
> >>> ß = 0.5 when R1 = R2.
> >>>
> >>> R1 and R2 operate as a resistance divider.
> >>>
> >> This is indeed confusing and hopefully only for this particular form
of
> >> feedback topology.
> >
> > What is confusing?
> >
> > ß is the fraction of the output signal fed back, period.
> >
>
> Indeed and by your calculation this is 0.5. This means when the open
> loop gain is large enough the closed loop gain tends to 1/ß which in
> your case is 2. In practice the gain is unity. That is what is
confusing.
Think of a REAL amp, gain 10,000, inverting, with R1 = R2, ß = 0.5.
If there is +10V ac at the output, -0.001V is needed between the two
inputs, one of which is grounded.
10.001V is across R2, and this is also across R1, so the input voltage
MUST be -10.002V
People will say Gain = R1 / R2, ie, 1.0, but they are WRONG.
Its 10.000V / 10.002V, which is LESS than unity, OK.
Now consider a triode with OLgain = 20.
R1 = R2, and ß = 0.5, but lets calculate from the real observations we
would make.
With +10V output, -0.5V is at the grid, and cathode is bypassed fully.
10.5V appears across R2, so the same V appears across R1, so
the input voltage = -11.0V.
Closed loop gain = 10V / 11V, = 0.90909, inverting.
How does anyone get confused and believe gain will somehow be 2?
The tube with the shunt NFB is called an anode follower,
because it has nearly unity gain and thus resembles a cathode follower.
But the CF has ß = 1.0, because ALL the output is in series with the
input.
If the tube above is used as a CF, and there is +10V at the cathode as
output,
there will still be +0.5V needed between grid and cathode to cause that
voltage change at the cathode
assuming the load is the same as when anode loaded in the anode follower
case.
So grid voltage must be +10.5V.
CLG = 10.5V / 10V = 0.95 approximately, or slightly MORE than in the
anode follower,
all because ß is different.
If you try to make ß = 1.0 with an shunt FB network, you must make R2
approach zero ohms and R1
approach infinity, so that ß is R1 / ( R1 + R2 ).
Clealy this is impossible to achieve. Closed loop gain gets very low as
ß approaches 1.0.
Nobody uses such an amp except in the case where you use shunt NFB with
a LINEAR
gain pot, and then you get an approximate logarithmic gain control,
or just what ppl want, and the noise of the control pot is reduced with
the FB action.
Crown used this idea on the preamps but many ancient
AM radios also used the idea extremely effectively to clean up the
N&D endemic in so many AM radios.
Now with the triode CF, if there is a distortion voltage Vdn at the
output that you measure,
then this is amplified to become -20Vdn at the output.
So the distortion that exists without the FB is +21Vdn, and the -20Vdn
subtracts from the +21Vnd
leaving what we measure and see on the CRO.
THD reduction factor = 1/21 = 1 / (1+A)
CF CLG, A' = A / ( 1 + [A x ß] ), like all series NFB apps.
But you can see because ß = 1, you get a simpler equation for CF.
With the anode follower with triode of OLgain = 20,
if +Vdn is at the output, only +0.5Vdn is at the grid, because ß = 0.5,
so there is -10Vdn produced which reuces what must have been present
without FB to give 1Vdn.
Obviously, +11Vd must have been present, so the THD reduction factor =
1/11, much less than the CF, 1/21.
Using the CLG equation above but for the AF,
A' = 20 / (1 + [20 x 0.5] )
= 20 / 11, or nearly 2.
But above we observed real AF gain was actually 10/11.
Clearly the equation for the series voltage NFB is incorrect when used
for shunt NFB.
Using a formula i wrote down years ago with a sample EL34 tube
with A = 20, A' = A / ( 1 + {[ A + 2 ] x ß } ),
A' = 20 / ( 1 + { [ 20 + 2 ] x 0.5 )
= 20 / ( 1 + 11 ) = 20/12, also quite wrong.
Its 3:06 am, and I'll leave it for another day to clear up the math
mess.
Patrick Turner.
>
> Cheers
>
> Ian
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