Ian Thompson-Bell wrote:
>
> Patrick Turner wrote:
> >
> > Ian Thompson-Bell wrote:
> >> I finally cracked it - or rather found the text with an explanation I
> >> could understand. Basically, with shunt applied feedback it is
currents
> >> that are summed/subtracted at the input. As the output is a voltage
the
> >> stage is a transconductance amplifier so the closed loop gain is in
the
> >> form of Vo/Ii which does turn out to be Ao/(1+ß.Ao) and ß is simply
1/Rf
> >> where Rf is the feedback resistor
> >
> > Indeed the the transconductance amp is one with extremely high input
> > resistance.
> > So there is no input current to the amp, so a tube grid or fet gate
> > input
> > works well with shunt FB.
> >
>
> Actually, the input current is not zero; if it were the stage gain would
> be infinite. There will always be an Rg from grid to cathode so an input
> current i into the grid causes a Vgk of i.Rg
What I meant just to make things clearer is that the word 'amp' in my
above sentence
means the tube on its own, or any amp with extremely high Rin.
Yes there *is* current in the R1, but that's the current in the feedback
network only.
Biasing a grid with shunt FB can be done from the signal input to
ground, and letting
whatever signal source provide the signal current into both the bias R
and FB network.
The shunt FB is more effective when biasing this way.
>
> > For ß to be equal to 1/Rf, then how come? where does the 1 come from?
> >
>
> The transconductance amp does not have Ri present. ß is 1/Rf when
> feeding current into the input (gri cct). If the open loop gain is high
> then the closed loop gain is 1/ß or Rf so the output voltage is just
i.Rf
I don't see you logic. Rf, or R2, could be any onld value so ß could be
anything.
If you said R1 was 1 ohm, i'd understand.
And I don't neglect the slight grid voltage at the input grid even if
the
ciruit is a high gain type.
?
>
> > What is it?
> >
> > How come you don't express ß as Ri / ( Ri + Rf) , or in my way, R1 / (
> > R1 + R2 ) ?
> >
> > The Ii or input current supplied by the voltage signal source flows
> > equally in R1 and R2.
> >
>
> no, some flows into Rg.
But if Rg is from source to 0V, then input current is only flowing in
the FB network.
If you place Rg from g to 0V, everything becomed much more complex, and
I like simplicity
but the answer after using an equation must agree with simple analysis
using gain and ohm's
law.
>
> > But you have to know what the grid voltage is and Vo to sum these to
get
> > the current in R2.
> >
>
> Yes, it is i.Rg
>
> > Or if you like, Ii = ( A + 1 ) / R2, where +A volts is at output and
-1
> > volt is at the grid.
> >
> > So the voltage across R1 = R1 x [ ( A + 1 ) / R2 ] = ( R1/R2 ) x ( A +
1
> > )
> > To get the input voltage you must add the R1 voltage to the grid
voltage
> > which is 1, so you get
> > Vin = ( R1/R2 ) x ( A + 1 ) + 1.
> >
> > closed loop A' = Vo / Vi. But Vo = +A volts, with -1V at the grid.
> >
> > So A' = A / [ ( R1/R2 ) x ( A+1 ) + 1 ]
> >
> > Consider a 1/2 6SN7, A = 16, R1 = 40k, R2 = 80k.
> >
> > Say = +16 is at anode, then -1V is at grid, so voltage across 80k =
17V,
> > and across 40k is 8.5V, so input voltage MUST be -9.5V, ok?
> >
> > A' = 16 / 9.5 .
> >
> > Using the formula above without ß in it, just R1 and R2,
> >
> > A' = 16 / [ 0.5 x (16 + 1) + 1 ] = 16 / 9.5
> >
> >
> > You should be able to prove whether your reasons and formulas work
> > with common types of triodes or pentodes,
> > and make sure other dumb folks will understand you.
> >
>
> That I will do as part of NFB101.
>
> >> The input resistor effectively inputs a current equal to Vi/Ri from
> >> which you can work out the voltage gain as Rf/Ri since the closed
loop
> >> transconductance is Rf. I'll include this in NFB101.
> >
> > The input current is not Vin / R1.
> >
> > At the grid there is a smaller signal of the same phase as the input
> > signal.
> >
>
> Indeed, that voltage is i.Rg
>
> > So the current in R1 = ( Vin - Vg ) / R1.
> >
>
> No. You can replace Vin and Ri with a Norton equivalent current
> generator of Vin/Ri and parallel resistance Ri.
>
Yes but nobody would understand that so you ned careful schematic
showing the basics.
They must actually work when someone tries them out.
> The point is that the transconductance amplifier sums currents not
voltages.
OK, so you have a triode with gm = 2.5mA/V
You supply 1mA signal current within an R1 to a grid. and you have R2.
So what?, does not this end up allowing you to find what gain you'd need
in the tube
and what R2.
Most ppl know what tube they wish to use, then what R1 they should like
to be a minimum
value so it doesn't load down a previous tube stage, then the only
thing they want is the value of R2 to get the required gain reduction.
One has to decide early,
Whadda we want?
Whadda we have?
How can we use it?
Patrick Turner.
>
> Cheers
>
> Ian
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