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
> 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
> 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 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.
The point is that the transconductance amplifier sums currents not
voltages.
Cheers
Ian
|
