Patrick Turner wrote:
>
> 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.
>
No, there is also the current from Vin.
> 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.
>
Indeed. Remember, we are considering the amp *without* Ri, when it is a
transconductance amp. It's gain is Vo/Ii which must have the dimensions
of resistance (or impedance).
> 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.
>
And Kirchoff's law too.
>>> 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.
>
Point taken. NFB101 *is* for relative beginners.
Cheers
Ian
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