NFB101 Crisis Averted

NFB101 Crisis Averted

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

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.

Cheers

Ian

Re: NFB101 Crisis Averted



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.

For ß to be equal to 1/Rf, then how come? where does the 1 come from?

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.

But you have to know what the grid voltage is and Vo to sum these to get
the current in R2.

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.

>
> 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.

So the current in R1 = ( Vin - Vg ) / R1.

Patrick Turner.
>
> Cheers
>
> Ian

Re: NFB101 Crisis Averted

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

Re: NFB101 Crisis Averted



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

Re: NFB101 Crisis Averted

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

Re: NFB101 Crisis Averted



Ian Thompson-Bell wrote:
>
> 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.

You are disagreeing when you don't need to.

The current supplied from a voltage source signal flows in R1 and R2
equally. There is in effect, current flow between the amp input terminal
and the tube anode.

This one current determines grid voltage and the FB action we see.


>
> > 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).

Are we considering an amp without R1?

I thought we were considering a tume or opamp ect always with R1 and R2
connected.

If you wish to share your problems then you MUST spell the set up we are
talking about.

Don't EVER assume I can read your mind. I'm a terrible mind reader.


>
> > 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 need careful schematic
> > showing the basics.
> >
>
> Point taken. NFB101 *is* for relative beginners.

Assume the very worst about everyone out there trying to
design a simple triode line stage amp.

You do not need to mention Kirchoff, Norton.

Just use basics like I do and they'll all understand.

If you're going to write a paper on NFB, include the necessary 101
things,
not just 50, and never assume people will know the rest. They won't,
and they'll soon click off your pages once you have them bamboozled, and
they'll never return.

Include links to Norton and Kirchoff but being too heavy about these
will bamboozle.
All you need to know about building FB amp can be learnt by basic
current flow and ohm's law, and knowing how gain is developed in tubes.
Just my 2c opinion.

Patrick Turner.






>
> Cheers
>
> Ian

Re: NFB101 Crisis Averted

Patrick Turner wrote:
>
> Are we considering an amp without R1?
>
> I thought we were considering a tume or opamp ect always with R1 and R2
> connected.
>
> If you wish to share your problems then you MUST spell the set up we are
> talking about.
>
> Don't EVER assume I can read your mind. I'm a terrible mind reader.
>
>

I thought I made that clear in my original post.

Cheers

Ian

Re: NFB101 Crisis Averted



Ian Thompson-Bell wrote:
>
> Patrick Turner wrote:
> >
> > Are we considering an amp without R1?
> >
> > I thought we were considering a tume or opamp ect always with R1 and R2
> > connected.
> >
> > If you wish to share your problems then you MUST spell the set up we are
> > talking about.
> >
> > Don't EVER assume I can read your mind. I'm a terrible mind reader.
> >
> >
>
> I thought I made that clear in my original post.

After several posts go by, the focus is lost.

I sure lose the focus if you don't spell things out MORE CLEARLY.

This means the schematic of what you ARE talking about MUST be constant
and well described so there is NO confusion.

People want to know WHAT you are talking about.
Not just a formula ot theory, but something real doingt something,
ie, a vacuum tube or tubes set up for work to do something wanted.

You must learn to treat all the other guys like dumb bastards but
without insult,
and assume they know nothing unless you tell them.

All the best teachers I ever had were like this.
They ran hot and sweaty because they tried hard, god luv 'em.

That might mean a lot more typing for you.

The other day I had a posted on "Air gap on SE output transformers etc."
There were no replies because there are so few ppl building any amps
these days
AND who also want to take part in discussions here.
Its surprising really, that we have a forum available to so many folks,
and so few take part.
It proves how uncommunicative most men like to remain, and probably how
fickle they are when they
see a few trashy spams re Chinese crap produce and other catfights
regarding off topic issues.

But I never regret spelling it all out for everyone and the archives, so
that when somebody
googles 'air gap output transformers' then my post will come up amoung
and lotsa junk that isn't useful.

Same with negative feedback. What comes up after a search should be
useful, clear, and focused.

I find that when I edit a long post 3 times before clicking 'send' I
have
made the issue clearer in my own mind at least. Teaching other here
is teaching yourself.

Patrick Turner.


>
> Cheers
>
> Ian

Re: NFB101 Crisis Averted

Patrick Turner wrote:
>
> Ian Thompson-Bell wrote:
>> Patrick Turner wrote:
>>> Are we considering an amp without R1?
>>>
>>> I thought we were considering a tume or opamp ect always with R1 and R2
>>> connected.
>>>
>>> If you wish to share your problems then you MUST spell the set up we are
>>> talking about.
>>>
>>> Don't EVER assume I can read your mind. I'm a terrible mind reader.
>>>
>>>
>> I thought I made that clear in my original post.
>
> After several posts go by, the focus is lost.
>
> I sure lose the focus if you don't spell things out MORE CLEARLY.
>
> This means the schematic of what you ARE talking about MUST be constant
> and well described so there is NO confusion.
>
> People want to know WHAT you are talking about.
> Not just a formula ot theory, but something real doingt something,
> ie, a vacuum tube or tubes set up for work to do something wanted.
>
> You must learn to treat all the other guys like dumb bastards but
> without insult,
> and assume they know nothing unless you tell them.
>
> All the best teachers I ever had were like this.
> They ran hot and sweaty because they tried hard, god luv 'em.
>
> That might mean a lot more typing for you.
>

That is precisely what I am aiming for in NFB101. However, the original
post was just meant to let people know I had found a solution to that
particular problem rather than to present the solution in full.
Cheers

Ian