NFB101 Crisis!

NFB101 Crisis!

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

Re: NFB101 Crisis!

On Apr 22, 1:09 pm, Ian Thompson-Bell <ruffreco...@yahoo.co.uk> wrote:
> 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

I can't help you with the math but when you need a break you can get a
giggle at "ZNFB: Religious imperative or manner of speaking?":
* members.lycos.co.uk/fiultra/KISS%20104%20by%20Andre%20Jute.htm
and at: "Why there is no negative feedback in The KISS Amp"
* members.lycos.co.uk/fiultra/KISS%20123%20by%20Andre%20Jute.htm

Enjoy.

Andre Jute
Visit Jute on Amps at * members.lycos.co.uk/fiultra/
"wonderfully well written and reasoned information
for the tube audio constructor"
John Broskie TubeCAD & GlassWare
"an unbelievably comprehensive web site
containing vital gems of wisdom"
Stuart Perry Hi-Fi News & Record Review

Re: NFB101 Crisis!

Ian Thompson-Bell wrote

> I have just been working through the math for shunt
> derived shunt applied NFB around an amp and Ican't get the
> expected result.

Is that the same as voltage derived, current applied? I can
never remember this stuff, sigh.

> 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?

Your maths is correct or we're both wrong.

I don't know if it helps to see it as Ao/(1+ß.(1+Ao))

You haven't explained why you believe there is an anomaly.
What led you to your erroneous expectation?

Ian

Re: NFB101 Crisis!

Ian Iveson wrote:
> Ian Thompson-Bell wrote
>
>> I have just been working through the math for shunt
>> derived shunt applied NFB around an amp and Ican't get the
>> expected result.
>
> Is that the same as voltage derived, current applied? I can
> never remember this stuff, sigh.
>

No, it's voltage derived, voltage applied. As a rule:

Shunt = Voltage
Series = Current


CHeers

Ian

Re: NFB101 Crisis!

Ian Thompson-Bell wrote:

> Ian Iveson wrote:
>> Ian Thompson-Bell wrote
>>
>>> I have just been working through the math for shunt
>>> derived shunt applied NFB around an amp and Ican't get
>>> the expected result.
>>
>> Is that the same as voltage derived, current applied? I
>> can never remember this stuff, sigh.
>>
>
> No, it's voltage derived, voltage applied. As a rule:
>
> Shunt = Voltage
> Series = Current


Thanks. Then how come your circuit employs "shunt derived"
feedback?

And perhaps you could answer the question you cut out from
my post, considering I was good enough to answer yours:

"Your maths is correct or we're both wrong.

"I don't know if it helps to see it as Ao/(1+ß.(1+Ao))

"You haven't explained why you believe there is an anomaly.
What led you to your erroneous expectation?"

Ian

Re: NFB101 Crisis!

Ian Iveson wrote:
> Ian Thompson-Bell wrote:
>
>> Ian Iveson wrote:
>>> Ian Thompson-Bell wrote
>>>
>>>> I have just been working through the math for shunt
>>>> derived shunt applied NFB around an amp and Ican't get
>>>> the expected result.
>>> Is that the same as voltage derived, current applied? I
>>> can never remember this stuff, sigh.
>>>
>> No, it's voltage derived, voltage applied. As a rule:
>>
>> Shunt = Voltage
>> Series = Current
>
>
> Thanks. Then how come your circuit employs "shunt derived"
> feedback?
>

Sorry, don't understand - the feedback network is fed from the output
voltage i.e. across or shunting the output. Is that what you mean?

> And perhaps you could answer the question you cut out from
> my post, considering I was good enough to answer yours:
>

My apologies, I must have missed it.

> "Your maths is correct or we're both wrong.
>
> "I don't know if it helps to see it as Ao/(1+ß.(1+Ao))
>

Yes, I had it as that to start with but changed it to 'show' the anomoly.

> "You haven't explained why you believe there is an anomaly.
> What led you to your erroneous expectation?"
>

I had expected *all* NFB forms (however derived or applied) to reduce to
Ao/(1+ß.Ao) but this one does not.

Cheers

ian

Re: NFB101 Crisis!


Ian Thompson-Bell wrote

>>>>> I have just been working through the math for shunt
>>>>> derived shunt applied NFB around an amp and Ican't get
>>>>> the expected result.
>>>> Is that the same as voltage derived, current applied? I
>>>> can never remember this stuff, sigh.
>>>>
>>> No, it's voltage derived, voltage applied. As a rule:
>>>
>>> Shunt = Voltage
>>> Series = Current
>>
>>
>> Thanks. Then how come your circuit employs "shunt
>> derived" feedback?
>>
>
> Sorry, don't understand - the feedback network is fed from
> the output voltage i.e. across or shunting the output. Is
> that what you mean?
>
>> And perhaps you could answer the question you cut out
>> from my post, considering I was good enough to answer
>> yours:
>>
>
> My apologies, I must have missed it.
>
>> "Your maths is correct or we're both wrong.
>>
>> "I don't know if it helps to see it as Ao/(1+ß.(1+Ao))
>>
>
> Yes, I had it as that to start with but changed it to
> 'show' the anomoly.
>
>> "You haven't explained why you believe there is an
>> anomaly.
>> What led you to your erroneous expectation?"
>>
>
> I had expected *all* NFB forms (however derived or
> applied) to reduce to Ao/(1+ß.Ao) but this one does not.


I find the four possible combinations of shunt/series
feedback can be difficult to interpret, so it may be just
me.

For shunt derived feedback I would expect to see a load and
means of sensing the current through that load. An example
would be the use of a small current-sensing resistor on the
ground side of a loudspeaker to derive the feedback signal.
Another example is the use of an unbypassed cathode
resistor, which is shunt derived, series applied.

If the load is purely resistive, I can't see the difference
between the two methods of derivation, because the current
will always be in fixed proportion to the voltage.

Where the only load is the feedback path itself, which in
your case is purely resistive, I can't see the difference
either. The most obvious way to me of seeing your circuit,
however, is that the feedback signal is derived from a
voltage divider between output and input. It seems to me
that is where the mysterious "1+" comes from...the fact that
the feedback is not derived from a voltage divider between
the output and ground...there's an extra Vin with respect to
that.

Ian

Re: NFB101 Crisis!

Ian Iveson wrote:
> Ian Thompson-Bell wrote
>
>>>>>> I have just been working through the math for shunt
>>>>>> derived shunt applied NFB around an amp and Ican't get
>>>>>> the expected result.
>>>>> Is that the same as voltage derived, current applied? I
>>>>> can never remember this stuff, sigh.
>>>>>
>>>> No, it's voltage derived, voltage applied. As a rule:
>>>>
>>>> Shunt = Voltage
>>>> Series = Current
>>>
>>> Thanks. Then how come your circuit employs "shunt
>>> derived" feedback?
>>>
>> Sorry, don't understand - the feedback network is fed from
>> the output voltage i.e. across or shunting the output. Is
>> that what you mean?
>>
>>> And perhaps you could answer the question you cut out
>>> from my post, considering I was good enough to answer
>>> yours:
>>>
>> My apologies, I must have missed it.
>>
>>> "Your maths is correct or we're both wrong.
>>>
>>> "I don't know if it helps to see it as Ao/(1+ß.(1+Ao))
>>>
>> Yes, I had it as that to start with but changed it to
>> 'show' the anomoly.
>>
>>> "You haven't explained why you believe there is an
>>> anomaly.
>>> What led you to your erroneous expectation?"
>>>
>> I had expected *all* NFB forms (however derived or
>> applied) to reduce to Ao/(1+ß.Ao) but this one does not.
>
>
> I find the four possible combinations of shunt/series
> feedback can be difficult to interpret, so it may be just
> me.
>
> For shunt derived feedback I would expect to see a load and
> means of sensing the current through that load. An example
> would be the use of a small current-sensing resistor on the
> ground side of a loudspeaker to derive the feedback signal.

No, that is series derived as you are connected in series with the
output and also called current derived because you are sensing current.
I agree it can be confusing.

> Another example is the use of an unbypassed cathode
> resistor, which is shunt derived, series applied.
>

Correct, shunt derived because we connect the feedback network directly
across the output (cathode resistor) and series applied because derived
voltage is applied in series with the input.

> If the load is purely resistive, I can't see the difference
> between the two methods of derivation, because the current
> will always be in fixed proportion to the voltage.
>
> Where the only load is the feedback path itself, which in
> your case is purely resistive, I can't see the difference
> either. The most obvious way to me of seeing your circuit,
> however, is that the feedback signal is derived from a
> voltage divider between output and input. It seems to me
> that is where the mysterious "1+" comes from...the fact that
> the feedback is not derived from a voltage divider between
> the output and ground...there's an extra Vin with respect to
> that.
>

Yes, I am still not sure about it. Needs further thought. At least it
seems my maths is OK.

Cheers

Ian

Re: NFB101 Crisis!


"Ian Thompson-Bell" <ruffrecords@yahoo.co.uk> wrote in
message news:fupk67$i5l$1@energise.enta . net ...
> Ian Iveson wrote:
>> Ian Thompson-Bell wrote
>>
>>>>>>> I have just been working through the math for shunt
>>>>>>> derived shunt applied NFB around an amp and Ican't
>>>>>>> get the expected result.
>>>>>> Is that the same as voltage derived, current applied?
>>>>>> I can never remember this stuff, sigh.
>>>>>>
>>>>> No, it's voltage derived, voltage applied. As a rule:
>>>>>
>>>>> Shunt = Voltage
>>>>> Series = Current
>>>>
>>>> Thanks. Then how come your circuit employs "shunt
>>>> derived" feedback?
>>>>
>>> Sorry, don't understand - the feedback network is fed
>>> from the output voltage i.e. across or shunting the
>>> output. Is that what you mean?
>>>
>>>> And perhaps you could answer the question you cut out
>>>> from my post, considering I was good enough to answer
>>>> yours:
>>>>
>>> My apologies, I must have missed it.
>>>
>>>> "Your maths is correct or we're both wrong.
>>>>
>>>> "I don't know if it helps to see it as Ao/(1+ß.(1+Ao))
>>>>
>>> Yes, I had it as that to start with but changed it to
>>> 'show' the anomoly.
>>>
>>>> "You haven't explained why you believe there is an
>>>> anomaly.
>>>> What led you to your erroneous expectation?"
>>>>
>>> I had expected *all* NFB forms (however derived or
>>> applied) to reduce to Ao/(1+ß.Ao) but this one does not.
>>
>>
>> I find the four possible combinations of shunt/series
>> feedback can be difficult to interpret, so it may be just
>> me.
>>
>> For shunt derived feedback I would expect to see a load
>> and means of sensing the current through that load. An
>> example would be the use of a small current-sensing
>> resistor on the ground side of a loudspeaker to derive
>> the feedback signal.
>
> No, that is series derived as you are connected in series
> with the output and also called current derived because
> you are sensing current. I agree it can be confusing.

I was applying your rule, I thought, but got it the wrong
way round, mostly because according to what you say now it
is wrong:

>>>>> Shunt = Voltage
>>>>> Series = Current

because...

>> Another example is the use of an unbypassed cathode
>> resistor, which is shunt derived, series applied.
>>
>
> Correct, shunt derived because we connect the feedback
> network directly across the output (cathode resistor) and
> series applied because derived voltage is applied in
> series with the input.
>
...here you seem to be saying that voltage applied is series
applied

Careful. The voltage output is at the anode. The cathode
resistor is a current sensor, hence current derived. The
practical difference is that current derived, voltage
applied negative feedback inreases output impedance, whereas
voltage derived decreases it.

I'll continue to think in terms of current and voltage, and
forget about this series/shunt mularky.

>> If the load is purely resistive, I can't see the
>> difference between the two methods of derivation, because
>> the current will always be in fixed proportion to the
>> voltage.
>>
>> Where the only load is the feedback path itself, which in
>> your case is purely resistive, I can't see the difference
>> either. The most obvious way to me of seeing your
>> circuit, however, is that the feedback signal is derived
>> from a voltage divider between output and input. It seems
>> to me that is where the mysterious "1+" comes from...the
>> fact that the feedback is not derived from a voltage
>> divider between the output and ground...there's an extra
>> Vin with respect to that.
>>
>
> Yes, I am still not sure about it. Needs further thought.
> At least it seems my maths is OK.

If you see it like this:

>>>> "I don't know if it helps to see it as Ao/(1+ß.(1+Ao))

then the question is "why is it 1+Ao instead of Ao?"

The answer is likely to be that the feedback, instead of
being some proportion of Ao.Vin, comes from (1+Ao).Vin,
because the voltage divider used to derive it, instead of
being between the output and 0, is between the output and
Vin. I wish I could put it more clearly.

Anyway, if you look through your maths around the point
where the "1+Ao" appeared, then you should be able to see
why. That should be more fruitful than looking for the extra
ß in your expression.

Ian

Re: NFB101 Crisis!



Ian Iveson wrote:
>
> "Ian Thompson-Bell" <ruffrecords@yahoo.co.uk> wrote in
> message news:fupk67$i5l$1@energise.enta . net ...
> > Ian Iveson wrote:
> >> Ian Thompson-Bell wrote
> >>
> >>>>>>> I have just been working through the math for shunt
> >>>>>>> derived shunt applied NFB around an amp and Ican't
> >>>>>>> get the expected result.
> >>>>>> Is that the same as voltage derived, current applied?
> >>>>>> I can never remember this stuff, sigh.
> >>>>>>
> >>>>> No, it's voltage derived, voltage applied. As a rule:
> >>>>>
> >>>>> Shunt = Voltage
> >>>>> Series = Current
> >>>>
> >>>> Thanks. Then how come your circuit employs "shunt
> >>>> derived" feedback?
> >>>>
> >>> Sorry, don't understand - the feedback network is fed
> >>> from the output voltage i.e. across or shunting the
> >>> output. Is that what you mean?
> >>>
> >>>> And perhaps you could answer the question you cut out
> >>>> from my post, considering I was good enough to answer
> >>>> yours:
> >>>>
> >>> My apologies, I must have missed it.
> >>>
> >>>> "Your maths is correct or we're both wrong.
> >>>>
> >>>> "I don't know if it helps to see it as Ao/(1+ß.(1+Ao))
> >>>>
> >>> Yes, I had it as that to start with but changed it to
> >>> 'show' the anomoly.
> >>>
> >>>> "You haven't explained why you believe there is an
> >>>> anomaly.
> >>>> What led you to your erroneous expectation?"
> >>>>
> >>> I had expected *all* NFB forms (however derived or
> >>> applied) to reduce to Ao/(1+ß.Ao) but this one does not.
> >>
> >>
> >> I find the four possible combinations of shunt/series
> >> feedback can be difficult to interpret, so it may be just
> >> me.
> >>
> >> For shunt derived feedback I would expect to see a load
> >> and means of sensing the current through that load. An
> >> example would be the use of a small current-sensing
> >> resistor on the ground side of a loudspeaker to derive
> >> the feedback signal.
> >
> > No, that is series derived as you are connected in series
> > with the output and also called current derived because
> > you are sensing current. I agree it can be confusing.
>
> I was applying your rule, I thought, but got it the wrong
> way round, mostly because according to what you say now it
> is wrong:
>
> >>>>> Shunt = Voltage
> >>>>> Series = Current
>
> because...
>
> >> Another example is the use of an unbypassed cathode
> >> resistor, which is shunt derived, series applied.
> >>
> >
> > Correct, shunt derived because we connect the feedback
> > network directly across the output (cathode resistor) and
> > series applied because derived voltage is applied in
> > series with the input.
> >
> ...here you seem to be saying that voltage applied is series
> applied
>
> Careful. The voltage output is at the anode. The cathode
> resistor is a current sensor, hence current derived. The
> practical difference is that current derived, voltage
> applied negative feedback inreases output impedance, whereas
> voltage derived decreases it.
>
> I'll continue to think in terms of current and voltage, and
> forget about this series/shunt mularky.

I wish you'd have another read of all that RDH4 says about feedback.

There are many types of FB including positive and negative feedback,
and there is shunt or series voltage or current FB.

All have a different effect around an amplifier.
>
> >> If the load is purely resistive, I can't see the
> >> difference between the two methods of derivation, because
> >> the current will always be in fixed proportion to the
> >> voltage.
> >>
> >> Where the only load is the feedback path itself, which in
> >> your case is purely resistive, I can't see the difference
> >> either. The most obvious way to me of seeing your
> >> circuit, however, is that the feedback signal is derived
> >> from a voltage divider between output and input. It seems
> >> to me that is where the mysterious "1+" comes from...the
> >> fact that the feedback is not derived from a voltage
> >> divider between the output and ground...there's an extra
> >> Vin with respect to that.
> >>
> >
> > Yes, I am still not sure about it. Needs further thought.
> > At least it seems my maths is OK.
>
> If you see it like this:
>
> >>>> "I don't know if it helps to see it as Ao/(1+ß.(1+Ao))
>
> then the question is "why is it 1+Ao instead of Ao?"
>
> The answer is likely to be that the feedback, instead of
> being some proportion of Ao.Vin, comes from (1+Ao).Vin,
> because the voltage divider used to derive it, instead of
> being between the output and 0, is between the output and
> Vin. I wish I could put it more clearly.
>
> Anyway, if you look through your maths around the point
> where the "1+Ao" appeared, then you should be able to see
> why. That should be more fruitful than looking for the extra
> ß in your expression.
>
> Ian

Build sample amps and damn well measure them.

Observe the reality of the feedback and gain.

Then you'd know.

Patrick Turner.

Re: NFB101 Crisis!

Ian Iveson wrote:
> "Ian Thompson-Bell" <ruffrecords@yahoo.co.uk> wrote in
> message news:fupk67$i5l$1@energise.enta . net ...
>> Ian Iveson wrote:
>>> Ian Thompson-Bell wrote
>>>
>>>>>>>> I have just been working through the math for shunt
>>>>>>>> derived shunt applied NFB around an amp and Ican't
>>>>>>>> get the expected result.
>>>>>>> Is that the same as voltage derived, current applied?
>>>>>>> I can never remember this stuff, sigh.
>>>>>>>
>>>>>> No, it's voltage derived, voltage applied. As a rule:
>>>>>>
>>>>>> Shunt = Voltage
>>>>>> Series = Current
>>>>> Thanks. Then how come your circuit employs "shunt
>>>>> derived" feedback?
>>>>>
>>>> Sorry, don't understand - the feedback network is fed
>>>> from the output voltage i.e. across or shunting the
>>>> output. Is that what you mean?
>>>>
>>>>> And perhaps you could answer the question you cut out
>>>>> from my post, considering I was good enough to answer
>>>>> yours:
>>>>>
>>>> My apologies, I must have missed it.
>>>>
>>>>> "Your maths is correct or we're both wrong.
>>>>>
>>>>> "I don't know if it helps to see it as Ao/(1+ß.(1+Ao))
>>>>>
>>>> Yes, I had it as that to start with but changed it to
>>>> 'show' the anomoly.
>>>>
>>>>> "You haven't explained why you believe there is an
>>>>> anomaly.
>>>>> What led you to your erroneous expectation?"
>>>>>
>>>> I had expected *all* NFB forms (however derived or
>>>> applied) to reduce to Ao/(1+ß.Ao) but this one does not.
>>>
>>> I find the four possible combinations of shunt/series
>>> feedback can be difficult to interpret, so it may be just
>>> me.
>>>
>>> For shunt derived feedback I would expect to see a load
>>> and means of sensing the current through that load. An
>>> example would be the use of a small current-sensing
>>> resistor on the ground side of a loudspeaker to derive
>>> the feedback signal.
>> No, that is series derived as you are connected in series
>> with the output and also called current derived because
>> you are sensing current. I agree it can be confusing.
>
> I was applying your rule, I thought, but got it the wrong
> way round, mostly because according to what you say now it
> is wrong:
>
>>>>>> Shunt = Voltage
>>>>>> Series = Current
>
> because...
>
>>> Another example is the use of an unbypassed cathode
>>> resistor, which is shunt derived, series applied.
>>>
>> Correct, shunt derived because we connect the feedback
>> network directly across the output (cathode resistor) and
>> series applied because derived voltage is applied in
>> series with the input.
>>
> ...here you seem to be saying that voltage applied is series
> applied
>
> Careful. The voltage output is at the anode. The cathode
> resistor is a current sensor, hence current derived.



Sorry, I thought you were talking about a cathode follower, but you in
fact are talking about a common cathode amp with an unbypassed cathode
resistor. This is Series derived because the cathode resistor across
which the feedback voltage is derived is in series with the load. It is
also current derived because it is the output current not voltage that
determines the amount of feedback. ß is Rl/Rk where Rl is the anode
resistor and Rk is the cathode resistor

The
> practical difference is that current derived, voltage
> applied negative feedback inreases output impedance, whereas
> voltage derived decreases it.

How the feedback is *derived* affects the output impedance. Shunt
derived feedback reduces output impedance and Series derived increases it.

How feedback is *applied* affects input resistance. Series applied
increases input resistance and shunt applied reduces it.

>
> I'll continue to think in terms of current and voltage, and
> forget about this series/shunt mularky.
>

It is complex and confusing and remember so far we are only talking
about voltage amplifiers; there's also current amplifiers,
transconductance amplifiers etc to further confuse the issue.

>>> If the load is purely resistive, I can't see the
>>> difference between the two methods of derivation, because
>>> the current will always be in fixed proportion to the
>>> voltage.
>>>
>>> Where the only load is the feedback path itself, which in
>>> your case is purely resistive, I can't see the difference
>>> either. The most obvious way to me of seeing your
>>> circuit, however, is that the feedback signal is derived
>>> from a voltage divider between output and input. It seems
>>> to me that is where the mysterious "1+" comes from...the
>>> fact that the feedback is not derived from a voltage
>>> divider between the output and ground...there's an extra
>>> Vin with respect to that.
>>>
>> Yes, I am still not sure about it. Needs further thought.
>> At least it seems my maths is OK.
>
> If you see it like this:
>
>>>>> "I don't know if it helps to see it as Ao/(1+ß.(1+Ao))
>
> then the question is "why is it 1+Ao instead of Ao?"
>
> The answer is likely to be that the feedback, instead of
> being some proportion of Ao.Vin, comes from (1+Ao).Vin,
> because the voltage divider used to derive it, instead of
> being between the output and 0, is between the output and
> Vin. I wish I could put it more clearly.
>

That's a good thought. Thanks for that.

> Anyway, if you look through your maths around the point
> where the "1+Ao" appeared, then you should be able to see
> why. That should be more fruitful than looking for the extra
> ß in your expression.
>
> Ian
>
>

Cheers

ian

Re: NFB101 Crisis!



Ian Thompson-Bell wrote:
>
> 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

I think I can explain all, because I ran into the same problem you have
about
10 years ago when I tried to teach myself, and I discovered silly text
books
limited to basic oppamp FB equations just didn't work with simple low
gain triodes.

But let's consider the terms to be used and let input to grid R = R1,
and anode output to grid R = R2, bcause its the most common way I have
seen the R shown
in a shunt FB loop.

Also, to be simple, consider the inverting open loop gain be a positive
number, even though
it really is negative.

And ß = R1 / (R1 + R2) , OK? and not R1 / R2 !

Let us consider an example using an EL34 with R load of 2k2 ohms, in
pentode mode,
and with fully bypassed cathode.
OLG A = 20, ie, 10Vac input gives 200Vac output. Phase is inverted.

Let R1 = 20k, and R2 = 420k.

ß = 20 / ( 20 + 420 ) = 0.04545.

CLG, closed loop gain A' = A / ( 1 + { [ A + 2 ] x ß } ).

In this case, A'= 20 / ( 1 + { [ 20 + 2 ] x 0.04545 } ),

= 10.0, so my notes from my book of tested formulas says.

The effective anode resistance Ra' with NFB = Ra / ( 1 + [ µ x ß ] )

So if Ra = 12k, and µ = 130,

then Ra' with ß = 0.04545

= 12,000 / ( 1 + [ 130 x 0.04545 ] )

= 1,737 ohms.

Since gain is halved with a 2k2 load with the above amount of NFB,
the amount of applied FB = 6dB, but only when RL = 2k2.

If RL was say 4k7, A is maybe 40, and the gain reduction is greater so
the amount of NFB applied is greater.
With a CCS load, but with the above R2 = 420k, the load is approximately
A/( A + 1 ) x 420k.
This is because the R2 has output voltage PLUS the grid voltage across
it.

Notice that Ra' is independant of A, A', or RL;
Ra' is the effective Ra after NFB is applied.
it is the output resistance of the stage, Rout.
If the RL of 2k2 is used as a permanent dc Ia current feeding load,
then Ra' would be the same, but Rout = Ra' // RL.

Notice also that it doesn't take much NFB to dramatically
change the effective Ra' from a uselessly high Ra = 12k down to 1k7.

Obviously, since its not too hard to produce a fairly clean 40Vrms
of drive signal, we could make R2 considerably lower R without loading
down the tube too much
and get a much larger reduction of Ra to below triode connection values,
and with less THD than a triode.

When calculating A with shunt FB applied, the load on the tube = the dc
carrying RL
in parallel with any following load such as a gain pot, AND ALSO
in parallel with the effective value of R2.

To confirm what you measure with what you calculate
you must set up the circuit with FB connected to find out what A is.

Just using the data figures for µ, gm, and Ra is a bullshit practice
with zero merit
for accuracy in practice because these 3 paramaters vary with Ia and
between various makes and versions of tubes.
So its another reason I never simulate.

The data figures are only ever going to give you an APPROXIMATE answer,
but
still useful though if being within 10% accurate is good enough for you.

A with NFB is simply Va / Vg, and A' = Va / Vin at input terminal.

The only valid thing is what you actually have in front of you.

In the example above, R2 = 420k, and is in parallel with 2k2,
and there is no cap coupled load following, thus RL really = 2k2//420k
which I think = 2,181 ohms, which BTW is in my figuring in my notebook.

But set up an EL34 to really find outabout.

Patrick Turner.

Re: NFB101 Crisis!


"Ian Thompson-Bell" <ruffrecords@yahoo.co.uk> wrote in message
news:fukkiv$qch$1@energise.enta . net ...
> 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?

Compare two cases:
1. Classic case of a non-inverting amplifier with voltage series feedback.
In this case
An = Ao = 0.01 (approx.)

2. Your case of the inverting amplifier: An = 0.005 -- another half of the
gain is lost!
This is absolutely correct since the signal is divided by two by Ri/Rf
divider before being applied to the amp input.

Your formulae give exactly that -- correct and consistent with reality!

Regards,
Alex

Re: NFB101 Crisis!



Alex wrote:
>
> "Ian Thompson-Bell" <ruffrecords@yahoo.co.uk> wrote in message
> news:fukkiv$qch$1@energise.enta . net ...
> > 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?

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.

Suppose you have a tube with OLG, ( open loop gain) = 20, and ß = 0.5

Suppose you measure noise = +Vn at output, with NFB connected.

+0.5Vn appears at the grid, and is amplified to become -10Vn, the
correction signal.
But you measured +1Vn, so Vn without NFB must have been 11Vn without NFB
connected,
and becomes 1Vn because you have 11Vn without NFB and the correction
signal reduces it to 1Vn.

So it is with distortions. So the amp will have CLG THD = 1/11 of the
OLG THD.

If +20V appears at the anode, -1V is needed at grid.
R1 = R2, so there MUST be 21V across each of R1 and R2,
so input voltage MUST be 22V.

Gain with NFB = 20 / 22. it is a negative number actually
because of the inverting amp.

Notice this A' is less than unity.

For opamps and other ams with OLG A = 50,000,
when R1 = R2, gain is very close to 1.0, always below 1.0 though.
The amp input terminal has such a tiny voltage upon it that its
regarded as a virtual earth, and CLG, A' = R2 / R1, a close enough
approximation.

You could have R1 much larger than R2, then ß approaches unity
and the amount of NFB applied approaches the maximum possible.
But its far more easily done using series NFB where ALL the output
voltage may
be fed back to a second terminal as in the case of a simple cathode
follower
where the output terminal, the cathode, is also a second input terminal.
The CF is THE classic example of a simple series voltage NFB
application.

Shunt NFB gives slightly different figures than series NFB.

Patrick Turner.



>
> Compare two cases:
> 1. Classic case of a non-inverting amplifier with voltage series feedback.
> In this case
> An = Ao = 0.01 (approx.)
>
> 2. Your case of the inverting amplifier: An = 0.005 -- another half of the
> gain is lost!
> This is absolutely correct since the signal is divided by two by Ri/Rf
> divider before being applied to the amp input.
>
> Your formulae give exactly that -- correct and consistent with reality!
>
> Regards,
> Alex

Re: NFB101 Crisis!

Patrick Turner wrote:
>
> Alex wrote:
>> "Ian Thompson-Bell" <ruffrecords@yahoo.co.uk> wrote in message
>> news:fukkiv$qch$1@energise.enta . net ...
>>> 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.

A lot depends on your definitions. R1, R2 do indeed act as a pot divider
and if they are equal you might expect ß to be 0.5 but that would mean
the closed loop gain would be 2 (1/ß) when we know it is one with R1=R2.
Thing is, by the same argument, R1,R2 act as a pot divider for the input
signal, thus halving its value at the amp input so you could say the
input is halved then amplified by X2 with an overall unity result.

The problem is the topology of the input because the summing point is
somewhat isolated from the input by the input resistor. It is
interesting that RDH takes a different approach with no input resistor
and instead shows how the overall gain depends of the resistance of the
signal source which amounts to the same thing.

Funny stuff the NFB.

Cheers

Ian