Ian Iveson wrote:
> "Ian Thompson-Bell" <ruffrecords@[EMAIL PROTECTED]
> wrote in
> message news:fupk67$i5l$1@[EMAIL PROTECTED]
>> 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 pro****tion 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 pro****tion 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
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