A Negative Feedback Loose End

[John_BS]

Mark is right: you need to find an explanation which works for you.

My personal favourite is to use "vectors", i.e draw a line whose length is proportional to the relevant voltage, (rather like force diagrams in mechanics). The amplifier "input" vector* is the difference between the fed-back output vector and the actual circuit input; it's length is equal to the output vector divided by the amplifier gain (i.e gain with no feedback). This method also works well when there is a phase-shift in the amplifier (usually the output lagging the input by 90 degrees or so). If you can't draw a proper triangle, you've not got it right.

John

* this is sometimes called the error signal, as in systems with a large amount of feedback and very high amplifer gain, this signal is orders of magnitude smaller than the other two vectors)

[Skywave]

Originally Posted by John_BS

Mark is right: you need to find an explanation which works for you.

Maybe - but there is a danger in such a simplistic approach. An 'explanation' which 'works for you' - or rather which appears to 'work for you' - will depend on your level of comprehension of the topic when you meet that explanation. Should that explanation contain errors, (or other significant deficiencies), your limited comprehension will not detect those, so your 'newly acquired understanding' will thus be defective. In a phrase, a classic 'catch-22' situation.

So what's an alternative solution? I suggest that reading text books that have an established pedigree of thoroughness and accuracy of detail. However, that does not necessarily equate to the classical theory-heavyweight books which are available. There are many publications available that cater for the beginner and which have acquired a reputation for quality and ease of comprehension.

I am not claiming that any of the above is unquestionable fact: it's based merely my experience, gained over 50+ years in radio & electronics.

Al.

[MajorWest]

Originally Posted by daviddeakin

To answer your question very quickly, no, you can't get the feedback signal to cancel exactly the input signal leaving you with nothing. If you did then there would be nothing left to amplify, and therefore no feedback signal either! Paradox!

Recently I came across "transit time" The figure given was a thousandth of a microsecond for electrons to travel from cathode to grid. So, it's basically impossible for the negative feedback to be simultaneous with the signal input. So, I had some kind of delay in mind. Hopefully it will get clearer.

[MajorWest]

Originally Posted by John_BS

Mark is right: you need to find an explanation which works for you.

Actually that is what I decided to do. I have actually been looking at other explanations. I always found Owen Bishop to be very good at explaining electronics so I had a look at his explanation of phase. The problem is the large American Amateur radio book I have is really good but needs filling out in areas.
Also, I should add my question may not have been as vague if I'd included the diagram of the feedback circuit.
Another thing I just found, though, deserving of its own post, is how hopelessly ill-prepared I was for the reality of fixing an actual set. I found my theory nowhere near as helpful as good observation and detective work. So, I got to thinking I may have raced ahead of myself a bit since theory may be better learned more thoroughly once actual service skills are acquired. I think in depth theory is for more advanced electronics people but, sure, it's interesting to read these books.

[Skywave]

Originally Posted by MajorWest

I think in-depth theory is for more advanced electronics people but, sure, it's interesting to read these books.

"In-depth" theory is relative. When I was a youngster with little understanding at any depth, (but not short on enthusiasm), the stuff that I understand today would have been way outside of my comprehension then. And even now, there are many topics that I'm either quite ignorant of or only have a sketchy knowledge. "Advanced electronics people" have usually entered that category on account of many years of experience. So - provided that your enthusiasm in maintained and is manifest by practical activities and studying theory - one day you may find others referring to you as being in that class of 'advanced electronics people'.

If your interest in radio & electronics doesn't wane, you'll have your successes and your failures, sure - haven't we all? But even failures can be of some benefit - provided you go to the trouble to find out why it was a failure. Often, doing such a post mortem can seem like a pointless exercise: it isn't: it's educational. At least it will show you where you need to concentrate your further reading and studying.

Al.

[John_BS]

Originally Posted by Skywave

Should that explanation contain errors, (or other significant deficiencies), your limited comprehension will not detect those

Yes, this comment brings to mind another quote, this time by one of my favourite engineers:

"On two occasions I have been asked, 'Pray, Mr. Babbage, if you put into the machine wrong figures, will the right answers come out?' ... I am not able rightly to apprehend the kind of confusion of ideas that could provoke such a question."

[kalee20]

Originally Posted by MajorWest

Recently I came across "transit time" The figure given was a thousandth of a microsecond for electrons to travel from cathode to grid. So, it's basically impossible for the negative feedback to be simultaneous with the signal input.

You are right, but in practice you can forget about such miniscule delays.

If you think about putting a screwdriver into a screw, watching what you are doing to make sure the screwdriver blade is in line with the screw slit (ie visual feedback), this sort of delay time is comparable with the time it takes light to get from the screw to your eye. You don't need to allow for it, it's negligible.

[MajorWest]

I think I more or less have this clarified now in a way I can make sense of it. First of all, the signal voltage on the grid is 180 degrees out of phase with respect to the anode and cathode voltage.

In some wiring configurations, there is a hefty DC voltage drop across the anode load resistor. Yet the voltage drop here is out of phase to the signal. Thus, when a strong positive signal arrives at the grid, the anode cathode voltage will drop to more negative.

With negative feedback, the diagram uses a grid circuit where another resistor is in series with the anode load resistor. The second resistor is between grid and a grid bias battery. Given the second resistor is in series with the anode load resistor, any voltage drop is carried over to the grid circuit.

Thus, what seems to happen is there is a DC voltage drop across the grid in opposite phase to the incoming signal.

Strange that the signal comes between a grid bias battery and grid in this circuit but it's an American vintage design.

[Herald1360]

The anode signal is 180deg out of phase with the grid, but the cathode signal is in phase.

A grid bias battery provides a negative dc voltage for the grid, but looks like a low impedance to signal (ac) voltages so inputting signal between bias battery and grid will work OK as long as the signal source is a reasonably low resistance at dc (a transformer winding is one possibility) so that the bias dc reaches the grid.

A sketch of your American circuit would help make sense of what happens there. A picture paints.....