Gain isn't everything. Noise and input current are also factors. The input current may be non-linear, so you want it to be small. Otherwise you might have to put a follower as a buffer before the amp proper. Then the follower has to have small input current, so you have simply moved the problem.
Well I'm not sure I understand it all myself, but remember this is current derived voltage feedback, and we (I) are mostly familiar with voltage derived voltage feedback. We know that emitter degeneration, while being a form of nfb, is not identical to, in it's effects, say overall loop feedback.
If you take the case of Re = 1/Gm, your voltage gain becomes 0.5 (see my earlier equation). You can consider 1/Gm as a kind of build-in output resistance that drops the no-load Vout if you connect an Re - a load for the emitter. Now if we increase Re we lose less and less voltage across the build-in 1/Gm Rout which gets us closer to Acl = 1.
I don't know how that should be considered in terms of Aol and loop gain, but I'm sure others do. Dave? John? Scott?
jan
On reflection I might have got that wrong. It might be possible to cancel any particular order by using the one below that. Easiest is cancelling third by using second. Above that the maths gets more messy, although not conceptually any harder. You just have to do the nth derivative of the gain expression, and look for a zero condition.DF96 said:
The easy way to consider follower voltage gain is to regard it as a potential divider, made of 1/gm and Re. Then it is obvious that increasing Re (i.e. reducing the load) gives more voltage.
To think of it in feedback terms it is best to treat it as a transconductance stage, followed by a load resistor. Then G=gmxRe/(gmxRe+1). As Re increases this tends to 1, because the increase in load resistor (numerator) always outweighs the increase in feedback (denominator).
Jan, maybe I'm misunderstanding the confusion and the confusion confuses me. As the resistor AND load get smaller, the OLG decreases, right? And as the resistor and load get larger, OLG increases, right? If we neglect base current, the load can be either in the collector or emitter, same OLG, right?
Yes sounds right. Cathodyne?
But OLG is strictly only available with open emitter, so I don't know if you can talk about changing OLG with load.
The load here doubles as feedback generator.
With open emitter, your OLG is 1. When you add Re, the CLG drops below the OLG and you only can get back to '1' by increasing Re, in the limit to infinity, which is the same as open emitter of course.
Edit: If I compare with a global nfb loop around an amp, increasing EF Re is equivalent to decreasing beta (nfb beta, not transistor beta).
In the limit, making Re infinite is the same as making beta zero (opening the fb loop). Both get you back to the OLG.
So in both cases feedback (degenerative or global) decreases CLG. In the global case, the min CLG is '1', for the EF, the min CLG = 0 and the max OLG = '1'. Same difference ;-)
jan
But OLG is strictly only available with open emitter, so I don't know if you can talk about changing OLG with load.
The load here doubles as feedback generator.
With open emitter, your OLG is 1. When you add Re, the CLG drops below the OLG and you only can get back to '1' by increasing Re, in the limit to infinity, which is the same as open emitter of course.
Edit: If I compare with a global nfb loop around an amp, increasing EF Re is equivalent to decreasing beta (nfb beta, not transistor beta).
In the limit, making Re infinite is the same as making beta zero (opening the fb loop). Both get you back to the OLG.
So in both cases feedback (degenerative or global) decreases CLG. In the global case, the min CLG is '1', for the EF, the min CLG = 0 and the max OLG = '1'. Same difference ;-)
jan
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It is always difficult to put yourself in the other person's shoes and work out what is confusing them. In this case I think it is the idea that more feedback leads to a greater closed-loop voltage gain. The solution is that we are not applying more feedback to the same amplifier, but more feedback to a different amplifier. Increasing Re both raises the feedback, and raises the open loop gain (if considered as a voltage amplifier).
The more general problem is that there are several different ways of analysing a given circuit. All of them must give the same result and, correctly applied, they do. Problems come when an attempt is made, usually innocently, to mix concepts or rules of thumb from one model to another.
The more general problem is that there are several different ways of analysing a given circuit. All of them must give the same result and, correctly applied, they do. Problems come when an attempt is made, usually innocently, to mix concepts or rules of thumb from one model to another.
Hmmmmmm, so lets say we load it with a current source and no other load.
We are close to OLG then, if I understood correctly. But in that case we have no current gain and voltage gain approaches one. And distortion is low in that case. Not a sign of open loop operation.
And in a earlier post i was told that a higher Re is more feedback.
The other case is a very small Re. Current gain is much higher compared to the above, voltage gain is a bit lower, but according to earlier explanations mostly because loading.
To me, not a sign of more feedback, which should decrease the gain.
I still have no idea how to derive loop gain from that, which would indicate how much feedback we throw at the device.
The longer i think about, the more I believe it is very small...
Wikipedia links do not help much in that regard, either.
And i am still confused.
Higher Re is more feedback.
I did a comparison once but can't find those files atm.
Draw an opamp symbol, with a build-out output R of value 1/Gm. Label the non-inverting input 'B', the output after the output resistor 'E' and connect 'E' to the inverting input.
Load 'E' with an Re. There's your emitter follower, complete with feedback and you can (easily? sorry Jacco) calculate all you want. I think.
jan
Ok fine, thanks for the patience. And what would be the loop gain in that case? Do i have to assume gain as infinite for the virtual op amp, and only reduced by build out resistor? Btw, in that simplified model, Re is outside the loop....its function is only to load the output, hence higher Re is higher output voltage.
Still searcing my loop gain you say; of what?
A single stage thing or a more complex amp?
Things are not that easy when you are analysing a complex amp.
I guess that your amp is not a single stage thing, but includes an input stage, a trans impedance stage and an output stage.
I think the best thing for you is to open a separate thread for your amp and see if some will give you some advices on that particular design.
Cheers
S
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