Steve Eddy said:
Well, if I could adequately describe it, I wouldn't be looking for a definitive answer as to whether or not the CFP has more feedback than the Darlington.
I'm just going on intuition here.
se
CFP has a bit more feedback than the Darlington and the reason is that in a Darlington the first transistor has a bit of a local negative feedback from the input impedance of the second transistor, reducing overall loop gain. In the CFP there is no such additional local feedback.
This difference is clear if you think about the gain of both combinations in a common emitter mode (i.e. "open loop") . It is easy to see that the transconductance of CFP in this case is higher than transconductance of a Darlington - in a first case (CFP) you need to increase the voltage only across one B-E junction, but for a Darlington the input voltage has to increase the voltage on both B-E junctions .
x-pro
x-pro said:
Hmmmm. Not quite sure what you mean here. I've been of the understanding that the two devices in the CFP are in a local feedback loop as well, which some say accounts for its tendency to be a bit less stable compared to the Darlington.
Sure, but the difference due to the two BE junctions is rather marginal.
Charles claimed that the Darlington has no feedback at all. In fact, not even a single device emitter follower has any feedback. They are both, according to him, "open loop" circuits.
In a Darlington pair, an essentially unity-gain open-loop emitter follower drives a second essentially unity-gain open-loop emitter follower.
So we're not talking just a marginal difference between the two.
se
Steve Eddy said:
In both cases you've got closed overall loop through the load. However CFP has two inverting stages with complete loop around these. That will require a proper compensation as in an op-amp. Usually this requirement fulfilled by internal capacitances but it could be a problem.
It is not a big difference, but it is there.
It is ONLY according to Charles
. "Unity-gain open-loop emitter follower" - is a nonsense. x-pro
x-pro said:
It is ONLY according to Charles
Not so fast there, big boy. Try listening to Burr-Brown designers and Maxim designers:
http://www.diyaudio.com/forums/showthread.php?postid=331335#post331335
Some of the best solid state designers in the business agree with my use of terminology. For example, please refer to the data sheet for the Burr-Brown BUF600. This is an "open loop follower" that basically consists of a complementary pair of emitter followers driving another complementary pair of Darlington emitter followers. In the data sheet it says:
"The BUF600 and BUF601 are 3-stage open-loop buffer amplifiers consisting of complementary emitter followers with a symmetrical class AB Darlington output stage.... The amplifiers use no feedback, so their low-frequency gain is slightly less than unity and somewhat dependent on loading." (emphasis mine)
or refer to the data sheet for the Maxim MAX4200 open loop buffer:
"The MAX4200–MAX4205 are ultra-high-speed, open loop buffers....Since these devices operate without negative feedback, there is no loop gain to transform the input impedance upward, as in closed-loop buffers." (emphasis mine)
Charles Hansen said:
OK, I admit not having read the datasheet, but an equally
reasonable intepretation seems to be that they mean the buffer
has no negative input to allow for an external feeback loop
around the buffer. It could thus mean that they do not refer
to internal local feedback at all.
Charles Hansen said:
I've seen even worse errors and mistakes in datasheets
. And even in application papers, like here for instance:http://www.maxim-ic.com/appnotes.cfm/appnote_number/2238/ln/en
It is a good candidate for an elaborate April Fool joke, however it's published in July
x-pro
Ultima Thule said:
Here are a few:
Electronics Engineers' Handbook, Second Edition, Fink/Christiansen, McGraw Hill, p.7-44:
"Common Collector. In the common-collector connection, the source voltage and the output voltage are in series and have opposing polarities. This is a negative feedback arrangement, which gives a high input impedance and approximately unity voltage gain. Current gain is about the same as that of the common-emitter connection."
From: http://www.williamson-labs.com/480_xtor.htm
"In the case of the emitter follower, as the base voltage is increased, there is a corresponding tracking of the base/emitter differential: the emitter rises to--or follows--the base's change. If the output follows the input, there can never be enough current drawn by the base to cause a voltage drop across the emitter which exceeds the input voltage--hence no voltage gain. This is an elegant case of (internal) negative feedback."
From: http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/negfeed.html
"The emitter follower can be seen to be an example of negative feedback when the relationship..."
"The gain of 1 can be seen as a result of all the output being fed back to the input."
From: http://www.geocities.com/researchtriangle/node/2356/afeede.html
"As a matter of fact, no amplifier can be built without negative feedback, because emitter-follower is a 100% negative feedback device."
From: http://www.stereophile.com/showarchives.cgi?70:7
"A cathode/emitter follower is an amplifier with 100% negative feedback."
From: http://www.etdv.ruhr-uni-bochum.de/dv/publications/icinas98/uhsm-owan/uhsm-owan.pdf
"Tansimpedance stages and emitter followers are feedback circuits. Both circuits have a 100% negative feedback."
se
Charles Hansen said:
it is very hard, if not downright impossible, to get everyone to agree to even the simplest matter. so it is not hard to find people who agree with your definition. That's expected.
However, I thought the concensus is probably not on your side. there are likely more people who would agree that a EF has 100% feedback.
Emitter followers give essentially a step function; that is, their output tracks the input, with the difference always being Vbe.
If the feedback were, by conventional standards, 100%, then you would expect this step function to be invariate.
However, it is not, and varies according to a mix of Ib*rb, Ic*re and Vce. As current through the DUT increases, so does Vbe, and it is not strictly linear. This introduces a distortive mechanism, minimized by making the collector current constant and cascoding the EF so Vce is a constant, but nonetheless non-linear.
I would have expected that this mechanism militates against the notion of feedback which, if it were 100%, should compensate this phenomenon completely since the feedback is so complete.
It seems an indirect argument, but if there were a feedback mechanism here of a voltage nature then Vbe would remain essentially constant. The fact that this distortion can be almost completely eliminated (albeit with an infinite load!) by use of a CCS and a flying cascode tells me that no such feedback is in operation. Ergo, any feedback on an EF is NOT voltage feedback, though it may be possible to argue something valid on current grounds.
Cheers,
Hugh
If the feedback were, by conventional standards, 100%, then you would expect this step function to be invariate.
However, it is not, and varies according to a mix of Ib*rb, Ic*re and Vce. As current through the DUT increases, so does Vbe, and it is not strictly linear. This introduces a distortive mechanism, minimized by making the collector current constant and cascoding the EF so Vce is a constant, but nonetheless non-linear.
I would have expected that this mechanism militates against the notion of feedback which, if it were 100%, should compensate this phenomenon completely since the feedback is so complete.
It seems an indirect argument, but if there were a feedback mechanism here of a voltage nature then Vbe would remain essentially constant. The fact that this distortion can be almost completely eliminated (albeit with an infinite load!) by use of a CCS and a flying cascode tells me that no such feedback is in operation. Ergo, any feedback on an EF is NOT voltage feedback, though it may be possible to argue something valid on current grounds.
Cheers,
Hugh
Thanks SE,
how about this....
In the case of the diode, as the anode voltage is increased, there is a corresponding tracking of the anode/cathode differential: the cathode rises to--or follows--the anode's change. If the output follows the input, there can never be enough current drawn by the anode to cause a voltage drop across the cathode which exceeds the input voltage--hence no voltage gain. This is an elegant case of (internal) negative feedback.
From.... with m.d.f.c.t.n.
BTW the last link page 5, 6.. was good, but......
Cheers!
how about this....
In the case of the diode, as the anode voltage is increased, there is a corresponding tracking of the anode/cathode differential: the cathode rises to--or follows--the anode's change. If the output follows the input, there can never be enough current drawn by the anode to cause a voltage drop across the cathode which exceeds the input voltage--hence no voltage gain. This is an elegant case of (internal) negative feedback.
From.... with m.d.f.c.t.n.
BTW the last link page 5, 6.. was good, but......
Cheers!
AKSA said:
Well, 100% feedback isn't the same as an infinite amount of feedback. If you've say 20dB of gain, with 100% feedback, you've got just 20dB of feedback.
But look what's happening when you try for that infinite load with the CCS.
If the gain of the EF is gmRL/(1+gmRL), then that high impedance CCS load increases the gain and therefore the amount of feedback.
Why would it not be voltage feedback?
se
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