Well, I tried breaking in by splitting the connection between the emitter and the rest of the circuitry. Even tried swapping LG1 and LG2. No go. Zero loop gain with the technique because there's no physical loop. It's just a mathematical abstraction.
As an interesting aside, Dr. Leach showed our class back in the '70s how cascading a simple PNP emitter follower with a simple NPN emitter follower gave less distortion with a high impedance load than just the NPN emitter follower alone. He was at a loss to explain. I suspect that the nonlinear transfer function of the PNP was an approximate inverse of that of the NPN.
Anyway, I think we'll either have to just use the calculated value of the loop gain for the simple EF first stage (gm * REeq) where REeq = (beta2 + 1) * RL.
As an interesting aside, Dr. Leach showed our class back in the '70s how cascading a simple PNP emitter follower with a simple NPN emitter follower gave less distortion with a high impedance load than just the NPN emitter follower alone. He was at a loss to explain. I suspect that the nonlinear transfer function of the PNP was an approximate inverse of that of the NPN.
Anyway, I think we'll either have to just use the calculated value of the loop gain for the simple EF first stage (gm * REeq) where REeq = (beta2 + 1) * RL.
Charles Hansen said:
Ok. Then by what mechanism does the Darlington increase its linearity over the simple (i.e. single device) emitter-follower?
se
Does this cross-posting mean that the moderators should re-merge the threads they split?
http://www.diyaudio.com/forums/showthread.php?postid=334032#post334032
http://www.diyaudio.com/forums/showthread.php?postid=334032#post334032
andy_c said:
That's what I'm thinking as well.
But the question is, why is the Darlington significantly more linear than a singe-device emitter-follower?
In the other thread, Charles quoted Douglas Self saying that the CFP was more linear than the single-device follower because adding the second transistor "increase
How can the Darlington also be significantly more linear than the single device follower except by way of the same mechanism?
If you just take a single device emitter-follower and drive a second single-device emitter-follower, their noise and distortion will simply add.
So it seems to me that the Darlington must also effectively increase the open-loop gain.
se
Charles Hansen said:
Ok, so now you're claiming that a Darlington pair is no more linear than a single-device emitter-follower?
se
Wasn't it Eddy who said "claims don't usually end with question marks"?
http://www.diyaudio.com/forums/showthread.php?postid=331365#post331365
http://www.diyaudio.com/forums/showthread.php?postid=331365#post331365
I can't imagine two "same sex" EFs cascaded would have less distortion than just one. But I have seen the distortion get smaller with an NPN-PNP cascade as I mentioned above. This speaks well for the classic diamond buffer.
andy_c said:
By "two 'same sex' EFs cascaded" are you meaning a Darlington pair or two distinct EFs in series?
se
Okay, can I assume you're defining "darlington pair" as two "same sex" transistors in which the emitter of the first is connected to the base of the second, and the collectors are connected together? In this case, for AC purposes, it should be equivalent to allow a constant current load from the emitter of the first to the negative supply, if only to allow the DC current of the first device to be greater than the base current of the second device.
andy_c said:
Yes.
Well, that's not quite what I was getting at. I was thinking more in terms of feedback.
In other words, with two independent same-sex EFs, one feeding the other, the linearity of the first EF is going to be determined by the gain of that EF's transistor. The same holds true for the second EF.
So, any distortion produced by the first EF will simply be seen as of it were signal by the second EF. The second EF won't do anything about the distortion from the first and will simply add its own distortion to the mix.
Now, the way I see it, a Darlington pair isn't simply one EF feeding another EF. It's two cascaded transistors forming a single EF.
And instead of having the gain of two transistors in two feedback loops, you've go the gain of two transistors in one feedback loop.
In other words, with the single transistor EF you've only got the gain of the one transistor within the loop whereas in the Darlington you've got the gain of two transistors within the loop.
The greater the gain within the loop, the greater the amount feedback and the greater the linearity.
Am I making sense?
se
Here's something from Douglas Self's comments on the CFP that might help explain why we're having such a hard time thinking in terms of "loops" when it comes to simple emitter-followers and Darlington pairs:
This also allows the stage to be configured to give voltage-gain, as the output and feedback point are no longer inherently the same.
In any case, the way I see it, a transistor doesn't lose its intrinsic gain just because it's being used as an emitter-follower. If it did, an emitter-follower stage would be no more linear than a common-emitter stage. But an emitter-follower stage is more linear than a common-emitter stage. And I suspect that it is precisely because the transistor's intrinsic gain becomes part of the feedback which is inherent in the emitter-follower.
se
This also allows the stage to be configured to give voltage-gain, as the output and feedback point are no longer inherently the same.
In any case, the way I see it, a transistor doesn't lose its intrinsic gain just because it's being used as an emitter-follower. If it did, an emitter-follower stage would be no more linear than a common-emitter stage. But an emitter-follower stage is more linear than a common-emitter stage. And I suspect that it is precisely because the transistor's intrinsic gain becomes part of the feedback which is inherent in the emitter-follower.
se
Steve Eddy said:
higher i/p Z = better linearity. less current is drawn from the previous stage. The linearity is lost in the previous stage because of current draw through the load it sees
don't necessarily need FB to increase linearity
look at the low frequency linearity of a single mosfet before the i/p C becomes a significant factor.
it's simple math ( as you guys say other there ) - you know, ratios
mike
mikelm said:
True enough.
But by the same token, I don't see a transistor's intrinsic gain simply disappearing just because it's configured as an emitter-follower.
se
Steve Eddy said:
plenty of current gain - yes...but no voltage gain...yes ?
the lack of voltage gain is not due to some inherent feedback mechanism.
it's just that, because of the reasonably fixed Vbe's, THERE IS JUST NO VOLTAGE GAIN.
The feedback that Charles was saying he has noticed changes the sound ( for the worse ), involves a POSSIBLE voltage gain ( that can be demonstated by removing the feedback ) that is controled / reduced by an external feedback cct.
You know, just like the three wise men, the feedback goes back by a different route. and also, significantly perhaps, at a ( slightly ) different time.
hope this helps
mike
mikelm said:
Well, no. Followers don't have voltage gain. The reason for that, according to a number of sources I've read, is because of feedback.
It's not? Why not?
Take a grounded emitter common-emitter stage. What happens to its voltage gain as you add some emitter resistance? The gain drops. Why? Again, according to a number of sources I've read, it's because of feedback.
Horowitz & Hill for example, in the second edition of The Art of Electronics, chapter 2, p. 84, "Emitter resistor as feedback."
Adding an external series resistor to the intrinsic emitter resistance re (emitter degeneration) improvesd many properties of the common-emitter amplifier, at the expense of gain. You will see the same thing happening in Chapters 4 and 5, when we discuess negative feedback, an important technique for improving amplifier characteristics by feeding back some of the output signal to reduce the effective input signal. The similarity here is no coincidence; the emitter-degenerated amplifier itself uses a form of negative feedback. Think of the transistor as a transconductance device, determining collector current (and therefore output voltage) according to the voltage applied between the base and emitter; but the input to the amplifier is the voltage from base to ground. So the voltage from base to emitter is the input voltage, minus a sample of the output (IERE). That's negative feedback, and that's why emitter degeneration improves most properties of the amplifier (improved linearity and stability and increased input impedance; also the output impedance would be reduced if the feedback were taken directly from the collector).
So why is the emitter resistor providing negative feedback in the case of the common-emitter amplifier but suddenly it stops providing negative feedback just because you've removed the collector resistor and take your output across that same emitter resistor?
Unless Horowitz & Hill are just talking out of their collective *****, it seems to me that removing the collector resistor simply results in the amount of feedback going from something less than 100% to 100%.
se
No!
You have got this thing in your head that transistors work on a feedback principle and you are struggling to force fit this theory to reality.
Two transistors in EF are no more a feedback system than two resistors in series. It doesn't matter that current gain is involved. Imagine a magic resistor that is just like a normal resistor except that the current flowing into it from the positive end is 1/10th of what it should be. The other 9/10ths come from somewhere else and it doesn't matter where. The current gain reduces the input current, that's all, it doesn't imply feedback.
You have mis-interpreted H&H. They are saying that the Ic will be more linearly related to Vb if you use an emitter-degeneration resistor Re. There is no magic here, this is simply sacrificing transconductance for linearity. The same will happen to the transconductance of an EF or darlington if you add a resistor in series, or indeed any device or component if you add a resistor for that matter.
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