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Old 12th December 2002, 11:17 AM   #1
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Question Complex Emitter follower

Hi all,

I am working on an amplifier (how suprising but I haven't decided wich
emitter follower structure to use. It will be a class A 10 W amplifier whith
symmetric design. Not a big deal at all and to be honest I do not like the
amount of NFB it will use, but I could not figure out how to decrease the
output resistance without strong NFB. (As I remember there was a thread
called "I would like to see your amp" or something like this, where HH
described sortly his amplifier and according those lines it is damn close to
what I imagine as a really good amplifier. (Just local NFB, DC coupled,
completly FET based...) But there is a secret about the output resistance
and damping factor wich I do not know yet.)

Anyway, the reason because I wrote that I need some help regarding this
complex emitter follower I found in JLH's amplifier design book. I wasn't
seelpingThe first BJP is a simple emitter follower I understand this. The DC offset
between its base and emitter is around 0.6 Volts or so and I can bias the
output by implementing a proper voltage across the 0.22 resistor. So the
bias would be lets say for 1A*0.22Ohms + 0.6 V = 820 mVolts. But what is the
function of the other transistor? In the book it was meantioned as an
alternate and superior version of the simple darlington structure, so I
assume the first transistor would take over the task of a first BJT in the
commercional darlington leaving the real work - providing the dissipation
for the 1A - to the second one. But how does this 1 Amp divide up between
the two transistors? The first transistor has a resistor between the power
supply and its colletor providing some voltage wich is - I think - definitly
0.6 Volts or so beacuse of the base-emitter drop voltage of the second
transistor, wich is paralell with this resistor. Also in the book both the
emitter and the collector resistance had the same 100 Ohms value. In this
way there should flow about 6mA (0.6/100) across the first BJT leaving 994
mA to the second BT. (Assuming that the emitter and collector current is
equal.)
to much last night beacuse I was wondering how this can work at
all, how does it the trick JLH meantioWhat I do not understand is:

This topology is implemented in many applications as the final stage wich
drives the speaker, so the whole signal voltage is apparent at the base of
both BJTs. I just can not see how could then the second BJT work with the
whole signal without any emitter voltage potential. (Emitter connected
directly to the power rail.) As I see it should cut one half of the signal.
Or the second BJT is not an emitter follower? Maybe they are grounded
emitters using each other as an active load? Anyway is it not class B? And
at least JLH meantioned that this type of topology provides less output
imedance as a simple emitter follower. (According to the book the output
impedance of the complex emitter follower equals the output impedance of a
simple emitter follower devided by the Hfe of the second BT.) Why and how?

So, please somebody tell me the working mechanism of this follower
structure...

Thanks anyway and sorry for the long mail.

bt




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Old 12th December 2002, 12:20 PM   #2
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The first transistor can be thought of as an error-amplifier. It responds to the potential difference between its base and emitter terminals.

You can think of the second transistor as a common-emitter stage, providing lots of gain.

So, you have an error amplifier, and a gain stage. A complete system employing negative feedback! Using only 4 components!

It works like this: Suppose you raise the input potential a bit... The first transistor will see an increased VBE, and will respond to this by increasing its collector-current. This collector-current will flow through the base-emitter junction of the second transistor - it also responds by increasing its collector-current.

Here's the clever bit: This increased collector current will raise the potential of the output of your circuit. However, if it raises it too much, what happens to the VBE of the first transistor?

That's right, the VBE seen by the first transistor shrinks, it stops passing current, the second transistor also stops passing current, and your output voltage will fall!

Of course, it won't fall far, because the first transistor will switch on again and cause the voltage to stabilise. In reality, you won't observe any "hunting", but I think it helps understanding if you break everything down and think in this simplistic way...

So, the circuit uses 100% negative feedback to achieve a gain of 1, a high input impedance, and low output impedance. It's neat

Cheers,

Mark
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Old 12th December 2002, 12:38 PM   #3
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Hmmm.... Thanks for the explanation mhennessy . I always have some problem when thinking about NFB circuits beacuse they are a bit unreal for me. (Modifying a signal wich have already passed through the network by coupling it back to the input...Somewhat not logical, even if all the elements are ultra fast in the circuit. If I try to imagine how such amplifiers work always some kind of oscillation falls into my mind and in the next minute I am completly confused about the whole theory. Anyway, it seems that I can not avoid using NFB and your post was a great help.

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Old 12th December 2002, 12:48 PM   #4
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Default aka Sziklai pair or Complementary Feedback Pair (CFP)

I think the second transistor is suposed to be a pnp?
http://jever.phys.ualberta.ca/~gingr...ml#ch5DarlSzik
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Old 12th December 2002, 12:51 PM   #5
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Yes, it's a pnp. Sorry.
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Old 12th December 2002, 05:35 PM   #6
PMA is offline PMA  Europe
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Default CFP

The circuit you show is a so called "complementary follower pair". It has a voltage gain of 1 (as emitter follower) and there is an additional current gain created by 2nd transistor (PNP). The circuit has strong local NFB, very good linearity and is better than Darlington pair. To see an example of amps with this circuit you might click on the www button below this message.
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Old 12th December 2002, 09:54 PM   #7
Jeff R is offline Jeff R  United States
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Default Re: CFP

Quote:
Originally posted by PMA
The circuit you show is a so called "complementary follower pair". It has a voltage gain of 1 (as emitter follower) and there is an additional current gain created by 2nd transistor (PNP). The circuit has strong local NFB, very good linearity and is better than Darlington pair. To see an example of amps with this circuit you might click on the www button below this message.
There seems to be a lot of controversy regarding the "best" BJT output stage. Prof. Leach discounts this approach and considers the Darlington, overall, to be superior. He claims that the use of local negative feedback can cause oscillations in the output stage and that the bandwidth is reduced vs. what a Darlington would provide due the Miller effect causing the junction capacitance to be much larger. I belive the Darlington also provides the lowest output impedance.

I think the complementary follower does have better linearity due to the local feedback, but Leach appears to think the price of this is too high. Regardless, some good amps have been built using both approaches, so it may be worth some more investigation.
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Old 13th December 2002, 07:13 AM   #8
PMA is offline PMA  Europe
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Default CFP stability

to Jeff R:
yes, the question is whether we speak about DIY approach without instruments or about a try to find optimal solution. With oscilloscope and simulated speaker load it is easily possible to make the CFP circuit stable. I also use capacitive load up to 10uF. Modern BJT devices, as 2SA1943 and 2SC5200 are superb in linearity and fast enough to enable application of frequency correction by a small cap. The amp (I speak about push-pull follower from my web page) is very very quick, no SR or fh limitation, several hundreds of kHz even with freq. compensation. I have tried, measured and listened both approachs and I do prefer CFP. Another big advantage is much better thermal stability, as there is only one Vbe.
Note: please take to the account that a frequency stability of Darlington is not any better than CFP :-)
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Old 13th December 2002, 08:51 AM   #9
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Yes, the thermal stability issue is good news - you only need to monitor the temperature of the drivers, so the VBE-multiplier transistor can be mounted on the much smaller driver heatsinks. No need to put it on the main heatsink with all the layout issues that go with it.

One problem with the CFP - you have to control the bias voltage much more accurately than you do with EF output. This is a reason to use the EF - you might get a slightly worse result, but it will be more consistent with time and temperature.

As a lot of you will know, Doug Self has lots of good information about this...

BTW, Banfi T. - you won't be able to avoid feedback in transistor circuits. Even the humble EF has 100% negative feedback
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Old 13th December 2002, 09:17 AM   #10
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I agree unequivocally with Professor Marshall Leach.

I have built two otherwise identical SS amplifiers, one with a CFP (Sziklai) output stage, the other with a conventional double emitter follower, a Self Type III.

The comparison is very obvious audibly, the only way that counts.

The problem appears to be oscillation in the CFP, particularly the negative rail block. This oscillation robs detail, and creates spurious intermodulation products which destroy any musicality. We can fix this by simply strapping a 100pF cap from base to collector of the driver, however. But then if you listen, the musical presentation lacks vitality, sounds 'drained'.

Leach is quite right; the price of securing stability with the CFP is so high sonically you might as well use an emitter follower, where Zout is actually not so low, but the sonics are far superior.

Go back to the Type III Self double emitter follower, and the vitality is immediately apparent. There is no comparison audibly. Vocalists no longer sound 'tired'.

Of course, to a technologist, the emitter follower lacks the appeal of the CFP, which just reeks of elegance, but sounds like !@#$.

Furthermore, the much touted advantage of superior thermal stability is overshadowed by very tetchy and low bias voltage requirements; it must be deadly accurately set, since with only two pn junctions across the bases of the drivers there is little room for error; the Vbe of a large transistor varies considerably with collector current and the tolerance setting bias is far less sensitive.

Another major problem the CFP has is the switch off behaviour; as the output comes off, the driver has difficulty with the extremely fine control of collector/emitter voltage because of the huge voltage gain of the output device, and this fudges the crossover event subliminally with spurious, short term oscillation whichever remedy you use to prevent it. This might make them suitable for Class A, but for Class AB, no dice.......

Sorry, long post......

Cheers,

Hugh

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