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Old 11th October 2014, 12:19 PM   #1
Stee is offline Stee  Italy
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Default cfp - variant --> current coupled

as you know
the Sziklai pair also known as a "complementary feedback pair" (CFP)
have a resistor
http://www.epanorama.net/sff/Misc/Am...dback_Pair.pdf

my scheme drastically reduces the instability
Click on the diagram
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Old 11th October 2014, 01:18 PM   #2
RNMarsh is offline RNMarsh  United States
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That looks good. Does it have any affect on linearity -- distortion? And, also as a push-pull version?

-THx RNMarsh
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Old 11th October 2014, 01:44 PM   #3
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Assuming transistors Q2 and Q4 are identical (figure attached), Q2's current must be less than Q4's current. They have the same base voltage but Q2's emitter voltage is less than Q4's emitter voltage, due to the IR drop across R5. Thus Q2's base-to-emitter voltage is smaller than Q4's base-to-emitter voltage. Thus Q2's current is smaller than Q4's current.

Therefore "current booster" Q2, can only increase the total output current thru R6, by a factor of two at most - because Ic2 < Ie1. In my opinion, this nullifies one of the big advantages of the CFP, namely its enormous current gain (Iout / Iin). In a standard CFP the current gain is proportional to the square of Beta, while in this circuit the current gain is merely proportional to Beta. In fact it's less than 2 x Q1_Beta.
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Old 11th October 2014, 08:20 PM   #4
Ketje is offline Ketje  Belgium
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Quote:
Originally Posted by Mark Johnson View Post
Assuming transistors Q2 and Q4 are identical (figure attached), Q2's current must be less than Q4's current. They have the same base voltage but Q2's emitter voltage is less than Q4's emitter voltage, due to the IR drop across R5. Thus Q2's base-to-emitter voltage is smaller than Q4's base-to-emitter voltage. Thus Q2's current is smaller than Q4's current.

Therefore "current booster" Q2, can only increase the total output current thru R6, by a factor of two at most - because Ic2 < Ie1. In my opinion, this nullifies one of the big advantages of the CFP, namely its enormous current gain (Iout / Iin). In a standard CFP the current gain is proportional to the square of Beta, while in this circuit the current gain is merely proportional to Beta. In fact it's less than 2 x Q1_Beta.
+1
And why is the OUT where it is drawn
Mona
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Old 15th October 2014, 04:42 AM   #5
Stee is offline Stee  Italy
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Default in other words

The advantage is stability
current don't need to transform itself in voltage and again in current
to control the unit gain
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Old 15th October 2014, 11:19 AM   #6
Stee is offline Stee  Italy
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Default also as a push-pull version

high speed
good for headphone amplifier
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Old 15th October 2014, 03:04 PM   #7
DF96 is online now DF96  England
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Quote:
Originally Posted by Mark Johnson
Assuming transistors Q2 and Q4 are identical (figure attached)
I would assume that Q2 is a power device and Q4 is similar to a driver device.

Quote:
Originally Posted by Stee
The advantage is stability
current don't need to transform itself in voltage and again in current
to control the unit gain
And there was I always thinking that stability problems were caused by unwanted parasitic feedback, rather than current-voltage transformations. If this version of CFP is more stable, then this arises from the reduced internal gain - which may have to made up elsewhere.

Roughly speaking, this is an emitter follower augmented by an asymetric current mirror. I am unclear what Q3 (in post 3) does - it will increase Miller effect in Q1.
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Old 15th October 2014, 03:58 PM   #8
Stee is offline Stee  Italy
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Q3 (in post 3) cut RF on Q1
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Old 15th October 2014, 04:05 PM   #9
DF96 is online now DF96  England
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How?
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Old Today, 10:56 AM   #10
Stee is offline Stee  Italy
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Default because

work as a diode @0.6V
so cut half wave

see the application of this superdevice on VAS stage
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