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Old 29th August 2011, 12:25 PM   #1
dinstam is offline dinstam  Greece
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Smile Need help please

Hello,
Will please somebody explain to me:
How do we calculate the intrinsic emitter resistor for a differential amplifier with current mirror.
If the tail current is (say) 2mA then for each leg: re’=26/1mA OR re’=26/2mA?
Thank you in advance
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Old 29th August 2011, 01:34 PM   #2
sreten is offline sreten  United Kingdom
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Hi, If its balanced the current each side is 1mA, rgds, sreten.
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Old 29th August 2011, 05:28 PM   #3
dinstam is offline dinstam  Greece
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Hi sreten and thank you.
The circuit is balanced. So re'=26R or maybe 13R since all of the tail current is available for each leg due to the current mirror? I have looked up this in textbooks and the net but I cannot find the answer.
Cheers
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Old 30th August 2011, 03:32 PM   #4
sreten is offline sreten  United Kingdom
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Hi,

No. a current mirror load will share the current between the legs,
in the linear region of operation. Degenerating the emitters helps.

rgds, sreten.
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There is nothing so practical as a really good theory - Ludwig Boltzmann
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Old 30th August 2011, 05:21 PM   #5
DF96 is offline DF96  England
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re is set by the current in each BJT. The transconductance of the stage comes from 2*re (because the signal sees two in series) but it can be doubled by using a current mirror at the output.

2mA total means 1mA per BJT, so re=25. If you took LTP output just from one collector then gm=1/(2*re). A current mirror doubles this to gm=1/re.
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Old 30th August 2011, 06:24 PM   #6
dinstam is offline dinstam  Greece
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Hello and thank you both sreten and DF96.
For the usual input differential stage in a power amp without current mirror gm=1/2*re. Addition of the current mirror doubles gm=1/re. Now, if I want to degenerate back to gm=1/2*re I must make my calculations with re=25. Correct?
Sorry for the naive approach but I am really lost. I thought that gm doubles because mirroring reduces re to re/2.
My best regards
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Old 30th August 2011, 08:34 PM   #7
DF96 is offline DF96  England
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Yes, use re=25. To degenerate back to half the gain, use R=25 in each emitter.

Mirroring does not alter re. The gain doubling comes from using signal current from both collectors.
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Old 31st August 2011, 12:44 AM   #8
sreten is offline sreten  United Kingdom
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Quote:
Originally Posted by DF96 View Post
The gain doubling comes from using signal current from both collectors.
Hi, you've lost me on that one without a differential VAS, rgds, sreten.

I thought the "standard" LTP, SE VAS and EFP was being discussed.
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There is nothing so practical as a really good theory - Ludwig Boltzmann
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Last edited by sreten; 31st August 2011 at 12:47 AM.
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Old 31st August 2011, 12:07 PM   #9
DF96 is offline DF96  England
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The current mirror in the collectors of the LTP adds together the signal currents from both LTP transistors, and sends the sum to the VAS input. It does not 'share current' as you said in post 4, it adds current.
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Old 31st August 2011, 12:28 PM   #10
dinstam is offline dinstam  Greece
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Thank you both for the knowledge which cannot be found in text books.
The amp is as simple as can be:
LTP (TR1 input, TR2 feedback), VAS, output transistors in Darlington arrangement.
Regards
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