Need help please

dinstam · 29th August 2011, 12:25 PM

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

sreten · 29th August 2011, 01:34 PM

Hi, If its balanced the current each side is 1mA, rgds, sreten.

dinstam · 29th August 2011, 05:28 PM

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

sreten · 30th August 2011, 03:32 PM

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.

DF96 · 30th August 2011, 05:21 PM

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.

dinstam · 30th August 2011, 06:24 PM

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

DF96 · 30th August 2011, 08:34 PM

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.

sreten · 12:47 AM

Quote:

Originally Posted by DF96

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.

DF96 · 31st August 2011, 12:07 PM

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.

dinstam · 31st August 2011, 12:28 PM

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

29th August 2011, 12:25 PM	#1
dinstam diyAudio Member Join Date: Dec 2005 Location: Serres	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

29th August 2011, 01:34 PM	#2
sreten diyAudio Member Join Date: Nov 2003 Location: Brighton UK	Hi, If its balanced the current each side is 1mA, rgds, sreten. __________________ There is nothing so practical as a really good theory - Ludwig Boltzmann When your only tool is a hammer, every problem looks like a nail - Abraham Maslow

30th August 2011, 03:32 PM	#4
sreten diyAudio Member Join Date: Nov 2003 Location: Brighton UK	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. __________________ There is nothing so practical as a really good theory - Ludwig Boltzmann When your only tool is a hammer, every problem looks like a nail - Abraham Maslow

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29th August 2011, 05:28 PM	#3
dinstam diyAudio Member Join Date: Dec 2005 Location: Serres	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

30th August 2011, 05:21 PM	#5
DF96 diyAudio Member Join Date: May 2007	re is set by the current in each BJT. The transconductance of the stage comes from 2re (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/(2re). A current mirror doubles this to gm=1/re.

30th August 2011, 06:24 PM	#6
dinstam diyAudio Member Join Date: Dec 2005 Location: Serres	Hello and thank you both sreten and DF96. For the usual input differential stage in a power amp without current mirror gm=1/2re. Addition of the current mirror doubles gm=1/re. Now, if I want to degenerate back to gm=1/2re 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

30th August 2011, 08:34 PM	#7
DF96 diyAudio Member Join Date: May 2007	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.

31st August 2011, 12:07 PM	#9
DF96 diyAudio Member Join Date: May 2007	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.

31st August 2011, 12:28 PM	#10
dinstam diyAudio Member Join Date: Dec 2005 Location: Serres	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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