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Old 12th September 2009, 05:47 AM   #1
Bitrex is offline Bitrex  United States
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Default Differential amplifier gain question

I know that the gain of a differential amplifier when using a differential output and differential input is basically Rc*Io*(Vid/2*Vt), where Rc is the collector load, Io is the current through the long tail, Vid is the differential input voltage, and Vt is the thermal voltage. This is a small signal simplification of the Taylor series expansion of the second term, which is actually tanh(Vid/2*Vd). My question is how can I calculate the gain when using a single ended input and a single ended output? I think that it will be half of the above, because the input only "sees" one half the signal swing? Also, what effect will using a current mirror load have instead of a collector resistor?
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Old 12th September 2009, 10:52 AM   #2
jony is offline jony  Poland
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The gain for asymmetric output is equal
Av_asym=Rc/2re=Rc*Io/(4*26mV)
And when you use current mirror load and asymmetric output,then the gain is equal
Av=Rc/re=Rc*Io/(2*26mV)
So using a current mirror load you get the same gain for asymmetric output
as for differential output.
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Old 14th September 2009, 02:10 AM   #3
Bitrex is offline Bitrex  United States
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Thank you!
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