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guest Post subject: tail current mirror Posted: Tue Jan 03, 2006
7:02 pm hi in IC's usually differential pairs are biased using
current mirror (to set current in the emitters of the differential pair
transistors).i'm reading in the book but i don't see how the ac voltage
wont' see the current mirror?my understanding is that ac will also go
through the current mirror and thus voltage across the differentail
pair will be reduced.i don't see any ac chokes, etc between current
mirror and differential pair. Top Guest Post
subject: Current MirrorPosted: Wed Jan 04, 2006 3:14 am The output
impedance of the current mirror is very high, though it does decrease
with frequency (the exact behavior is device specific). That means it
acts like an RF choke. The underlying issue you seem to have
is distinguishing between voltage and current. The voltage at the junction
of the two emitters and the collector of the current mirror does in
fact go up and down with the base voltages of the diff-pair. The current
through the current source (= current mirror) does not change, because
it's determined (primarily) by beta times the base current - which is
not connected to a signal point. So "AC goes through the current mirror"
is not correct - the current stays essentially constant. But there IS
an AC voltage at the junction. (English can be such a confusing language
- "AC voltage" really has nothing to do with Current, but we say it
anyway!) Good Luck! Top guest Post subject:
Posted: Thu Jan 05, 2006 6:17 am thanks...makes sense; appreciate
it Top guest Post subject: Posted: Sat Jan 07,
2006 1:24 am i said it makes sense but i gain started looking at
it and i again have this question: if as you said if current mirror
acts like a RF/ac choke then the emitter of my actual transistor needs
to be grounded in AC operation....if my current mirror is a choke then
my emitter is floating...please reply thanks Top
guest Post subject: Posted: Sat Jan 07, 2006 2:49 am i just
asked the question again....in answering assume that i'm biasing a single
transistor and not a differential pair....the way i'm understanding
is that if you bias a single transistor then you will have to provide
a dc block at the emitter to ground if you want your ac signal to couple
to ground at the emitter.....however, if you use differential pair then
because of what is known as ac virtual ground at the two emitter connection
then you don't need a coupling cap for your ac to couple to ground
is my understanding correct please tell me or explain what is the
right way. Top Guest Post subject: Posted: Sat
Jan 07, 2006 11:34 am Hi You're right. For some reason if
you want to bias a single transistor with a current mirror in its emitter
then you would need a coupling cap. from emitter to ground if your emitter
is supposed to be at ac ground. But in case of differential pair
you don't need a coupling cap. at the connected emitters. You can think
of it two ways. In small signal ac, there is a virtual ground at the
emitter node. Also, you can look at it from impedance point of view,
as earlier post suggested. The impedance looking into the current mirror
is quite high, Ro = early voltage/IC and so the current instead of going
to the current mirror instead goes to the "other side" (transistor)---meaning
no current goes to current mirror. However, as I said, if you're
using just a single transistor biased with a current mirro then you
need a coupling cap because there is no virtual ground or for current
to any where else. Hope this helps.
Posted 11/12/2012
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