Post subject: tail current mirror Posted: Tue Jan 03,
2006 7:02 pm
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.
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!)
Post subject: Posted: Thu Jan 05, 2006 6:17 am
thanks...makes sense; appreciate it
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
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.
Post subject: Posted: Sat Jan 07, 2006 11:34 am
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
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.