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tail current mirror - RF Cafe Forums

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Below are all of the forum threads, including all the responses to the original posts.


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.


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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!


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guest
Post subject: Posted: Thu Jan 05, 2006 6:17 am
thanks...makes sense; appreciate it


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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


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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.


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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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