Two-Stage Transistor Phase Splitter

daviddeakin

Joined Aug 6, 2009
207

hobbyist

Joined Aug 10, 2008
892
I have searched the internet and found some useful stuff, I had an idea and started trying to solve.
I have been trying to calculate and find, an equation that shows that the gain is related to rf/rs, but I can't. Can you help me out?

Hi,

I found this in my course books,
I'm rereading it again to try to get a grasp at what there saying,

I just breadboarded a quick CE amp, with colloector feedback,
and got a volt. gain approx. = to (RC /RE).
Just as a series current mode feedback config. works out at.

When I get a better understanding about this myself, I'll post what I found.
 

The Electrician

Joined Oct 9, 2007
2,971
If you are driving the input with a voltage source (zero output impedance ideally) and if you don't have a series resistor, Rs, then you don't have collector feedback. The feedback through Rf will be shorted to ground by the zero output impedance of the (ideal) signal source.

If the emitter resistor is unbypassed, and if you do have an Rs resistor between the input signal source and the base, plus an Rf collector feedback resistor, then the gain will depend on both the collector feedback and the unbypassed emitter resistor.
 

Thread Starter

cebrax

Joined Jan 30, 2010
26
The feedback through Rf will be shorted to ground by the zero output impedance of the (ideal) signal source.
So, base of the transistor will be grounded too, then? What about the coupling capacitor?
 

The Electrician

Joined Oct 9, 2007
2,971
The coupling capacitor will have an effect, of course, but its value would usually be chosen to cause its effect to be below the audio passband.

Given a transistor gain stage with resistors (biasing resistors excluded) Rc, Rf, re (intrinsic emitter resistance) and a fully bypassed external emitter resistance RE, the gain function for the stage is:

\(-\frac{Rc(Rf-re)}{re(Rc+Rf)}\)

That (Rf-re) term in the numerator means that the gain can be made zero (if Rf=re), and can even change sign (if Rf < re). When there is no Rs, there is no feedback (when the signal source has zero output impedance), but there is feedforward from the base to the collector through Rf; that's how the (Rf-re) term comes about.

It should be possible to choose resistors to give a gain of -1.

If an input resistor Rs is included, the gain function becomes:

\(-\frac{Rc(Rf-re)}{re(Rc+Rf)+Rs(Rc+re)}\)
 

hobbyist

Joined Aug 10, 2008
892
Hi, cebrax

After rereading it, it seems to be under certain conditions, when the rf/rs holds true, but mainly the gain is givien as (RC / RE) just as a current feedback amp works at.

After "The Electrician", posted I pretty much gave up any further pursuit of this thread, because that person has more knowledge in the electronics field then I do, and I felt that he was better qualified to answer your question, than I was.

I must say I hardly ever designed any transistor circuits using collector feedback, it has always been emitter feedback.
 

Thread Starter

cebrax

Joined Jan 30, 2010
26
Oh, damn I am so stupid and sorry :(
Really, I feel so bad that I omitted your post The Electrician.

Can you please explain how you derived the gain equation of the circuit with Rs? Did you make only AC analysis?
 
I hope it's clear just what is the circuit I analyzed. It's the circuit surrounding Q2 in figure I-28 on this page:

http://www.tpub.com/content/neets/14180/css/14180_41.htm

with Rs = R4, Rf = R5 and Rc = R7; RE = R6, which is considered to be fully bypassed at the frequency of operation. The intrinsic emitter resistance is re, which is, of course, determined by the emitter current. Biasing arrangements are not considered, so the analysis is AC only.

The method of analysis is that which I described in this thread:

http://forum.allaboutcircuits.com/showthread.php?t=26710

Especially have a look at post #43, et. seq., for the details.

Posts #48 and #49 are particularly relevant to your question.

I allowed β->∞ to get the expressions you're asking about.
 
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