Micpre of Graham
in [Design]
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From: John Larkin on 24 Mar 2007 13:53 On Sat, 24 Mar 2007 11:47:04 GMT, "Kevin Aylward" <kevin_aylward(a)ntlworld.com> wrote: > >That is, the collecter shot noise is directly at the internal ce nodes. This >internal noise current is a constant set by ICDC, irespective of any >external circuitary. Now..if there is an emitter resister, a simple >calculation on the equivelent circuit will show that the noise actually >generated into the external collecter load is: > >in_load = icn_shotout/(1+Re/re), because the controlled source swips some >of the current away. > >So, in an actual circuit, the output noise current, dropped across the load >resister, is reduced by Re feedback. However, again, this noise is not a >function of the nature of the driving source or resistance, in addition this >reduced output noise does not imply a better S/N ratio, as the signal also >gets reduced by Re. Of course, if Re generates additional noise, then this >will have to be added into the calculation. That was my point. If the emitter current source is noise-free, it reduces the shot noise in the collector load; it must, since Ie = Ic + Ib. Even if Ib has full shot noise, Ib is 1/b of Ie, so its "diversion" noise current is small. In the case of the differential mic preamp, if the transistors are biased by dc drops across big resistors, no-signal emitter shot noise is nil. The AC gain isn't correspondingly reduced, because AC-coupled emitter-emitter resistance sets AC gain, and that contributes no shot noise current (although it does add gain to the equivalent emitter noise.) I suppose Ib has full shot noise because there's insufficient charge interaction path in the base of a transistor to get the smoothing effects seen in metallic conductors, and maybe because holes are clumsier than electrons, and maybe because the carriers are essentially thermally driven. Or something. OK, next question, again for the simple emitter follower with a stiff base supply and Rbias from emitter to ground. Assume a forced Ie that is noiseless. Suppose we have some DC base current, say 1/100 of Ie. Now suppose that base current transiently increases by d above the normal current because of shot noise. That pumps 100*d current into the emitter. That raises emitter voltage by 100*d*Rbias, reducing b-e junction voltage, reducing base current. Since Rbias can be arbitrarily large, the effective gain here can be large. Since this is a feedback gain that reduces the base current, it sounds like we can reduce base shot current by something like beta, 100:1. If not, why not? John
From: Kevin Aylward on 24 Mar 2007 14:24 John Larkin wrote: > On Sat, 24 Mar 2007 11:47:04 GMT, "Kevin Aylward" > <kevin_aylward(a)ntlworld.com> wrote: > >> >> That is, the collecter shot noise is directly at the internal ce >> nodes. This internal noise current is a constant set by ICDC, >> irespective of any external circuitary. Now..if there is an emitter >> resister, a simple calculation on the equivelent circuit will show >> that the noise actually generated into the external collecter load >> is: >> >> in_load = icn_shotout/(1+Re/re), because the controlled source >> swips some of the current away. >> >> So, in an actual circuit, the output noise current, dropped across >> the load resister, is reduced by Re feedback. However, again, this >> noise is not a function of the nature of the driving source or >> resistance, in addition this reduced output noise does not imply a >> better S/N ratio, as the signal also gets reduced by Re. Of course, >> if Re generates additional noise, then this will have to be added >> into the calculation. > > > That was my point. If the emitter current source is noise-free, it > reduces the shot noise in the collector load; This is so obvious that wouldn't, and I didn't, think for a moment that you are referring to this external to the transistor noise, as shot noise. Resisters don't have shot noise, so clearly, one is going to ignore them as shot noise generators. >it must, since Ie = Ic + > Ib. Even if Ib has full shot noise, Ib is 1/b of Ie, so its > "diversion" noise current is small. In the case of the differential > mic preamp, if the transistors are biased by dc drops across big > resistors, no-signal emitter shot noise is nil. Yeah, we really have a terminology problem here. "Emitter (or Collector) shot noise", just means the noise internal to the transiser to me. >The AC gain isn't > correspondingly reduced, because AC-coupled emitter-emitter resistance > sets AC gain, and that contributes no shot noise current (although it > does add gain to the equivalent emitter noise.) > > I suppose Ib has full shot noise because there's insufficient charge > interaction path in the base of a transistor to get the smoothing > effects seen in metallic conductors, and maybe because holes are > clumsier than electrons, and maybe because the carriers are > essentially thermally driven. Or something. I just don't understand what you are trying to say here. > > OK, next question, again for the simple emitter follower with a stiff > base supply and Rbias from emitter to ground. Assume a forced Ie that > is noiseless. Suppose we have some DC base current, say 1/100 of Ie. > Now suppose that base current transiently increases by d above the > normal current because of shot noise. That pumps 100*d current into > the emitter. That raises emitter voltage by 100*d*Rbias, reducing b-e > junction voltage, reducing base current. Since Rbias can be > arbitrarily large, the effective gain here can be large. Since this is > a feedback gain that reduces the base current, it sounds like we can > reduce base shot current by something like beta, 100:1. > > If not, why not? All this transient forcing is all darkness to me. Look, draw the small signal equivalent circuit. Put in i_base_noise generated noise across r_pi (hfe.re), put i_collector across ce. put in the emitter resister. Add in any externally generated noise, and churn the handle. If you dont want to do that, do it is spice. You can even use ideal transistors, which still generate shot noise. You can even get spice to list out contributions from each component separately. To get thermally noiseless resisters, you can use a VCCC connected to itself. Assuming a stiff base supply, i.e. Rs=o, and neglecting, rbb' as well, In is the base current noise, I get: i_out_noise = In. Re.hfe/((1+hfe)(Re + re)) So, for the case of Re=0, we get i_out_noise = 0, as expected, to wit, ib is s/c directly on the input, so no vbe.gm. For the general case, with hfe >>1: i_out_noise = In.Re/(Re +re) So, for large Re, i_out_noise ~ In, where In is due to noise from sqrt(Ic/hfe.q) Interestingly, Re allows for Ib to be feed directly into the output circuit, essentially because ib no longer has a s/c across it. -- Kevin Aylward ka(a)anasoftEXTRACT.co.uk www.anasoft.co.uk SuperSpice
From: Michael A. Terrell on 25 Mar 2007 00:39 Fred Bartoli wrote: > > Isn't PSD *power* spectral density? In Graham's case its "Petty, Snide Donkey" ;-) -- Service to my country? Been there, Done that, and I've got my DD214 to prove it. Member of DAV #85. Michael A. Terrell Central Florida
From: Eeyore on 25 Mar 2007 01:00 "Michael A. Terrell" wrote: > Fred Bartoli wrote: > > > > Isn't PSD *power* spectral density? > > In Graham's case its "Petty, Snide Donkey" ;-) IDIOT
From: Fred Bloggs on 25 Mar 2007 09:47
Michael A. Terrell wrote: > Fred Bartoli wrote: > >>Isn't PSD *power* spectral density? > > > > In Graham's case its "Petty, Snide Donkey" ;-) > > No- it is Petty Simpleton's Demise... |