Multiple current mirrors
in [Design]
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From: C Egernet on 7 Jan 2010 10:51 > > So what does the common-base BJT do? Act as a > > current-follower, pinning the voltage of the emitters of the > > differential pair? > > No, it's the other way up: low impedance in, high impedance out--the > collector goes to the diff pair emitters. You want the current not to > depend on the voltage at the emitters. I think I get it now but just to reinforce my understanding: The common-base transistor pins the voltage of the end of the resistor at the BJT emitter, right? I can understand that because the voltage drop over the resistor is now well-defined and so will the current be. The common-base BJT collector just faithfully sends the current through to the differential pair emitters (assuming large beta, alpha unity). That leaves me with what you said originally, Phil: > Also, make sure you use a diode-connected transistor > (CB shorted) for the compensation, because the BE diode doesn't have the > same characteristics as the transistor in normal bias. Does this imply that a common-base BJT behaves a bit like a diode- connected transistor? about the non-linearity: > The nonlinear emitter impedance and temperature drift of the common-base device Can I think of it this way: We drive the feedback path of the TIA, resistor and diode-connected BJT, by a current source. There will be a Vbe voltage drop and a drop over the resistor. This is the voltage signal out of the TIA, shared to the noise cancellers. At the noise cancellers, the current is regenerated by taking the voltage and sending it through a similar resistor and the common-base BJT to ground. Since the majority of the voltage drop is over the resistor, the current into (out of, really) the emitter is roughly the same as photocurrent so the Vbe is correct too. Because Vbe is correct, the current ends up being not just roughly right but very accurate. Best regards, Chris Egernet
From: Phil Hobbs on 7 Jan 2010 12:26 On 1/7/2010 10:51 AM, C Egernet wrote: >>> So what does the common-base BJT do? Act as a >>> current-follower, pinning the voltage of the emitters of the >>> differential pair? >> >> No, it's the other way up: low impedance in, high impedance out--the >> collector goes to the diff pair emitters. You want the current not to >> depend on the voltage at the emitters. > > I think I get it now but just to reinforce my understanding: > > The common-base transistor pins the voltage of the end of the resistor > at the BJT emitter, right? > Right. > I can understand that because the voltage drop over the resistor is > now well-defined and so will the current be. > > The common-base BJT collector just faithfully sends the current > through to the differential pair emitters (assuming large beta, alpha > unity). > Yes. > That leaves me with what you said originally, Phil: > >> Also, make sure you use a diode-connected transistor >> (CB shorted) for the compensation, because the BE diode doesn't have the >> same characteristics as the transistor in normal bias. > > Does this imply that a common-base BJT behaves a bit like a diode- > connected transistor? A diode connected transistor is actually in its normal bias region--it doesn't saturate until the collector drops ~200-300 mV below the base. That's why it makes a good compensator for the BJT. Early effect will make a bit of difference, as Jon pointed out, but it'll mainly be a slight gain error--the tempco and Vbe nonlinearity will track pretty well. You don't have to worry too much about dissipation-induced thermals with only a few microamps, but if that were ever to be a problem, just bias the CB stage's base at -0.8V or so compared with the diff pair's bases. > > about the non-linearity: >> The nonlinear emitter impedance and temperature drift of the common-base device > > Can I think of it this way: > > We drive the feedback path of the TIA, resistor and diode-connected > BJT, by a current source. There will be a Vbe voltage drop and a drop > over the resistor. This is the voltage signal out of the TIA, shared > to the noise cancellers. > > At the noise cancellers, the current is regenerated by taking the > voltage and sending it through a similar resistor and the common-base > BJT to ground. Since the majority of the voltage drop is over the > resistor, the current into (out of, really) the emitter is roughly the > same as photocurrent so the Vbe is correct too. Because Vbe is > correct, the current ends up being not just roughly right but very > accurate. Right. Cheers Phil Hobbs -- Dr Philip C D Hobbs Principal ElectroOptical Innovations 55 Orchard Rd Briarcliff Manor NY 10510 845-480-2058 email: hobbs at electrooptical dot net http://electrooptical.net
From: Jon Kirwan on 7 Jan 2010 17:58 On Thu, 7 Jan 2010 08:37:34 -0600, "Tim Williams" <tmoranwms(a)charter.net> wrote: >"Jon Kirwan" <jonk(a)infinitefactors.org> wrote in message >news:iasak5pia917e27tnjh5bs601reh86qvpe(a)4ax.com... >> Can you describe more about how you did your measurement of >> Va? > >Constant Ib. I can't imagine trying to take such a measurement at constant >Vbe, think of the runaway. However, I did not track change in Vbe vs. Vce, >so if it's defined in terms of constant Vbe, my data may be incorrect. > >As for the slope, I measured Ic for Vce = 0...10V and got the >slope-intercept for points from 1-10V. Okay. As I understand things constant Vbe works without runaway (say, for example, 600mV for Si types.) Since Ic moves rapidly with small Vbe changes, though, it seems using a constant current at the base is indeed easier and probably better. So I think I'm with you on this. >> In any case, it would seem also that calculations generating >> large Va figures are often fraught with inaccuracies (not >> that it matters for large Va values, since that means the >> effect is nearing a negligible value, anyway.) > >That could very well be. Hey, I didn't get a negative number, so that's one >thing! Well, Va is on the negative side of the Vce axis. The main thing, of course, is that you set up your equations right and perform measurements appropriately. >> I'm curious if you tried this at a number of different Vbe's >> to see if there was a basewidth modulation effect, too. > >The Early effect is explained in terms of base width modulation from the B-C >depletion region, and has nothing to do with the B-E junction itself. I >don't know why Vbe might change (except to secondary effects like Ie and >improved beta). (Actually, I was thinking of modulating Vbe by changing your Ib.) But there is a width in the emitter-base space charge layer that is normally assumed to be negligible, but which may be measurable for you. It's called the "Late Effect." Va captures the Vbc basewidth modulation, and since that is the topic here it should be obvious that there _might_ be such a Late Effect, right? Anyway, the best way to verify, of course, is to sit down and _do_ it and see. If you get two different Va for different Ib/Vbe settings, then it's certainly an indication, right? I had wanted to someday try and play with both. Jon
From: Tim Williams on 7 Jan 2010 19:39 "Jon Kirwan" <jonk(a)infinitefactors.org> wrote in message news:stock513nq6g9lr6g9uuqj9dchf5hqopi9(a)4ax.com... > (Actually, I was thinking of modulating Vbe by changing your > Ib.) But there is a width in the emitter-base space charge > layer that is normally assumed to be negligible, but which > may be measurable for you. It's called the "Late Effect." :-p That's true, even a forward-biased junction still has some V across it and therefore some depletion region thickness. (It's dastardly difficult to shove enough current density through silicon to "unbias" the junction -- IIRC, kiloamperes for a 1N914.) Tim P.S. Yeah, but was it named for Mr. Late? Is it actually Lat� and we're saying it wrong? ;-) -- Deep Friar: a very philosophical monk. Website: http://webpages.charter.net/dawill/tmoranwms
From: Jon Kirwan on 7 Jan 2010 20:03
On Thu, 7 Jan 2010 18:39:17 -0600, "Tim Williams" <tmoranwms(a)charter.net> wrote: >"Jon Kirwan" <jonk(a)infinitefactors.org> wrote in message >news:stock513nq6g9lr6g9uuqj9dchf5hqopi9(a)4ax.com... >> (Actually, I was thinking of modulating Vbe by changing your >> Ib.) But there is a width in the emitter-base space charge >> layer that is normally assumed to be negligible, but which >> may be measurable for you. It's called the "Late Effect." > >:-p > >That's true, even a forward-biased junction still has some V across it and >therefore some depletion region thickness. (It's dastardly difficult to >shove enough current density through silicon to "unbias" the junction -- >IIRC, kiloamperes for a 1N914.) The procedure I've seen for seeing if Vb (or Spice's Var) is important for the "normal" region is to plot ln(Ic) vs the venerable (q/kT)*Vbe and see if the slope varies much from 1. If it does depart, Var is important in the normal region. >Tim > >P.S. Yeah, but was it named for Mr. Late? Is it actually Lat� and we're >saying it wrong? ;-) hehe. Early's 1952 paper was probably just later 'turned around' on him! Jon |