Multiple current mirrors

43 noise cancellers in one box? Cool.

Noise cancelling nanoamps isn't impossible, but it's hard to do well. Manyfold current mirroring at megahertz bandwidths with only 100 nA is going to be very tough. I'd suggest using a TIA plus and 43 biggish resistors--big enough to drop a few volts at your comparison photocurrent value--each one driving the emitter of a small RF BJT such as a BFT25A, whose collectors goe to the noise canceller diff pairs. Putting a similar BJT in series with the feedback resistor of the TIA will get you a nice first-order temperature compensation and reduce the nonlinearity. Keep the transistor dissipation down to something reasonable--well below 1 mW--so that the temperature tracking doesn't get screwed up. 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.

As far as the noise cancellers go, the bandwidth of the integrating loop needs to be wide enough to cover all the important modulation of the photocurrent. This is because of the pronounced nonlinearity of the logarithmic Ic vs Vbe characteristic of the BJTs. If you don't need to be too close to the shot noise, you can use an LM13700 instead of the diff pair--by using the linearizing diodes (see the data sheet), you can get rid of the loop nonlinearity and have a straight ratio output rather than a log ratio, which reduces the effect by a lot. That'll cost you probably 8-10 dB in SNR, but you may not mind that--you'll still be within shouting distance of the shot noise.

It's possible to do this other ways and get better performance, but it takes real work. (I've been designing a follow-on to the original noise canceller over the last 6 months or so, so I've been reminded of it.)

The temperature tracking of the mirror devices is the biggie. If you keep their Ic and Vce the same, and mount them identically, it will help, and of course you aren't going to get much heating down at a few microwatts of dissipation.

If you don't need to noise-cancel all the photocurrents them at once, you might want to look at multiplexing them into a single noise canceller.

The MAT04 that I used in my published circuits was discontinued earlier this year, unfortunately.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058
hobbs at electrooptical dot net
http://electrooptical.net

When I first read about the noise canceller idea, I knew that I had my hammer. I just needed to find a suitable nail.

In fairness, I would mirror the several uA current but still. Noise cancelling in a much wider bandwidth than I plan to measure in is maybe also the wrong approach? It is essentially a DC signal I am after.

This is a simple, idea even I can understand. Seems to make a lot of sense.

The temperature compensation sounds very neat, however, which nonlinearity are you referring to?

But getting a linear signal rather than a logarithmic signal will move problems to the A/D-conversion. I am just as interested in knowing whether the signal from the 23rd photodiode, say, is 1% or 1.01% as I am in knowing whether the 37th photodiode gives 100% or 101% of the reference.

Exactly. And the instrument is going to live in a carefully air conditioned environment with plenty of airflow. I plan to keep power dissipation down.

Best regards,

Chris Egernet

The nonlinear emitter impedance and temperature drift of the common-base device that you need between the resistor and the emitters of the noise canceller's diff pair. The feedback network of the TIA would be a resistor plus a diode connected transistor, which would match those of each of the split outputs. There would be some small offset voltage if the currents weren't identical in the feedback and output branches, but if you're really making 43 1:1 mirrors with 44 identical resistors and

44 transistors, they'd match very well, as long as you're dropping a few volts in the resistors.

Terrific. Most folks don't like the log output because the gain depends on the signal level, but your application is very good. Watch out for base current errors, though--you need transistors with really good beta linearity for that job. Have a look at the HFA3046 array.

Lots of airflow is going to produce a certain amount of low frequency noise, from microphonics and temperature variations due to turbulence. I'd want to put a bit of insulation around the transistors--a 1 mK fluctuation will give you 2 uV of drift, which is ~100 ppm in collector current--not particularly subtle.

Please keep me posted on how it's going--that's the largest noise canceller instrument I've heard of.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058
hobbs at electrooptical dot net
http://electrooptical.net

Okay, so it is the action of the common-base transistor I don't understand:

The TIA will convert the comparison photodiode current into a voltage which is shared among the 43 noise cancellers, right? The large-ish resistors will then convert that voltage into currents for each noise canceller, right? So what does the common-base BJT do? Act as a current-follower, pinning the voltage of the emitters of the differential pair?

Sorry for asking elementary questions, but I'd very much like to understand this clearly.

It's good that you mention it. It seems that in physical science, it is always easy to make a thermometer. The hard part is making something that is either _not_ a thermometer or _only_ a thermometer.

So far I am only toying with the concept so I am increasingly convinced that this is the way to go.

Best regards,

Chris Egernet

Okay, so it is the action of the common-base transistor I don't understand:

The TIA will convert the comparison photodiode current into a voltage which is shared among the 43 noise cancellers, right? The large-ish resistors will then convert that voltage into currents for each noise canceller, right? So what does the common-base BJT do? Act as a current-follower, pinning the voltage of the emitters of the differential pair?

Sorry for asking elementary questions, but I'd very much like to understand this clearly.

It's good that you mention it. It seems that in physical science, it is always easy to make a thermometer. The hard part is making something that is either _not_ a thermometer or _only_ a thermometer.

So far I am only toying with the concept so I am increasingly convinced that this is the way to go.

Best regards,

Chris Egernet

You need to give the diff pair of the canceller a good quality current, because otherwise the cancellation won't work as well--changing Vbe will change the total current as well as the splitting ratio, which is Bad News.

The easiest way to do that is to stick the current into the emitter of a transistor, bias the base someplace reasonable--somewhere _very_ quiet,

1 nV/sqrt(Hz) or less--and come out the collector...i.e. a common base buffer.

Yup.

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.

Yup.

Good luck!

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058
hobbs at electrooptical dot net
http://electrooptical.net

What about Early effect here? With the common base arranged collector feeding the diff pair emitters that are moving around? That would seem to move the collector and thus some difference due to the Early effect, to me. What am I missing?

Jon

The magnitude of the change. In general purpose NPNs, the Early voltage is ~100V, whereas the emitters only move 50 mV at most, and the resistor current sources won't be dropping anywhere near 100V. The OP can probably use BF862 JFETs just as well, but I wouldn't lose too much sleep over it, at least at the 60 dB cancellation level. If he needs more than that, it'll start to matter, but so will a lot of other things.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058
hobbs at electrooptical dot net
http://electrooptical.net

Got it. The diff pair bases can't be driven around that much so their shared emitters won't move much, etc.

By the way, 100V seems to be kind of a "blind default" case (unmeasured, it seems to me, but estimated by some gross sweep of the hand) for modeling small signal NPNs like the

2N3904 or 2N2222. But I don't have much experience actually measuring the parameter -- something I should remedy. The specific NPNs the OP might use for the common base design might have significantly smaller magnitude Early voltage, though. Regardless, your point stands. I can't recall ever seeing a model (as I said, I've not measured so I'm lacking experience there) with an Early voltage less than 20V. And even there movement in the tens of millivolts shouldn't be a serious problem. (Maybe before I say that, I should see what 50mV movement does on a VAF=20V device against the 60dB level, though. At my level of understanding, this is kind of interesting to think about.)

Thanks, Jon

I looked at the 2N3904, which has a typical Early voltage of 140V as measured by Google. ;)

Quicker things like the HFA3046 are down in the 20V range, which can be very inconvenient...I had a beautiful noise canceller front end blown out of the water by that problem.

It used a non-monolithic diff pair, and the temperature tracking was done by changing V_CE of one half of the diff pair so that the power dissipation in the two sides was identical, regardless of the current split ratio. It would have worked great, except that V_A was about

12V.... (John L. pointed that out to me, the rotter. ;) )

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058
hobbs at electrooptical dot net
http://electrooptical.net

I just measured it the other day, as a matter of fact. (I'm finally to the point in the curriculum where we're supposed to investigate and use transistors... how droll!) I got 220V for this particular 2N3904 at 3.8mA Ic. In contrast, the Multisim v10 Spice model has the unusually low value of 74V.

Tim

--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms

Well, the LTSpice 2n3904 (Philips, it says) shows VAF=100. YMMV, of course.

Okay. So it _could_ be a problem. Glad to see I'm not totally confused.

Egads. Maybe I should be glad I picked up some monolithic pairs of both polarities. (I haven't used them for diff pairs, as just current mirrors.)

Thanks, Jon

Can you describe more about how you did your measurement of Va? I'm guessing you set Vbe to some voltage and Vce to about the same and measured Ic at 3.8mA and then changed Vce to some other higher voltage and measured Ic again, and then computed the slope and divided that into 3.8mA to get your figure. (Or multiplied, depending upon whether you computed the slope as a resistance or conductance.) But I'm curious if any of that is right. Could you describe the details?

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.)

I'm curious if you tried this at a number of different Vbe's to see if there was a basewidth modulation effect, too.

Thanks, Jon

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.

That could very well be. Hey, I didn't get a negative number, so that's one thing!

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).

Tim

--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms

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:

Does this imply that a common-base BJT behaves a bit like a diode- connected transistor?

about the non-linearity:

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

Right.

Yes.

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.

device

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

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.

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.

(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

:-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 Laté and we're saying it wrong? ;-)

--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms

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.

hehe. Early's 1952 paper was probably just later 'turned around' on him!

Jon