Vbe stuff

What's wrong with 59mV per decade??

Jon

I am miles ahead of you, and this using the most elemental analysis of the diode Is*(exp(V/VT)-1) diode equation. For low level injection, which is certainly the case in the example measurements, the volts-per-decade characteristic derives precisely from Vt and nothing else. The measurement deviations are not about non-linearity due to Is dependence on injection level or anything like that. So the explanation of the behavior goes directly to the estimation of THE POTENTIAL DROP ACROSS THE TRANSITION REGION AND THE ACCOMPANYING MINORITY CARRIER CHARGE DENSITIES AT THE BOUNDARIES! Charge neutrality in steady state requires that the majority carrier densities increase by the same amount as the minority density at the boundaries, and this creates the dependence of each minority carrier density upon the other. The simplified model of say n-sub-P being in ratio to n-sub-N by the factor exp(V/Vt) no longer holds when the doping densities in the two complementary regions are of the same order. Anything having to due with surface diffusion, recombination rates, and/or diffusion lengths all influence an Is non-linearity and having nothing to do with transition region fields and potentials. The transition region is the purest structure in the device, and it is this that determines the volts-per-decade current dependence.

Fred, Thank you for the quote from some textbook, but an engineer finds it more useful to think in terms of current crowding (IKF) and dirty junctions (recombination, ISE and associated terms).

The "most elemental" equation doesn't demonstrate those conditions.

I've posted those two pages of equations before. If you missed it I will post them again.

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.

"DCT" mode was used, as i mentioned - that is, the base was connected to the collector during measurement.

Nothing. But note the measured Vbe is significantly lower than what Mr. Pease mentioned. Hence the query.

Oh. I see. That's just Is.

Jon

Hmm, is it Is or is it T? Glancing at the discussion of logarithmic circuits in the Analog Devices Nonlinear Circuits Handbook (highly recommended) the firat thing I noticed is that (kT/q)ln(10) = 59 at

24.21 C and 60 at 29.25 C, resulting in 59mV per decade at the lower temp and 60mV per dacade at the higher temp regardless of Is. What temp was the measurement made at?

He mentioned some temps. But in any case, I assumed (I think reasonably) that the change in temperature between 10uA and 100uA and the time between the measurements all conspired to little effect.

You are now mentioning the delta-V, which I had assumed was the earlier question, but I've now been corrected to understand that he was talking about comparing the absolute magnitudes of the Vbe. Which is what brings me back to Is. I don't think T is the issue.

Jon

Temp seemed to be about 21C (two thermometers).

At 21C the ideal diode equation predicts about 58.4mV per decade of current. The reason you were expecting 60mV per decade has escaped me and I don't have time to go back and read the thread to figure it out now, but I would bet that the answers to most of your logarithmic circuit questions can be found in the Nonlinear Circuits handbook; which goes into good detail about the behavior of diode connected transistors, and why that behavior requires accurate log circuits to be temperature compensated log ratio circuits. IIRC the discussion includes everything Pease has published on the subject in greater detail, except for the method of improving low current frequency response.

My question had little to do about the log slope; i was wondering why measurement of a transistor VBE was lower than what Mr Pease stated (as fact).

I had first misunderstood you, because you were talking about the differential Vbe change when speaking about 100uA and 10uA and that made me focus on that aspect. Later, you clarified the question and I now take it that you are mostly wondering about the Vbe differences between parts at the same current, but that the reason you ALSO mentioned the finite Vbe difference, too, at two different currents is that you thought this has something important to do with it, too.

For example, you show that at 10uA: 1N4006: 387mV at 10uA 474mV at 100uA MPSA42: 505mV at 10uA 564mV at 100uA Pease: 640mV at 10uA 700mV at 100uA

From the Shockley, I=Is*(e^(Vd/(nKT/q))-1) equation and KT/q = 25-26mV at roughly room temp, assuming e^(Vd/(nKT/q)) >> 1, you get:

Vd = nKT/q * ln(I/Is), or Vd = n * 25-26mV * ln(I/Is)

The value for 'n' depends and it is the mean value for the emission coefficient. There are several different currents involved and they will usually have n=1 or n=2 in their equation. However, depending on which dominate, the mean value when stuffed into a single exponential is usually between 1 and 2 and not necessarily an integer. The upshot of these is that for diode-connected transistors you will usually find that "n=1" is very close to the mark. For diodes, it may vary upwards towards 2 and even past that, or below 1 even. In other words, expect a lot more difference there.

The 60mV per decade part actually comes directly from that 25-26mV figure. The difference is ln(10)*25-26mV, which is from about 58mV to about 60mV when you multiply that out. A change by 'e' in the current would be 25-26mV. But a change by '10' would be ln(10) times larger in volts.

So you could just write:

Vd = n * 59mV * log10(I/Is)

That's where Bob's rough figure is coming from (assuming n=1 here.) One needs to have some "one point calibration" to set the curve. So on some parts, this will be say, .6V at 10uA. On others, it might be .55V at 10uA. From those points, it should go up and down at about that 59mV per decade rate, though.

So you really have two questions. One is why the rate of change in Vd is larger than 59mV per decade, from your measurements, in the 1N4006. The other is why the Vd is different between a diode-connected MPSA42 in one case and a 1N4006 in another case, both running at the same currents.

So two questions. Two answers. And keep in mind I'm not an expert like others here. I'm just a lowly hobby type. So take it all with a grain of salt, eh?

The difference in the absolute magnitude is due to at least two things in these simple models at once -- one is 'n', which for your 1N4006 must be about 87mV/59mV or about 1.47 and for your MPSA42 is 1. The other is the value of Is, which is probably rather much higher than it is for the MPSA42.

Let's calibrate some guesses here. You can then try it out and see how well it predicts. Recall these:

1N4006: 387mV at 10uA 474mV at 100uA MPSA42: 505mV at 10uA 564mV at 100uA Pease: 640mV at 10uA 700mV at 100uA

Vd = n * 59mV * log10(I/Is), or Is = I * 10^[-((Vd/59mV)/n)]

We get: n Is KT/q*ln(10) ------------------------------- 1N4006 87/59 3.56nA 59mV MPSA42 1 27.6fA 59mV Pease: 1 215aA 60mV

The reason I used 60mV for Bob's entry is simply because that's the estimate he was using. 59mV is about what you were getting for the MPSA42, so I used that instead for the real-world cases.

So the new equations would be:

1N4006 Vd = 87mV * log10( I / 3.56nA ) MPSA42 Vd = 59mV * log10( I / 27.6fA ) Pease Vd = 60mV * log10( I / 215aA )

Those may match much more nicely. Try it out and let us know.

Of course, now you want to know why the effective 'n' is 1 in the case of the MPSA42 and why it is some other value in the case of the

1N4006. And also why the Is is different in different cases.

I could try my hand at this, but I'd rather let others here provide some good information. There are several effects involved and it's probably better that others put them in context as I'm likely to make some mistakes along the way.

Jon

By the way, I dropped out the "- 1" part of the Shockley equation in forming the approximations. That works for currents well above the Is current (by a couple of orders of magnitude or more.)

Jon

Well done, Professor Jon. (I mean that as a compliment!)

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.

Thanks, Jim. I don't know if any of it helped.

Jon

I understood what you were saying, and I've been doing Vbe things for

44 years... Tom Frederiksen, when he was at Motorola, used to call me "Vbe Thompson" ;-)

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.

Mr Pease did not mention that Is was a factor at all. Now you are saying, in effect, fiddle with the presumed or guessed value for Is until it makes the equation "work".

Eh?

"IS" is NOT a factor in the DELTA VBE... it drops out of the equations.

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.

He didn't need to, for the case he was discussing. And I've not even bothered to read what you read, since you didn't provide a link. But I'm pretty certain he knows what he's talking about and I accept your assurance that he didn't mention Is in what you read. So the easy conclusion is that it wasn't needed for what he was talking about -- namely the change of Vbe vs current.

Let's look at the equation:

Vd(I) = nKT/q * ln(I/Is)

But this is also just:

Vd(I) = nKT/q * ( ln(I) - ln(Is) )

For a given part, Is is a constant. So ln(Is) will also be a constant. I don't need to tell you that when you take the derivative of a constant, it goes to zero. So Is drops out. And if you take the derivative with respect to the ln(I) (not I, but the log of it) so as to look at decade changes and not linear magnitude changes, even the ln(I) just becomes 1. So you get:

d Vd(I) ------- = nKT/q d ln(I)

So you can notice that this is a fixed value for a part. That's why Mr. Pease didn't need to discuss Is in whatever it is that you read.

Is is a estimated parameter. Take any part you want and try and set fixed currents through it. Measure the voltage. Or set a fixed voltage across it and measure the current. Either way. Plot the values on a log graph. You will see that it pretty much makes for a nice, straight line with a fixed slope to it. However, while you cannot actually go about measuring what the current actually is at 0V across the junction, it is also the case that if you put a ruler up to the curve and draw the line back to the intercept you will see that it does NOT cross at exactly 0A. (In fact, your log chart would have a little problem there because log(0) is a problem.) Instead, it actually intercepts the current axis elsewhere. That spot is Is.

An estimation for Is can be had by looking at the device geometry and doping, etc. I didn't want to pursue that here because I'm kind of ignorant and there are really good educational papers and slide shows on the subject and you should go read those. At least, they provide some nice 1D approaches to "get the idea" in mind. In real parts, made in three dimensions, you would need to do 3D spatial integrals to get there and that's not very pretty. I've tried my hand at some numerical methods to approximating this in order to reverse engineer out the dopant levels of a Hamamatsu detector (I asked and they refused to tell me) and with some success at then predicting its responses over a much wider temperature range than their data sheets covered, but it was a serious pain in the ass to do.

In any case, Is in the models is usually a model parameter input and not a derived value. It's something you estimate from measurements. In transistors, it is actually _defined a priori_ by a reciprocity expression found in the EM model, with the base and collector at the same potential (which is exactly what you were doing with your 'diode' connection) and the transistor in the active region, otherwise.

Typical values in IC transistors is close to 100-200aA, so Pease's estimated values at certain currents were pretty good, from my modest experience.

You can do a one-point calculation of Is, just as I did earlier. That works and is simple, but it may not be enough at times. You can do better than this by doing curve-tracing of Ic versus Vce with a range of fixed Vbe's set up. Then take each fixed Vbe that you used and find it on the Vce (horizontal) axis and mark them. Take a vertical line up from there and mark its intercept with the associated Vbe curve you traced out. Do that for each Vbe you used. That will give you points where Vbc=0. (Not done like you did with B and C shorted.) This allows for a more accurate estimation of Is, drawing the line back to its intercept with the Ic axis.

Jon