J
Jon Kirwan
- Jan 1, 1970
- 0
Not in my experience. (but my experience is fiarly limited... a few
transitors tested.)
I always got a number that was a bit off ~0.3%, so about 1 degree at
room temp. I always assumed the error was due to the transistor
beta... Since the current is Ic and Ib. (I think I got a temperature
that was always a bit high, but I'd have to check my notebook.) You
could add some beta 'fudge factor'.... but then beta changes with
temperature too.
It also depended a bit on the collector current. (1 uA to 10uA were
'nice' currents)
Hi, George. I posted up a link to Linear's AN45 elsewhere
under this topic. See page 7 there. But I take your
experiences here seriously and wanted to think about this,
not at the 'charged gas' theory level but at the higher (and
more usual for an EE) device modeling level.
The two-current pulse method, using say 1X and 10X currents,
depends upon the following:
dV = (k/q) * ln( 1+Ic/Is ) dT
Although k and q are known, the entire factor that includes
the ln( 1+Ic/Is ) part isn't knowable in advance. But if one
assumes that the +1 term is negligible then the two pulsed
currents results in:
dV1/dT = (k/q) * ln( Ic1 ) - (k/q) * ln( Is )
dV2/dT = (k/q) * ln( Ic2 ) - (k/q) * ln( Is )
Subtracting dV1/dT from dV2/dT yields:
dV1 - dV1 = (k/q) * ln( Ic2/Ic1 ) ) dT
And if the ratio of Ic2/Ic1 is known a priori then the entire
factor, (k/q) * ln( Ic2/Ic1 ) ), is also known. And as a
consequence could be used to measure temperature without
having to calibrate the system. Or so it seems at first
blush.
But it's also the case that the value of the saturation
current, Is, is itself a rather complex function of T:
Is(T) = Is(Tn) * (T/Tn)^3 * e^( -(q*Eg/k) * (1/T-1/Tn) )
Where Tn is some chosen T(nominal).
In fact, this particular component is what overwhelms the
first equation (which is positive vs temperature) and yields
the usually quoted -2mV/K figure (very approximately.) So, in
fact, Is(T) is the dominant factor in Vbe change over T and
in no possible way is it a simple function of T!
(Even the above Is(T) equation itself is a simplification.
The power ((T/Tn)^3) for example is an approximation and not
strictly true in practice. Same with Eg, which itself is also
taken as a single approximation value.)
Just as a guess, the idea of ln( Is ) being cancelled
entirely out of the equation by ratiometry, even assuming
that the die is at thermal equilibrium, would make me worry a
little. (I accept that pulsing the 1X/10X current change fast
enough or that using low enough currents, like the 1uA and
10uA you mention, would yield a near-equilibrium state.) I'm
just not sure that at this level of modeling, that _Is_
remains dead stable as a modeling parameter when facing a 10X
current change. There is a lot of linearity over orders of
magnitude change, as a broad statement. But exactly how
linear is it when provided a two point ratio a decade apart,
vs device variation?
I wonder that some temperature error is swept under this
Is(T) rug and hidden from the analysis, so to speak. Even
assuming thermal equilibrium. Because it may really be that
Is(V,T), not Is(T), as both the power (^3) and Eg are taken
as simple constants for simplification when they aren't, in
fact, invariant at this level of modeling.
You mention base currents as a possible error. I've ignored
that so far. The equation:
dV = (k/q) * ln( 1+Ic/Is ) dT
in the diode connected case refers to Ic. The currents
through it, on whole, are (beta+1)/beta times as much. If you
cobble up precision current sources at exactly 1X and 10X,
the ratio of Ic2/Ic2 would still be 10, even though you are
driving Ie, I think. However, beta itself changes vs Ic. So
there is that to account for, if you were only using a beta
level model. But the:
dV1 - dV1 = (k/q) * ln( Ic2/Ic1 ) ) dT
method doesn't use or rely upon beta. So I'm not imagining a
problem there because (1) the ratio is still 10X and (2) beta
isn't used in the analysis method.
Interesting problem getting past a certain level of accuracy,
though. There must be several papers that go beyond the AN45
app note I'd posted up earlier. I haven't seen one, yet.
Jon