[sdiy] Temperature compensation results

[Ian Fritz]

Hi all --

At 01:59 PM 6/12/2003, René Schmitz wrote:

>My initial lack of understanding has stumbled me upon the following 
>reasoning: If the offset must be applied after the TC-scaling to get a 
>"absolute drift" for compensation, then what happens if the offset is 
>introduced before the TC and scaled along. (I'd prefer calling this 
>"absolute drift" "tune drift" instead, because thats what it effectively 
>is for our application)

"Tuning drift" is a good term -- I'm going to start using it 
also.  Actually, the way I define thing "absolute" drift is the log 
derivative of the frequency and so it includes both, drift from the scale 
factor (proportional to Vin) and tuning drift.

>Also for the following recall the definition of offset voltage: Its a 
>voltage that must be externally *applied* to get the same collector 
>current in a differential pair. As such it is a circuit property not a 
>transistor property.
>
>Usual case: OP, long tail pair etc.
>
>So in that case we have Ic1 = Ic2, Vb2 = Vb1+Voff (To say the transistors 
>aren't similar. And I apply a correction voltage to the second transistor.)
>
>In the following I am using the Ebers Moll Transistor Model.
>
>So this comes down to:
>
>Ic1 = Is1 * exp (Vb1 * q /(k*T)) -1
>Ic2 = Is2 * exp (Vb1+Voff * q /(k*T)) -1


Picky minor note: a pair of () is missing here.  (Just so you know someone 
is actually reading your eqns!)


>Now Ic >> Is
>
>Ic1 = Is1 * exp (Vb1 * q /(k*T))
>Ic2 = Is2 * exp (Vb1+Voff * q /(k*T))
>
>If Ic1 = Ic2 we have:
>
>Is1 * exp (Vb1 * q /(kT)) = Is1 * exp (Vb1+Voff * q /(kT))
>
>-> Is1/Is2 = exp (Vb1+Voff * q /(kT)) / exp (Vb1 * q /(kT))
>= exp (Voff * q /(kT))   (1)
>
>Or Voff = (ln(Is1/Is2) * (k*T)) / q
>
>So if Is1 = Is2 (i.e. Transistors are matched) then Voff = 0.
>if Is1 != Is2 the offset Voltage must rise with T to keep Ic1=Ic2.
>this shows that the better the matching, the better the drift will be.
>
>But this is only for the normal case of a long tail pair or a current 
>mirror (Is1/Is2 is called mirror gain).
>
>Now to the humble exponentiator:
>
>Ic2 != Ic1; Iref := Ic2; Iout := Ic1;
>
>Iout = Iref * Is1/Is2 * exp (dVbe q / (kT))
>= Iref exp ((dVbe + Voff) q / (kT))
>
>
>So if our offset is scaled along with T there is no harm done if the 
>transistors are dissimilar.
>
>If Voff is independant of T, then Iref is effectively linearly
>varied by T. Which can be used to compensate for a tune drift.

This argument is fine as long as the mismatch offset is independent of 
T.  If it comes from different T dependences in Is1 and Is2 then the 
situation is much more complicated.

>>I *assume* that a non local compensation approach (i.e. compensation
>>circuit not near to physical thermo process that you want to compensate,
>>and perhaps physically not even related to the bunch of residual tempcos
>>in the circuit) will suffer from such temperature gradients.
>>I could imagine that even the sign of compensation may be wrong
>>under certain circumstances.
>
>Indeed. I think it is somewhat difficult to archeive that on a board. 
>There are influences like TC of summing resistors, the cap, even things 
>like offset drifts of OPs, that all boil down to a single drift figure.
>It would mean to ovenize or insulate the whole board. (Maybe not the worst 
>idea, but more effort.)

Gradients in themself may not be a show stopper, as long as they are 
reproducible and non-fluctuating.  Turbulent air flow and drafts are what 
need to be controlled, I think.

   Ian






>Cheers,
>  René
>
>--
>uzs159 at uni-bonn.de
>http://www.uni-bonn.de/~uzs159
>
>