Tempco calcs

Ian Fritz ijfritz at earthlink.net
Wed Nov 10 04:46:22 CET 1999


Hi all --

I had time today to work out some analytic relations to help design
alternate tempco schemes.

To start, I think it is useful to derive an analytic relation describing
the temperature stability of an expo converter. Since the output current
(or the module's frequency) is proportional to exp(qV/kT), we need the
temperature derivative of this quantity to be zero. Differentiating and
rearranging yields an important general equation:

dV/dT = V/T (1)

Enforcing (1) at room temperature (300K) ensures compensation at that
temperature.

Whatever the relation between V and T is, it may be linearized at any
temperature. If we write the linearization at 300K as

V = AT + B 

and then apply (1), we find that temperature stability requires the
constant term B to be zero. This is fairly obvious, because qV/kT in
this case has a temperature dependent contribution of qB/kT.

Now let's look at some examples.

A.)  Tempco resistor R(T) in op amp feed back path. Here V(T) is
proportional to R(T), so applying (1) recovers the familiar result

(1/R)(dR/dT) = 1/T. 

So at room temperature the tempco needs to be C(300) = 1/300 or 0.33%/K.

B.)  Tempco resistor R2(T) in series with non-tempco R1 in op amp
feedback loop. V is now proportional to R1 + R2(T), so (1) becomes

[1/(R1 + R2)](dR2/dT) = 1/T, or

[1/(1 + R1/R2)]C(300) = 1/300,

where C(300) is the tempco of R2 at 300K. Solving gives

R1 = R2[300C(300) - 1].

This is the value of R1 needed to produce a composite with the correct
tempco if the tempco of R2 is larger than .33%/K. As an example, the U.
S. Sensor Corp. PTC thermistors have C(300) = .7%/K. So R1 = 1.1 R2.

C.) Martin's voltage-divider arrangement. Ru is the upper-leg,
non-tempco resistor and R1(T) is the thermistor in the lower leg. V is
now proportional to R1/[R1 + Ru]. The math takes a couple of extra
steps, but it's not too bad. I get

1/T = [Ru/(R1+Ru)][(1/R1)(dR1/dT)], or

1/T = [1/(1 + R1/Ru)]C(T), or

Ru = R1/[300C(300)-1],

where C(T) is the tempco of the thermistor. For the C(300) = .7%/K
case,we get Ru = R1/1.1.

I think these cover the main interesting cases, but other networks can
be easily calculated with the same general technique.

  Ian



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