Temperature compensation resistor value for oscillator circuit

Question

I'm reading the CEM3340 datasheet, section "operation of the temperature compensation circuitry" over and over and still can't understand how do you choose value of the tempco resistor. In the typical application diagram the value is 5.6k. This exact value is hard to find, so I want to understand how do I choose the resistor value for another supply voltage, and what do I have to modify if I choose another tempco resistor value.

I understand why this resistor even needed: the transistor pair in the oscillator IC has temperature drift, so the resistor with 3500ppm drift would compensate by providing changing voltage (or current?).

The supply voltage in the typical application is ±15V, but I'm going to do ±12V.

Answer 1

It should not be hard to find a 5.6K resistor- that is a standard E24 (5%) value. You could use 5.62K 1% too.

Rt/Rz are not 'temperature compensating' resistors with a special tempco, they are just normal resistors that bias the internal junctions to produce a voltage related to absolute temperature (Vt).. that way the temperature compensation circuitry is measuring the temperature directly on the die- it actually is there to compensate for the logging transistors which are also nearby on the same die.

The values of Rt and Rz don't change with supply voltage because the voltage at pin 3 is regulated to -6.5V by the internal Zener. Change the value of Ree to maintain that same current from your negative supply if the latter is not -15V.

Answer 2

I figured this out: this chip does not even need a tempco resistor. It contains the temperature compensating generator inside which cancels the temperature drift of the oscillator transistors. Rt on the circuit diagram is just normal resistor which sets the working mode of the temperature compensating generator.