A very technical promblem on the log shaper circuit

[vaualbus]

In realation to my function generator project I've found this usefull document: http://www-inst.eecs.berkeley.edu/~ee100/su06/lab/lab10-ProjectFinish/TheDifferentialPairAsTriangleSineWaveConverter.pdf

where it is explained how to use and design the log shaper, metioned in the AN-263 document: http://www.ti.com/lit/an/snoa665c/snoa665c.pdf

This circuit is used to transform a triangle waveform into a sine wave.
Altough the equations are almost easy I don't get how to find the Vt paramater.
Any body can understand how to optained that paramater.
For transistors I'm using a LM394 supermatched transistor paired.
My input triangle wave havw Vpp of 10V on  frequency range from 1hz to 1Mhz.
Any body have some else ideas to convert the triangle wave into sine wave.
How you optaine q, T (I think this is the ambient temperature) and k metioned on the definition of Vt. (Vt = (kT)/q)

I've seen circuits using the old style and opsolate AD639 circuit but I want to not use it (24$ on ebay).

Best regards, Alberto.

PS:
Happy Christmas and happy new year to everybody.

[SeanB]

Please note. First link will not load, trailing comma in URL. Second a trailing period.

[vaualbus]

now the link work, I've done a stupid error.

[vaualbus]

I found waht paramaters are:
k -> Boltzaman constant
q -> Electron Constant
T -> Like I suspect is the temperature.

[mrflibble]

I found waht paramaters are:
k -> Boltzaman constant
q -> Electron Constant
T -> Like I suspect is the temperature.

Given your original post ... note that T is the absolute temperature in Kelvin.

[vaualbus]

Of course.

[Icarus]

The key is that the input voltage of log shaper must be lower 100mv (or 500mV something like that) reference to virtual ground. So we can put bjts in exponential region.
http://www.ti.com/lit/an/snoa665c/snoa665c.pdf
Look at figure 14.
LM311, LF356 and all those clamping diodes are just a triangle gen.
Wave-Shape Trim-pot and 200 ohm resistor network form up an attenuator