Exponential Control Voltage for VCF using LOG102?

[ioplex]

Hello,

Hopefully someone can help me with this, I've been beating myself up over the last few days with this ...

I need to compute an exponential voltage to control a VCF. I thought I would use a Burr Brown LOG102 chip to do the work for me but I don't understand how this chip can be used to generate an exponential output.

The LOG102 chip outputs a voltage that is the log of the ratio of two currents. Internally this is basically the same as any logarithmic circuit but with everything on the chip and matched and temperature compensated and so on. My understanding was that if you connect Vout to I1 with a smallish resistor then you should get an antilog output or in other words an exponentiator. But the output is still completely log AFAIKT. As I increase I2, Vout's slope gets smaller, not greater. I tried inverting it and DC shifting it up so that it was at least exponential but this seems very wrong to me. I am doubting that this chip can be used at all. Can I use this LOG102 chip to generate a the Vf = 2**CV exponent I need?

The following are detailed measured (blue diamonds) and computed values (pink squares) of the filter's required control voltages:

http://173.230.137.181/~miallen/FilterCv2012.pdf

You can see that starting with 27.5Hz the voltage is 13.6mV and then doubles each octave. When plotted this is very much a straight line (but the chart shown has log x-axis). Meaning the filter can be precisely (albeit perhaps not accurately) controlled with a predictably voltage.

So the exponential function I need is:

Vf = 2 ** CV * 0.01375

where Vf is the voltage required by the filter to give you the frequency corresponding to the control voltage CV (using the standard 0-10V with 1V per octave).

What do I have to do to generate the necessary control voltages for this filter? Can I use this LOG102 chip or do I need to built a discrete exponentiator with a thermistor and matched transistors and so on?

Mike

[ioplex]

Answering my own post ....

This page explains pretty clearly how to get a good 1V/octave exponential output:

https://plus.google.com/107231411209962227230/posts/C8iVtXKzvet#107231411209962227230/posts/C8iVtXKzvet

I think maybe the LOG102 cannot be used for a 1V/octave exponentiator because the voltage divider in front of Q1 is hard-wired for log base 10 or at least something that cannot be adjusted.

Mike