Squarerooting transistors.

[Christopher List]

>>   What I'm wondering is this: You can use two transistors to create an
>>   exponential current source from a linear input voltage. Is there a way
>>   to wire them in reverse so that you have a square-root (reverse log?)
>>   voltage source from a linear input current?  Just a point in the
>>   proper direction would be helpful.

> Why would you want "square-root"?  And "reverse log" is exponential,
> which you've already got.
> 
> What I think you're really asking for is a log amp, the circuit isn't
> too difficult, and can be found in most published collections of opamp
> circuits.  Basically a transistor feedback around an opamp.

Doh! I swear I was in A.P. calc in high school! Yeah, I don't know how I got my 
terminology screwed up. Wait a sec, for the exp. converter, isn't out=in^2? - 
That's why I thought I needed square rooting. No, wait, it's out=2^in, OK, I'll 
be quiet now....
 
> However... there's likely to be a far simpler solution.  I don't have
> the data sheets for the SSM2164 on me, but most devices like this have
> a diode junction at their control input, which means if you drive the
> device with a low-impedance voltage you get exponential and if you
> drive it with a high-impedance current you get linear control.  The
> data sheet will tell you for sure.
>
>  -- Don

Mmmm, I don't think so for this one. It's resistors to transistor bases at the 
control input, but then again I've been known to make mistakes in the past :).

Anyway, thanks, Don - you're right, I've got schems for log amps, so I'll try 
it both ways...

- Chris