SV: Re: [sdiy] Expo converters, lin log etc.

[Magnus Danielson]

From: Karl Ekdahl <elektrodwarf at yahoo.se>
Subject: SV: Re: [sdiy] Expo converters, lin log etc.
Date: Wed, 31 Jan 2007 23:06:00 +0100 (CET)
Message-ID: <804964.79259.qm at web26204.mail.ukl.yahoo.com>

> Hi all,
> 
> > Not really. A "normal" VCO is exponential and
> > therefore an exponential function
> > is needed. A freq-to-CV converter is logarithmic in
> > its frequency to CV
> > conversion for a "normal" VCO.
> 
> So, i've got two answers, "normal" is linear and
> normal is exponential.
> 
> Basically, if i input 1V and get 100hZ, i then input
> 3V and get 500hZ it'd be assumed that inputing 2V
> should give 300hZ - that's a linear response if i'm
> not mistaken, am i correct in my thinking?

Indeed. You get 200 Hz per Volt, and you have a -100 Hz offset.
Usually you attempt to be at 0 Hz offset.

> > For some reason the terms is widely missused.
> 
> I figured there was something fishy...

Indeed there is! ;O) You where right to figure something was fishy!

> >        U
> > f = f 2
> >      0
> > 
> > where U is the input frequence. The base 2 is
> > natural since you want to double
> > the frequency for the rise in voltage of 1 V. f0 is
> > the frequency of the
> > oscillator with the net sum of 0 V.
> 
> Okay awesome, what if i want to do some
> reverse-engineering; basically i'm experimenting with
> this oscillator which i'm unsure of what kind of
> response it has, i've measured a couple of voltages
> and the resulting frequencies and i'm unsure wheter
> the response is in fact exponential or what it is.

Simple. Raise the voltage 1 Volt, does it double the frequency? OK, if it did,
will another 1 Volt increase also double the frequency? If it did, you have an
exponential responce of 1 V. If you think there is another exponential response
just raise the voltage until you double the frequency, then raise the voltage
by the same amount and if you doubled frequency again you do have an
exponential response and you also know by how much.

You can naturally do this with more finess and precission, but it should work
well enought to make the distinction.

> > The transistor has the "right" properties, i.e. the
> > exponential function but
> > with incorrect values. For most practical matters it
> > is:
> >          qV
> >         (--)
> > I = Is e kT
> 
> I gave this a shot a while ago, but i've forgotten
> most of the math skills (i once had) necessary to
> understand this. 

In that case just see the transistors (when rigged up correctly) as some
suitable function y = f(x) and you just need to scale your input to x and your
y to your output and the magic just happends.

> > It is not too hard to have those strange transistor
> > to have some 15-20 octaves
> > of nice exponential response. It is still a valid
> > method even if it may feel
> > strange. Infact, I'm not fully sure that I
> > understand all that happends in
> > there, but for all practical reasons it does the
> > right thing and that is what
> > matters in the end.
> 
> That's my problem; i have a hard time accepting "black
> boxes" until i understand the underlying functions -
> at that point i might as well just forget about it
> because then it is "valid" in my head.

Well, that is how I am too. Most of the time, but I tend to recall the overall
plot. :)

There are a number of application notes which you could read which dwelves into
this, but the formula I gave above is correct except for a -Is term which will
be neglible for all practical purposes in our synths.

The PMI (later Analog Devices) app-note used to be my favorite.

Cheers,
Magnus