[sdiy] true VCOs, expo ?

[René Schmitz]

Hi Tony, Aaron and all,

Aaron Lanterman wrote:
> 
> Well... the distinction gets a bit pedantic... You could replace that 
> 100k resistance with some kind of other current source and happily go 
> along your merry way.

I beg to differ. The circuit as such is voltage controlled, and doesn't 
lend itself to easy current control.

The divider at the + input establishes a Uc/2 node voltage.
The opamp tries to maintain equal node voltages at both inputs. (Opamp 
golden rules.)

Hence when the transistor is switched off, the current running into the 
- node (and hence getting integrated) is
I1 = (Uc-Uc/2)/100k=(Uc/2)/ 100k.

If the transistor is on, you can say that an additional
I2 = (Uc/2) / 50k

is pulled out of the - input (which again is held at Uc/2), with the net 
result that the total current is then I1-I2, which is -I1 (since the 
resistors have 1:2 ratio, the currents are I and 2I).

What you really need to control this via an expo is two expo circuits, 
one that delivers twice as much current as the other, and one that sinks 
(the one with the larger current) while the other one sources current. 
Then you could switch the larger one in and out. Watch out for 
compliance of the sources. The + input of the opamp then can be tied to 
a constant potential.

Maintaining precise 1:2 relation could be difficult, so symmetry might 
shift somewhat, also you have more components you need to thermally 
couple for temperature compensation. But then its far easier to use an 
OTA to drive the integrator, using it as a "switchable current mirror". 
I.e. the standard OTA triangle core.

> If you think about most voltage-to-voltage op amp circuits with the + 
> terminal at some fixed voltage, you see that the input impedance more or 
> less is turning a voltage to a current, and the feedback impedance is 
> turning that current back into a voltage.

Yes, but that isn't the case here.

> I like to tell my students that while the most commonly known op amp 
> circuits represent quantities as voltages, they do the actual 
> computation via currents. Hence, you'll almost never see me use 
> mesh-current analysis in my class. It's all node-voltage analysis, since 
> I think more insight is to be gained that way.

I think the important thing is to know when to apply which view on a 
particular problem. Some things are easier if you think of currents, 
some aren't.

Cheers,
  René

-- 
uzs159 at uni-bonn.de
http://www.uni-bonn.de/~uzs159