[sdiy] Re: Leapfrog

[jhaible at debitel.net]

Hi Magnus


> Thus, the amplitude responce (when having excessive deviations) can give you
> a
> hint that this is going on, but since you can have various compensations by
> other defects, it may also not be able to tell you about the problem. So,
> when
> the tools are lacking (if they are lacking, it may be in the operator not
> "asking the right questions") the ear has to take up and listen to it with
> some
> set of reference material (music, test-CDs or whatever).


Yes, and even before listening tests, one should analyse if any inner node
has an excessive signal that might be distorted, even when the output 
looks "flat" for small signal analysis. (Is this what you meant?)

Of course you can also do it the other way round, design some soft
clipping on inner nodes (where the clipping is frequency dependent),
while the overall small signal response doesn't betray what's going
on inside. That's the secret of many good-sounding filters, from
Moog to SSM to many others. But I disgress.

But as for the integrator analysis, I wanted to add this method,
which avoids a lot of the big equations.

Here's how I calculated it (using your 
variable names, for convenience):

Negative input of opamp is 1/2 output because
of equal resistor divider.
Positive input of opamp is the same, asuming
ideal infinite gain.
So we only have one variable, V4.
(The capacitor voltage is 1/2 V4.)

Sum of currents thru input resistor and positive
feedback resistor must be equal to current thru
capacitor and grounded second input resistor.

I_in + I_posfeedback = 
      = I_capacitor + I_grounded_resistor

(V1 - 1/2 V4) 1/(2Ro) + (V4 - 1/2 V4) 1/Ro =
              = 1/2 V4 (jwCo + 1/(2Ro))

Multiply with Ro and voila: 

V1 = V4 * jwRoCo

Best Regards,

JH.

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