[sdiy] AC coupling caps on MS20 clone
lanterma at ece.gatech.edu
lanterma at ece.gatech.edu
Sun Aug 8 00:19:51 CEST 2010
I think you just want R1 in your first sentence?
Sent from my iPhone
On Aug 7, 2010, at 5:10 PM, "David G. Dixon"
<dixon at interchange.ubc.ca> wrote:
>> If you want switch in front panel, this solution implies connect a
>> cable
>> in a high impedance part -the inverting opamp input- when ideal
>> design pcb
>> wants short traces at that points.
>> I don't see the need.
>>
>> It would be better place the capacitor on the input or between two
>> stages.
>> So I would say between the sum inverting amplifier and the 30K input
>> ssm2164 resistor. (I'm supposing the schematic)
>
> I believe that the impedance at the inverting input of an inverting
> opamp is
> R1 || R2 (where R1 is the input resistor and R2 is the feedback
> resistor).
> Hence, if a blocking capacitor C is inserted between R1 and the
> inverting
> input, then this impedance becomes (-j/wC + R1) || R2 which, at DC
> (w = 0),
> becomes simply R2.
>
> If this were a non-inverting opamp, then you'd be spot on. (Indeed,
> that's
> the issue with a sample-and-hold cap-and-follower, where the high-
> impedance
> trace should be kept as short as possible, and even surrounded with
> a guard
> trace). In this case, a resistor to ground would be required to
> shunt off
> any current, and the impedance would then be that of the shunt
> resistor in
> parallel with the cap.
>
> You are confusing the impedance into the inverting input (which is
> very
> large indeed) with the impedance of the trace connected to the
> inverting
> input (which is a parallel combination of all of the impedances
> leading away
> from the trace: the input resistor (or cap), the feedback resistor,
> and the
> inverting input (which can be ignored because it is so large
> compared to the
> other two).
>
> (If I'm smoking crack here, then someone please correct me!)
>
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