[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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