freq= 1/(2x pi x R x C) or 1/(6.28 x R C)Kingston said:I've a 100Mhz scope with 10x probes, Rigol DS1052E that got universal praise at eevblog upgraded to 100Mhz by their recommendation. What size resistor should there be in series with the opamp output and probe? I've been measuring the output pin directly of course, didn't know any better...
As far as me playing around with exotic opamps that began with blind swapping, it seems I'm finally asking the right questions. I also suspect it was the better decoupling implementation that gave me these improvements, little to do with opamps themselves. 0.001% THD is still several zeroes away from what the better opamps are capable of in ideal environments. Perhaps after this thread I'll gravitate towards new designs where I will achieve that.
Samuel Groner said:Capacitive loading at the inverting input. Cancel with a feedback capacitor; for general audio work, I'd size the time constant with the corresponding feedback resistor for about 300 kHz (this assumes a unity-gain stable opamp).
Is this the compensation/integration cap parallel to the feedback resistor? How do I calculate the time constant and the corresponding cap size?
solving for C
C= (1/freq) /(6.28xR)
note: as Sam mentioned this is only applicable for unity-gain stable opamps, since it could actually cause issues with decompensated opamps. In theory you can add a cap across a decomp opamp feedback R also but that cap needs to be smaller than the cap to ground at - input in the proper ratio to satisfy the minimum gain requirement to insure stability, so generally not worth the effort, unless you have unusually large input pin capacitance.
Probing an output pin with a scope probe is generally not a problem unless the circuit is really marginal. The x10 probe has less capacitance. A small resistor (couple hundred ohms) in series should be adequate. As Ricardo mentioned too large of a resistor could form a LPF pole with the probe capacitance and hide very HF oscillation.
JR