Scott Dattalo wrote:
>
>PSRR - is the Power Supply Rejection Ratio. Right? If you have NO noise on your
>power supply then you can live with such a small value. But it's one of those
>theory vs. practice things. So the question I have is, can you reduce the noise
>on your power supply and (this is equally important) reduce the power supply
>source impedance? The former is probably obvious, however the latter is
>important because the output stage of your opamp will cause current to flow
>through the power supply. If the power supply has a high impedance, this
current
>will be converted to a voltage and cause the supply to ripple synchronously to
>the opamp's output. If there's a lot of gain in the circuit, then you'll get an
>oscillator! (Don't ask how I know :).
>
>It's fairly straight forward to calculate the noise voltage that would be
>induced on your power supply leads due to the opamp switching.
.....

Hi Scott, thanks for your input. Yes, to all of your questions - re
PSRR and P/S source impedance. However, I wouldn't think the opamps
themselves [at least in my case] are responsible for creating much
P/S buss ripple, since they move a relatively small amount of current
- a few milliamps.

In general, putting .1 uF and 10-20 uF caps on v.reg high and low
sides is the usual way to lower P/S impedance. However, you are also
right about the "theory vs practice" thing. Caps are not ideal
devices, they have series R and L. And all leads/traces have
inductance. This is why they always say put the bypass caps as
close as possible to the Vcc/Vss pins.

But no matter how much bypassing you seem to have, and how small
1/(2*PI*F*C) may seem to be, you can always measure tons of hash
on digital busses - at least in my experience. And if the analog
busses pick up some of this noise, because they are on the same
pcb with the digital, run for inches and are inductive, and the
bypass caps aren't ideal, then you need good opAmp PSRR to help
deal with whatever pickup there is. PSRR = 20 dB [as in most std
opAmps at 1 Mhz] is only 10:1 rejection, so a mere 20 mV of buss
noise would translate into 2 mV of opAmp noise.

This is how I see the problem.
=============

>It's fairly straight forward to calculate the noise voltage that would be
>induced on your power supply leads due to the opamp switching. The 'filter
>capacitance' along with equivalent series resistance (esr) can be used to
>approximate the filter's impedance. If you look at the max current swing in
your
>output and say that it's a sinusoidal current source driving (or sinking) the
>filter then the voltage swing will be approximately:
>
>  V ~ I * sqrt(R^2 + (1/(2*pi*C))^2)
>
>Which says you want zero R and infinite C...
>
>But, even if you did this filtering perfectly, you need to also look at the
>other side of the filter. Ground is not always ground. If the ground to which
>this filter is referenced is bouncing around like a trampoline, then it will
>couple right into the opamp's supplies. Bummer.
>

As mentioned above, filtering is never perfect, since real caps aren't
perfect, plus all leads have some inductance. And correct, gnd is not
always gnd. So I guess the main solution here is to use wide traces,
physically separate analog and digital subsystems, don't track digital
lines over analog traces and gnds, and return the analog and digital
gnds separately to the power source with only a 1-point connection
between the two.

This last is interesting because some people say the 1-point
connection should be made at the power entry point, while others
say it should be directly under the A/D converter - ie, the place
where analog and digital meet - and that the analog currents should
be conducted thru the digital gnd plane, but via a physically
separate route than the high-frequency/high-current digital signals.
These things seem to be a little difficult to get a handle on.

best regards,
- Dan Michaels
Oricom Technologies
===================