On Thu, 27 Apr 2000, Dan Michaels wrote:

> Russell McMahon wrote:
>
> >1.    It's vital that your op-amps have good gain at the common mode
> >frequency.
> >ie - even if you are amplifying a dc or low frequency signal, if you have
> >digital (eg clock, uP data, SPI signals etc) appearing as common mode
> >signals then your op-amps MUST be able to reject these actively . ie you
> >should be using amps with a 10MHz plus gain bandwidth and really the more
> >the better! A low gbw opamp will pass the common mode signals as noise in an
> >uncontrolled way.
> >
> >2.    Consider using parts like the AD7730 or the newer AD7731 with internal
> .....
> >
>
> My present opamp, TLE072, has 10 Mhz GBW but only 20 dB PSRR at 1 Mhz,
> and I've been looking for a better drop-in replacement. Natl has the
> LM61x2 series, going to 75 Mhz GBW, but amazingly, the PSRR on every
> one of these goes into the mud (ie, <= 20 dB) right near 1 Mhz <-- this
> seems to be a magical number.

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. 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.

Scott