Zero-drift amplifiers with negative PSRR

I'm having that problem too, on a 1 volt supply for an FPGA core. I'm trying to get the kids to spin up a bunch of MACs or something, to use more current and stop the switcher from burping. I'm getting 6 ps RMS jitter and I want less; I blame the power supply.

At these current levels, use a linear reg! Or RC filter the amp supply hard.

Looks like that amp chops at 1 KHz, so it could be beating with the supply. But then, the CMRR (fig 21) is all-by-itself terrible at even

1 KHz.

We use some opamps with PSRR that has actual gain at some frequencies.

I think people tend to design elegant low-noise circuits, and then power them from noisy supplies.

--

John Larkin         Highland Technology, Inc 

lunatic fringe electronics

I bodged in a 270uH inductor for the moment, which does a surprisingly good job. But yes, there's an LDO in this circuit's future.

That's an excellent point. Since it's a unity-gain buffer, the problem here is really CMRR, not PSRR.

No kidding. Elsewhere, this design was using the ripply 3.3v rail as Vref!

Cheers, James

Do you really need the 0.65uA supply current? A higher supply current op-amp would work better.

--
 Thanks, 
    - Win

No, I don't. In fact I don't even need the buffers -- I'm not sure why the previous gent felt we needed them. They buffer against a little kick-out charge from the a/d, but an existing filtering cap should suppress that error just fine. Replacing the buffer amps with jumpers would eliminate this one PSRR problem completely. (And save 3 x 650 nA! :)

But boards not in my possession have to be installed and in service in a few days, so changes are inconvenient. The inductor easily drops in to replace a two-pin power-supply-isolating shunt that the previous designer thoughtfully provided.

The real solution, of course, is an LDO.

Yep.

Cheers, James Arthur

snipped-for-privacy@highlandsniptechnology.com wrote: ...

At your current levels, you could filter the supply with an RC with a

1 second time constant.

Depends on the precise definition of CMRR. At 1 KHz, the CMRR of that part is about 70 dB worse than the PSRR.

It's unfortunate that opamps don't have ground pins. All the internal guts are hung off the power supplies.

--

John Larkin         Highland Technology, Inc 

lunatic fringe electronics

I can't. These buffers run all the time and draw nothing. But other loads intermittently draw ~120-ish mA off the same rail.

270uH with 20uF makes a 2kHz filter. That knocked most of the piss out of it.

Easiest and most accurate is just to remove the buffers.

They're really unnecessary, too slow and hi-z to compensate the a/d's charge kick-out anyhow. But the same opamp might be handy elsewhere, when I revise the board.

|\ >----R1----+----+------|+\ | C1 | | >-+--> R2 --- .--|-/ | | --- | |/ | === | '-------' ===

Well, if a unity-gain follower's rail bounces and one input stays fixed, the part sees the rail bounce as a common mode input, right?

Any chopper that's alternately sampling its inputs will see a slewing common mode voltage as a differential mode signal.

Yet another proof for Larkin's Law of Op Amps: Always invert. (Except I can't, here.)

Cheers, James Arthur

Alas, I've got other beasties to feed of this rail, and they're much hungrier. Using an r-c- might apply at revision time, when I could separate those few devices from the power plane and filter them specifically. But I'm more likely to take them out completely, saving power, parts, and improving accuracy too.

Cheers, James Arthir

PSR is input-referred, of course, and so is much worse at higher gains.

Cheers

Phil Hobbs

Right, so if I'd had an inverting unity-gain amplifier with zero CMV, 160mV on the rail suppressed by 70dB PSRR should've made 50uV on the output.

Instead I'm getting 180mV, thus diverting attention to the most-worse CMRR, instead.

Cheers, James Arthur