From: email@example.com (Tom Bruhns)
Subject: Re: linear power supply noise questions
Date: 13 Nov 2002 12:29:19 -0800
NNTP-Posting-Date: 13 Nov 2002 20:29:19 GMT
Yes, I think you've analyzed it correctly. It's really a bridge
circuit; if the resistor ratios are correct, you will "perfectly"
cancel the noise in the output. Beware of errors in the op amp gain
and phase, though. You need very accurate gain+phase to do very
accurate cancellation. Beware of the effects of AC coupling at low
frequencies, and of the op amp gain rolloff at high frequencies.
In the circuit I just did, I used an 0.1 ohm resistor. To take out 1
millivolt peak noise, you then need 10 milliamps in that resistor.
Many op amps by themselves can do five times that, and with an
emitter-follower output, they can do even more. Your net power supply
impedance may not be much less than that without the noise-cleaner
You are right: returning your feedback to the rail is also bad
because of things your circuit puts on that rail, not just because of
what the power supply puts on it. That's true of anything else that
rail supplies, too! You may even get positive feedback that way. All
in all, it's a bad idea. But...you COULD do the following to avoid
the use of a single non-polar electrolytic: use two polarized ones,
one to your feedback return and one to ground, and use a large
resistance to the (-) rail to bias them. You can even just let them
self-bias. And if the input to the amplifier is AC coupled, do you
really need to bias the cap anyway? Won't the voltage across it be
low enough without a bias? Reverse bias on a tantalum is bad, but I'm
under the impression that a few millivolts is OK. (I'm sure I'll be
soundly thrashed if it's not!) Finally, I can conceive of a circuit
(as above, with one more R to common) using a couple lower-voltage
caps with a small bias, and ending up with as much or more net
capacitance with the two in series than you would with a
higher-voltage one of same total (cost,size) to the rails, if you must
avoid any possibility of even tiny reverse biases.
Walter Harley wrote in message news:...
> In article <firstname.lastname@example.org>,
> email@example.com says...
> > The best way, if you want _really_ low noise, is to use a circuit like
> > shown at http://www.wenzel.com/documents/finesse.html.
> That's a neat idea! When I first looked at the circuit I thought it was
> sensing the voltage drop across the series resistor, which would have
> meant that there was positive feedback actively impairing the load
> regulation. But I see that's not the case; the series resistor is there
> just so that there's something to shunt "against," and the sense input
> is simply the high-pass-filtered supply rail, not the voltage drop. But
> the load regulation will still be impaired by however much the series
> resistor is, correct? Or perhaps by slightly less, since there's
> actually a little bit of negative feedback due to the bias being taken
> after the series resistor (the more the gain, the less the negative
> feedback, if I understand).
> In my own case, the net effect would still be kind of bad, since my
> output signal would then show up on the power supply line and cause
> asymmetric gain reduction as noted by another poster; considerably more
> so than at present because of the much-higher source impedance of the
> supply with this system. Still, I might be able to use this idea
> elsewhere. Seems like very clever engineering.