This seems to be the optimal way of reducing capacitor size in power supply filtering but I'm unsure how to apply this to a high voltage system of several hundred volts.
Right now I'm looking at using a capacitor multiplier circuit to achieve low ripple but the kanner cap seems better suited. I'm unsure of how to apply this to a high voltage supply though. Maybe a few of you guys know how?
This article shows some analysis:
Regards, Allan
It might be a good time to review active noise compensation methods in general. The function is that of a 'nullor' and is used in compound amplifiers and regulators where the bandwidth is within the capabilities of the active circuit elements.
Passive methods are also possible, where the noise source is known and can be harnessed to produce inverted signals for passive cancellation at higher bandwidths.
High voltage nodes can coupled capacitively to the correction circuitry; correction circuits can even 'drive' the low voltage end of the main filter capacitor, over it's limited compliance range.
RL
e w
This doesn't seem to have anything to do with the kanner circuit.
The problem I'm having with high voltage is that almost all IC's are for low voltage of around a few voltage to at most 20V with a few opamps reaching as much as 80V.
The Kanner circuit is a nullor that forces current into the regulated output in order to reduce ripple. The error amplifier and driver are low-voltage circuitry tied to ground. Input and output coupling to the load is largely capacitive.
Note that if 1mA of ripple is present at the regulation point, and the ratio between filter capacitance and coupling capacitance is 10:1, then the driver of the coupling capacitor will have to produce 10x the normal output ripple voltage at it's end of the coupling capacitor, in order to compensate for the same current variation.
Do you have access to a spice simulator, like LTSpice4?
This is not a low loss circuit and is only suited to the compensation of already filtered nodes. Linear amps driving totally reactive loads suffer from the same losses that they would driving a short circuit, due to the reactive phase difference of voltage and current in the load.
RL
Both circuits involve reducing noise on a supply rail by AC coupling the noise of the rail into an inverting amplifier, then coupling the output of the amplifier back into the supply rail. The only significant differences that I can see are:
- the kanner circuit uses AC coupling on the output of the amplifier
- the wenzel circuit uses a series resistor on the output of the rail; the kanner circuit doesn't have an explicit resistor here - it relies on the output impedance of the previous stage.
In any case, you might find more information here:
Regards, Allan
My favourite method is to lift the cold end of the supply from ground, drive it with an op amp, and use a cap from the hot end to the summing junction (suitably protected, and with suitable DC feedback for stability). That way the op amp jiggles the bottom of the supply to keep the top still. Works amazingly well.
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal ElectroOptical Innovations 55 Orchard Rd Briarcliff Manor NY 10510 845-480-2058 hobbs at electrooptical dot net http://electrooptical.net
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