understanding the cutoff frequency(Pwr supply LPF)&resistor position

[KORGULL]

I have a couple questions in reference to the power supply filtering procedure that is in Anderton's Projects for guitarists book (and has probably been discussed on the forum alot).
This is the deal where you put a 1000mF cap across the pos and neg lines of a wall wart and also put in a 100ohm series resistor.
I understand this to be a low pass filter and when I calculate the cut-off frequency (F=159,000/RC) (F=159,000/100*10000) ...I get F=1.59Hz. Correct??

1)Is this such a low freq because it is designed to cut everything (as close as possible) that is not pure DC?
When I first started thinking about this filter I thought the frequency would be higher - like somewhere above 50-60Hz.
*I think I may have figured out question #1 as I was typing it, but I'll leave it here anyway in case I'm wrong about anything, or someone can add more details.

2) In Anderton's book, he shows two diagrams for the filter - one for wall warts with positive tip plugs and one for negative tip.
For each diagram, the cap polarity is switched around to follow the polarity of the tip/sleeve. Cap positive always goes to power supply positive wire. No problem so far - just what I would expect to avoid popping the polarized capacitor.
What I don't get is why the resistor always remains on the wire going to the plug's tip - regardless of the polarity.
In other words, in one diagram the resistor sits on the positive wire and also connects with the cap +, and in the other diagram it is on the negative wire and connects to the cap neg.
I thought the resistor would always go on the positive wire and connect with the cap positive - whether or not the pos wire and cap pos were on the plug's tip or sleeve.

One minute I think I have answers to these questions, but then I get confused/unsure again.

I hope this makes sense.

[R.G.]

This is the deal where you put a 1000mF cap across the pos and neg lines of a wall wart and also put in a 100ohm series resistor.
I understand this to be a low pass filter and when I calculate the cut-off frequency (F=159,000/RC) (F=159,000/100*10000) ...I get F=1.59Hz. Correct??

Correct - as far as it goes. What's missing here is an accurate concept of what "cutoff frequency" means for filters. Cut-off frequency is defined as the frequency where the signal being filtered is at half power, 6DBv down or 3db power down. What it does after that point is not specified. The rate at which frequencies are cut after the cutofff depends on the order of the filter - literally, how many reactance components are in there. A single RC filter response slopes down at -6db / octave, -20db/decade after the cutoff.

1)Is this such a low freq because it is designed to cut everything (as close as possible) that is not pure DC?
When I first started thinking about this filter I thought the frequency would be higher - like somewhere above 50-60Hz.
*I think I may have figured out question #1 as I was typing it, but I'll leave it here anyway in case I'm wrong about anything, or someone can add more details.

And you've probably answered your own question. With a -20db/decade slope, the noted 1.59Hz filter is 26db down at 15.9hz, and -40db at 159Hz. Since what you're trying to do is to filter out (probably) 100 Hz or 120Hz power supply ripple, the -36 to -40db of attenuation at the power supply ripple frequency is what you're after.

A single RC filter is a voltage divider, but one that varies its attenuation with frequency as the impedance of the capacitor changes.

So to get good removal of power supply ripple, you have to park the cutoff frequency of the filter way down from the frequency you want to filter.

2) In Anderton's book, he shows two diagrams for the filter - one for wall warts with positive tip plugs and one for negative tip.
For each diagram, the cap polarity is switched around to follow the polarity of the tip/sleeve. Cap positive always goes to power supply positive wire. No problem so far - just what I would expect to avoid popping the polarized capacitor. What I don't get is why the resistor always remains on the wire going to the plug's tip - regardless of the polarity.
In other words, in one diagram the resistor sits on the positive wire and also connects with the cap +, and in the other diagram it is on the negative wire and connects to the cap neg. I thought the resistor would always go on the positive wire and connect with the cap positive - whether or not the pos wire and cap pos were on the plug's tip or sleeve.

What's missing there is that as long as the effect is seeing only the voltage across the capacitor, it doesn't matter which power supply lead the resistor is in series with. The wall wart is acting like a voltage source - an ugly one, but a voltage source - in series with that resistor. To the capacitor, it doesn't matter which lead the resistor is in series with. It works equally well each way. And frankly, it would also work as well to split the resistor into two half-value resistors, one in series with each side.

Here's a tricky one for you, and one that's hard to believe in at first. If you have a series string of passive components, as long as you look at voltages and currents only at the ends of the string, not the middle connections, the order of the components makes no difference at all. The internal parts can be arranged in any order at all and you'll get the same voltage across the string and current through it. Obviously if you look at the voltages at the middle points, that's order sensitive, but at the ends, it makes no difference.

[KORGULL]

Thanks a million for filling in the blanks for me R.G., these things are always a little more complicated than I think. I've got some new things to study & memorize here.

Here's a tricky one for you, and one that's hard to believe in at first. If you have a series string of passive components, as long as you look at voltages and currents only at the ends of the string, not the middle connections, the order of the components makes no difference at all. The internal parts can be arranged in any order at all and you'll get the same voltage across the string and current through it. Obviously if you look at the voltages at the middle points, that's order sensitive, but at the ends, it makes no difference.

Yup, I already got my head around that one when I first started really studying electronics (not very long ago ) - and it definitely was not how I first assumed things worked. I was just too dense and inexperienced to actually think to apply the concept in this instance I guess.