47uf vs. 100uf-470uf power supply filtering.

[panterafanatic]

2. I = C*dV/dT (the current into/out of a capacitor is equal to the capacitor times the rate of change of voltage in the cap in volts/second)

who knew calc class would actually teach me something. i hated learning derivatives

[slideman82]

Dereivatives are not so difficult... integration is really a PITA!

[aziltz]

When in doubt, integrate by parts

or punt.

[slideman82]

That's because you don't know Fourier or Laplace transformation with integration... maths is such a huge plain...

[Cliff Schecht]

That's because you don't know Fourier or Laplace transformation with integration... maths is such a huge plain...

The Fourier transform is a function of the integral from -infinity to infinity of f(x)e^(j*w*t) where f(x) is the function being analyzed (sin, cos, etc..), w is in radians per second and t is what you're integrating over (-infinity to infinity). The Fourier transform is just a way to represent a periodic signal (any repeating signal) in terms of frequency vs. amplitude instead of time vs. amplitude. You can look at a Fourier series as the time-independent representation of a repeating signal. The math can get a bit hairy yes, but it is essentil to being able to understand what the Fourier transform is showing you. Conversely, the math doesn't mean anything if you don't understand what is being modeled by the math.

[Guitar_Ninja]

So all other things being equal, is there an ideal corner frequency you should be aiming for when attempting to filter out power supply noise?

Also, when measuring the current draw of a pedal to make these calculations, should you include the added current draw of an LED if you're going to use one, or just the basic circuit itself? I've been attaching my LEDs directly to the power supply before the current limiting resistor leading into the circuit proper. So will that affect anything?

[tempus]

Also, when measuring the current draw of a pedal to make these calculations, should you include the added current draw of an LED if you're going to use one, or just the basic circuit itself?

Yes. If the LED is part of the pedal circuit (which I assume it is), then it too will draw current when it's on.

[R.G.]

So all other things being equal, is there an ideal corner frequency you should be aiming for when attempting to filter out power supply noise?

No, unfortunately not.

Well, OK, I said that wrong. The ideal corner frequency is zero Hz. This is, of course, impossible.

The corner frequency of a filter is the frequency where the signal drop to half of it's power, 0.7071 of its voltage, which are two ways of saying the same thing. A simple R-C filter will then have a falloff beyond the corner frequency of -6db/octave for each added rolloff section. So two RCs fall at -12db/octave, three at -18db/octave, etc. Since what you're trying to do is reduce (at least!) rectifier ripple to inaudible levels, you're generally looking for at least -60db of reduction below your signal level in the circuits powered, or at least that much after the power supply rejection ratio of the circuits. Since the raw signal has a peak-to-peak voltage of the entire size of the DC produced by the rectifier/filter, and since the power supply rejection of simple transistor stages may be 0db, you are trying to get all the suppression you can get, and then praying for more. Since the falloff of the skirts simply goes down at the same rate forever (for the purists: neglecting the rise in parasitics at high frequencies that happens in the Real World), but never reaches infinite attenuation, it follows that you never suppress it all. You can only get to "good enough".

There is a class of LC (not R-C or R-L) filters and active filters that either mimic or implement the same thing, depending on how you look at it, that have infinite attenuation at some specific frequencies in their stop band. The Cauer/Elliptical filters do this. It's considered elegant to put one of these infinite-rejection points right on top of the frequency you're most wanting to suppress. This works well with high frequencies where you can afford either the inductors or the circuits. At power line frequencies, the inductors are too big, non-standard values, and expensive to do. You can do this with active filters, of course, but then you're trying to produce high power active filters. It's much simpler to just build a power supply regulator, or change your powered circuit to give you some kind of power supply rejection.

Also, when measuring the current draw of a pedal to make these calculations, should you include the added current draw of an LED if you're going to use one, or just the basic circuit itself? I've been attaching my LEDs directly to the power supply before the current limiting resistor leading into the circuit proper. So will that affect anything?

If it draws DC out of the power supply, it enters into making power supply ripple. Connecting LEDs directly to the power supply is another way of saying "putting a filter between the LED and the circuit", which is good. How good? As good as the power supply noise suppression of your filtering.

[Ripthorn]

I really like elliptic filters for the reason R.G. mentioned.  A similar thing can be done with a simple Chebyshev filter, though you are limited to only one or two notches.  Of course, puting one of the notches over your frequency will only do a good job if your noise has a higher Q than the notch in your filter.

As for using inductors, you would simply stick the inductor in series with the power supply as opposed to parallel, i.e., instead of shunting the cap to ground, the +V is fed through the inductor.  Of course, the problem with inductors is that they are big, expensive, and have a time required to "charge up" which means that if you have a sudden spike in the current draw, there will be a time lag and probably some sagging in your circuit performance.  I personally really like trying to use inductors in circuits, but one way around the very large size, etc. is to use an inductance mimicking ciruit that uses OTA's.

Short and long of it is much like everyone else has said, there are two main reasons bigger caps are better:

1. Greater ripple rejection
2. Better "storage" for current demand spikes

[m_charles]

So going by RG's first post, For a 47uf cap, ideally a 200 ohm resist would do the job because I'd want to raise the resistor if lowering the cap?
As opposed to a 100.

[Guitar_Ninja]

Thanks R.G. I think that makes sense. 

So I've been looking at this online calculator to make my calculations on some current builds. It's for tube amp power filters, but it should be the same for pedals, should it not?

http://www.pentodepress.com/calculator/RC-ripple-filter.html

[Paul Marossy]

I just drew up a schematic for the BBE Green Screamer the other day in order to properly mod one. The first thing I noticed when I pulled off the bottom cover is the 1000uF filter cap! And there is no series resistor, just that big cap.

There is more than one way to skin a cat, as they say. 

[tempus]

I suppose we also have to keep in mind that that series resistor is going to limit the amount of current that can be drawn from the supply.

Everything comes with its price...

[R.G.]

I suppose we also have to keep in mind that that series resistor is going to limit the amount of current that can be drawn from the supply.
Everything comes with its price...

Yep. Like I said:

1. V = I*R (that is, the voltage dropped across a resistor is the current times the resistor)
...
If your pedal pulls a current I of 20 ma, and you put a resistor in front of it between it and the power supply then the voltage the pedal sees is going to be less. It will be V  = 20ma times R less. A 100 ohm resistor will make for two volts less! A 10 ohm will make for 0.2V less. And if your pedal pulls 1ma with the bypass LED off and 20 ma with the bypass LED on, then a 100 ohm resistor will change the power supply from 8.9V to 7V when you switch the bypass. A 10 ohm will make it change from 8.99 to 8.8 when you switch.

[lowend]

This is the best thread I've found on the topic so please excuse the update. I don't pretend to understand the concept so here's the thing:

There's no resistor possible with the valvecaster as it's a high current draw. What's a good cap to put across that will filter any power supply hiss or hum out?

[caress]

I suppose we also have to keep in mind that that series resistor is going to limit the amount of current that can be drawn from the supply.
Everything comes with its price...

Yep. Like I said:

1. V = I*R (that is, the voltage dropped across a resistor is the current times the resistor)
...
If your pedal pulls a current I of 20 ma, and you put a resistor in front of it between it and the power supply then the voltage the pedal sees is going to be less. It will be V  = 20ma times R less. A 100 ohm resistor will make for two volts less! A 10 ohm will make for 0.2V less. And if your pedal pulls 1ma with the bypass LED off and 20 ma with the bypass LED on, then a 100 ohm resistor will change the power supply from 8.9V to 7V when you switch the bypass. A 10 ohm will make it change from 8.99 to 8.8 when you switch.

that formula is very useful indeed.

[lowend]

Stuff it. Something between 200 and 500uF will be going on the order. Off to go shopping.

[amptramp]

I have been waiting for R.G.'s patient explanations, but I use one equation that seems to work that I haven't seen yet:

C * V = Q = I * T

where C = capacitance in farads
          V = voltage change in volts
          Q = charge in coulombs
          I = current in amperes
          T = time in seconds

The rectifiers only top up the voltage on the capacitor when their voltage is above that of the capacitor since the diodes only conduct current in one direction.  At 60 Hz with fullwave rectification the capacitor is topped up every 8.333 milliseconds, which becomes the value of T.  Let's say the effect takes a constant current of 20 mA and the capacitor is 220 uF.  Rearranging the equation, the voltage ripple due to hum is:

(I * T)/C

or (2E-2 * 8.333E-3) / 2.2E-4

=7.575E-1 or 0.7575 volts

or you could simply use:

http://www.duncanamps.com/psud2/index.html

This level of ripple may be tolerable for some circuitry using op amps with high power supply rejection.  This would possibly be the case for an equalizer but certainly not for something like a fuzz face.  If the wall wart has a capacitor of, say 900 uF and the effect has a capacitance of 100 uF, one tenth of the pulse current from the diodes will go down the line to the effect if there is no series resistor.  The pulse current is high because it exists for a short time but must carry the average current.  The resistor separating the wall wart and the internal capacitor reduces the peak of this current.

[lowend]

Thanks. I'm not quite following but I have to get to work. I'll just point out the valvecaster takes 300mA for the valve heaters and some more for the rest. It doesn't have a rectifier so I'm lost.

[merlinb]

Thanks. I'm not quite following but I have to get to work. I'll just point out the valvecaster takes 300mA for the valve heaters and some more for the rest. It doesn't have a rectifier so I'm lost.

The rectifier is in the wall wart / power brick. The valvecaster demands so much current that you would probably need a huge capacitor to reduce hum to negligible levels. You could try 1000uF perhaps, provided the wall wart is rated for more than 500mA.

Ultimately though, for a project like that, you should buy a voltage-regulated wall wart. That will give you a clean DC supply without worrying about huge smoothing caps (a small cap, say 100uF, could still be added for peace of mind)


Incidentally, if you built this version of the Valvecaster (page 1)
http://www.aronnelson.com/DIYFiles/up/12AU7-6111_Valve_Caster_Summary_Rev002.pdf
then the current demand is only 120mA,  not 300mA.