AMZ Power Supply Filter alternative part question

[bushidov]

Hi All (and possibly Jack),

I was looking at the AMZ Power Filter Supply page (http://www.muzique.com/lab/filter.htm), which I realize was written back almost 20 years ago (wow!) and it mentioned using a Radio Shack choke, then going into a 220uF electrolytic capacitor. As Radio Shack has long been dead, I was looking at DigiKey for similar parts.

I was wondering if this Bourns 7108-RC would work as an alternative for a guitar pedal power supply filter:
https://www.digikey.com/en/products/detail/bourns-inc/7108-RC/3782691

I know it isn't rated for 4A, but as my whole pedal board doesn't even reach 1A, I figured that would still be good enough. The impedance is a hair lower (1.7mH instead of 1.9mH) at 1kHz, and the max DC resistance is a smidge less at 60mOhms instead of 87.2mOhms, but would this be a good alternative? What things in a common mode choke should I be specifically looking for in this task?

Thanks again for all your time and posts,
Erik

[Vivek]

Are Low Dropout Voltage Regulators better than chokes these days, in terms of weight, cost, ease, ripple rejection ?

[antonis]

What things in a common mode choke should I be specifically looking for in this task?

Other than current rating you can estimate inductance via f = 0.159/SQR(L*C), where f = cut-off frequency (the lower the better), C= shunt cap capacitance (in Farads) and L= series choke inductance (in Henries)

But, as Vivek said, chokes are no more in use for practical reasons..

[amz-fx]

This LC filter is not going to remove 60Hz line noise since the corner frequency of the filter is about 259 Hz, but it can remove higher noise frequencies. It was useful in blocking power supply noise that came from AC motors that switched on/off that cause interference, such as the air con.

You would have to add (a lot) more capacitance to filter the fundamental line frequency. If you are dealing with low currents, then a simple active power supply filter is a better choice for your pedal power needs.

Best regards, Jack

[bushidov]

You would have to add (a lot) more capacitance to filter the fundamental line frequency. If you are dealing with low currents, then a simple active power supply filter is a better choice for your pedal power needs.

I figured as much, as when I did the corner cut math, I got around 260 Hz as well and was thinking I was doing something wrong, as it wasn't 60 Hz.

I was skittish about using an LDO voltage regulator as a means of removing noise, as I've seen that backfire badly before as they suck at rejecting large amounts of input ripple as demonstrated here:

[antonis]

when I did the corner cut math, I got around 260 Hz as well and was thinking I was doing something wrong, as it wasn't 60 Hz.

It always depends on how much attenuation you wish for a given frequency...

e.g. rectified 60Hz mains exhibit 120Hz ripple so a LPF of 60Hz cut-off frequency attenuates that ripple by -9dB (-3dB plus -6dB per octave)
That may or may not be adequate for particular use while is additionaly dependent on reservoir capacitor value & load..

[Rob Strand]

I was skittish about using an LDO voltage regulator as a means of removing noise, as I've seen that backfire badly before as they suck at rejecting large amounts of input ripple as demonstrated here:

In order for any active device to reject ripple the output voltage must be lower than the *dips* of the input ripple.   A linear regulator cannot boost the output back up.   In fact you need to set the output sufficiently low such that,

                 Vout <= Vin_min + Vdropout + Vmargin
                      Vout <= Vin_min - Vdropout - Vmargin                  ; corrected, see Vivek's post below
where      Vin_min is the lowest dips of the ripple
                  Vdropout is a maximum dropout
                  Vmargin is a safety margin above dropout to ensure the regulator maintains performance
                  to ensure enough ripple is rejected.   The performance of all regulators drops as you
                  approach dropout and the specs aren't always consistent.

Believe it or not some regulators are not good at rejecting ripple, particularly low power devices.
They regulate but rely in the input also being regulated.

The active filters, like the one in the video image, rely on the output being lower than the  input voltage in order to reject ripply.   As soon as the dips of the ripple drop below the output they will not longer filter.    The Boss pedals use single transistor version of the active filter.   If you want to reject large amounts of ripple you need to force the circuit to have more drop by placing a second resistor from the base+cap node to ground.

The advantage of the active filters is they don't care about the output voltage like a regulator.  They will reject ripple when the input voltage drops to low values.

Here you can see how the high input ripple causes the first active filter to drop out and the ripple increases.

The second active filter has the output forced to a lower voltage using R3 and handles the higher input ripple
without dropping out.  The ripple remains low.

[Vivek]

Thanks Rob

Great spice simulation!


Is this correct :

Vout <= Vin_min - Vdropout - Vmargin

[Rob Strand]

Vout <= Vin_min - Vdropout - Vmargin

Thanks for checking.     You really made me think about it.  Your equation is correct.
[Corrected above post.]

It might make more sense as,

Vout + Vdropout + Vmargin   <=  Vin_min
--------------------------------------
All positive quantities adding up from the output back towards the input.
Then the actual input must be larger than that so as to keep above the "bad zone".

[Vivek]

So now we can marry these two ideas

Capacitance multiplier
LDO regulator or zener

And get one circuit that does

Great ripple rejection
Regulated supply
High output current
Adjustable output

All at the same time!

(Always assuming we are working in the safe zone specified by Rob's equation)

[Vivek]

e.g. rectified 60Hz mains exhibit 120Hz ripple so a LPF of 60Hz cut-off frequency attenuates that ripple by -9dB (-3dB plus -6dB per octave)

I feel that the attenuation at 2Fc (Twice the cutoff frequency) is -6.9897dB, which can normally be rounded to -7dB

It's a factor of 20.log(Sqrt(1/5))

[Rob Strand]

Capacitance multiplier
LDO regulator or zener
...
(Always assuming we are working in the safe zone specified by Rob's equation)

Whatever the method you have to choose an output voltage low enough to keep out of dropout.

In theory a low dropout design will let you have the highest output voltage.

You can put RC filters before or after a zener regulator.    There's a couple of other tricks,
- Feeding a zener with a current source, which gives low ripple without a cap.
  Low noise designs might filter off the zener noise.
https://i.stack.imgur.com/Joxqz.png
The main point about this version which uses a transistor current source is the dropout
can be kept relatively low without the current source pooping out.

- There's a self bias zener circuit,
https://www.eeweb.com/stable-zener-reference/
You can add an output transistor for more current.
It's possible to reg-jig this for lower drop-out but you need to consider
start-up issues.

Notice how the zener and feedback keep the zener current constant
and so the zener voltage is independent of input ripple.
Where things creep in are PSRR of the opamp.

- There's a version which is the two previous cases in one
  Called the "Ring of two" self biased reference.  Discussed in
  Wireless-world magazine in the 1960's.

  You can see a version of it here,

It's not really set-up for low dropout.

These are all old school designs.

The LDO is the modern day solution, you just need to find one which gives good ripple rejection and some of that comes down to dropping the output voltage low enough.

This post has the links to the old wireless world stuff
https://www.diystompboxes.com/smfforum/index.php?topic=122929.msg1161238#msg1161238

Williams, WW Sept 1966, p456, Letters to the Editor ; *** The first appearance of the "ring of two" idea.
Williams, WW July 1967, p318, Ring of 2 Voltage Reference,  Article

https://worldradiohistory.com/Wireless_World_Magazine.htm
https://worldradiohistory.com/UK/Wireless-World/60s/Wireless-World-1966-09.pdf
https://worldradiohistory.com/UK/Wireless-World/60s/Wireless-World-1967-07.pdf

[Vivek]

I'm sure this is an ancient idea :




(Or use LDO with an adjust pin or feedback pin and use emitter follower/capacitance multiplier with it)
( Or use a Zener instead of a LDO regulator)

Caveat : Always operate in safe zone as per Rob's formula

[antonis]

I feel that the attenuation at 2Fc (Twice the cutoff frequency) is -6.9897dB, which can normally be rounded to -7dB

Can't get your feeling..

At cut-off frequency, attenuation already is -3dB (by definition) with a slope of -6dB per octave..

[Vivek]

I feel that the attenuation at 2Fc (Twice the cutoff frequency) is -6.9897dB, which can normally be rounded to -7dB

Can't get your feeling..

At cut-off frequency, attenuation already is -3dB (by definition) with a slope of -6dB per octave..

Here is a graph of RC filter with Corner Frequency of 60Hz (1uF Cap and 2.65392781316 KΩ resistor)



We can see by the cursor measurements:

-3 db down at 60 Hz  = Fc
-7 db down at 120 Hz = 2.Fc

Here is another view



Here is an interactive Bode Plot generator :
https://www.multisim.com/content/i5giQfciZBMpaNyK82mtXZ/bode-plot-circuit/
https://www.multisim.com/content/i5giQfciZBMpaNyK82mtXZ/bode-plot-circuit/open/



It shows for 5K ohms and 0.1uf = 318.47Hz
-3db at 318Hz at Fc
-7db at 643Hz = 2.Fc


The basic math for first order filters also predict that at twice the cutoff frequency, the attenuation is 20.log(Sqrt(1/5)) ie -7 dB

[ElectricDruid]

I think what's going on here is that the cutoff is a curve that approaches a slope of -6dB/Oct. If you're reasonably far from the cutoff frequency, approximating the curve as a straight line of -6dB/Oct slope works well enough. But close to the cutoff frequency, it's clear that the approximation breaks down, since we move from the passband (slope of 0) to the stopband (slope of -6dB) and the line curves smoothly from one to the other.

[Vivek]

For a low pass, we can empirically consider slope =0  from 0 up till Fc/3

And slope = -6db/octave from 3Fc onwards.

In between Fc/3 till 3Fc, it's better to calculate / simulate the response

[m4268588]

Accurate formula?

log(0.5/sqrt(1+0.5))*20
log(0.5/sqrt(0.5^2+1))*20

[amz-fx]

Here is a quick and simple power supply filter that I drew up for someone not long ago:



If your pedal doesn't need much current then TR1 can be most any NPN bipolar transistor. You don't need R2 either for pedal applications.

This will perform as good or better than linear regulators for filtering out hum/noise.

regards, Jack

[Vivek]

Shouldn't there be a voltage divider near C2, such that Vce is maybe around 0.5V at least, or maybe 1V ?

If Vce is zero, maybe we cannot handle down swings on the ripple

Hope I understood this correctly.