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Thread: Power supply filtering for ripple attenuation but also as HPF

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    Power supply filtering for ripple attenuation but also as HPF

    Greetings all,

    I think I understand pretty much everything in here:
    The Valve Wizard

    But one thing is unclear: what is the point of calculating the corner frequencies of the RC (or LC) filters that emerge? Obviously if the capacitors are so small that these frequencies become too large (>50Hz for example), then we have attenuation of bass... But in the text, it says "the only frequency we really want to pass is 0Hz or DC, so we just make the cut-off frequency as low as possible, often below 1HZ."

    Is this the same as saying we want to attenuate the ripple as much as possible?

    Because I think that having several RC (or LC) filters in series, the attenuation will be multiplicative (cumulative), and i don't see why the corner frequencies in question should be below 1Hz, as long as they are below audio frequencies.

    So is there some other meaning to what ValveWizard wrote?

    And as a further question, perhaps many times asked, what amount of ripple in mV are you guys aiming for? I know that the sensitivity of the speaker, OT and whether we have PP or SE power amp all affect the answer.

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    Senior Member Malcolm Irving's Avatar
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    Quote Originally Posted by Habbe View Post
    Is this the same as saying we want to attenuate the ripple as much as possible?
    Basically, yes.
    The lowest frequency component of the ripple is at 100Hz (or 120Hz depending on the supply frequency) so I suppose the factor that is really of greatest interest is the attenuation at that frequency. The corner frequency gives us a guide to that.
    Each amplification stage amplifies any ripple present in the B+ of the previous stage, and we want the final ripple at the speaker to be inaudible. So it’s efficient to progressively attenuate the ripple as we work back from the power stage to the first pre-amp stage.

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    Last edited by Malcolm Irving; 11-10-2016 at 10:13 AM.

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    Quote Originally Posted by Malcolm Irving View Post
    Basically, yes.
    The lowest frequency component of the ripple is at 100Hz (or 120Hz depending on the supply frequency) so I suppose the factor that is really of greatest interest is the attenuation at that frequency. The corner frequency gives us a guide to that.
    Right, but we can also compute directly the attenuation?
    https://www.ampbooks.com/mobile/ampl...ripple-filter/
    I don't know the formula though (but I have ordered the book which it references). This calculator also computes the corner frequency, which I guess is only used to check that it is smaller than 20Hz (or whichever we want, but not necessarily below 1Hz).

    Quote Originally Posted by Malcolm Irving View Post
    Each amplification stage amplifies any ripple present in the B+ of the previous stage, and we want the final ripple at the speaker to be inaudible. So itís efficient to progressively attenuate the ripple as we progress back from the power stage to the first pre-amp stage.
    Right. I think that for SE power amps the B+ ripple goes to the speaker directly trough the output transformer so it should be small in mV already there.

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    Senior Member Malcolm Irving's Avatar
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    Agreed.

    The other thing that the filter caps provide is decoupling between gain stages to prevent instability. To achieve decoupling, we need the attenuation of the power supply filter stage to be greater than the gain (in relation to B+ disturbances) of the associated amplification stage(s) at any frequency.
    This is not usually a problem at audio or higher frequencies, where the filter provides plenty of attenuation, but can be an issue at the sub-sonic frequencies associated with motor-boating.

    EDIT: To be more precise we need to think about phase change as well as gain of course.

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    Last edited by Malcolm Irving; 11-10-2016 at 11:02 AM.

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    Check the list of several reasons the VW gives. Then think about how much the lowest audio frequency of interest is a factor in each of them.

    Quote Originally Posted by Habbe View Post
    Greetings all,


    Because I think that having several RC (or LC) filters in series, the attenuation will be multiplicative (cumulative), and i don't see why the corner frequencies in question should be below 1Hz, as long as they are below audio frequencies.

    So is there some other meaning to what ValveWizard wrote?

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    Quote Originally Posted by Habbe View Post
    Right, but we can also compute directly the attenuation?
    A single RC will attenuate at a rate approaching 6dB per octave or 20dB per decade above the corner frequency. If we make it easy and chose 3.3k and 47u the corner frequency is 1Hz. 1Hz is two decades below 100Hz so the attenuation at 100Hz will be about 40dB which is a factor of 100. In other words if you want to attenuate by a factor of 100 the corner frequency must be 1Hz (or lower)

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    Last edited by Dave H; 11-10-2016 at 07:05 PM.

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    Quote Originally Posted by Habbe View Post
    So is there some other meaning to what ValveWizard wrote?
    1Hz is sometimes a very convenient number because it 'disappears' from equations. For example:

    C = 1 / (2 * pi * f * R)

    becomes:

    C = 1 / (2 * pi * 1* R)
    C = 1 / (2 * pi * R)

    So if you're going to pick some arbitrary 'below audio' frequency, then 1Hz is a good place to start.

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    Quote Originally Posted by Merlinb View Post
    1Hz is sometimes a very convenient number because it 'disappears' from equations. For example:

    C = 1 / (2 * pi * f * R)

    becomes:

    C = 1 / (2 * pi * 1* R)
    C = 1 / (2 * pi * R)

    So if you're going to pick some arbitrary 'below audio' frequency, then 1Hz is a good place to start.
    Or if you just love simplicity and very good filtering, just use C = 1./R. Then make C a bit smaller so that you can afford it!

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