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-30 dB cut box - stereo signal

Discussion in 'Electronic Repair Modifications DIY' started by info@soundcomposer.dk, Aug 17, 2011.

  1. info@soundcomposer.dk

    info@soundcomposer.dk Active Member

    Hey all!

    Recently I got a diagram from an audio technician at a company telling me how to make my own -10 dB cut and -20 dB cut stereo box. But now that I made them did I realize that I need one single box with around -25/-30 dB cut. I can see from the numbers I got from the technician that it's not just doubling the ohm numbers to go 10 dB extra.

    So my question is - how do I calculate to know exactly which resistors I need to make either a -25 dB or -30 dB cut?

    Here are the numbers from the technician:

    -10 dB - 820 Ohm + 270 Ohm
    -20 dB - 5,6K Ohm + 560 Ohm
    -25 dB - ???
    -30 dB - ???

    And the diagram for the box:


    Thanks all!! :)
  2. bouldersound

    bouldersound Real guitars are for old people. Well-Known Member

    Hint: note the ratio between the parallel and series resistors. It's 3:1 for -10dB and 10:1 for -20dB. I'd be shooting for something like 30:1. Also notice how the parallel value doubles going from the -10 to the -20. Just off the top of my head, assuming the values you give are correct, I would try a 27M on the series and an 870ohm on the parallel for -30dB.
  3. MrEase

    MrEase Active Member

    What a mess this lot is. None of the figures given actually take into account what the load impedance on the driver should be. The tech's figures for a 10dB pad would give ~12dB attenuation with an input impedance of just over 1kOhm while the figures for the 20dB pad would give ~21dB with an input impedance of just over 6kOhm. Both these calculations are unloaded figures.

    Bouldersounds figures would actually give you ~90dB with an input impedance of 27MOhm. I think he meant 27kOhm which would be much much closer! :<)

    Let's use maths instead of intuition for a more reliable result...

    Lets target an input impedance of around 10k so make Rseries = 10k and Rshunt = 330Ohm. With a reasonably high input impedance on the output (say 10k or more), then attenuation is 29.9dB and input impedance is 10.33 kOhm with no load.

    For 25dB, same 10k and 560Ohm will give 25.5dB at 10.56kOhm unloaded. Maybe change the 560 to 620 for lower impedance loads.

    For completeness for 20dB I would use 9.1kOhm and 1kOhm for a 10.1kOhm input impedance and consider changing 1kOhm to 820Ohm.

    For 10dB, use 6k2 and 3k6 to get 8.7dB at 9.8kOhm input but with a 10kOhm load, this would become 10.5dB with an input impedance of 8.8kOhm.

    Note that with higher shunt values, load impedances are more significant hence actual attenuation and input impedance values will vary more. Larger attenuations are more immune to this effect due to the lower shunt values. This may account for the lower values provided by the tech. but this also results in a very low input impedance - in this case possibly too low for some outputs. For instance it would NEVER work on a passive guitar output. Then again neither would my values as the input impedance is still far too low!
  4. MrEase

    MrEase Active Member

    Aha, I didn't answer the original question!

    For a quick unloaded calculation, take the shunt resistor value, divide it by the sum of both resistors, take the logarithm and then multiply by 20.

    I.e. for 30dB, using my values of 10k and 330Ohm, you get 330/10330 = 0.03194.... Log of this is -1.4955... Times 20 = -29.9117... which is approximately -30dB.

    You can of course reverse these calculations to work out values needed for any particular attenuation.
  5. info@soundcomposer.dk

    info@soundcomposer.dk Active Member

    Thanks MrEase!!

    I know nothing of the mathematics of electronics so that's why I ask here. I only know how to use a soldering iron and build a sound studio :)

    Thanks for your very precise answer. I'm though a bit confused as how to use it in my case. What I have is a condenser mic going into a field mixer that I connect with a video camera with a stereo-minijack input. The dB cut box is supposed to take the signal from the field mixer output, lower the level, and send it further to the camera. This is because the mixer has too high an output to be able to calibrate the input in the camera gain.

    On the specifications of the mixer it says so:

    Azden FMX-20 Field Mixer
    Main Output
    0.5dBu with 600 ohm load 0dBu ref @ 0.775Vrms
    0.5dBu with 1000 ohm load 0dBu ref @ 0.775Vrms

    More info here

    I don't understand why they specify two ohm loads and how should I use this to calculate my dB cut box?

    Thanks for your help and time! :)
  6. Boswell

    Boswell Moderator Distinguished Member

    That series of Azden field mixers has a -50dBu stereo output on a 3.5mm jack designed for feeding the stereo mic input of a video camera. You don't need attenuators with that - just use a minijack stereo cable.

    From what passes for a user manual for the FMX-20:
  7. bouldersound

    bouldersound Real guitars are for old people. Well-Known Member

    Right k, not M.
  8. MrEase

    MrEase Active Member

    Hey, no problem from me but others may not know the difference so I guessed it should be pointed out! This type of slip is so easily done - I did it myself in the clock jitter thread and Boswell kindly pointed out that error.

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