SuperQ;161297 Wrote:
> Yea, single value attenuators are just a resistor network. Would you
> mind posting your circuit and other info to the wiki?
>
> I've had a hard time googling for pre-made balanced attenuators, most
> of what I find is for microphone inputs.
I don't think I should post anything to the wiki before others have had
a chance to review what I've done, so I'll describe it here.
A theoretically ideal resistor network to attenuate a simple
(unbalanced) signal uses three resistors. The circuit is shown in the
attached diagram ("Ideal Attenuation Network"). In principle it should
be possible to juggle the values of R1, R2 and R3 so as to achive the
required attenuation AND maintain the source and destination
impedances. However, solving the simultaneous equations is fiendishly
difficult for my mathematically feeble brain, and my understanding is
that in many cases the solution could call for negative resistances in
one or more places.
So an ideal network isn't really practical. Fortunately for us, we're
dealing with audio signals (ie. very low frequencies), so maintaining
the impedances isn't critical. We can instead use the circuit shown in
the second diagram ("Basic Attenuation Network"). This does not
maintain the source and destination impedances, but provided the values
for R1 and R2 are chosen sensibly, the effect will be insignificant.
Let S be the output impedance of the source component, and D be the
input impedance of the destination component. After inserting the
resistor network, the resulting source impedance seen by the
destination device will be:
1 / ( 1/(S+R1) + 1/R2 )
The destination impedance seen by the source device will be:
R1 + 1 / ( 1/D + 1/R2 )
And the attenuation in dB will be:
20 * LOG ( R1 / ( 1 / ( 1/D + 1/R2 ) ) )
For anyone interested in building their own attenuators, it is simple
to set up an Excel spreadsheet with the formulae above so you can plug
in values for S, D, R1 and R2 to find the resulting new impedances and
attenuation.
Provided D (the destination input impedance) is fairly high - which it
is for all modern audio gear, the attenuation factor is primarily
defined by the ratio of R1 to R2. A ratio of 4:1 gives about 12dB of
attenuation. Provided R2 is kept reasonably small, the resulting source
impedance will not get too high. But don't make R2 too small, or R1 will
also have to be small, and the resulting destination impedance will be
too low.
For example, in my own setup, the Transporter has an output impedance
of 100 Ohms, and the power amp has an input impedance of 10 kOhm. Using
R1 = 3k and R2 = 680, we get an attenuation of 13.5dB, with a new source
impedance of 557 Ohm and destination impedance of 3.6 kOhm. If the input
impedance of the power amp was 100kOhm, the attenuation would then be
12.9dB. As you can see, the actual attenuation isn't affected that much
by the destination input impedance. I've never seen a power amp with an
input impedance below 10kOhm.
Finally, for a balanced signal, you simply need to duplicate the
circuit for both +ve and -ve signal lines, as shown in the final
diagram ("Attenuation for Balanced Signal"). The two R1s and two R2s
should be matched as closely as possible. It's probably worth buying a
handful of the required resistors and measuring them to find the best
matched pairs. I use 1% 0.6W metal film resistors.
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|Filename: balanced_pad.gif |
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--
cliveb
Performers -> dozens of mixers and effects -> clipped/hypercompressed
mastering -> you think a few extra ps of jitter matters?
------------------------------------------------------------------------
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