Hi all,
I need some attentuation between my MiniDSP 10x10 and 3eAudio TPA3251 (4 channel) amp. The MiniDSP is very noisy and the drivers in my active setup are sensitive (10" pro driver and compression driver). I have my MiniDSP going balanced straight in to the TPA chip bypassing the onboard op-amps. MiniDSP is fed by SPDIF. The amp is dead quiet.
The MiniDSP input gains are down 18db, I have a 10db L-pad on my compression drivers and yet I still have my windows volume level hovering between 10%-30%. Something must give.
I would like to put some passive attenution on the balanced line betwen MiniDSP and amp and having read everything I could find here would someone mind looking over my sums, please?
MiniDSP output = 560 ohms
10db attenuation K factor = 3.16
Rshunt = 560ohms
Rseries = 560 * (K-1) 2.16 = 1209 ohms
Rseries/2 = 604.5 ohms
R1/2 ≈ 600 ohms
R2 = 560 ohms
I plan on using 1w metal film resistors.
Cheers! 🙂
I need some attentuation between my MiniDSP 10x10 and 3eAudio TPA3251 (4 channel) amp. The MiniDSP is very noisy and the drivers in my active setup are sensitive (10" pro driver and compression driver). I have my MiniDSP going balanced straight in to the TPA chip bypassing the onboard op-amps. MiniDSP is fed by SPDIF. The amp is dead quiet.
The MiniDSP input gains are down 18db, I have a 10db L-pad on my compression drivers and yet I still have my windows volume level hovering between 10%-30%. Something must give.
I would like to put some passive attenution on the balanced line betwen MiniDSP and amp and having read everything I could find here would someone mind looking over my sums, please?
MiniDSP output = 560 ohms
10db attenuation K factor = 3.16
Rshunt = 560ohms
Rseries = 560 * (K-1) 2.16 = 1209 ohms
Rseries/2 = 604.5 ohms
R1/2 ≈ 600 ohms
R2 = 560 ohms
I plan on using 1w metal film resistors.
Cheers! 🙂
Remember to include the source output impedance (560R) in the calculation.
Without the attenuator the source impedance has little effect due to the high input impedance of the amplifier,
but this attenuator will load the source heavily. You can just subtract 560/2 from each of the two series resistors
to compensate and include the loading effect. Then each would be 320R.
Without the attenuator the source impedance has little effect due to the high input impedance of the amplifier,
but this attenuator will load the source heavily. You can just subtract 560/2 from each of the two series resistors
to compensate and include the loading effect. Then each would be 320R.
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A source impedance usually varies with frequency, especially at the extremes of its intended bandwidth. It's reasonable to make the load seen by the source at least ten times the nominal source impedance to minimise frequency response aberrations that result.
The normal purpose of balanced signal lines is to allow for the rejection of common mode noise induced on the interconnect cable, however unless the interconnect cables used are technically incongruous, there's no benefit where the connection distance is small away from a hostile EMI environment, such as in a domestic setting.
If the mini-DSP has a properly balanced output with no common mode noise, you can drive the amplifier single ended which will give you 6dB gain reduction at the amplifier input with no quality or fidelity penalty, or even better, drive it single ended through an L-pad. A pad that has 5k6Ω input & 560Ω shunt resistors will satisfy the first requirement of keeping the load >10 times the source impedance, and the source impedance seen by the amp will be the same as before. Combined with a single ended input configuration that will give you 26dB of noise and distortion free attenuation.
The normal purpose of balanced signal lines is to allow for the rejection of common mode noise induced on the interconnect cable, however unless the interconnect cables used are technically incongruous, there's no benefit where the connection distance is small away from a hostile EMI environment, such as in a domestic setting.
If the mini-DSP has a properly balanced output with no common mode noise, you can drive the amplifier single ended which will give you 6dB gain reduction at the amplifier input with no quality or fidelity penalty, or even better, drive it single ended through an L-pad. A pad that has 5k6Ω input & 560Ω shunt resistors will satisfy the first requirement of keeping the load >10 times the source impedance, and the source impedance seen by the amp will be the same as before. Combined with a single ended input configuration that will give you 26dB of noise and distortion free attenuation.
Take 2: It isn't necessary (unless you are driving a transmission line*) to match the input impedance of the attenuator to the source impedance, in fact much better not to as I pointed out above. Make R/2 2k7Ω, and then make R2 1k8Ω. That will give you close to 10dB whilst keeping the load ~10 times the source impedance, and keep the source impedance seen by the amplifier ~1/10 of the amplifier input impedance.
*At audio frequencies transmission line effects occur in kilometres of cable, not meters.
*At audio frequencies transmission line effects occur in kilometres of cable, not meters.
Thanks man, where did I go wrong with my working? I was thinking to make a set of 5/10/15db boards to drop in to a mini chassis should I need to go up or down in the future so it would be great to know where I messed up 🙂
Cheers for the input! Yes, I have issues with USB hash and being a multiway active system it can be a magnet for noise!
Would you mind telling me the quation that got to you the "R/2 2k7Ω, and then make R2 1k8Ω" values, please? It would be great to understand in full 🙂
Actually it turns out that the MiniDSP output impedance in balanced mode is 1.12k
But they don't say what the minimum load impedance is. Let's assume it is 2k
However, before building anything check with the mfr and see what it is.
First you scale the two series resistors plus the shunt resistor so their sum totals 2k
Then adjust the relative values of R1 and R2 to attenuate by the desired amount, 1/3.16
So first we have: R1+ R2 = 2k (neglecting the amplifier input impedance)
Also the attenuation is: R2 / (R2 + R1 + 1.12k) = 1/3.16
Solving these two equations gives:
R1 = 1013R
R2 = 987R
If you want to account for the loading due to the amplifier input impedance (roughly 1dB),
just increase the value of R2, so the parallel combination of R2 and the input impedance is 987R.
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Below is the equivalent schematic of your DAC output / attenuator / amplifier input. It is symmetrical so we can redraw it in a simplified version as underneath. No let's say your DAC outputs 0dB. Without any attenuator in series the proportion of the DAC's voltage across the amplifier input is forming a potential divider with the output impedance of the DAC.
We can work out the attenuation as 20*log10(20,000÷(20,000+560)) = 20*log10(20,000÷20,560) = 20*log10(0.973) = -0.24dB. There is already a small amount of attenuation.
Now if we put 5k4Ω in series with the 560Ω output impedance of the DAC, the total 'output impedance' is now 5k96Ω. If we put 1k8Ω in // with the 20kΩ input impedance of the amp, the load seen by the DAC + series r = (1,800 x 20,000)÷(1,800+20,000) = 1k651Ω.
So now the attenuation at the amplifier input is 20*log10(1,561÷(5,960+1,561) = 20*log10(1,562 ÷ 7,521) = 20*log10(0.208) = -13.65dB. The actual attenuation is the difference between padded and not padded = -13.65 + 0.24 = -13.41dB.
I have guesstimated a little over the correct amount by 3.14dB - doh!
The voltage across the load needs to be 31.6% of the DAC output. Therefore the output impedance + series Rs has to be the other 68.4%. From that we can work out that the input load has to be ((Rsource ÷ 0.684) x 0.316) = 2k729Ω.
Using the R48 preferred value of 3k010Ω in parallel with 20kΩ = 2k616Ω, and the attenuation now works out as 20*log10(2,616÷(5,960+2,616) = -10.31dB.
Then subtract the original .24dB attenuation without the pad and you have added and additional -10.07dB attenuation.
We can work out the attenuation as 20*log10(20,000÷(20,000+560)) = 20*log10(20,000÷20,560) = 20*log10(0.973) = -0.24dB. There is already a small amount of attenuation.
Now if we put 5k4Ω in series with the 560Ω output impedance of the DAC, the total 'output impedance' is now 5k96Ω. If we put 1k8Ω in // with the 20kΩ input impedance of the amp, the load seen by the DAC + series r = (1,800 x 20,000)÷(1,800+20,000) = 1k651Ω.
So now the attenuation at the amplifier input is 20*log10(1,561÷(5,960+1,561) = 20*log10(1,562 ÷ 7,521) = 20*log10(0.208) = -13.65dB. The actual attenuation is the difference between padded and not padded = -13.65 + 0.24 = -13.41dB.
I have guesstimated a little over the correct amount by 3.14dB - doh!
The voltage across the load needs to be 31.6% of the DAC output. Therefore the output impedance + series Rs has to be the other 68.4%. From that we can work out that the input load has to be ((Rsource ÷ 0.684) x 0.316) = 2k729Ω.
Using the R48 preferred value of 3k010Ω in parallel with 20kΩ = 2k616Ω, and the attenuation now works out as 20*log10(2,616÷(5,960+2,616) = -10.31dB.
Then subtract the original .24dB attenuation without the pad and you have added and additional -10.07dB attenuation.
Thanks, I need to do figure this out now!
Yes, the MiniDSP has a 1.12kohm out so I need to figure out how to insert that in to the maths 🙂
Yes, the MiniDSP has a 1.12kohm out so I need to figure out how to insert that in to the maths 🙂
Just subtract half of it (560R), from each of the two series resistors to include the source impedance.
So, looking at your equation R1 + R2 must equal 1.12k and then the balance between the two resistors forms the atenuation. I don't understand how 1/3.16 (or 0.316) influences the balanced or distribution of values.
Is this correct?
R1 = 766ohm
R2 = 350ohm
Sorry for my difficulty understanding this 🤢
Is this correct?
R1 = 766ohm
R2 = 350ohm
Sorry for my difficulty understanding this 🤢
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No problem. For those values the attenuation is R2/(R1 + R2 + Rsource) = 350/(766 + 350 +1120) = 0.156 = -16dB.
You wanted 1/(3.16) = 0.316 which is -10dB.
If we choose R1 + R2 = 2k as a minimum load
and attenuation 0.316 = R2 / ( R1 + R2 + 1120 )
then 0.316 = R2 / (2000 + 1120 )
so 0.316 = R2 / 3120 and R2 = 0.316 x 3120 = 986
then R1 = 2000 - 986 = 1014
Check the answers: R2/(R1 + R2 + Rsource) = 986/(1014 + 986 + 1120) = 0.316 which is what you wanted.
Of course you could choose R1 + R2 to be a higher value, such as 5k or 10k, but then you will have to include
the power amp input impedance, since that will now have much more of an effect.
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Thanks for being so patient! I think I'm beginning to get it, sorry for the late reply, I've been super busy!
So, R1/2 = 507 and R2 = 986?
Or is R1/2 still 1014 in the u-pad config?
So, R1/2 = 507 and R2 = 986?
Or is R1/2 still 1014 in the u-pad config?
Whilst I wait for my basket to fill up so I don't have tp spend £15 for a handful of peny resistors I got wondering how these passive units work:
Behringer Monitor1
https://www.gear4music.com/Recordin...ME7DhB9JFJs16dwn-ncfMEvv20xEBzMhoCkwwQAvD_BwE
I don't understand how they are able to pick seemingly arbitrary input and output resistances and how they are maintained. I can't find a teardown for this kind of unit with a detailed assessment so it's hard to know!
Any thoughts, anyone? 🙂
Behringer Monitor1
https://www.gear4music.com/Recordin...ME7DhB9JFJs16dwn-ncfMEvv20xEBzMhoCkwwQAvD_BwE
I don't understand how they are able to pick seemingly arbitrary input and output resistances and how they are maintained. I can't find a teardown for this kind of unit with a detailed assessment so it's hard to know!
Any thoughts, anyone? 🙂
Here's the schematic of a similar one, probably the same as the Monitor1. Has a pair of ganged 5k pots per channel.
Being passive, the resistances are not maintained.
Being passive, the resistances are not maintained.
They can type any number they want in the chart. Behringer 'specs' are at best plagiarized from the product they are copy-catting, and often just made-up.