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-   -   Lowest resistor needed to sum to ADC (https://www.gearslutz.com/board/geekslutz-forum/1265139-lowest-resistor-needed-sum-adc.html)

whippoorwill 16th May 2019 02:29 PM

Lowest resistor needed to sum to ADC
 
I am trying to sum 2 stereo signals from two different preamps to a single stereo signal.
L1 + L2 = L
R1 + R2 = R

Signal is going straight to an ADC.
I have a huge amount of gain available from the mic pre's so a small loss isn't an issue, but I want to keep losses at a minimum.
What minimum resistor values do I need?
Would even 100 Ohms be enough?

Murky Waters 16th May 2019 05:24 PM

Use a mixer.

DrewE 16th May 2019 06:27 PM

A resistor network is a simple passive mixer.

The size answers really depend on the output impedance of your preamplifiers and the input impedance of the ADCs. For many line level applications, the input impedance (perhaps 10 kOhm) is much, much greater than the output impedance of the drivers (perhaps 150 Ohms). 100 Ohms is probably lower than you want; 1 kOhm resistors might be a reasonable starting point. Trial and (possible) error are perhaps the best way to figure these things in practice.

If you're only ever using one at a time, you can instead have a switch or patch bay or even just move cables around. That would generally be preferable to a passive combiner circuit.

Jay Rose 16th May 2019 06:31 PM

1 Attachment(s)
The point of the resistor is to prevent source 2 from loading source 1 too much, and vice-versa.

It works like this: Source 1 sees the parallel load of Destination's input impedance, in parallel with a series combination of Source 2's output impedance and the resistor. So to find an appropriate R, you need to know the load S1 has been designed to feed, and the actual impedance of S2's output, and the actual impedance of D's input.

You don't say what your impedances are, or even if you're working in a mid-Z unbalanced or low-Z balanced environment. I'm going to assume the ADC doesn't have an input transformer, but that's not necessarily a given either.

On the other hand: if you're not working in a transformer environment, the actual currents involved will be very low. That means the resistor can be fairly high without producing too much loss.

As a rule of thumb, with modern transformerless balanced gear I usually use about 1k for each combining R. I could probably go all the way to 2k7 if I wanted to, without any major effect.

BTW, don't overdrive your mic pre to make up any missing gain (there won't be much), unless you're absolutely sure you're not causing distortion in its output stage (or you like that kind of distortion). Usually better to make it up at the ADC's input.

mskala 16th May 2019 06:55 PM

There's a heck of a leap between "I don't want much loss" and "I want a small resistor value," so that the original question as asked cannot be meaningfully answered. Passive mixers just don't operate the way this question implies.

Your preamps are going to be fighting with each other through those resistors. Lower resistor value means the preamps see lower impedance; too low and they won't be able to produce the necessary current and it will lead to distortion (conceivably even damage to the preamps, though probably not a serious risk in practice). On the other hand, with too high a resistance the averaged signal won't be able to drive whatever input it's being connected to.

I would suggest finding out what impedance your preamps are designed to drive. (That means the BEST impedance, not the LOWEST impedance!) Use that as the starting resistor value; calculate half of it and compare the result to the input impedance of the ADC; and if the ADC's input impedance is not significantly more, then you're not going to get good results in any case and need to use a proper active mixer. Some ADCs do require a surprisingly low input impedance.

whippoorwill 16th May 2019 07:12 PM

Quote:

Originally Posted by Murky Waters (Post 13986605)
Use a mixer.

So many stages, My preamp signals sound incredible, and most mixers sound gross, muddy, indistinct in direct comparison.
Even the very fancy ones.

whippoorwill 16th May 2019 07:16 PM

Excellent advice given, and I have learnt something here and had a fundamental misunderstanding of the way the circuit operates.

The preamps involved are the Sonodore MPB-502, output impedence 56 Ohms and the Sonosax Sx-m2, impedence unknown.
https://www.sonosax.ch/product/sx-m1m2/

ADC is the RME ADI-2 Pro and, at other times, the Sony Wm-d6c and other analog capturing devices.

Jay Rose 16th May 2019 11:53 PM

If you can't be sure of the ADC's input load (not the impedance it's designed for, but its load... which will be a lot higher)... then add a gain stage. Or an active combiner.

Or try one of the suggested values and see if it works. Unless you make a mistake in the wiring, adding a 1k or higher resistor in series is not going to damage your pre's output stage. You might not like the sound, but heck... it's worth a resistor to find out.

cbc6403 17th May 2019 01:45 AM

You mention that you are concerned about loss, but at a minimum, there will be a 6 dB loss at each input. If the ADC input impedance is significantly lower than ten times the summing resistors, the loss will be higher than 6 dB. It doesn't matter how low you make the summing resistors, there has to be at least 6 dB of loss.

Jay's suggestion is a very good starting point, and likely will be the end point of the experiment too.

Geoff

Radardoug 17th May 2019 02:46 AM

Perhaps the OP could tell us why he wants to do this? Then we could give him better advice.

whippoorwill 17th May 2019 02:05 PM

I’m trying to not have any sonic quality loss if possible. I want to sum two stereo signals as simply as possible with as little quality difference as possible, partially because I only have one stereo input on my RME/analog recorder and it’s the best AD stage I have, also because I like to mixdown to 2 track and also because I do 100% non studio recording so carrying another box to a location is bulky.
1k seems like the winner.

bitman 17th May 2019 02:39 PM

I would go with 4.7k.