I built a passive attenuator to sit between my mixing desk and my PC. The output from the mixer was way too high for the PC, and this has solved the problem in a way. However, the sound is a little dull, and I'm lead to believe that may be something to do with "impedance mismatching". I can boost the treble after recording, but I'd prefer a better signal going in to begin with, as I suspect it will then sound even better.

I looked at the specs for the mixer, and all the outputs say they are "600ohm min load, min +4dBU, max 22dBU"

I looked at the specs for the recording interface, and the inputs say they are "min >20k impedance, -10dBV"

To make life easy, all inputs and outputs are unbalanced, so essentially I've contructed an "L-Pad" attenuator.

The attenuation itself obviously takes care of the whole +22dBu -> -10dBV issue, but I'm sure if I choose better pot I'll get a better result, whether that's in terms of quality of the pot, or a more suitable value.

At the moment I have a 100k log (audio) pot. across the "mixer output" (attenuator's input) between signal and ground. I then have the taper from the pot going to the signal line of the "recording interface input" (attenuator's output) - the ground of both is tied together.

As far as I understand, I've got two directions I can go:
1 - Just get a different value Pot... in which case, what value? 10k? 20k? 50k?
2 - Get a rotary switch, and then use a "series and shunt" resistor on each position... but again, what should the resistance add up to? 10k? 20k? 50k? 100k?

I'd really appreciate some advice, particularly with regard to how to work out what total resistance should be.

 

If you use a 2K pot you will be comfortably above 600 ohms load and comfortably below the 20K input impedance of the next stage. This lower impedance will keep the capacitance of the cables and the input stage from quenching the high frequencies. It's not, "impedance matching", but it should work.

Smarter people than I lurk here and some might draw a circuit for you that includes impedance matching.

 

I would connect the taper from the any value potentiometer direct to the PC input and forget about your L pad. Give just a tiny signal adjust to the potentiometer. If happy, the potentiometer could be replaced by a couple of resistors as voltage divider.
What is that PC input? Microphone or 'line' level ?

Why would you want to match impedances ?

 

I'm not sure I need to match impedance exactly, but I think it's fair to say I'm widely enough away from it for it to adversely affect the sound quality.

The recording interface is -10dBV, which I believe means it is consumer line level, rather than +4dBU which is professional line level, hence the need for attenuation.

 

" This lower impedance will keep the capacitance of the cables and the input stage from quenching the high frequencies."

That sounds exactly like what is happening! So, I should be looking to go somewhere between the two impedances. If I can find good quality pots somewhere between 1k and 20k without breaking the bank, it should be improved.

Am I right in thinking that the higher I go, the less "signal drop" I would get on the desk? I'm actually tapping into the FX insert loop for each channel... not ideal, but it saves a fortune on extra equipment.

 

Am I right in thinking that the higher I go, the less "signal drop" I would get on the desk? I'm actually tapping into the FX insert loop for each channel... not ideal, but it saves a fortune on extra equipment.

You've already tried, "the higher I go, the less signal drop" and the results suck. Besides, the whole point of this is to get the signal to drop to a lower amplitude. If you want quality sound reproduction, get in the right range, 1K to 10K. Guess what's right in the middle of that on a log scale? 3.1K, but you can't buy one of those. 2K or 5K is the best area if you can get them in audio taper. If not, I bet you can get a 10k audio taper.

 

Impedances are matched to provide maximum power transfer.
Your console can put out +22dBu into 600 ohm, that is 150 miliwatts. Waaaay too many miliwatts for a PC signal recording input. You can drive a speaker with that.

You want the up to +22 dBu = +22dBm = 158mW = 9.7VRMS = 13.8Vp = 27Vpp signal to drop level. A lot. Like 60 dB.

 

You can use a standard linear 5k or 10k pot. For a quasi-log response just connect a resistor from the pot wiper to ground of about 20% of the pot value (1k ohms for a 5k pot for example). This will give a low-impedance output, as desired to minimize the effects of cable cable capacitance, that is reasonably close to the log output of an audio-taper pot for the purposes of audio signal level adjustment.

 

In that case, I would go with a 10K linear pot and a 2.2K resistor.

 

"You've already tried, "the higher I go, the less signal drop" and the results suck. "

Well yes, but that's because I've gone far too high it seems.

What I mean is, that the FX loop has a send and return - what I'm doing it tapping off of that. I would have thought, if I go towards the low end (600 ohms) then I'll find the signal on the desk that then goes to the speakers will drop more than if I went higher to, say 20k... or am I misunderstanding how it would work? I thought that by tapping off of the FX loop I'd be stealing a bit of the signal... I don't want to steal too much of it so that the mixer then doesn't give a good signal to the main outputs.

 

You can use a standard linear 5k or 10k pot. For a quasi-log response just connect a resistor from the pot wiper to ground of about 20% of the pot value (1k ohms for a 5k pot for example). This will give a low-impedance output, as desired to minimize the effects of cable cable capacitance, that is reasonably close to the log output of an audio-taper pot for the purposes of audio signal level adjustment.

Is there any advantage to that approach, rather than just buying a log pot? Cost doesn't seem to be any different really, as what you save in a linear pot you lose in then buying an extra resistor...

 

Are you assuming that the Fx output doesn't have a driver of its own or have you seen a signal loss when attaching to it? My Fx loops have their own drivers so you can short them dead and have no effect on the main amplitude.

PS, It's difficult to buy log pots in every size. That's why we're giving you options.

 

Are you assuming that the Fx output doesn't have a driver of its own or have you seen a signal loss when attaching to it?

It's an insert for the channel, so I know the signal path...
input (line/mic) -> Preamp -> FX Send -> FX return -> on-board EQ -> channel allocation/mixer (main outs, etc.)

If you plug into the FX Send, then the channel is effectively muted unless you return the signal on FX return, hence the need to effectively short the Send and Return and take a tap off of the signal line.

 

I've' not noticed a signal drop... but then with 100k impedence on the tapped signal, I doubt the drop would have been noticeable.

 

Thank you. The plug throws a switch that demands an FX return. Didn't we already go through this where you said the impedance of the source was 600 ohms? Didn't we already say a pot in the 2K to 10K range would satisfy both ends of the loop?

 

Thank you. The plug throws a switch that demands an FX return. Didn't we already go through this where you said the impedance of the source was 600 ohms? Didn't we already say a pot in the 2K to 10K range would satisfy both ends of the loop?

I did say I was getting confused... lol

What I've got happening is this...

FX Send -> SPLIT --> Attenuator (currently 100k impedance, which we're changing to around the 2k-10k range) -> computer recording interface
........................... --> FX return -> on-board EQ -> channel allocation/mixer (main outs, etc.)

I've literally soldered the Send and return together on the plug.

I guess I'll just have to try it and see what happens.

 

Let's quit beating this horse and connect some parts!

You might just try a 22K resistor on your 100K pot.
Oh, wait, it's already a log pot.
Well, try it anyway.

If the sound suddenly gets cleaner, you will have your answers!

 

Nah, that's way too obvious and easy...

Yeah I do feel like an idiot for not thinking of just doing that, lol!

 

Is there any advantage to that approach, rather than just buying a log pot? Cost doesn't seem to be any different really, as what you save in a linear pot you lose in then buying an extra resistor...

No advantage. It's just a workaround if you have a problem finding a low impedance log pot or happen to have a linear pot handy but not a log pot.