How to tame a high voltage output

[Loquah]

I have a DAC with a 6.8V XLR output.

I'd really like to make use of this output as a second option alongside the RCA outs, but at 6.8V it's pretty hot and doesn't really work with my Crack, S.E.X. or Quickie. Is there any way (i.e. inline resistors, etc.) I can tame the 6.8V somehow or will any such mods alter the sound?

[Paul Joppa]

You can L-pad the inputs of anything with two resistors per channel. Yes it will affect the sound, but most of that effect is that it will stop distorting if you reduce the input enough, and it will become noisy if you reduce the input too much.

It's very useful to look at and try to optimize the gain structure of a system, though the tools and specifications to do so are usually hard to locate and use. It's a pet peeve of mine  :^)  check my "signals and noise" white paper, linked off the Community page, for more details than anyone really wants. (Sorry everyone, I would simplify it if I could!)

[Loquah]

Sorry Paul, what do you mean by "L-pad"?

[Mike B]

It's 2 resistors added to the input that drops the voltage but preserves the impedance.

Lets say the input impedance is 100K.  You add a 100K resistor in parallel which makes the input impedance 50K.

Then you add 50K in series.

So the inpedance is still 100K, but only half the voltage appears at the input.  Half across the 50K series and the other half across the 2 parallel 100K's which is 50K.

[Loquah]

Thanks Mike. I'm still learning so let me see if I understand...

• The 50k resistor goes in line with the active lead of the interconnect (i.e. from the end of the active lead to the centre pin of the RCA lead)
• The 100k resistor bridges the active and ground / cold leads (in a single-ended setup)

Is that correct?

[Mike B]

Yup, that's it.

By playing with the values, you can reduce the signal however much you want.

[Loquah]

Thanks so much Mike!

Does the parallel one have to match the input impedance or do I just have to keep the 1:2 ratio between the 2 resistors?

[Mike B]

No baby, you have to do the math. 

Let's say you want 1/4 the voltage.  That means you have to drop 3X on the series and X on the input.

If the input resistance is to remain the same that means you need 75K on the series resistor and the input needs to be 25K.

The amp has 100K, so you tell me the value needed to put in parallel to make 25K.

What, you thought this was easy and you don't have to work?

I gave you the series resistor value, you give me the parallel.

[Loquah]

Happy to do the work, but wanted to make sure I understood first. 

Let me go get some exact specs for the DAC and come back with the maths for a couple of options to see if I've got it all right.

Thanks again for your help Mike!

[Loquah]

OK, so I'm not entirely sure that I understand your challenge, but here's what I've come up with...

I think if I'm using a 25k resistor in parallel (with 100k input) I would need an 80k resistor in series. Is that correct?

I've noticed that the resistor values I can buy aren't in round amounts (for example I can get a 56k resistor, not a 50k). How exact do I have to be? Do I need to play with the calculations until I find 2 available resistors of exactly the correct values?

[Paul Birkeland]

You don't need to be that accurate.  There's also nothing that magical about the 100K input impedance, just keep it between 50K and 250K and everything should be peachy. 

For 6.8V, you may want to go down to 1.7V, which is 1/4 of what you started with.  Putting a 27K across the pot and a 75K from input to pot should get you pretty close (this is an on-the-fly estimate).

[Mike B]

That will get you 1/5.

For 1/4 you would need 75K and 33K.

You got it!

[Loquah]

Eeek! I'm not sure I follow! 

I'm not sure of the equation for the series resistor value. Based on PB's post about the 6.8V out, my maths was almost correct because I came to 80k and 25k for that one, but I'm worried it was dumb luck. Let me share my current reasoning to see if I'm somewhere close to correct.

1. Identify desired impedance (X) using V=IR
2. Use the following equation to determine resistor values: X = 1 / (1/(100k + series resistor)) + 1/parallel resistor)

Is that correct? Please forgive me if I'm way off the mark - it's >15 years since I left school and I never got the practical application of these equations back then so it takes me a little while to fully understand them now.

[Mike B]

I know you don't understand, that's why you came up with the wrong answer (the hard part) and the right impedance (the easy part)

OK.  You need to find the correct resistance to parallel with 100K to get 25K.  The parallel combo is R1 * R2/ R1 + R2.

25 = 100X / 100 + X

2500 +25X = 100X

100 + X = 4X

100 = 3X

X = 33

I am here and I will help anyone wanting education.  If you want to learn stuff you have to work at it.  There is a reason they spent all those years in school teaching math.  If you want to do anything technical or scientific, you need to know math.

I get a big kick out of all the people wanting to be educated re vacuum tube theory when they don't even know how to use and apply  Ohm's Law.

I will be happy to help, but I ain't gonna give ya the answer w/o you doing at least some of the work.

It's just like school.  If you don't do your homework you ain't gonna pass the test.

[Loquah]

Mike, I appreciate the help and am willing to put in some work (hence why I am sharing my process thus far for feedback). Please don't mistake my asking for guidance for laziness. I am a trainer and facilitator and fully understand adult learning. I WANT to work this out and play with examples in order to learn how and why this works, not just how to punch numbers into a formula in a spreadsheet.

Thank you for expanding on the maths of it. My issue was using an incorrect equation for the process. I Googled "how to determine the resistance of two resistors in parallel" and it sent me down the wrong path for the task at hand (i.e. the equation was good, but not the right one for this purpose).

My understanding now is this:

1. I determine the required impedance to create the desired voltage (Y)
2. I use the desired voltage (Y) to calculate the parallel resistor value (X) --> Y = 100X / (100 + X)

My remaining question then is should the series resistor value be 100-Y or  something else? (Again, I'm not asking for the answer on a platter, but the means to calculate it and understand the "why" of it)