Bottlehead Forum
General Category => Technical topics => Topic started by: Loquah on October 21, 2013, 02:08:55 PM
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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?
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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!)
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Sorry Paul, what do you mean by "L-pad"?
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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.
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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?
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Yup, that's it.
By playing with the values, you can reduce the signal however much you want.
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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?
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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.
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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!
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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?
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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).
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That will get you 1/5.
For 1/4 you would need 75K and 33K.
You got it!
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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.
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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.
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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)
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What you are asking is how do I get the voltage I need from what I have.
An LPad is a way to create any voltage you want from what you have while maintaining the same load impedance.
If you measured the voltage you have now in a loaded condition (source connected to load of known impedance) the LPad calcs we just did apply nicely. They don't apply if you measure open circuit voltage. Every source has an impedance too.
You could just add series resistance and that would drop the voltage too, but the load impedance would change.
The why of it is just Ohms Law. And Kirchoff's current and voltage laws.
Voltage Law - the sum of the voltage drops in any closed loop is zero.
Current Law - The current leaving any junction is equal to the current entering.
Tough stuff eh?
If you have 6 volts and you need 1.5, then the 1/4 stuff we did is good. If you have 6 volts and you need 3, then the 1/2 stuff is good.
If you need something else, you figure it out and I will review your findings.
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That will get you 1/5.
For 1/4 you would need 75K and 33K.
You got it!
You are correct, the 33K and 100K of the pot will make 25K.
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For those who need more clarification, I would refer you to the following resource:
http://www.goldpt.com/mods.html (http://www.goldpt.com/mods.html)
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Can someone please confirm the equation to calculate the value of the series resistor (i.e. what's its mathematical relationship to the input impedance, parallel resistor and / or target impedance)?
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The place to start is how many dB you'd like to pad at the input. This determines how much voltage is to be attenuated (20Log(Vout/Vin))=dB.
If we take 6dB of attenuation as an example, then we would need to attenuate half the voltage. If we didn't care about impedance, we could use a pair of 100K resistors in series with the pot to attenuate the signal. If we wanted to keep the 100K input impedance, then we would want a pair of 50K resistors in series with the pot and a pair of resistors across the pot which would make that parallel combination 50K (100K resistors will do that).
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Take a look here:
http://www.siteswithstyle.com/VoltSecond/12_posistion_shunt/12_Position_Pure_Shunt.html
There is a lot of information on the page but what you want is Figure 2.2.1 Trick 1.
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Thanks for the great link Grainger.
Can someone answer the question of how the series resistor value is calculated? Is the series resistor just there to keep the load on the source the same?
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The series resistor both reduces the voltage and helps set the impedance.
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I just googled L-pad calculator and found a nice fill in the blank calculator that came to this solution. Series resister 75k, shunt resister 33k in parrallel with 100k pot. This circuit will attenuate 75% of your input signal.
Because the 33k resister is in parrallel with 100k pot their combined resistance is about 25k. This circuit is a voltage divider network with a series resister of 75k, which is labeled R1 and a shunt resistance of 25k, which is labeled R2 and an overall imput impedance of 100k or R1+R2 . The voltage divider rule is: voltage out= voltage in x R2/R1+R2. In your case 6.8volts x 25k/100k=1.7v
To use the calculator you will have to know the level of db reduction you want. In your case you want a 12 db reduction which is a signal reduction of 75%.
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I just googled L-pad calculator and found a nice fill in the blank calculator that came to this solution. Series resister 75k, shunt resister 33k in parrallel with 100k pot. This circuit will attenuate 75% of your input signal.
Because the 33k resister is in parrallel with 100k pot their combined resistance is about 25k. This circuit is a voltage divider network with a series resister of 75k, which is labeled R1 and a shunt resistance of 25k, which is labeled R2 and an overall imput impedance of 100k or R1+R2 . The voltage divider rule is: voltage out= voltage in x R2/R1+R2. In your case 6.8volts x 25k/100k=1.7v
To use the calculator you will have to know the level of db reduction you want. In your case you want a 12 db reduction which is a signal reduction of 75%.
I think that was the piece of the puzzle I was trying to grasp - thank you. I understood the general concept and also found various calculators, but I was trying to understand the exact maths that went into the value for the series resistor. I was under the impression that it was the difference between the starting input impedance (e.g. 100K) and the resultant impedance of the input impedance + parallel resistor. It seems it's slightly more complex so I'll just rest with the formulas provided.
Thanks everyone for your help!
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One thing that makes this odd is that we don't hear voltage reduction in a linear way. Cutting it in half then cutting to 1/4 doesn't sound the same. Our volume pots are logarithmic not linear.