help required with a signal pad

[mr_doyle]

hello,

i need to build a simple "U" signal pad of the type found in this site: http://www.uneeda-audio.com/pads/



problem is that i don't understand how should i choose resistors value

source is 100 ohm (balanced out from a processor) and need 6 db of attenuation to enter balanced input on receiving device (another processor, 1 M ohm input)

i don't understand the relation between impedances on out, in and the pad

tia for your help,
D.

[ashcat_lt]

For 6db attenuation, you want the series resistors to equal each other and be half of the value of the parallel.  Since you've got such a big difference in Z's between the source and the load, there's a bit of leeway as to actual values.  For safety's sake I'd try to follow the (at least) 10 to 1 rule (of thumb).  Make the total of the three resistors at least 10x the output Z of the source and less than 1/10 the Z of the load.  So anywhere between 1K and 100K total should do fine.

[mr_doyle]

Thank you for your reply!

So, if i understand:

att = 6db
k = 1.9953

standard 1% resistor values for my application could be:

R1/2 = 14K3
R2 = 28K7

or

R1/2 = 15K4
R2 = 30K9

or

R1/2 = 15K
R2 = 30K1

Did i get it right?

THank you,
D.

[PRR]

Your numbers will work. I'd pick them a bit different.

BTW: do you need 6dB (K=1.9953) or 6.02dB (K=2)? Probably you can be happy with 5dB to 7dB of loss, maybe 10dB between line-level sources. Unless you are the kind of tech who sleeps with a calibration set. And then you'd already know your pads.

So don't use too sharp a pencil. 5% values are good enough for most audio. Stereo pairs should be matched 5% worst-case, so you use 2% resistors or hand-select 2% pairs from a 5% lot.

> source is 100 ohm

That does not mean it will drive 100 ohms. Studio gear usualy has minimum load of 600 ohms, lightweight gear may be happier at 2K and up.

> 1 M ohm input

That may be in the spec. It is an astonishingly high number. Even if "this" box is 1Meg, many are 10K or 22K. And it is good to have a pad where you don't have to look-up specs every time you change boxes.

You also have some length of cable which has capacitance. A capacitor is infinite ohms at zero frequency and zero ohms at infinite frequency. For audio we like to look at/past 20KHz. I remember that 300 feet shielded cable (happens sometimes in large studios and live sound) is below 1K ohms at 20KHz. (This is why big studios aim at a 600 ohm standard.) 30 feet is 10K and 3 feet is 100K at 20KHz (actually 16KHz but we'll leave some margin).

So see what you can do with >2K input and >10K nominal load.

The unbalanced pad could be 1K+1K with load tapped from one of the 1K resistors. If the load is infinite then the input is 2K and the attenuation is 2:1 or 6.02dB (ignoring the 100 ohms for now). The output impedance is (100+1K)||1K) or 498 ohms; will easily drive 300 feet cable. The input impedance is 2K, right at our spec. Adding a 10K load and including the 100 ohms gives (1K||10K)/(100+1K+(1K||10K)) or 6.07dB attenuation.

Point: if the cable is pinched the load falls to 1K. In small work, a short stops the take and you fix it, in some large organizations there are multiple loads and the show must go on even if one is shorted. Will 1K load on processor degrade the sound? Maybe not. But what if we double the pad values, 2K+2K? Now the minimum is 2K (safe), the output is about 1K (low), the unloaded or 1Meg-load loss is still 6.02dB and the loss with 10K load is about 7dB (check my math) which is hardly noticeable from 6dB.

If the unbalanced L-pad is 2K+2K then the balanced U-pad is 1K+2K+1K.

Looking in my box I find 1K and 2.2K. This gives no-load loss of 5.6dB. If you really have things aligned so exact that a 0.4dB excess will hurt, find some 1.2K and the 2.2K for 6.4dB loss un-loaded, 7.4dB loss in 10K load. You can spend the morning picking more-closer values, but 1dB "errors" usually don't matter, and between line-level boxes a dB more loss can usually be made-up trivially.

Your answer arrives about 15 times higher impedance. For the long-cable problem this means you must stay 15 times shorter, about 20 feet. This is fine in most small studios. You also have an output impedance near 15K. In 1Meg load this will drop 0.13dB. But if you switch the box for a 10K box, it drops 8dB! We usually like to have sources under 1K so that loads from infinity to 10K cause "negligible" added loss.

[mr_doyle]

Thanks for your in depth answer!

Try to give you some answers:

BTW: do you need 6dB (K=1.9953) or 6.02dB (K=2)? Probably you can be happy with 5dB to 7dB of loss, maybe 10dB between line-level sources.

No chances at all that my ears would get the difference between 6 and 6.02 db ;-)

I've been asked by a friend to build a pad to connect an fx device to his guitar router.
I have no details about the fx but "it's output impedance is low, 100 Ohm", but i know for sure that his router is a Switchblade by SoundSculpture, and that thing has 1MOhm impedance on inputs.
The Switchblade can also cut 6db on inputs if signals are too hot: that's what the guy did (and it worked but he wants some external pad to do the job), and here's why i'm talking about 6db of attenuation.

Once i get the right way to calculate resistor values then it would be no problems at all building both a 6db and a 10db pad and let him decide what he prefers.

You also have some length of cable which has capacitance.

As this is going to be used in his guitar rig i think that cable won't be longer than a couple of meters.

Now, back to what you wrote, let's see if i got it:

But what if we double the pad values, 2K+2K? Now the minimum is 2K (safe), the output is about 1K (low), the unloaded or 1Meg-load loss is still 6.02dB and the loss with 10K load is about 7dB (check my math) which is hardly noticeable from 6dB.

If the unbalanced L-pad is 2K+2K then the balanced U-pad is 1K+2K+1K.

So, these values would result in a lower impedance so that the pad could be places between a wider range of devices, right?


Thank you so much for your help!
D.

[ashcat_lt]

Is this actually balanced to balanced?  It's unusual for guitar gear to be balanced, and the 1M input leads me to believe that it's unbalanced intended to take a guitar input.  This might also explain the need for a pad - feeding a line level output to an input with guitar level headroom.

If it's a balanced>unbalanced deal you can usually get a 6db drop just by dropping the negative side of the balanced signal.  Depending on the device, you would either short that to ground (by using a TS cable) or by leaving it open (using a custom TRS>TS cable).  Of course, if I'm right, he's been unbalancing the signal one way or another and needs yet another 6db.

I'd like to point out, too, that PRR explained the reason - in this instance - the (at least) 10 to 1 rule (of thumb).

[ashcat_lt]

Is this actually balanced to balanced?  It's unusual for guitar gear to be balanced, and the 1M input leads me to believe that it's unbalanced intended to take a guitar input.  This might also explain the need for a pad - feeding a line level output to an input with guitar level headroom.

If it's a balanced>unbalanced deal you can usually get a 6db drop just by dropping the negative side of the balanced signal.  Depending on the device, you would either short that to ground (by using a TS cable) or by leaving it open (using a custom TRS>TS cable).  Of course, if I'm right, he's been unbalancing the signal one way or another and needs yet another 6db.

I'd like to point out, too, that PRR explained the reason - in this instance - for the (at least) 10 to 1 rule (of thumb).

[mr_doyle]

Is this actually balanced to balanced?

Yes!

[PRR]

> both a 6db and a 10db pad

10dB is about "one third", near enuff.

Using the 1:10 rule from rated 100 ohms we start with 1K. Since the 1Meg load is much-much higher, we may cheat this up. Use a 2K or 5K total string of carbon.

For "half" we tap 1K or 2.5K out of the total.

For "third" we tap 0.66K or 1.66K out of the total.

For "tenth" (20dB) we tap 100 or 500 ohms out of the total.

For balanced, we tap it out of the middle.

A simple 10dB balanced pad with >2K input is three 1K resistors. Ignoring source and load Z it is 9.54dB. With 100r source and 1Meg load it is 9.8dB. With 10K load it is 10.6dB.