Suhr ISO Line Out Box: DIY-able? Speaker Jack to Line Out conversion!

This is an indispensable tool for guitar rigs, and I'm trying to figure out how it works and if it's possible to achieve a speaker jack to line out conversion some other crafty (read cheaper!) DIY way.  I've googled some other methods that use a couple resistors in a simple circuit, but they're a bit too quick and dirty = not true line out.  Before I shell out 180$ I figured I'd submit this to the hive mind.  Something else to consider is that, for my particular application, one amp is a Bassman head that has 2 external speaker jacks in parallel and wants to see a total 4 ohm load.  The other amp is a MusicMan that also has 2 external speaker jacks in parallel but has a 4 and an 8 ohm tap. 
Thanks for helpin'!

This is the best link I've found so far.

http://www.diyguitarist.com/GuitarAmps/LineOutBox.htm

It indicates a "general" attenuation of 10:1, depending on the amp. So does the Suhr box simply use a little transformer to accomplish this versus the voltage divider circuit shown? What is the benefit of doing that and what would the layout for that look like? What transformer would be suitable for this I'm guessing pretty low power purpose?  Something I can pick up at radio shack?!?!?! 

...So many questions! sux being dumb!!!

As usual Jensen Transformers has a solution.

see: http://www.jensen-transformers.com/as/as061.pdf

regards,
Jeff

> a bit too quick and dirty = not true line out.

Why would Q&D == untrue?

You need 0.001 Watts. You have 10W-100W. No great trick to tap 0.1% of signal.

What is wrong with this?

there's also this if you want balanced...

From tedweber.com:

Not to speak of a prompt reply, you guys are frigging radical with the nice info!  This is just what I needed to save some dough, damn boutique box makers and their plagarism.  And (heaven forbid) I'm learning something!!!  That Jensen site is ridiculous, too. 

I guess quick and dirty works just fine, PRR. The transformer is only used for the purpose of balancing then, no other benefit? Probably not neccessary for my use - thru wet f/x to slave a power amp... But definitely something I'll do eventually anyway.

***SO*** if I understand correctly (dubious), either version presents NO load on the speaker jack?  Therefore, as long as the other speaker jack feeds a 4 ohm cab, the Fender is happy? (Or with the series version feeding 4 ohms)

correct, the amp will still see something very close to 4 ohms, which will be in parallel with the high input Z of the line out.

Thanks for all the responses. I'm experimenting with some different values for R1, R2 because I'm unsure of the math and principles. 
Bassman (50W @ 4 ohms) = 14V
Vout = R2/R1+R2 * Vin
... something or other
???

So Line level voltage is 0.775VAC. Well thats pretty easy math. Ok, nevermind!

> presents NO load on the speaker jack?

Work it out. Simplify to a Worst-Case. Some fool shorts the line. There is 2K2 across the speaker.

4 in parallel with 2200 is 3.9927 ohms.

There isn't a Fender alive (yes, they are alive) which knows 3.9927 ohms from 4.0000 ohms.

(Anyway all audio is approximate. Your "4 ohms" is never 4.0000 ohms. If it were, and you started playing, the heat in the copper raises its resistance. Up to double for a highly-abused PA woofer, though we try to stay below such extremes.)

"Square root of (Watts RMS * Speaker impedance) gives you VAC at the OT secondary.
(VAC @ OT secondary -0.775VAC)/0.775VAC gives you the voltage drop ratio
So R1/drop ratio gives you the value for R2.

E.g. ...

Sq. root of (50W*8ohms) = 20VAC
20VAC - 0.775VAC = 19.225VAC
19.225VAC/0.775VAC = 24.8
2200ohms/24.8 = 88.7ohms
Therefore R2 = 88.7ohms ...nearest common value is 91ohms. "

> 19.225VAC

We do not know an audio-amp's output to 5-place precision.

Be more lazy. 19V/0.8V = 24:1

> R1/drop ratio gives you the value for R2.

"drop-ratio +plus+ one"

In this case, 24.8 or 25.8 _do_ give the "same" R2 after rounding to Standard Value. 82 or 91 ohms, whichever you have more of.

In the case of a flea-watt amp with 1.5V output, we want 2:1 ratio. Your simplified formula might give 2200:1100. But that is a 3:1 division. A righter answer might be 2200:2200. Or since this is a little high impedance on the line side, perhaps 1K:1K.

looks like the ol' quick and dirty isn't so dirty after all!