Why does a DI work?

[ashcatlt]

Wasn't sure if this fit here or in Amps or what. I'm sure Sumbody will be around to move it if necessary.

We're always worried about the impedance to which we're connecting our guitars. As understand it, an impedance too low is bad for 2 reasons:

1) Pickup loading - pickups can't produce enough current to properly drive the device it's plugged into

2) The cutoff of the lo pass filter created by the capacitance of the cable is reduced, causing loss of high register harmonic content.

With that in mind, I often find myself doing, and advising others, to always buffer the guitar's signal before performing any splitting. Splitting to 2 x 1M inputs presents a 500K load to the pickups. So buffer with a known input Z first, to get that good old 1M, and let the extremely low Z output of the buffer device drive the lower Z from the parallel inputs.

Great.

But then we whack the old passive DI in there. It's often used on bass and sometimes used on guitar to split the signal between a lo-Z input (usually a mic pre) and the hi-Z input on an instrument amp.

From what I can tell, the transformer on the low-Z output is in parallel with the amp input via the "thru" or "amplifier" jack.

Here's the actual question I've got: Why doesn't this cause the issues I mentioned above?

I've done a little digging and found that the transformer in a Whirlwind DI should present an input Z of about 133 times whatever is on the other side of it. Looking around I find quotes for typical mic pre input Zs in the range between 100-2K. So, the transformer will be showing somewhere between 13.3K-266K.

Shouldn't that give us a parallel impedance of somewhere between 13.1K and 210k? If 500K is too low then surely this abyssmal, no?

On top of the overall Z being too low, I think I understand that it also creates something like a voltage divider - causing major voltage mismatch between the too outputs. I'm not real clear on this point, but based on what i know of our good old volume pot it seems like the lion's share of the voltage would be heading toward ground through the transformer, "starving" the amp input.

But nobody ever complains about the passive DI sucking tone. It's pretty standard procedure in both live and studio situations.

Why?

Thanks

edit – minor grammar and spelling tweaks
edit2 - too many sentences starting with "So"

[sumgai]

SO riddle me this........

(SOrry, couldn't resist! )


ash,

There's more to the story, but let me cut to the chase here...... A DI box does more than just match impedance for a given device and load, it also transforms a signal from unbalanced to balanced. In certain electro-circles, we call that a 'balun', short for balanced-unbalanced. Obviously, as a transformer, it can (and does) work both ways, so we don't call it 'unbal' - that just makes a guy quiver needlessly.

The transformer is actually in series with the signal chain, not in parallel. In point of fact, every tranformer serves at least two functions, one of which is always isolation. It's fair to say that when the chips are down, the electrical signal from the pickups ends at the transformer - the xfmr (short-hand, OK?) is the proper load for the pickups, and all is well. A new signal is generated magnetically by the xfmr, and that's what goes down the line to the load (the input of whatever device). The two signals are isolated from each other electrically, they share only a common magnetic property.

Clear as mud, eh?

More to the point, the DI's input is one side of the xfmr, the primary. It's really nothing more than a simple coil of wire, so it can't divide the signal, thus the pickup circuitry is happy, the volume control acts normally, all is well in Axe-Thumper land.

The secondary however, that's a new ball of wax. Watch this: (But don't you dare try to hold my beer! ;D)

The secondary winding is indeed tapped at the center point. That center point is referenced wired to the amp's chassis ground, and the other two lines are the signal plus and minus. Note strongly that the signal minus is NOT ground. Doing so would reduce the signal strength by more than half, for reasons we'll not get into here, just suffice it to say, "don't do that".

This 'dual' signal line is fed into a pair of inputs (most often pins 2 and 3 of an XLR connector (see WIKI Page - XLR), but it could as easily be a TRS ¼" connector), and amplified accordingly. The reason for all this? When the amp circuitry can deal with both halves of a signal, independently of ground, then it can reject all manner of noise, no sweat. The fact that the xfmr's secondary winding is very low impedance (just as you stated) is also beneficial - the voltage of the signal suffers very little loss, even over a long cable run.

And there you have it, the reasons why a passive DI box can, and does, act beneficially even though it's unpowered. A powered DI box can drive several loads at once, they usually have both hi and low impedance outputs available. (Most of those derive phantom power from the mixer, very handy indeed. )

Now, to address your first two points:

1) Pickup loading - they can't reproduce enough current to properly drive the device they're plugged into
Strictly speaking, that's a misleading statement - pickups don't generate current beyond miniscule amounts.... they generate voltage. Voltage transfer requires a high impedance ratio to be the most effective. If you feed a low impedance source into a high impedance load, you'll have the best voltage transfer (albeit, the worst current transfer). I assume you have heard of the Les Paul Recording model, no? Or perhaps the Langcaster Low-Impedance pups? Sadly, they're much more expensive, so there's little incentive to experiment amongst us, the unwashed masses.

2) The cutoff of the lo pass filter created by the capacitance of the cable is reduced, causing loss of high register harmonic content.
How's that again? Low impedance (at the input) lowers the frequency cut-off point, thanks to cable capacitance? If that were true, then yes, we'd experience very undesirable loss of high frequency content. But consider: Do you think the above referenced Les Paul would have ever seen the light of day, were that true? Or more succinctly (and much closer to home for my failing memory), have they changed the formulas for calculating how capacitive reactance (a component of impedance) since I went to engineering school? Let's not get into a math discussion here, but the short of it is, if you increase the source impedance, you increase the overall reactance value of the capacitor - that will lower the freqency cut-off point, not raise it. This is exactly why John installs a buffer circuit - it has a much lower output impedance than the pickups themselves. And you've never heard John complain about loss of harmonic content in this context, have you?

I think I've overstayed my welcome.

HTH




sumgai

[ashcatlt]

Okay, I do appreciate the discussion on how a DI works, but I guess I buried my real question under a bunch of stuff. So let me re-phrase:

Is ~200K (let alone 13K) an acceptable load for typical passive pickups?

Now, I don't mean to argue, but I guess some of this needs clarification.

I do understand that the xformer is in series with the signal on its way to the mic pre (or other low-Z input). It is, unless I'm just completely nuts, in parallel with the Hi-Z amplifier input connected to the "thru" or "amplifier" jack of the DI. AFAIK, that jack is basically there for convenience. It eliminates the need to buy a Y-cable in the (relatively common) case where you'd like to connect to both the DI and an amplifier. That jack is always on "the guitar side" of the transformer, wired in parallel with the one marked "input" or "instrument".

1) But isn't the reason we need a high-Z for a meaningful V output precisely the fact that the current is so small? If V=IR, and I is small, then R needs to be that much bigger so that V can be a practical value. No?

2) Doesn't it have something to do with the ratio of the output (source) impedance and the input (load) impedance? It was my understanding that the buffer John installs was there to present a consistent, and sufficiently high, impedance to the pickups. The output side of that buffer is low-enough Z to interface with much lower impedance inputs than would make the pickups happy.

As for what's changed since you left EE: I've never been myself (which is why I'm always bugging you guys). But, did you hear that scientists are saying that the world is round, and revolves around the sun?

So, like WTF'n'F, O? Why are we always telling these kids they'd best buffer their signals before sending them down long runs of cable or splitting them out to multiple amplifiers?

[ashcatlt]

Okay, still haven't got a god's answer on the "ideal impedance" part of the question.

I did run across something which seemed to show that lowering the load impedance actually just lowers the amplitude of the resonant peak, while leaving the cutoff in the same place.

Am I to assume that this happens because the "cable filter cutoff" rises, to the point of becoming irrelevant compared to the rest of the circuit?

[JohnH]

ash - this is both interesting and easy to test using my 5spice model of a PAF pickup with 500k pots:



C1 represents the capacitance of about a 10' cord and the input of the amp, while R3 is the amps input impedance, nominally 1000k for starters.

The graphs show the output frequency response as the amp input impedance R3 is reduced, first from 1000k to 100k:



and then from 100k down to 10k:



It would seem, that with these values, the first graph is showing mainly a damping of the resonant peak, then as input impedance is reduced further, it is cutting deeper into the treble as the resistance load interacts with the pup inductance, acting as a low pass filter.

John

[ashcatlt]

Thanks John!

That first graph is pretty much exactly what I had found on some random google link which I didn't bother to save so I could point to it.

Looking at that, it seems that the value of ~200K has a pretty profound effect. I would think that to most of us "damping of the resonant peak" would sound a whole lot like reducing high-frequency content. Especially since it sits right there at the extreme high end of the (guitar) spectrum. That's where all the bite and zing and "tone" is, no? Seems this would probably be just fine - might even be desirable - for most things we do with a bass guitar, but far from ideal for the six-stringers.

I think that 2K is probably a bit on the high end of mic preamp input-Z. From what I'm seeing in your second graph, one of these extremely low-Z inputs would make things that much worse!

This thing started when a guy on another forum was asking whether it would be okay to split his guitar signal using the mult output on a patchbay, which is essentially just a Y-cable. He wanted to connect one side to his Pod and the other to an amp. Then he was talking about using a DI to split between the Lo-Z in on the Pod and the amp. Seems to me like he'd be better off with the 500K total from the 2 x 1M in parallel than the (at best) 210K from the transformer||1M. Best, though, would still be to buffer the signal before any splitting happens.